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Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3

I. Select the correct option from the given alternatives.

Question 1.
The value of sin(n + 1) A sin(n + 2) A + cos(n + 1) A cos(n + 2) A is equal to
(a) sin A
(b) cos A
(c) -cos A
(d) sin 2A
Answer:
(b) cos A
Hint:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let (n + 2)A = a and (n + 1)A = b … (i)
∴ L.H.S. = cos a . cos b + sin a . sin b
= cos (a – b)
= cos [(n + 2)A – (n + 1)A] ……..[From (i)]
= cos [(n + 2 – n – 1)A]
= cos A
= R.H.S.

Question 2.
If tan A – tan B = x and cot B – cot A = y, then cot (A – B) = ________
(a) \(\frac{1}{y}-\frac{1}{x}\)
(b) \(\frac{1}{x}-\frac{1}{y}\)
(c) \(\frac{1}{x}+\frac{1}{y}\)
(d) \(\frac{x y}{x-y}\)
Answer:
(c) \(\frac{1}{x}+\frac{1}{y}\)
Hint:

Question 3.
If sin θ = n sin(θ + 2α), then tan(θ + α) is equal to
(a) \(\frac{1+n}{2-n}\) tan α
(b) \(\frac{1-n}{1+n}\) tan α
(c) tan α
(d) \(\frac{1+n}{1-n}\) tan α
Answer:
(d) \(\frac{1+n}{1-n}\) tan α
Hint:

Question 4.
The value of \(\frac{\cos \theta}{1+\sin \theta}\) is equal to ________
(a) \(\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)\)
(b) \(\tan \left(-\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(d) \(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\)
Answer:
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
Hint:

Question 5.
The value of cos A cos (60° – A) cos (60° + A) is equal to ________
(a) \(\frac{1}{2}\) cos 3A
(b) cos 3A
(c) \(\frac{1}{4}\) cos 3A
(d) 4cos 3A
Answer:
(c) \(\frac{1}{4}\) cos 3A
Hint:

Question 6.
The value of \(\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}\) is ________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{64}\)
(c) \(\frac{1}{128}\)
(d) \(\frac{1}{256}\)
Answer:
(b) \(\frac{1}{64}\)
Hint:

Question 7.
If α + β + γ = π, then the value of sin² α + sin² β – sin² γ is equal to ________
(a) 2 sin α
(b) 2 sin α cos β sin γ
(c) 2 sin α sin β cos γ
(d) 2 sin α sin β sin γ
Answer:
(c) 2 sin α sin β cos γ
Hint:
sin² α + sin² β – sin² γ
= \(\frac{1-\cos 2 \alpha}{2}+\frac{1-\cos 2 \beta}{2}-\sin ^{2} \gamma\)
= 1 – \(\frac{1}{2}\) (cos 2α + cos 2β) – 1 + cos² γ
= \(\frac{-1}{2}\) × 2 cos(α + β) cos(α – β) + cos² γ
= cos γ cos (α – β) + cos² γ …..[∵ α + β + γ = π]
= cos γ [cos (α – β) + cos γ]
= cos γ [cos (α – β) – cos (α + β)]
= 2 sin α sin β cos γ

Question 8.
Let 0 < A, B < \(\frac{\pi}{2}\) satisfying the equation 3sin² A + 2sin² B = 1 and 3sin 2A – 2sin 2B = 0, then A + 2B is equal to ________
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) 2π
Answer:
(b) \(\frac{\pi}{2}\)
Hint:
3 sin 2A – 2sin 2B = 0
sin 2B = \(\frac{3}{2}\) sin 2A …….(i)
3 sin² A + 2 sin² B = 1
3 sin² A = 1 – 2 sin² B
3 sin² A = cos 2B ……(ii)
cos(A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin² A) – sin A (\(\frac{3}{2}\) sin 2A) …..[From (i) and (ii)]
= 3 cos A sin² A – \(\frac{3}{2}\) (sin A) (2 sin A cos A)
= 3 cos A sin² A – 3 sin² A cos A
= 0
= cos \(\frac{\pi}{2}\)
∴ A + 2B = \(\frac{\pi}{2}\) ……..[∵ 0 < A + 2B < \(\frac{3 \pi}{2}\)]

Question 9.
In ∆ABC if cot A cot B cot C > 0, then the triangle is ________
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles right-angled
Answer:
(a) acute angled
Hint:
cot A cot B cot C > 0
Case I:
cot A, cot B, cot C > 0
∴ cot A > 0, cot B > 0, cot C > 0
∴ 0 < A < \(\frac{\pi}{2}\), 0 < B < \(\frac{\pi}{2}\), 0 < C < \(\frac{\pi}{2}\)
∴ ∆ABC is an acute angled triangle.
Case II:
Two of cot A, cot B, cot C < 0
0 < A, B, C < π and two of cot A, cot B, cot C < 0
∴ Two angles A, B, C are in the 2nd quadrant which is not possible.

Question 10.
The numerical value of tan 20° tan 80° cot 50° is equal to ________
(a) √3
(b) \(\frac{1}{\sqrt{3}}\)
(c) 2√3
(d) \(\frac{1}{2 \sqrt{3}}\)
Answer:
(a) √3
Hint:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

II. Prove the following.

Question 1.
tan 20° tan 80° cot 50° = √3
Solution:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

Question 2.
If sin α sin β – cos α cos β + 1 = 0, then prove that cot α tan β = -1.
Solution:
sin α sin β – cos α cos β + 1 = 0
∴ cos α cos β – sin α sin β = 1
∴ cos (α + β) = 1
∴ α + β = 0 ……[∵ cos 0 = 1]
∴ β = -α
L.H.S. = cot α tan β
= cot α tan(-α)
= -cot α tan α
= -1
= R.H.S.

Question 3.
\(\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}\)
Solution:

Question 4.
\(\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}\)
Solution:

Question 5.
cos 12° + cos 84° + cos 156° + cos 132° = \(-\frac{1}{2}\)
Solution:

Question 6.
\(\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x\)
Solution:

Question 7.
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Solution:

Question 8.
sin² 6x – sin² 4x = sin 2x sin 10x
Solution:

Question 9.
cos² 2x – cos² 6x = sin 4x sin 8x
Solution:

Question 10.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:

Question 11.
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Solution:

Question 12.
If sin 2A = λ sin 2B, then prove that \(\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}\)
Solution:

Question 13.
\(\frac{2 \cos 2 A+1}{2 \cos 2 A-1}\) = tan (60° + A) tan (60° – A)
Solution:

Question 14.
tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A
Solution:

Question 15.
3 tan⁶ 10° – 27 tan⁴ 10° + 33 tan² 10° = 1
Solution:

Question 16.
cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0
Solution:
L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384°
= cosec 48° + cosec (180° – 84°) + cosec (180° + 12°) + cosec (360° + 24°)
= cosec 48° + cosec 84° + cosec (-12°) + cosec 24°

Question 17.
3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x) = 13
Solution:
(sin x – cos x)⁴
= [(sin x – cos x)²]²
= (sin² x + cos² x – 2 sin x cos x)²
= (1 – 2 sin x cosx)²
= 1 – 4 sin x cos x + 4 sin² x cos² x
(sin x + cos x)² = sin² x + cos² x + 2 sin x cos x = 1 + 2 sin x cos x
sin⁶ x + cos⁶ x
= (sin² x)³ + (cos² x)³
= (sin² x + cos² x)³ – 3 sin² x cos² x (sin² x + cos² x) …..[∵ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3 sin² x cos² x (1)
= 1 – 3 sin² x cos² x
L.H.S. = 3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x)
= 3(1 – 4 sin x cos x + 4 sin² x cos² x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin² x cos² x)
= 3 – 12 sin x cos x + 12 sin² x cos² x + 6 + 12 sin x cos x + 4 – 12 sin² x cos² x
= 13
= R.H.S.

Question 18.
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution:
We have to prove that,
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
i.e., to prove,
cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0

∴ cot θ – tan θ = 2 cot 2θ …..(i)
L.H.S. = cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A
= 2 cot 2A – 2 tan 2A – 4 tan 4A – 8 cot 8A …..[From (i)]
= 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8A
= 2 × 2 cot 2(2A) – 4 tan 4A – 8 cot 8A ……[From (i)]
= 4(cot 4A – tan 4A) – 8 cot 8A
= 4 × 2 cot 2(4A) – 8 cot 8A ……[From (i)]
= 8 cot 8A – 8 cot 8A = 0
= R.H.S.
Alternate Method:

Question 19.
If A + B + C = \(\frac{3 \pi}{2}\), then cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C
Solution:

Question 20.
In any triangle ABC, sin A – cos B = cos C. Show that ∠B = \(\frac{\pi}{2}\).
Solution:
sin A – cos B = cos C
∴ sin A = cos B + cos C

A = B – C ………(i)
In ∆ABC,
A + B + C = π
∴ B – C + B + C = π
∴ 2B = π
∴ B = \(\frac{\pi}{2}\)

Question 21.
\(\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}\) = sec x cosec x – 2 sin x cos x
Solution:

Question 22.
sin 20° sin 40° sin 80° = \(\frac{\sqrt{3}}{8}\)
Solution:
L.H.S. = sin 20°. sin 40°. sin 80°
= sin 20°. sin 40°. sin 80°
= \(\frac{1}{2}\) (2 . sin 40°. sin 20°) . sin 80°
= \(\frac{1}{2}\) [cos(40° – 20°) – cos (40° + 20°)] . sin 80°
= \(\frac{1}{2}\) (cos 20° – cos 60°) sin 80°
= \(\frac{1}{2}\) . cos 20° . sin 80° – \(\frac{1}{2}\) . cos 60° . sin 80°
= \(\frac{1}{2 \times 2}\) (2 sin 80° . cos 20°) – \(\frac{1}{2 \times 2}\) . sin 80°
= \(\frac{1}{4}\) [sin(80° + 20°) + sin (80° – 20°)] – \(\frac{1}{2}\) . sin 80°

Question 23.
sin 18° = \(\frac{\sqrt{5}-1}{4}\)
Solution:
Let θ = 18°
∴ 5θ = 90°
∴ 2θ + 3θ = 90°
∴ 2θ = 90° – 3θ
∴ sin 2θ = sin (90° – 3θ)
∴ sin 2θ = cos 3θ
∴ 2 sin θ cos θ = 4 cos³ θ – 3 cos θ
∴ 2 sin θ = 4 cos² θ – 3 …..[∵ cos θ ≠ 0]
∴ 2 sin θ = 4 (1 – sin² θ) – 3
∴ 2 sin θ = 1 – 4 sin² θ
∴ 4 sin² θ + 2 sin θ – 1 = 0
∴ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{8}\)
= \(\frac{-2 \pm 2 \sqrt{5}}{8}\)
= \(\frac{-1 \pm \sqrt{5}}{4}\)
Since, sin 18° > 0
∴ sin 18°= \(\frac{\sqrt{5}-1}{4}\)

Question 24.
cos 36° = \(\frac{\sqrt{5}+1}{4}\)
Solution:
We know that,
cos 2θ = 1 – 2 sin² θ
cos 36° = cos 2(18°)
= 1 – 2 sin² 18°

∴ cos 36° = \(\frac{\sqrt{5}+1}{4}\)

Question 25.
sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\)
Solution:
We know that, sin² θ = 1 – cos² θ
sin² 36° = 1 – cos² 36°
= 1 – \(\left(\frac{\sqrt{5}+1}{4}\right)^{2}\)
= \(\frac{16-(5+1+2 \sqrt{5})}{16}\)
= \(\frac{10-2 \sqrt{5}}{16}\)
∴ sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\) ……[∵ sin 36° is positive]

Question 26.
\(\sin \frac{\pi^{c}}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}\)
Solution:

Question 27.
tan \(\frac{\pi}{8}\) = √2 – 1
Solution:

Question 28.
tan 6° tan 42° tan 66° tan 78° = 1
Solution:

Question 29.
sin 47° + sin 61° – sin 11° – sin 25° = cos 7°
Solution:

Question 30.
√3 cosec 20° – sec 20° = 4
Solution:

Question 31.
In ∆ABC, ∠C = \(\frac{2 \pi}{3}\), then prove that cos² A + cos² B – cos A cos B = \(\frac{3}{4}\).
Solution:

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.4

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines, i. 2x + 3y-6 = 0 ii. 3x-y-9 = 0 iii. x + 2y = 0
Solution:
i. Given equation of the line is 2x + 3y – 6 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 2, b = 3, c = -6

ii. Given equation of the line is 3x – y – 9 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 3, b = – 1, c = – 9

iii. Given equation of the line is x + 2y = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 1, b = 2, c = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
i. y = 2x – 4
ii. y = 4
iii. \(\frac{x}{2}+\frac{y}{4}=1\)
iv. \(\frac{x}{3}-\frac{y}{2}=0\)
i. y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

ii. y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

iii. \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}\)
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

iv. \(\frac{x}{3}-\frac{y}{2}=0\)
∴ 2x – 3y = 0
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.
[Note: Answer given in the textbook is ‘2x – 3y – 6 = 0’. However, as per our calculation it is ‘2x-3y + 0 = 0’.]

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 15 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 15 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{-4}=\frac{1}{2}\)
Since m₁ = m₂
the given lines are parallel to each other.

Question 4.
Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x + y + 1 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{1}=-2\)
Since m₁ x m₁ = \(\frac{1}{2}\) x (- 2) = -1,
the given lines are perpendicular to each other. Consider,
x – 2y – 7 = 0 …(i)
2x + y + 1 =0 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y + 2 = 0 …(iii)
Adding equations (i) and (iii), we get
5x – 5 = 0
∴ x = 1
Substituting x = 1 in equation (ii), we get
2 + y + 1 = 0
∴ y = – 3
∴ The point of intersection of the given lines is (1,-3).

Question 5.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y axes at points A and B respectively.
3x + 4y = p

This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A (a, 0) ≡ (\(\frac{p}{3}\), 0) and B ≡ (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A (∆OAB) = 24 sq. units
∴ \(\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=24\)
∴ \(\left|\frac{1}{2} \times \frac{\mathrm{p}}{3} \times \frac{\mathrm{p}}{4}\right|=24\)
∴ p² = 576
∴ p = ± 24

Question 6.
Find the co-ordinates of the foot of the perpendicular drawn from the point A(- 2,3) to the line 3x-y -1 = 0.
Solution:
Let M be the foot of perpendicular drawn from
point A(- 2,3) to the line
3x-y- 1 = 0 …(i)
Slope of the line 3x-y – 1 = 0 is \(\frac{-3}{-1}\) =3.
Since AM ⊥ to line (i),
slope of AM = \(\frac{-1}{3}\)
∴ Equation of AM is
y – 3 = \(\frac{-1}{3}\)(x + 2)

∴ 3(y – 3) = – 1(x + 2)
∴ 3y – 9 = -x – 2
∴ x + 3y – 7 = 0 …………(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).
By (i) x 3 + (ii), we get 10x -10 = 0
∴ x = 1
Substituting x = 1 in (ii), we get
1 + 3y – 7 = 0
∴ 3y = 6
∴ y = 2
∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

Question 7.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(- 2, 3), B(6, -1), C(4,3),e
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of AABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC respectively.

Since FD passes through (5, 1) and has slope 1/2 equation of FD is
y – 1 = \(\frac{1}{2}\)(x-5)
∴ 2 (y – 1) = x – 5
∴ 2y – 2 = x – 5
∴ x – 2y – 3 = 0 …(i)
Since both the points A and C have same y co-ordinates i.e. 3,
the given points lie on the line y = 3.
Since the equation FE passes through E(1, 3),
the equation of FE is x = 1. .. .(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y -3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F ≡ (1, – 1).

Question 8.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(3, – 2), B(7,6), C (-1,2).
Solution:
Let O be the orthocentre of ∆ABC.
Let AD and BE be the altitudes on the sides BC and AC respectively.

Slope of side BC = \(\frac{2-6}{-1-7}=\frac{-4}{-8}=\frac{1}{2}\)
∴ Slope of AD = – 2 [∵ AD ⊥ BC]
∴ Equation of line AD is
y – (-2) = (- 2) (x – 3)
∴ y + 2 = -2x + 6
∴ 2x + y -4 = 0 …(i)
Slope of side AC = \(\frac{-2-2}{3-(-1)}=\frac{-4}{4}\) = -1
∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]
∴ Equation of line BE is
y – 6 = 1(x – 7)
∴ y – 6 = x – 1
∴ x = y + 1 …(ii)
Substituting x = y + 1 in (i), we get
2(y + 1) + y – 4 = 0
∴ 2y + 2 + y – 4 = 0
∴ 3y – 2 = 0
∴ y = \(\frac{2}{3} in (ii), we get
Substituting y = [latex]\frac{2}{3}\) in (ii), we get
x = \(\frac{2}{3}+1=\frac{5}{3}\)
∴ Co-ordinates of orthocentre, O = \(\left(\frac{5}{3}, \frac{2}{3}\right)\)

Question 9.
Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.
Solution:
The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are
3x – 4y + 5 = 0 …(i)
7x-8y + 5 = 0 …(ii)
4x + 5y – 45 = 0 …(iii)
By (i) x 2 – (ii), we get
– x + 5 = 0
∴ x = 5
Substituting x = 5 in (i), we get
3(5) – 4y + 5 = 0
∴ -4y = – 20
∴ y = 5
∴ The point of intersection of lines (i) and (ii) is given by (5, 5).
Substituting x = 5 and y = 5 in L.H.S. of (iii), we get
L.H.S. = 4(5) + 5(5) – 45
= 20 + 25 – 45
= 0
= R.H.S.
∴ Line (iii) also passes through (5, 5).
Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

Question 10.
Find the equation of the line whose x-intercept is 3 and which ¡s perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ Slope of the required line perpendicular to
3x – y + 23 = 0 is \(\frac{-1}{3}\)
Since the x-intercept of the required line is 3, it passes through (3, 0).
∴ The equation of the required line is ‘
y – 0 = \(\frac{-1}{3}\)(x – 3)
∴ 3y = x + 3
∴ x + 3y = 3

Question 11.
Find the distance of the origin from the line 7x + 24y – 50 = 0.
Solution:
Let p be the perpendicular distance of origin
fromtheline7x + 24y – 50 = 0
Here, a = 7, b = 24, c = -50

Question 12.
Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = – 5, c = – 13, x₁ = -2, y₁ = 3

Question 13.
Find the distance between parallel lines 4x – 3y + 5 = 0 and 4xr – 3y + 7 = 0.
Solution:
Equations of the given parallel lines are 4x – 3y + 5 = 0 and 4x – 3y + 1 = 0
Here, a = 4, b = – 3, c₁ = 5 and c₂ = 7
∴ Distance between the parallel lines

Question 14.
Find the distance between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
Equations of the given parallel lines are 3x + 2y + 6 = 0 and
9x + 6y – 1 = 0 i.e., 3x + 2y – \(\frac{7}{3}\) =0
Here, a = 3, b = 2, c₁ = 6 and c₂ = \(\frac{-7}{3}\)
∴ Distance between the parallel lines

Question 15.
Find the points on the line x + y – 4 = 0 which are at a unit distance from the line 4JC + 3y = 10.
Solution:
Let P(x₁, y₁) be a point on the line x + y – 4 = o.
∴ x₁ + y₁ – 4 = 0
∴ y₁ = 4 – x₁ …(i)
Also, distance of P from the line 4x + 3y- 10 = 0 is 1

∴ 5 = | x₁ + 2 |
∴ x₁ + 2 = ± 5
∴ x₁ + 2 = 5 or x₁ + 2 = – 5
∴ x₁ = 3 or x₁ = – 7
From (i), when x₁ = 3, y₁ = 1
and when x₁ = -7, y₁ = 11
∴ The required points are (3, 1) and (-7, 11).
[Note: The question has been modified]

Question 16.
Find the equation of the line parallel to the X-axis and passing through the point of intersection of lines x + y – 2 = 0 and 4x + 3y = 10.
Solution:
Let u = x + y – 2 = 0 and v = 4x + 3y – 10 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)
∴ x + y – 2 + 4kx + 3ky – 10k = 0
∴ x + 4kx + y + 3ky – 2 – 10k = 0
∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0
But, this line is parallel to X-axis.
∴ Its slope = 0
∴ \(\frac{-(1+4 k)}{1+3 k}\) = 0
∴ 1 + 4k = 0
∴ k = \(\frac{-1}{4}\)
Substituting the value of k in (i), we get
(x + y – 2) + (4x + 3y – 10) = 0
∴ 4(x +y – 2) – (4x + 3y -10 ) = 0
∴ 4x + 4y – 8 – 4x – 3y + 10 = 0
∴ y + 2 = 0, which is the equation of the required line.
[Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

Question 17.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2xr – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Let u ≡ x + y – 2 = 0 and v ≡ 2x – 3y + 4 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)
But, x-intercept of line is 3.
∴ It passes through (3, 0).
Substituting x = 3 and y = 0 in (i), we get
(3 + 0 – 2) + k(6 – 0 + 4) = 0
∴ 1 + 10k = 0
k = \(\frac{-1}{10}\)
Substituting the value of k in (i), we get (x + y – 2) + \(\left(\frac{-1}{10}\right)\) (2x – 3y + 4) = 0
∴ 10(x + y – 2) – (2x – 3y + 4) = 0
∴ 10x + 10y -20 — 2x + 3y-4 = 0
∴ 8x + 13y – 24 = 0, which is the equation of the required line.

Question 18.
If A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC, then find the equation of bisector of angle BAC.
Solution:
Let the bisector of ∠ BAC meets BC at point D.
∴ Point D divides seg BC in the ratio l(AB) : l(AC)

∴ 18 (y – 3) = 6 (x – 4)
∴ 3(y – 3) = x – 4
∴ 3y – 9 = x – 4
∴ x – 3y + 5 = 0

Question 19.
D(- 1, 8), E(4, – 2), F(- 5, – 3) are midpoints of sides BC, CA and AB of AABC. Find
i. equations of sides of ΔABC.
ii. co-ordinates of the circumcentre of ΔABC.
Solution:
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

∴ x₁ + x₂ = -10 …………. (v)
and y₁ + y₂ = – 6 …………(vi)
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = – 4
∴ x₁ + x₂ + x₃ = -2 …………..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = – 10
Solving (v) and (vii), we get x₃ = 8

For y-coordinates:
Adding (ii), (iv) and (vi), we get 2y₁ + 2y₂ + 2y₃ = 6
y₁ + y₂ + y₃ = 3 …….(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of AABC are A(0, – 13), B(- 10, 7), C(8, 9)

∴ 8(y + 13) = 22x
∴ 4(y + 13) = 11x
∴ 11x – 4y – 52 = 0

ii. Here, A(0, – 13), B(- 10, 7), C(8, 9) are the vertices of ΔABC.
Let F be the circumcentre of AABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
D and E are the midpoints of side BC and AC.

∴ Slope of FD = -9 … [ ∵ FD ⊥ BC]
Since FD passes through (-1, 8) and has slope -9, equation of FD is
y – 8 = -9 (x +1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) [ ∵ FE ⊥ AC]
Since FE passes through (4, -2) and has slope -4
\(\frac{-4}{11}\), equation of FE is
(y + 2) = \(\frac{-4}{11}\) (x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = – 4x + 16
∴ 4x + 11y = -6 …………(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value ofy in (ii), we get
4x + 11(-9x- 1) = – 6
∴ 4x – 99x -11 = – 6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F ≡ \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

Question 20.
0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.
Solution:
Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E
∴ Point D divides seg AB in the ratio l(OA): l(OB)
and point E divides seg BO in the ratio l(AB): l(AO)
Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8
i.e. 3 :4

∴ y = x …(i)
Now, by distance formula,
l(AB) = \(\begin{aligned}
&=\sqrt{(6-0)^{2}+(0-8)^{2}} \\
&=\sqrt{36+64}=10
\end{aligned}\)
l(AO) = \(\sqrt{(6-0)^{2}+(0-0)^{2}}\) = 6
∴ Point E divides BO internally in 10 : 6 i.e. 5:3

∴ -2y = x – 6
∴ x + 2y = 6 …(ii)
To find co-ordinates of incentre, we have to solve equations (i) and (ii).
Substituting y = x in (ii), we get
x + 2x = 6
∴ x = 2
Substituting the value of x in (i), we get
y = 2
∴ Co-ordinates of incentre I ≡ (2, 2).

Alternate Method:
Let I be the incentre.
I lies in the 1st quadrant.
OPIR is a square having side length r.
Since OA = 6, OP = r,
PA = 6 – r
Since PA = AQ,
AQ = 6 – r …(i)
Since OB = 8, OR = r,
BR = 8 – r
∴ BR = BQ
∴ BQ = 8 – r …(ii)

AB = BQ + AQ
Also, AB = \(\begin{aligned}
&=\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}} \\
&=\sqrt{6^{2}+8^{2}} \\
&=\sqrt{100}=10
\end{aligned}\)
∴ BQ + AQ= 10
∴ (8 – r) + (6 – r) = 10
∴ 2r = 14- 10 = 4
∴ r = 2
∴ I = (2,2)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.3

Question 1.
Write the equation of the line:
i. parallel to the X-axis and at a distance of 5 units from it and above it.
ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).
Solution:
i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above X-axis, k = 5
∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to Y-axis is x = h. Since the line is at a distance of 5 units to the left of Y-axis, h = -5
∴ The equation of the required line is x = -5.
[Note: Answer given in the textbook is ‘y = -5
However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since the line is at a distance of 4 units from the point (- 2, 3),
k = 4 + 3 = 7 or k = 3- 4 = -1
∴ The equation of the required line is y = 1 or y = – 1.

Question 2.
Obtain the equation of the line:
i. parallel to the X-axis and making an intercept of 3 units on the Y-axis.
ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution:
i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k.
Here, y-intercept = 3
∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ The equation of the required line is x = 4.

Question 3.
Obtain the equation of the line containing the point:
i. A(2, – 3) and parallel to the Y-axis.
ii. B(4, – 3) and parallel to the X-axis.
Solution:
i. Equation of a line parallel to Y-axis is of the form x = h.
Since the line passes through A(2, – 3), h = 2
∴ The equation of the required line is x = 2.

ii. Equation of a line parallel to X-axis is of the formy = k.
Since the line passes through B(4, – 3), k = -3
∴ The equation of the required line is y = – 3.

Question 4.
Find the equation of the line:
i. passing through the points A(2, 0) and B(3,4)
ii. passing through the points P(2, 1) and Q(2,-1)
Solution:
i. The required line passes through the points A(2, 0) and B(3,4).
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x₁y₁) = (2,0) and (x₁,y₂) = (3,4)
∴ The equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)
∴ \(\frac{y}{4}=\frac{x-2}{1}\)
∴ y = 4(x – 2)
∴ y = 4x – 8
∴ 4x – y – 8 = 0

ii. The required line passes through the points P(2, 1) and Q(2,-1).
Since both the given points have same
x co-ordinates i.e. 2,
the given points lie on the line x = 2.
∴ The equation of the required line is x = 2.

Question 5.
Find the equation of the line:
i. containing the origin and having inclination 60°.
ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7).
iii. having slope 1/2 and containing the point (3, -2)
iv. containing the point A(3, 5) and having slope 2/3
v. containing the point A(4, 3) and having inclination 120°.
vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.
Solution:
i. Given, Inclination of line = θ = 60°
Slope of the line (m) = tan θ = tan 60°
= \(\sqrt{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
.‘. The equation of the required line is y = \(\sqrt{3}\) x

ii. Given, A (2, 4) and B (1, 7)
Slope of AB = \(\frac{7-4}{1-2}\) = -3 1-2
Since the required line is parallel to line AB, slope of required line (m) = slope of AB
∴ m = – 3 and the required line passes through the origin.
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = \(=\frac{1}{2}\) and the line passes through (3, – 2).
Equation of the line in slope point form is
y-y ₁= m(x-x₁)
∴ The equation of the required line is
[y-(- 2)]=\(\frac{1}{2}\)(x-3)
∴ 2(y + 2)=x – 3
∴ 2y + 4 = x – 3
∴ x – 2y – 7 = 0

iv. Given, slope(m) = \(\frac{2}{3}\) and the line passes through (3, 5).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is y – 5 = \(\frac{2}{3}\)(x-3)
∴ 3 (y – 5) = 2 (x – 3)
∴ 3y – 15 = 2x – 6
∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°
Slope of the line (m) = tan θ = tan 120°
= tan (90° + 30°)
= – cot 30°
= – \(\sqrt{3}\)
and the line passes through A(4, 3).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is
y- 3 = –\(\sqrt{3}\)(x-4)
∴ y – 3 = –\(\sqrt{3}\) x + 4\(\sqrt{3}\)
∴ \(\sqrt{3}\)x + y – 3 -4\(\sqrt{3}\) = 0

vi.

Given equation of the line is 3x +y = 6.
∴ \(\frac{x}{2}+\frac{y}{6}=1\)
This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}\) = 1,
where a = 2, b = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.
∴ Midpoint of AB = ( \(\frac{2+0}{2}, \frac{0+6}{2}\) ) = (1,3)
∴ Required line passes through (0, 0) and (1,3).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{3-0}=\frac{x-0}{1-0}\)
\(\frac{y}{3}=\frac{x}{1}\)
∴ y = 3x
∴ 3x – y = 0

Alternate Method:
Given equation of the line is 3x + y = 6 …(i)
Substitute y = 0 in (i) to get a point on X-axis.
∴ 3x + 0 = 6
∴ x = 2
Substitute x = 0 in (i) to get a point on Y-axis.
∴ 3(0) + 7 = 6
∴ y = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.
Let M be the midpoint of AB.
M = \(\left(\frac{2+0}{2}, \frac{0+6}{2}\right)\) = (1,3)
Slope of OM (m) = \(\frac{3-0}{1-0}\) = 3
Equation of OM is of the formy = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 6.
Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.
Solution:
Given, A(2, 1) and B(3,2)
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-1}{2-1}=\frac{x-2}{3-2}\)
∴ \(\frac{y-1}{1}=\frac{x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1
Comparing this equation with y = mx + c, we get
m = 1 and c = – 1

Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1

Question 7.
Find the equation of the line having inclination 135° and making x-intercept 7.
Solution:
Given, Inclination of line = 0 = 135°
∴ Slope of the line (m) = tan 0 = tan 135°
= tan (90° + 45°)
= – cot 45° = – 1 x-intercept of the required line is 7.
∴ The line passes through (7, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ The equation of the required line is y — 0 = – 1 (x – 7)
∴ y = -x + 7
∴ x + y – 7 = 0

Question 8.
The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containing
i. side BC
ii. the median AD
iii. the midpoints of sides AB and BC.
Solution:
Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6).
i. Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the side BC is
\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\)
\(\frac{y}{6}=\frac{x-2}{-3}\)
∴ – 3y = 6x – 12
∴ 6x + 3y – 12 = 0
∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)
The median AD passes through the points
A(3,4) and D( \(\frac{1}{2}\) , 3)

∴ The equation of the median AD is
\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)
\(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)
\(\frac{5}{2}\)(y-4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

iii. Let D and E be the midpoints of side AB and side BC respectively.
The equation of the line DE is

∴ -4(y-2) = 2x-5
∴ 2x + 4y – 13 = 0

Question 9.
Find the x and y-intercepts of the following lines:
i. \(\frac{x}{3}+\frac{y}{2}=1\)
ii. \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
iii. 2x – 3y + 12 = 0
Solution:
i. Given equation of the line is latex]\frac{x}{3}+\frac{y}{2}=1[/latex]
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = 3, y-intercept = 2

ii. Given equation of the line is \(\frac{3 x}{2}+\frac{2 y}{3}\) = 1
∴ \(\frac{x}{\left(\frac{2}{3}\right)}+\frac{y}{\left(\frac{3}{2}\right)}\) = 1
This is of the form = \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = \(\frac{2}{3}\) and y-intercept = \(\frac{3}{2}\)

iii. Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = – 12
∴ \(\frac{2 x}{(-12)}-\frac{3 y}{(-12)}=1\)
∴ \(\frac{x}{-6}+\frac{y}{4}=1\)
This is of the form \(\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = – 6 and y-intercept = 4

Question 10.
Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be
\(\frac{x}{a}+\frac{y}{b}=1\) ………..(i)
Since, the sum of the intercepts of the line is zero.
∴ a + b = 0
∴ b = – a
Substituting b = – a in (i), we get
\(\frac{x}{a}+\frac{y}{(-a)}=1\)
x – y = a .. .(ii)
Since, the line passes through A(1, 3).
∴ 1 – 3 = a
∴ a = – 2
Substituting the value of a in (ii), equation of the required line is
∴ x – y = – 2,
∴ x – y + 2 = 0

Case II: Line passing through origin.
Slope of line passing through origin and
A(1, 3) is m = \(\frac{3-0}{1-0}\) = 3
∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 11.
Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …………(i)
This line passes through A(3, 4).
∴ \(\frac{3}{a}+\frac{4}{b}=1\)……………..(ii)
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b …(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{4}{a}=1\)
∴ \(\frac{7}{a}=1\)
∴ a = 7
∴ b = 7 …[From (iii)]
Substituting the values of a and b in (i), equation of the required line is
\(\frac{x}{7}+\frac{y}{7}=1\) = 1
∴ x + y = 7

Case II: Line passing through origin.
Slope of line passing through origin and A(3,4) is m = \(=\frac{4-0}{3-0}=\frac{4}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is 4
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0

Question 12.
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).
Solution:

A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
∴ Slope of AD = -5 …[∵AD ⊥ BC]
Since altitude AD passes through (2, 5) and has slope – 5,
equation of the altitude AD is y – 5 = -5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x +y -15 = 0
Now, slope of AC = \(\frac{-3-5}{-4-2}=\frac{-8}{-6}=\frac{4}{3}\)
Slope of BE = \(\frac{-3}{4}\)
…[∵ BE ⊥ AC]
Since altitude BE passes through (6,-1) and has slope \(\frac{-3}{4}\),
equation of the altitude BE is
y-(-1) = \(\frac{-3}{4}\) (x – 6)
∴ 4 (y + 1) = – 3 (x – 6)
∴ 4y + 4 =-3x+ 18
∴ 3x + 4y – 14 = 0
Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)
∴ Slope of CF = \({2}{3}\) ….[∵ CF ⊥ AB]
Since altitude CF passes through (- 4, – 3) and has slope , \(\frac{2}{3}\)
equation of the altitude CF is
y-(-3) = \(\frac{2}{3}\)[x-(-4)]
∴ 3 (y + 3) = 2 (x + 4)
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0

Question 13.
Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).
Solution:

Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR.
A is the midpoint of side PQ.

Slope of perpendicular bisector of PQ is \(\frac{1}{2}\) and it passes through (\(\frac{3}{2}\)), 3).
Equation of the perpendicular bisector of side PQ is
y – 3 = \(\frac{1}{2}\)(x – \(\frac{3}{2}\))
y – 3 = (\(\frac{1}{2}\left(\frac{2 x-3}{2}\right)\))
∴ 4(y – 3) = 2x – 3
∴ 4y – 12 = 2x – 3
∴ 2x – 4y + 9 = 0
B is the midpoint of side QR
∴ B = \(\left(\frac{4-5}{2}, \frac{-2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-5}{2}\right)\)
Slope of side QR = \(\frac{-3-(-2)}{-5-4}=\frac{-1}{-9}=\frac{1}{9}\)
∴ Slope of perpendicular bisector of QR is -9 and it passes through \(\left(-\frac{1}{2},-\frac{5}{2}\right)\)
∴ Equation of the perpendicular bisector of side QR is

∴ 2y + 5 = -18x – 9
∴ 18x + 2y + 14 = 0
∴ 9x + y + 7 = 0
C is the midpoint of side PR.

Equation of the perpendicular bisector of PR is \(y-\frac{5}{2}=-\frac{4}{11}(x+3)\)
∴ \(11\left(\frac{2 y-5}{2}\right)\) =-4(x + 3)
∴ 11(2y – 5) = – 8 (x + 3)
∴ 22y – 55 = – 8x – 24
∴ 8x + 22y -31 = 0

Question 14.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).
Solution:
Let O be the orthocentre of AABC.
Let AM and BN be the altitudes of sides BC and AC respectively.
Now, slope of BC = \(\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\)
Slope of AM = -2 ,..[∵ AM ⊥ BC]
Since AM passes through (2, – 2) and has slope -2,
equation of the altitude AM is y – (- 2) = – 2 (x – 2)
∴ y + 2 = -2x + 4
∴ 2x + y – 2 = 0 …(i)

Also, slope of AC = \(\frac{0-(-2)}{-1-2}=\frac{2}{-3}\)
∴ Slope of BN = \(\frac{3}{2}\) …[∵ BN ⊥ AC]
Since BN passes through (1,1) and has slope \(\frac{3}{2}\), equation of the altitude BN is
y – 1 = \(\frac{3}{2}\)(x-1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0 …(ii)
To find co-ordinates of orthocentre, we have to solve equations (i) and (ii).
By (i) x 2 + (ii), we get
7x – 5 = 0
∴ x = \(\frac{5}{7}\)
substituting x = \(\frac{5}{7}\) in eq (i), we get
2(\(\frac{5}{7}\)) + y – 2 = 0
∴ y = -2(\(\frac{5}{7}\)) + 2
∴ y = \(\frac{-10+14}{7}=\frac{4}{7}\)
∴ Coordinates of orthocentre O = \(\left(\frac{5}{7}, \frac{4}{7}\right)\)

Question 15.
N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.
Solution:

Slope of ON = \(\frac{-4-0}{3-0}=\frac{-4}{3}\)
Since line L ⊥ ON,
slope of the line L is \(\frac{3}{4}\) and it passes through point N(3, -4).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line L is
y-(-4) = \(\frac{3}{4}\)(x-3)
∴ 4(y + 4) = 3(x – 3)
∴ 4y + 16 = 3x – 9
∴ 3x – 4y – 25 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.1

Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3), B(2, 1) and
PA = PB
∴ PA² = PB²
∴ (x – 1)² + ( y – 3)² = (x – 2)² + (y – 1)²
∴ x² – 2x + 1 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 1
-2x – 6y + 10 = -4x – 2y + 5
∴ 2x – 4y + 5 = 0
∴ The required equation of locus is 2x – 4y + 5 = 0.

Question 2.
A(- 5,2) and B(4,1). Find the equation of the locus of point P, which is equidistant from A and B.
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(- 5, 2) and B(4, 1).
∴ PA = PB
∴ PA² = PB²
∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)²
∴ x² + 10x + 25 + y² — 4y + 4
= x² – 8x + 16 + y² – 2y + 1
∴ 10x – 4y + 29 = -8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
The required equation of locus is 9x -y + 6 = 0.

Question 3.
If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
∴ AP² = 4BP²
∴ (x – 2)² + (y – 0)² = 4[(x – 0)² + (y – 3)²]
∴ x² – 4x + 4 + y² = 4(x² + y² – 6y + 9)
x² – 4x + 4 + y² = 4x² + 4y² – 24y + 36
∴ 3x² + 3 y² + 4x – 24y + 32 = 0
∴ The required equation of locus is
3x² + 3y² + 4x – 24y + 32 = 0.
[Note: Answer given in the textbook , is
‘3x² + 3y² + 4x + 24y + 32 = O’.
However, as per our calculation it is ‘3x² + 3y² + 4x – 24y + 32 = 0’.]

Question 4.
If A(4,1) and B(5,4), find the equation of the locus of point P such that PA² = 3PB².
Solution:
Let P(x, y) be any point on the required locus. Given, A(4,1), B(5,4) and PA2 = 3PB2
∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)² + (y – 4)²]
∴ x² – 8x + 16 + y² – 2y + 1 = 3(x² – 10x + 25 + y² – 8y + 16)
∴ x² – 8x + y² – 2y + 17 = 3x² -30x + 75 + 3y² – 24y + 48
∴ 2x² + 2y² – 22x – 22y + 106 = 0
∴ x² + y² – 11x – 11y + 53 = 0
∴ The required equation of locus is
x² + y² – 11x – 11y + 53 = 0.

Question 5.
A(2, 4) and B(5, 8), find the equation of the
locus of point P such that PA² – PB² = 13.
Solution:
Let P(x, y) be any point on the required locus. Given, A(2,4), B(5, 8) and PA²-PB² = 13
∴ [(x -2)² + (y – 4)²] – [(x -5)² + (y- 8)²] = 13
∴ (x² – 4x + 4 + y² – 8y + 16) – (x² – 10x + 25 + y² – 16y + 64) =13
∴ x² – 4x+ y² – 8y + 20 – x² + 10x – y² + 16y – 89 = 13
∴ 6x + 8y- 69 = 13
∴ 6x + 8y – 82 = 0
∴ 3x + 4y – 41 = 0
∴ The required equation of locus is 3x + 4y- 41 = 0.

Question 6.
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends right angle at P. (∠APB = 90°)
Solution:
Let P(x, y) be any point on the required locus. Given,
A(l, 6) and B(3, 5),
∠APB = 90°
∴ ΔAPB is a right angled triangle,
By Pythagoras theorem,

AP² + PB² = AB² P (x,y)
∴ [(x – 1)² + (y – 6)²] + [(x – 3)² + (y – 5)²] = (1 – 3)² + (6 -5)²
∴ x² — 2x + 1 + y² — 12y + 36 + x² – 6x + 9 + y² – 10y + 25 = 4 + 1
∴ 2x² + 2y² – 8x – 22y + 66 = 0
∴ x² + y² – 4x – 11y + 33 = 0
∴ The required equation of locus is x² + y² – 4x – 11y + 33 = 0.
[Note: Answer given in the textbook is
‘3x² + 4y² – 4x – 11y + 33 = 0’.
However, as per our calculation it is ‘x² + y² – 4x – 11y + 33 = O

Question 7.
If the origin is shifted to the point 0′(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points
i. A(1, 3) ii. B(2,5)
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + handy = Y + k
x = X + 2 andy = Y + 3 …(i)

i. Given, A(x, y) = A( 1, 3)
x = X + 2 andy = Y + 3 …[From(i)]
∴ 1 = X + 2 and 3 = Y + 3 X = – 1 and Y = 0
∴ The new co-ordinates of point A are (- 1,0).

ii. Given, B(x, y) = B(2, 5)
x = X + 2 and y = Y + 3 …[From(i)]
∴ 2 = X + 2 and 5 = Y + 3
∴ X = 0 and Y = 2
∴ The new co-ordinates of point B are (0, 2).

Question 8.
If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points
i. C(5,4) ii. D(3,3)
Solution:
Origin is shifted to (1,3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + h andy = Y + k
∴ x = X+1 andy = Y + 3 …(i)

i. Given, C(X, Y) = C(5, 4)
x = X +1 andy = Y + 3 …[From(i)]
∴ x = 5 + 1 = 6 andy = 4 + 3 = 7
∴ The old co-ordinates of point C are (6, 7).

ii. Given, D(X, Y) = D(3, 3)
x = X + 1 andy = Y + 3 …[From(i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
∴ The old co-ordinates of point D are (4, 6).

Question 9.
If the co-ordinates A(5, 14) change to B(8, 3) by shift of origin, find the co-ordinates of the point, where the origin is shifted.
Solution:
Let the origin be shifted to (h, k).
Given, A(x, y) = A(5,14), B(X, Y) = B(8, 3)
Since x = X + h andy = Y + k,
5 = 8 + hand 14 = 3 + k ,
∴ h = – 3 and k = 11
The co-ordinates of the point, where the origin is shifted are (- 3, 11).

Question 10.
Obtain the new equations of the following loci if the origin is shifted to the point 0′(2,2), the direction of axes remaining the same:
i. 3x-y + 2 = 0
11. x²+y²-3x = 7
iii. xy – 2x – 2y + 4 = 0
iv. y² – 4x – 4y + 12 = 0
Solution:
Given, (h,k) = (2,2)
Let (X, Y) be the new co-ordinates of the point (x,y).
∴ x = X + handy = Y + k
∴ x = X + 2 andy = Y + 2
i. Substituting the values of x and y in the equation 3x -y + 2 = 0, we get
3(X + 2) – (Y + 2) + 2 = 0
∴ 3X + 6-Y-2 + 2 = 0
∴ 3 X – Y + 6 = 0, which is the new equation of locus.

ii. Substituting the values of x and y in the equation
x² + y² – 3x = 7, we get
(X + 2)² + (Y + 2)² – 3(X + 2) = 7
∴ X² + 4X + 4 + Y² + 4Y + 4 – 3X – 6 = 7
∴ X² + Y² + X + 4Y – 5 = 0, which is the new
equation of locus.

iii; Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get
(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0
∴ XY + 2X + 2Y + 4 – 2X – 4-2Y- 4 + 4 = 0
∴ XY = 0, which is the new equation of locus.

iv. Substituting the values of x and y in the equation y² – 4x – 4y + 12 = 0, we get
(Y + 2)² – 4(X + 2) – 4(Y + 2) + 12 = 0
∴ Y² + 4Y + 4 – 4X – 8 – 4Y -8 + 12 = 0
∴ Y² – 4X = 0, which is the new equation of locus.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.2

Question 1.
Find the derivative of the function y = f (x) using the derivative of the inverse function x = f^-1( y) in the following
(i) y = \(\sqrt {x}\)
Solution:
y = \(\sqrt {x}\) … (1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
y² = x ∴ x = y²

(ii) y = \(\sqrt{2-\sqrt{x}}\)
Solution:
y = \(\sqrt{2-\sqrt{x}}\) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(iii) y = \(\sqrt[3]{x-2}\)
Solution:
y = \(\sqrt[3]{x-2}\) ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(iv) y = log (2x – 1)
Solution:
y = log (2x – 1) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(v) y = 2x + 3
Solution:
y = 2x + 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(vi) y = e^x – 3
Solution:
y = e^x – 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
e^x = y + 3
∴ x = log(y + 3)
∴ x = f^-1(y) = log(y + 3)

(vii) y = e^{2x – 3}
Solution:
y = e^{2x – 3} ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
2x – 3 = log y ∴ 2x = log y + 3

(viii) y = log₂\(\left(\frac{x}{2}\right)\)
Solution:
y = log₂\(\left(\frac{x}{2}\right)\) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
\(\frac{x}{2}\) = 2^y ∴ x = 2∙2^y = 2^y+1
∴ x = f^-1(y) = 2^y+1

Question 2.
Find the derivative of the inverse function of
the following
(i) y = x²·e^x
Solution:
y = x²·e^x
Differentiating w.r.t. x, we get

(ii) y = x cos x
Solution:
y = x cos x
Differentiating w.r.t. x, we get

(iii) y = x·7^x
Solution:
y = x·7^x
Differentiating w.r.t. x, we get

(iv) y = x² + logx
Solution:
y = x² + logx
Differentiating w.r.t. x, we get

(v) y = x logx
Solution:
y = x logx
Differentiating w.r.t. x, we get

Question 3.
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them.
(i) y = x⁵ + 2x³ + 3x, at x = 1
Solution:
y = x⁵ + 2x³ + 3x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x⁵ + 2x³ + 3x)
= 5x⁴ + 2 × 3x² + 3 × 1
= 5x⁴ + 6x² + 3
The derivative of inverse function of y = f(x) is given by

(ii) y = e^x + 3x + 2, at x = 0
Solution:
y = e^x + 3x + 2
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(e^x + 3x + 2)
The derivative of inverse function of y = f(x) is given by

(iii) y = 3x² + 2 log x³, at x = 1
Solution:
y = 3x² + 2 log x³
= 3x² + 6 log x
Differentiating w.r.t. x, we get

The derivative of inverse function of y = f(x) is given by

(iv) y = sin (x – 2) + x², at x = 2
Solution:
y = sin (x – 2) + x²
Differentiating w.r.t. x, we get

Question 4.
If f(x) = x³ + x – 2, find (f^-1)’ (0).
Question is modified.
If f(x) = x³ + x – 2, find (f^-1)’ (-2).
Solution:
f(x) = x³ + x – 2 ….(1)
Differentiating w.r.t. x, we get

Question 5.
Using derivative prove
(i) tan^-1x + cot^-1x = \(\frac{\pi}{2}\)
Solution:
let f(x) = tan^-1x + cot^-1x
Differentiating w.r.t. x, we get

Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k
Let x = 0.
Then f(0) = k ….(2)
From (1), f(0) = tan^-1(0) + cot^-1(0)
= 0 + \(\frac{\pi}{2}=\frac{\pi}{2}\)

(ii) sec^-1x + cosec^-1x = \(\frac{\pi}{2}\) . . . [for |x| ≥ 1]
Solution:
Let f(x) = sec^-1x + cosec^-1x for |x| ≥ 1 ….(1)
Differentiating w.r.t. x, we get

Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k, where |x| > 1
Let x = 2.
Then, f(2) = k ……(2)

Question 6.
Diffrentiate the following w. r. t. x.
(i) tan^-1(log x)
Solution:
Let y = tan^-1(log x)
Differentiating w.r.t. x, we get

(ii) cosec^-1(e^-x)
Solution:
Let y = cosec^-1(e^-x)
Differentiating w.r.t. x, we get

(iii) cot^-1(x³)
Solution:
Let y = cot^-1(x³)
Differentiating w.r.t. x, we get

(iv) cot^-1(4^x
Solution:
Let y = cot^-1(4^x
Differentiating w.r.t. x, we get

(v) tan^-1(\(\sqrt {x}\))
Solution:
Let y = tan^-1(\(\sqrt {x}\))
Differentiating w.r.t. x, we get

(vi) sin^-1\(\left(\sqrt{\frac{1+x^{2}}{2}}\right)\)
Solution:
Let y = sin^-1\(\left(\sqrt{\frac{1+x^{2}}{2}}\right)\)
Differentiating w.r.t. x, we get

(vii) cos^-1(1 – x²)
Solution:
Let y = cos^-1(1 – x²)
Differentiating w.r.t. x, we get

(viii) sin^-1\(\left(x^{\frac{3}{2}}\right)\)
Solution:
Let y = sin^-1\(\left(x^{\frac{3}{2}}\right)\)
Differentiating w.r.t. x, we get

(ix) cos³[cos^-1(x³)]
Solution:
Let y = cos³[cos^-1(x³)]
= [cos(cos^-1x³)]³
= (x³)³ = x⁹
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x⁹) = 9x⁸.

(x) sin⁴[sin^-1(\(\sqrt {x}\))]
Solution:
Let y = sin⁴[sin^-1(\(\sqrt {x}\))]
= {sin[sin^-1(\(\sqrt {x}\))]}⁸
= (\(\sqrt {x}\))⁴ = x²
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x²) = 2x.

Question 7.
Diffrentiate the following w. r. t. x.
(i) cot^-1[cot (e^x2)]
Solution:
Let y = cot^-1[cot (e^x2)] = e^x2

(ii) cosec^-1\(\left(\frac{1}{\cos \left(5^{x}\right)}\right)\)
Solution:

(iii) cos^-1\(\left(\sqrt{\frac{1+\cos x}{2}}\right)\)
Solution:

(iv) cos^-1\(\left(\sqrt{\frac{1-\cos \left(x^{2}\right)}{2}}\right)\)
Solution:

(v) tan^-1\(\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)\)
Solution:

(vi) cosec^-1\(\left(\frac{1}{4 \cos ^{3} 2 x-3 \cos 2 x}\right)\)
Solution:

(vii) tan^-1\(\left(\frac{1+\cos \left(\frac{x}{3}\right)}{\sin \left(\frac{x}{3}\right)}\right)\)
Solution:

(viii) cot^-1\(\left(\frac{\sin 3 x}{1+\cos 3 x}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{\sin 3 x}{1+\cos 3 x}\right)\)

(ix) tan^-1\(\left(\frac{\cos 7 x}{1+\sin 7 x}\right)\)
Solution:

(x) tan^-1\(\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\)
Solution:
Let y = tan^-1\(\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\)

(xi) tan^-1(cosec x + cot x)
Solution:
Let y = tan^-1(cosec x + cot x)

(xii) cot^-1\(\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right)\)
Solution:

Question 8.
(i)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:


= e^x.

(vi)
Solution:

y = sin^-1[sin(2^x)∙cosα – cos(2^x)∙sinα]
= sin^–[sin(2^x – α)]
= 2^x – α, where α is a constant
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(2^x – α)
= \(\frac{d}{d x}\)(2^x) – \(\frac{d}{d x}\)(α)
= 2^x∙log2 – 0
= 2^x∙log2

Question 9.
Diffrentiate the following w. r. t. x.
(i) cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:

(ii) tan^-1\(\left(\frac{2 x}{1-x^{2}}\right)\)
Solution:

(iii) sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:

(iv) sin^-1(2x\(\sqrt{1-x^{2}}\))
Solution:

(v) cos^-1(3x – 4x³)
Solution:

(vi) cos^-1\(\left(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right)\)
Solution:

(vii) cos^-1\(\left(\frac{1-9^{x}}{1+9^{x}}\right)\)
Solution:

(viii) sin^-1\(\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)\)
Solution:

(ix) sin^-1\(\left(\frac{1-25 x^{2}}{1+25 x^{2}}\right)\)
Solution:

(x) sin^-1\(\left(\frac{1-x^{3}}{1+x^{3}}\right)\)
Solution:

(xi) tan^-1\(\left(\frac{2 x^{\frac{5}{2}}}{1-x^{5}}\right)\)

(xii) cot^-1\(\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\)

Question 10.
Diffrentiate the following w. r. t. x.
(i) tan^-1\(\left(\frac{8 x}{1-15 x^{2}}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{8 x}{1-15 x^{2}}\right)\)

(ii) cot^-1\(\left(\frac{1+35 x^{2}}{2 x}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{1+35 x^{2}}{2 x}\right)\)

(iii) tan^-1\(\left(\frac{2 \sqrt{x}}{1+3 x}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{2 \sqrt{x}}{1+3 x}\right)\)

(iv) tan^-1\(\left(\frac{2^{x+2}}{1-3\left(4^{x}\right)}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{2^{x+2}}{1-3\left(4^{x}\right)}\right)\)

(v) tan^-1\(\left(\frac{2^{x}}{1+2^{2 x+1}}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{2^{x}}{1+2^{2 x+1}}\right)\)

(vi) cot^-1\(\left(\frac{a^{2}-6 x^{2}}{5 a x}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{a^{2}-6 x^{2}}{5 a x}\right)\)

(vii) tan^-1\(\left(\frac{a+b \tan x}{b-a \tan x}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{a+b \tan x}{b-a \tan x}\right)\)

(viii) tan^-1\(\left(\frac{5-x}{6 x^{2}-5 x-3}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{5-x}{6 x^{2}-5 x-3}\right)\)
= tan^-1\(\left[\frac{5-x}{1+\left(6 x^{2}-5 x-4\right)}\right]\)

(ix) cot^-1\(\left(\frac{4-x-2 x^{2}}{3 x+2}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{4-x-2 x^{2}}{3 x+2}\right)\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.1

Question 1.
Differentiate the following w.r.t. x :
(i) (x³ – 2x – 1)⁵
Solution:
Method 1:
Let y = (x³ – 2x – 1)⁵
Put u = x³ – 2x – 1. Then y = u⁵

Method 2:
Let y = (x³ – 2x – 1)⁵
Differentiating w.r.t. x, we get

(ii) \(\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}\)
Solution:
Let y = \(\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}\)
Differentiating w.r.t. x, we get

(iii) \(\sqrt{x^{2}+4 x-7}\)
Solution:

(iv) \(\sqrt{x^{2}+\sqrt{x^{2}+1}}\)
Solution:
Let y = \(\sqrt{x^{2}+\sqrt{x^{2}+1}}\)
Differentiating w.r.t. x, we get

(v) \(\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}\)
Solution:
Let y = \(\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}\)
Differentiating w.r.t. x, we get

(vi) \(\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}\)
Solution:
Let y = \(\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}\)
Differentiating w.r.t. x, we get

Question 2.
Diffrentiate the following w.r.t. x
(i) cos(x² + a²)
Solution:
Let y = cos(x² + a²)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)[cos(x² + a²)]
= -sin(x² + a²)∙\(\frac{d}{d x}\)x² + a²)
= -sin(x² + a²)∙(2x + 0)
= -2xsin(x² + a²)

(ii) \(\sqrt{e^{(3 x+2)}+5}\)
Solution:
Let y = \(\sqrt{e^{(3 x+2)}+5}\)
Differentiating w.r.t. x, we get

(iii) log[tan(\(\frac{x}{2}\))]
Solution:
Let y = log[tan(\(\frac{x}{2}\))]
Differentiating w.r.t. x, we get

(iv) \(\sqrt{\tan \sqrt{x}}\)
Solution:
Let y = \(\sqrt{\tan \sqrt{x}}\)
Differentiating w.r.t. x, we get

(v) cot³[log (x³)]
Solution:
Let y = cot³[log (x³)]
Differentiating w.r.t. x, we get

(vi) 5^{sin3x+ 3}
Solution:
Let y = 5^{sin3x+ 3}
Differentiating w.r.t. x, we get

(vii) cosec (\(\sqrt{\cos X}\))
Solution:
Let y = cosec (\(\sqrt{\cos X}\))
Differentiating w.r.t. x, we get

(viii) log[cos (x³ – 5)]
Solution:
Let y = log[cos (x³ – 5)]
Differentiating w.r.t. x, we get

(ix) e^{3 sin2x – 2 cos2x}
Solution:
Let y = e^{3 sin2x – 2 cos2x}
Differentiating w.r.t. x, we get

(x) cos²[log (x²+ 7)]
Solution:
Let y = cos²[log (x²+ 7)]
Differentiating w.r.t. x, we get

(xi) tan[cos (sinx)]
Solution:
Let y = tan[cos (sinx)]
Differentiating w.r.t. x, we get

(xii) sec[tan (x⁴ + 4)]
Solution:
Let y = sec[tan (x⁴ + 4)]
Differentiating w.r.t. x, we get

= sec[tan(x⁴ + 4)]∙tan[tan(x⁴ + 4)]∙sec²(x⁴ + 4)(4x³ + 0)
= 4x³sec²(x⁴ + 4)∙sec[tan(x⁴ + 4)]∙tan[tan(x⁴ + 4)].

(xiii) e^{log[(logx)2 – logx2]}
Solution:
Let y = e^{log[(logx)2 – logx2]}
= (log x)² – log x² …[∵ e^{log x} = x]
Differentiating w.r.t. x, we get

(xiv) sin\(\sqrt{\sin \sqrt{x}}\)
Solution:
Let y = sin\(\sqrt{\sin \sqrt{x}}\)
Differentiating w.r.t. x, we get

(xv) log[sec(e^x2)]
Solution:
Let y = log[sec(e^x2)]
Differentiating w.r.t. x, we get

(xvi) log_e²(logx)
Solution:
Let y = log_e²(logx) = \(\frac{\log (\log x)}{\log e^{2}}\)

(xvii) [log{log(logx)}]²
Solution:
let y = [log{log(logx)}]²
Differentiating w.r.t. x, we get

(xviii) sin²x² – cos²x²
Solution:
Let y = sin²x² – cos²x²
Differentiating w.r.t. x, we get

= 2sinx²∙cosx² × 2x + 2sinx²∙cosx² × 2x
= 4x(2sinx²∙cosx²)
= 4xsin(2x²).

Question 3.
Diffrentiate the following w.r.t. x
(i) (x² + 4x + 1)³ + (x³ – 5x – 2)⁴
Solution:
Let y = (x² + 4x + 1)³ + (x³ – 5x – 2)⁴
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)[(x² + 4x + 1)³ + (x³ – 5x – 2)⁴]
= \(\frac{d}{d x}\) = (x² + 4x + 1)³ + \(\frac{d}{d x}\)(x³ – 5x – 2)⁴
= 3(x² + 4x + 1)²∙\(\frac{d}{d x}\)(x² + 4x + 1) + 4(x³ – 5x – 2)⁴∙\(\frac{d}{d x}\)(x³ – 5x – 2)
= 3(x² + 4x + 1)³∙(2x + 4 × 1 + 0) + 4(x³ – 5x – 2)³∙(3x² – 5 × 1 – 0)
= 6 (x + 2)(x² + 4x + 1)² + 4 (3x² – 5)(x³ – 5x – 2)³.
(ii) (1 + 4x)⁵(3 + x − x²)⁸
Solution:
Let y = (1 + 4x)⁵(3 + x − x²)⁸
Differentiating w.r.t. x, we get

= 8 (1 + 4x)⁵ (3 + x – x²)⁷∙(0 + 1 – 2x) + 5 (1 + 4x)⁴ (3 + x – x²)⁸∙(0 + 4 × 1)
= 8 (1 – 2x)(1 + 4x)⁵(3 + x – x²)⁷ + 20(1 + 4x)⁴(3 + x – x²)⁸.

(iii) \(\frac{x}{\sqrt{7-3 x}}\)
Solution:
Let y = \(\frac{x}{\sqrt{7-3 x}}\)
Differentiating w.r.t. x, we get

(iv) \(\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\)
Solution:
Let y = \(\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\left[\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\right]\)

(v) (1 + sin²x)²(1 + cos²x)³
Solution:
Let y = (1 + sin²x)²(1 + cos²x)³
Differentiating w.r.t. x, we get

= 3(1 + sin²x)² (1 + cos²x)²∙[2cosx(-sinx)] + 2 (1 + sin²x)(1 + cos²x)³∙[2sinx-cosx]
= 3 (1 + sin²x)² (1 + cos²x)² (-sin 2x) + 2(1 + sin²x)(1 + cos²x)³(sin 2x)
= sin2x (1 + sin²x) (1 + cos²x)² [-3(1 + sin²x) + 2(1 + cos²x)]
= sin2x (1 + sin²x)(1 + cos²x)²(-3 – 3sin²x + 2 + 2cos²x)
= sin2x (1 + sin²x)(1 + cos²x)² [-1 – 3 sin²x + 2 (1 – sin²x)]
= sin 2x(1 + sin²x)(1 + cos²x)² (-1 – 3 sin²x + 2 – 2 sin²x)
= sin2x (1 + sin²x)(1 + cos²x)²(1 – 5 sin²x).

(vi) \(\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}\)
Solution:
Let y = \(\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}[\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}]\)

(vii) log(sec 3x+ tan 3x)
Solution:
Let y = log(sec 3x+ tan 3x)
Differentiating w.r.t. x, we get

(viii) \(\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}\)
Solution:

(ix) cot\(\left(\frac{\log x}{2}\right)\) – log\(\left(\frac{\cot x}{2}\right)\)
Solution:
Let y = cot\(\left(\frac{\log x}{2}\right)\) – log\(\left(\frac{\cot x}{2}\right)\)
Differentiating w.r.t. x, we get

(x) \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution:

(xi) \(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\)
Solution:
let y = \(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\)
Differentiating w.r.t. x, we get

(xii) log[tan³x·sin⁴x·(x² + 7)⁷]
Solution:
Let y = log [tan³x·sin⁴x·(x² + 7)⁷]
= log tan³x + log sin⁴x + log (x² + 7)⁷
= 3 log tan x + 4 log sin x + 7 log (x² + 7)
Differentiating w.r.t. x, we get

= 6cosec2x + 4 cotx + \(\frac{14 x}{x^{2}+7}\)

(xiii) log\(\left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)\)
Solution:

(xiv) log\(\left(\sqrt{\left.\frac{1+\cos \left(\frac{5 x}{2}\right)}{1-\cos \left(\frac{5 x}{2}\right)}\right)}\right.\)
Solution:
Using log\(\left(\frac{a}{b}\right)\) = log a – log b
log a^b = b log a


\(-\frac{5}{2}\)cosec\(\left(\frac{5 x}{2}\right)\)

(xv) log\(\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)\)
Solution:

(xvi) log\(\left[4^{2 x}\left(\frac{x^{2}+5}{\sqrt{2 x^{3}-4}}\right)^{\frac{3}{2}}\right]\)
Solution:

(xvii) log\(\left[\frac{e^{x^{2}}(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right]\)
Solution:

(xviii) log\(\left[\frac{a^{\cos x}}{\left(x^{2}-3\right)^{3} \log x}\right]\)
Solution:

(xix) y= (25)^log₅(secx) − (16)^log₄(tanx)
Solution:
y = (25)^log₅(secx) − (16)^log₄(tanx)
= 5^{2log₅(secx)} – 4^{2log₄(tanx)}
= 5^{log₅(sec5x)} – 4^{log₄(tan2x)}
= sec²x – tan²x … [∵ = x]
∴ y = 1
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(1) = 0

(xx) \(\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}\)
Solution:
Let y = \(\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}\)
Differentiating w.r.t. x, we get

Question 4.
A table of values of f, g, f ‘ and g’ is given

(i) If r(x) = f [g(x)] find r’ (2).
Solution:
r(x) = f[g(x)]
∴ r'(x) = \(\frac{d}{d x}\)f[g(x)]
= f'[g(x)]∙\(\frac{d}{d x}\)[g(x)]
= f'[g(x)∙[g'(x)]
∴ r'(2) = f'[g(2)]∙g'(2)
= f'(6)∙g'(2) … [∵ g(x) = 6, when x = 2]
= -4 × 4 … [From the table]
= -16.

(ii) If R(x) = g[3 + f(x)] find R’ (4).
Solution:
R(x) = g[3 + f(x)]
∴ R'(x) = \(\frac{d}{d x}\){g[3+f(x)]}
= g'[3 + f(x)]∙\(\frac{d}{d x}\)[3 + f(x)]
= g'[3 +f(x)]∙[0 + f'(x)]
= g'[3 + f(x)]∙f'(x)
∴ R'(4) = g'[3 + f(4)]∙f'(4)
= g'[3 + 3]∙f'(4) … [∵ f(x) = 3, when x = 4]
= g'(6)∙f'(4)
= 7 × 5 … [From the table]
= 35.

(iii) If s(x) = f[9− f(x)] find s’ (4).
Solution:
s(x) = f[9− f(x)]
∴ s'(x) = \(\frac{d}{d x}\){f[9 – f(x)]}
= f'[9 – f(x)]∙\(\frac{d}{d x}\)[0 – f(x)]
= f'[9 – f(x)]∙[0 – f'(x)]
= -f'[9 – f(x)] – f'(x)
∴ s'(4) = -f'[9 – f(4)] – f'(4)
= -f'[9 – 3] – f'(4) … [∵ f(x) = 3, when x = 4]
= -f'(6) – f'(4)
= -(-4)(5) … [From the table]
= 20.

(iv) If S(x) = g[g(x)] find S’ (6)
Solution:
S(x) = g[g(x)]
∴ S'(x) = \(\frac{d}{d x}\)g[g(x)]
= g'[g(x)]∙\(\frac{d}{d x}\)[g(x)]
= g'[g(x)]∙g'(x)
∴ S ‘(6) = g'[g'(6)]∙g'(6)
= g'(2)∙g'(6) … [∵ g (x) = 2, when x = 6]
= 4 × 7 … [From the table]
= 28.

Question 5.
Assume that f ‘(3) = -1, g'(2) = 5, g(2) = 3 and y = f[g(x)] then \(\left[\frac{d y}{d x}\right]_{x=2}\) = ?
Solution:
y = f[g(x)]
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\){[g(x)]}

Question 6.
If h(x) = \(\sqrt{4 f(x)+3 g(x)}\), f(1) = 4, g(1) = 3, f ‘(1) = 3, g'(1) = 4 find h'(1).
Solution:
Given f(1) = 4, g(1) = 3, f ‘(1) = 3, g'(1) = 4 …..(1)

Question 7.
Find the x co-ordinates of all the points on the curve y = sin 2x – 2 sin x, 0 ≤ x < 2π where \(\frac{d y}{d x}\) = 0.
Solution:
y = sin 2x – 2 sin x, 0 ≤ x < 2π

= cos2x × 2 – 2cosx
= 2 (2 cos²x – 1) – 2 cosx
= 4 cos²x – 2 – 2 cos x
= 4 cos²x – 2 cos x – 2
If \(\frac{d y}{d x}\) = 0, then 4 cos²x – 2 cos x – 2 = 0
∴ 4cos²x – 4cosx + 2cosx – 2 = 0
∴ 4 cosx (cosx – 1) + 2 (cosx – 1) = 0
∴ (cosx – 1)(4cosx + 2) = 0
∴ cosx – 1 = 0 or 4cosx + 2 = 0
∴ cos x = 1 or cos x = \(-\frac{1}{2}\)
∴ cos x = cos 0

Question 8.
Select the appropriate hint from the hint basket and fill up the blank spaces in the following paragraph. [Activity]
“Let f (x) = x² + 5 and g(x) = e^x + 3 then
f [g(x)] = _ _ _ _ _ _ _ _ and g [f(x)] =_ _ _ _ _ _ _ _.
Now f ‘(x) = _ _ _ _ _ _ _ _ and g'(x) = _ _ _ _ _ _ _ _.
The derivative off [g (x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ _ _.
Therefore \(\frac{d}{d x}\)[f[g(x)]] = _ _ _ _ _ _ _ _ _ and [\(\frac{d}{d x}\)[f[g(x)]]]_{x = 0} = _ _ _ _ _ _ _ _ _ _ _.
The derivative of g[f(x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ __ _ _ _ _.
Therefore \(\frac{d}{d x}\)[g[f(x)]] = _ _ _ _ _ _ _ _ _ and [\(\frac{d}{d x}\)[g[f(x)]]]_{x = 1} = _ _ _ _ _ _ _ _ _ _ _.”
Hint basket : { f ‘[g(x)]·g'(x), 2e^2x + 6e^x, 8, g'[f(x)]·f ‘(x), 2xe^{x2 + 5}, -2e⁶, e^2x + 6e^x + 14, e^{x2 + 5} + 3, 2x, e^x}
Solution:
f[g(x)] = e^2x + 6e^x + 14
g[f(x)] = e^{x2 + 5} + 3
f'(x) = 2x, g’f(x) = e^x
The derivative of f[g(x)] w.r.t. x in terms of and g is f'[g(x)]∙g'(x).
∴ \(\frac{d}{d x}\){f[g(x)]} = 2e^2x + 6e^x and \(\frac{d}{d x}\){f[g(x)]}_{x = 0} = 8
The derivative of g[f(x)] w.r.t. x in terms of f and g is g’f(x)]∙f'(x).
∴ \(\frac{d}{d x}\){g[(f(x)]} = 2xe^{x2 + 5} and
\(\frac{d}{d x}\){g[(f(x)]}_{x = -1} = -2e⁶.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Miscellaneous Exercise 7

I) Select the appropriate alternatives for each of the following :
Question 1.
The value of objective function is maximum under linear constraints _______.
(A) at the centre of feasible region
(B) at (0, 0)
(C) at a vertex of feasible region
(D) the vertex which is of maximum distance from (0, 0)
Solution:
(C) at a vertex of feasible region

Question 2.
Which of the following is correct _______.
(A) every L.P.P. has an optimal solution
(B) a L.P.P. has unique optimal solution
(C) if L.P.P. has two optimal solutions then it has infinite number of optimal solutions
(D) the set of all feasible solution of L.P.P. may not be convex set
Solution:
(C) if L.P.P. has two optimal solutions then it has infinite number of optimal solutions

Question 3.
Objective function of L.P.P. is _______.
(A) a constraint
(B) a function to be maximized or minimized
(C) a relation between the decision variables
(D) equation of a straight line
Solution:
(B) a function to be maximized or minimized

Question 4.
The maximum value of z = 5x + 3y subjected to the constraints 3x + 5y ≤ 15, 5x + 2y ≤ 10, x, y≥ 0 is _______.
(A) 235
(B) \(\frac{235}{9}\)
(C) \(\frac{235}{19}\)
(D) \(\frac{235}{3}\)
Solution:
(C) \(\frac{235}{19}\)

Question 5.
The maximum value of z = 10x + 6y subjected to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y≥ 0. _______.
(A) 56
(B) 65
(C) 55
(D) 66
Solution:
(A) 56

Question 6.
The point at which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is obtained at _______.
(A) (30, 25)
(B) (20, 35)
(C) (35, 20)
(D) (40, 15)
Solution:
(D) (40, 15)

Question 7.
Of all the points of the feasible region, the optimal value ofz obtained at the point lies _______.
(A) inside the feasible region
(B) at the boundary of the feasible region
(C) at vertex of feasible region
(D) outside the feasible region
Solution:
(C) at vertex of feasible region

Question 8.
Feasible region is the set of points which satisfy _______.
(A) the objective function
(B) all of the given constraints
(C) some of the given constraints
(D) only one constraint
Solution:
(B) all of the given constraints

Question 9.
Solution of L.P.P. to minimize z = 2x + 3y such that x ≥ 0, y ≥ 0, 1 ≤ x + 2y ≤ 10 is _______.
(A) x = 0, y = \(\frac{1}{2}\)
(B) x = \(\frac{1}{2}\), y = 0
(C) x = 1, y = 2
(D) x = \(\frac{1}{2}\), y = \(\frac{1}{2}\)
Solution:
(A) x = 0, y = \(\frac{1}{2}\)

Question 10.
The corner points of the feasible solution given by the inequation x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0 are _______.
(A) (0, 0), (4, 0), (7, 1), (0, 4)
(B) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 4)
(C) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 7)
(D) (0, 0), (4, 0), (3, 1), (0, 7)
Solution:
(B) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 4)

Question 11.
The corner points of the feasible solution are (0, 0), (2, 0), (\(\frac{12}{7}\), \(\frac{3}{7}\)), (0, 1). Then z = 7x + y is maximum at _______.
(A) (0, 0)
(B) (2, 0)
(C) (\(\frac{12}{7}\), \(\frac{3}{7}\))
(D) (0, 1)
Solution:
(B) (2, 0)

Question 12.
If the corner points of the feasible solution are (0, 0), (3, 0), (2, 1) and (0, \(\frac{7}{3}\) ), the maximum value of z = 4x + 5y is _______.
(A) 12
(B) 13
(C) 35
(D) 0
Solution:
(B) 13

Question 13.
If the corner points of the feasible solution are (0, 10), (2, 2) and (4, 0) then the point of minimum z = 3x + 2y is _______.
(A) (2, 2)
(B) (0, 10)
(C) (4, 0)
(D) (3 ,4)
Solution:
(A) (2, 2)

Question 14.
The half plane represented by 3x + 2y < 8 contains the point _______.
(A) (1, \(\frac{5}{2}\))
(B) (2, 1)
(C) (0, 0)
(D) (5, 1)
Solution:
(C) (0, 0)

Question 15.
The half plane represented by 4x + 3y > 14 contains the point _______.
(A) (0, 0)
(B) (2, 2)
(C) (3, 4)
(D) (1, 1)
Solution:
(C) (3, 4)

II) Solve the following :
Question 1.
Solve each of the following inequations graphically using X Y plane.
(i) 4x – 18 ≥ 0
Solution:
Consider the line whose equation is 4x – 18 ≥ 0 i.e. x = \(\frac{18}{4}=\frac{9}{2}\) = 4.5
This represents a line parallel to Y-axis passing3through the point (4.5, 0)
Draw the line x = 4.5
To find the solution set we have to check the position of the origin (0, 0).
When x = 0, 4x – 18 = 4 × 0 – 18 = -18 > 0
∴ the coordinates of the origin does not satisfy thegiven inequality.
∴ the solution set consists of the line x = 4.5 and the non-origin side of the line which is shaded in the graph.

(ii) -11x – 55 ≤ 0
Solution:
Consider the line whose equation is -11x – 55 ≤ 0 i.e. x = -5
This represents a line parallel to Y-axis passing3through the point (-5, 0)
Draw the line x = – 5
To find the solution set we have to check the position of the origin (0, 0).
When x = 0, -11x – 55 = – 11(0) – 55 = -55 > 0
∴ the coordinates of the origin does not satisfy thegiven inequality.
∴ the solution set consists of the line x = -5 and the non-origin side of the line which is shaded in the graph.

(iii) 5y – 12 ≥ 0
Solution:
Consider the line whose equation is 5y – 12 ≥ 0 i.e. y = \(\frac{12}{5}\)
This represents a line parallel to X-axis passing through the point (o, \(\frac{12}{5}\))
Draw the line y = \(\frac{12}{5}\)
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 5y – 12 = 5(0) – 12 = -12 > 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{12}{5}\) and the non-origin side of the line which is shaded in the graph.

(iv) y ≤ -3.5
Solution:
Consider the line whose equation is y ≤ – 3.5 i.e. y = – 3.5
This represents a line parallel to X-axis passing3through the point (0, -3.5)
Draw the line y = – 3.5
To find the solution set, we have to check the position of the origin (0, 0).
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = – 3.5 and the non-origin side of the line which is shaded in the graph.

Question 2.
Sketch the graph of each of following inequations in XOY co-ordinate system.
(i) x ≥ 5y
Solution:

(ii) x + y≤ 0
Solution:

(iii) 2y – 5x ≥ 0
Solution:

(iv) |x + 5| ≤ y
Solution:
|x + 5| ≤ y
∴ -y ≤ x + 5 ≤ y
∴ -y ≤ x + 5 and x + 5 ≤ y
∴ x + y ≥ -5 and x – y ≤ -5
First we draw the lines AB and AC whose equations are
x + y= -5 and x – y = -5 respectively.

The graph of |x + 5| ≤ y is as below:

Question 3.
Find graphical solution for each of the following system of linear inequation.
(i) 2x + y ≥ 2, x – y ≤ 1
Solution:
First we draw the lines AB and AC whose equations are 2x + y = 2 and x – y = 1 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(ii) x + 2y ≥ 4, 2x – y ≤ 6
Solution:

(iii) 3x + 4y ≤ 12, x – 2y ≥ 2, y ≥ -1
Solution:
First we draw the lines AB, CD and ED whose equations are 3x + 4y = 12, x – 2y = 2 and y = -1 respectively.


The solution set of given system of inequation is shaded in the graph.

Question 4.
Find feasible solution for each of the following system of linear inequations graphically.
(i) 2x + 3y ≤ 12, 2x + y ≤ 8, x ≥ 0, y ≥ 0
Solution:

The feasible solution is OCPBO.

(ii) 3x + 4y ≥ 12, 4x + 7y ≤ 28, x ≥ 0, y ≥ 0
Solution:

The feasible solution is ACDBA.

Question 5.
Solve each of the following L.P.P.
(i) Maximize z = 5x₁ + 6x₂ subject to 2x₁ + 3x₂ ≤ 18, 2x₁ + x₂ ≤ 12, x₁ ≥ 0, x₂ ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 2x₁ + 3x₂ = 18 and 2x₁ + x₂ = 12 respectively.


The feasible region is OCPBO which is shaded in the graph. The vertices of the feasible region are O(0, 0), C(6, 0), P and B (0,6).
P is the point of intersection of the lines
2x₁ + 3x₂ = 18 ….(1)
and 2x₁ + x₂ = 12
On subtracting, we get
2x₂ = 6 ∴ x₂ = 3
Substituting x₂ = 3 in (2), we get
2x₁ + 3 = 12 ∴ x₂ = 9
∴ P is (\(\frac{9}{2}\), 3)
The values of objective function z = 5x₁ + 6x₂ at these vertices are
z(O) = 5(0) + 6(0) = 0 + 0 = 0
z(C) = 5(6) + 6(0) = 30 + 0 = 30
z(P) = 5(\(\frac{9}{2}\)) + 6(3) = \(\frac{45}{2}\) + 18 = \(\frac{45+36}{2}=\frac{81}{2}\) = 40.5
z(B) = 5(0) + 6(3) = 0 + 18 = 18
Maximum value of z is 40.5 when x₁ = 9/2, y = 3.

(ii) Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21
Question is modified.
Maximize z = 4x + 2y subject to 3x + y ≤ 27, x + y ≤ 21, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + y = 27 and x + y = 21 respectively.


The feasible region is OAPDO which is shaded region in the graph. The vertices of the feasible region are 0(0, 0), A (9, 0), P and D(0, 21). P is the point of intersection of lines
3x + y = 27 … (1)
and x + y = 21 … (2)
On substracting, we get 2x = 6 ∴ x = 3
Substituting x = 3 in equation (1), we get
9 + y = 27 ∴ y = 18
∴ P = (3, 18)
The values of the objective function z = 4x + 2y at these vertices are
z(O) = 4(0) + 2(0) = 0 + 0 = 0
z(a) = 4(9) + 2(0) = 36 + 0 = 36
z(P) = 4(3) + 2(18) = 12 + 36 = 48
z (D) = 4(0) + 2(21) = 0 + 42 = 42
∴ 2 has minimum value 48 when x = 3, y = 18.

(iii) Maximize z = 6x + 10y subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 10 and 5x + 3y = 15 respectively.


The feasible region is OCPBD which is shaded in the graph.
The vertices of the feasible region are 0(0, 0), C(3, 0), P and B (0, 2).
P is the point of intersection of the lines
3x + 5y = 10 … (1)
and 5x + 3y = 15 … (2)
Multiplying equation (1) by 5 and equation (2) by 3, we get
15x + 25y = 50
15x + 9y = 45
On subtracting, we get
16y = 5 ∴ y = \(\frac{5}{16}\)
Substituting y = \(\frac{5}{16}\) in equation (1), we get
3x + \(\frac{25}{16}\) = 10 ∴ 3x = 10 – \(\frac{25}{16}=\frac{135}{16}\)
∴ x = \(\frac{45}{16}\) ∴ P ≡ \(\left(\frac{45}{16}, \frac{5}{16}\right)\)
The values of objective function z = 6x + 10y at these vertices are
z(O) = 6(0) + 10(0) = 0 + 0 = 0
z(C) = 6(3) + 10(0) = 18 + 0 = 18
z(P) = 6\(\left(\frac{45}{16}\right)\) + 10\(\left(\frac{5}{10}\right)\) = \(\frac{270}{16}+\frac{50}{16}=\frac{320}{16}\) = 20
z(B) = 6(0) + 10(2) = 0 + 20 = 20
The maximum value of z is 20 at P\(\left(\frac{45}{16}, \frac{5}{16}\right)\) and B (0, 2) two consecutive vertices.
∴ z has maximum value 20 at each point of line segment PB where B is (0, 2) and P is \(\left(\frac{45}{16}, \frac{5}{16}\right)\).
Hence, there are infinite number of optimum solutions.

(iv) Maximize z = 2x + 3y subject to x – y ≥ 3, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB whose equation is x – y = 3.


The feasible region is shaded which is unbounded.
Therefore, the value of objective function can be in- j creased indefinitely. Hence, this LPP has unbounded solution.

Question 6.
Solve each of the following L.P.P.
(i) Maximize z = 4x₁ + 3x₂ subject to 3x₁ + x₂ ≤ 15, 3x₁ + 4x₂ ≤ 24, x₁ ≥ 0, x₂ ≥ 0
Solution:
We first draw the lines AB and CD whose equations are 3x₁ + x₂ = 15 and 3x₁ + 4x₂ = 24 respectively.


The feasible region is OAPDO which is shaded in the graph.
The Vertices of the feasible region are 0(0, 0), A(5, 0), P and D(0, 6).
P is the point of intersection of lines.
3x₁ + 4x₂ = 24 … (1)
and 3x₁ + x₂ = 15 … (2)
On subtracting, we get
3x₂ = 9 ∴ x₂ = 3
Substituting x₂ = 3 in (2), we get
3x₁ + 3 = 15
∴ 3x₁ = 12 ∴ x₁ = 4 ∴ P is (4, 3)
The values of objective function z = 4x₁ + 3x₂ at these vertices are
z(O) = 4(0) + 3(0) = 0 + 0 = 0
z(a) = 4(5) + 3(0) = 20 + 0 = 20
z(P) = 4(4) + 3(3) = 16 + 9 = 25
z(D) = 4(0) + 3(6) = 0 + 18 = 18
∴ z has maximum value 25 when x = 4 and y = 3.

(ii) Maximize z = 60x + 50y subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Solution:
We first draw the lines AB and CD whose equations are x + 2y = 40 and 3x + 2y = 60 respectively.


The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (20, 0), P and B (0, 20).
P is the point of intersection of the lines.
3x + 2y = 60 … (1)
and x + 2y = 40 … (2)
On subtracting, we get
2x = 20 ∴ x = 10
Substituting x = 10 in (2), we get
10 + 2y = 40
∴ 2y = 30 ∴ y = 15 ∴ P is (10, 15)
The values of the objective function z = 60x + 50y at these vertices are
z(O) = 60(0) + 50(0) = 0 + 0 = 0
z(C) = 60(20) + 50(0) = 1200 + 0 = 1200
z(P) = 60(10) + 50(15) = 600 + 750 = 1350
z(B) = 60(0) + 50(20) = 0 + 1000 = 1000 .
∴ z has maximum value 1350 at x = 10, y = 15.

(iii) Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30; x ≥ 0, y ≥ 0
Solution:
We first draw the lines AB, CD and EF whose equations are 3x + y = 27, x + y = 21, x + 2y = 30 respectively.


The feasible region is XEPQBY which is shaded in the graph.
The vertices of the feasible region are E (30,0), P, Q and B (0,27).
P is the point of intersection of the lines
x + 2y = 30 … (1)
and x + y = 21 … (2)
On subtracting, we get
y = 9
Substituting y = 9 in (2), we get
x + 9 = 21 ∴ x = 12
∴ P is (12, 9)
Q is the point of intersection of the lines
x + y = 21 … (2)
and 3x + y = 27 … (3)
On subtracting, we get
2x = 6 ∴ x = 3
Substituting x = 3 in (2), we get
3 + y = 21 ∴ y = 18
∴ Q is (3, 18).
The values of the objective function z = 4x + 2y at these vertices are
z(E) = 4(30) + 2(0) = 120 + 0 = 120
z(P) = 4(12) + 2(9) = 48 + 18 = 66
z(Q) = 4(3) + 2(18) = 12 + 36 = 48
z(B) = 4(0) + 2(27) = 0 + 54 = 54
∴ z has minimum value 48, when x = 3 and y = 18.

Question 7.
A carpenter makes chairs and tables. Profits are ₹140/- per chair and ₹ 210/- per table. Both products are processed on three machines : Assembling, Finishing and Polishing. The time required for each product in hours and availability of each machine is given by following table:

Formulate the above problem as L.P.P. Solve it graphically to get maximum profit.
Solution:
Let the number of chairs and tables made by the carpenter be x and y respectively.
The profits are ₹ 140 per chair and ₹ 210 per table.
∴ total profit z = ₹ (140x + 210y) This is the objective function which is to be maximized.
The constraints are as per the following table :

From the table, the constraints are
3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60.
The number of chairs and tables cannot be negative.
∴ x ≥ 0, y ≥ 0
Hence, the mathematical formulation of given LPP is :
Maximize z = 140x + 210y, subject to
3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.
We first draw the lines AB, CD and EF whose equations are 3x + 3y = 36, 5x + 2y = 50 and 2x + 6y = 60 respectively.


The feasible region is OCPQFO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (10, 0), P, Q and F (0, 10).
P is the point of intersection of the lines
5x + 2y = 50 … (1)
and 3x + 3y = 36 … (2)
Multiplying equation (1) by 3 and equation (2) by 2, we get
15x + 6y = 150
6x + 6y = 72
On subtracting, we get 26
9x = 78 ∴ x = \(\frac{26}{3}\)
Substituting x = \(\frac{26}{3}\) in (2), we get
3\(\left(\frac{26}{3}\right)\) + 3y = 36
3y = 1o y = \(\frac{10}{3}\)
Q is the point of intersection of the lines
3x + 3y = 36 … (2)
and 2x + 6y = 60 … (3)
Multiplying equation (2) by 2, we get
6x + 6 y = 72
Subtracting equation (3) from this equation, we get
4x = 12 ∴ x = 3
Substituting x = 3 in (2), we get
3(3) + 3y = 36
∴ 3y = 27 ∴ y = 9
∴ Q is (3, 9).
Hence, the vertices of the feasible region are O (0, 0),
C(10, 0), P\(\left(\frac{26}{3}, \frac{10}{3}\right)\), Q(3, 9) and F(0, 10).
The values of the objective function z = 140x + 210y at these vertices are
z(O) = 140(0) + 210(0) = 0 + 0 = 0
z(C) = 140 (10) + 210(0) = 1400 + 0 = 1400
z(P) = 140\(\left(\frac{26}{3}\right)\) + 210\(\left(\frac{10}{3}\right)\) = \(\frac{3640+2100}{3}=\frac{5740}{3}\) = 1913.33
z(Q) = 140(3) + 210(9) = 420 + 1890 = 2310
z(F) = 140(0) + 210(10) = 0 + 2100 = 2100
∴ z has maximum value 2310 when x = 3 and y = 9 Hence, the carpenter should make 3 chairs and 9 tables to get the maximum profit of ₹ 2310.

Question 8.
A company manufactures bicycles and tricycles, each of which must be processed through two machines A and B. Maximum availability of Machine A and B is respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on Machine A and 3 hours on Machine B. Manufacturing a tricycles requires 4 hours on Machine A and 10 hours on Machine B. If profits are ₹180/- for a bicycle and ₹220/- for a tricycle. Determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.
Solution:
Let x bicycles and y tricycles are to be manu¬factured. Then the total profit is z = ₹ (180x + 220y)
This is a linear function which is to be maximized. Hence, it is the objective function. The constraints are as per the following table :

From the table, the constraints are
6x + 4y ≤ 120, 3x +10y ≤ 180
Also, the number of bicycles and tricycles cannot be i negative.
∴ x ≥ 0, y ≥ 0.
Hence, the mathematical formulation of given LPP is :
Maximize z = 180x + 220y, subject to
6x + 4y ≤ 120, 3x + 10y ≤ 180, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are 6x + 4y = 120 and 3x + 10y = 180 respectively.


The feasible region is OAPDO which is shaded in the graph.
The vertices of the feasible region are O(0, 0), A(20, 0) P and D(0, 18).
P is the point of intersection of the lines
3x + 10y = 180 … (1)
and 6x + 4y = 120 … (2)
Multiplying equation (1) by 2, we get
6x + 20y = 360
Subtracting equation (2) from this equation, we get
16y = 240 ∴ y = 15
∴ from (1), 3x + 10(15) = 180
∴ 3x = 30 ∴ x = 10
∴ P = (10, 15)
The values of the objective function z = 180x + 220y at these vertices are
z(O) = 180(0) + 220(0) = 0 + 0 = 0
z(a) = 180(20) + 220(0) = 3600 + 0 = 3600
z(P) = 180(10) + 220(15) = 1800 + 3300 = 5100
z(D) = 180(0) +220(18) = 3960
∴ the maximum value of z is 5100 at the point (10, 15).
Hence, 10 bicycles and 15 tricycles should be manufactured in order to have the maximum profit of ₹ 5100.

Question 9.
A factory produced two types of chemicals A and B. The following table gives the units of ingredients P and Q (per kg) of chemicals A and B as well as minimum requirements of P and Q and also cost per kg. chemicals A and B :

Find the number of units of chemicals A and B should be produced so as to minimize the cost.
Solution:
Let the factory produce x units of chemical A and y units of chemical B. Then the total cost is z = ₹ (4x + 6y). This is the objective function which is to be minimized.
From the given table, the constraints are
x + 2y ≥ 80, 3x + y ≥ 75.
Also, the number of units x and y of chemicals A and B cannot be negative.
∴ x ≥ 0, y ≥ 0.
∴ the mathematical formulation of given LPP is
Minimize z = 4x + 6y, subject to
x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are x + 2y = 80 and 3x + y = 75 respectively.


The feasible region is shaded in the graph.
The vertices of the feasible region are A (80, 0), P and D (0, 75).
P is the point of intersection of the lines
x + 2y = 80 … (1)
and 3x + y = 75 … (2)
Multiplying equation (2) by 2, we get
6x + 2 y = 150
Subtracting equation (1) from this equation, we get
5x = 70 ∴ x = 14
∴ from (2), 3(14) + y = 75
∴ 42 + y = 75 ∴ y = 33
∴ P = (14, 33)
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(80)+ 6(0) =320 + 0 = 320
z(P) = 4(14)+ 6(33) = 56+ 198 = 254
z(D) = 4(0) + 6(75) = 0 + 450 = 450
∴ the minimum value of z is 254 at the point (14, 33).
Hence, 14 units of chemical A and 33 units of chemical B are to be produced in order to have the j minimum cost of ₹ 254.

Question 10.
A company produces mixers and food processors. Profit on selling one mixer and one food processor is ₹ 2,000/- and ₹ 3,000/- respectively. Both the products are processed through three Machines A, B, C. The time required in hours by each product and total time available in hours per week on each machine are as follows :

How many mixers and food processors should be produced to maximize the profit?
Solution:
Let the company produce x mixers and y food processors.
Then the total profit is z = ₹ (2000x + 3000y)
This is the objective function which is to be maximized. From the given table in the problem, the constraints are 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60
Also, the number of mixers and food processors cannot be negative,
∴ x ≥ 0, y ≥ 0.
∴ the mathematical formulation of given LPP is
Maximize z = 2000x + 3000y, subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤60, x ≥ 0, y ≥ 0.
First we draw the lines AB, CD and EF whose equations are 3x + 3y = 36, 5x + 2y = 50 and 2x + 6y = 60 respectively.


The feasible region is OCPQFO which is shaded in the graph.
The vertices of the feasible region are O(0, 0), C(10, 0), P, Q and F(0,10).
P is the point of intersection of the lines
3x + 3y = 36 … (1)
and 5x + 2y = 50 … (2)
Multiplying equation (1) by 2 and equation (2) by 3, we get
6x + 6y = 72
15x + 6y = 150
On subtracting, we get
9x = 78
∴ x = \(\frac{26}{3}\)
∴ from (1), 3\(\left(\frac{26}{3}\right)\) + 3y = 36
∴ 3y = 10
∴ y = \(\frac{10}{3}\)
∴ P = \(\left(\frac{26}{3}, \frac{10}{3}\right)\)
Q is the point of intersection of the lines
3x + 3y = 36 … (1)
and 2x + 6y = 60 … (3)
Multiplying equation (1) by 2, we get
6x + 6y = 72
Subtracting equation (3), from this equation, we get
4x = 12
∴ x = 3
∴ from (1), 3(3) + 3y = 36
∴ 3y = 27
∴ y = 9
∴ Q = (3, 9)
The values of the objective function z = 2000x + 3000y at these vertices are
z(O) = 2000(0) + 3000(0) = 0 + 0 = 0
z(C) = 2000(10) + 3000(0) = 20000 + 0 = 20000
z(P) = 2000\(\left(\frac{26}{3}\right)\) + 3000\(\left(\frac{10}{3}\right)\) = \(\frac{52000}{3}+\frac{30000}{3}=\frac{82000}{3}\)
z(Q) = 2000(3) + 3000(9) = 6000 + 27000 = 33000
z(F) = 2000(0) + 3000(10) = 30000 + 0 = 30000
∴ the maximum value of z is 33000 at the point (3, 9).
Hence, 3 mixers and 9 food processors should be produced in order to get the maximum profit of ₹ 33,000.

Question 11.
A chemical company produces a chemical containing three basic elements A, B, C so that it has at least 16 liters of A, 24 liters of B and 18 liters of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 liters of A, 12 liters of B, 2 liters of C. Each unit of compound II has 2 liters of A, 2 liters of B and 6 liters of C. The cost per unit of compound I is ₹ 800/- and that of compound II is ₹ 640/-. Formulate the problem as L.P.P. and solve it to minimize the cost.
Solution:
Let the company buy x units of compound I and y units of compound II.
Then the total cost is z = ₹(800x + 640y).
This is the objective function which is to be minimized.
The constraints are as per the following table :

From the table, the constraints are
4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18.
Also, the number of units of compound I and compound II cannot be negative.
∴ x ≥ 0, y ≥ 0.
∴ the mathematical formulation of given LPP is
Minimize z = 800x + 640y, subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x ≥ 0, y ≥ 0.
First we draw the lines AB, CD and EF whose equations are 4x + 2y = 16, 12x + 2y = 24 and 2x + 6y = 18


The feasible region is shaded in the graph.
The vertices of the feasible region are E(9, 0), P, Q, and D(0, 12).
P is the point of intersection of the lines
2x + 6y = 18 … (1)
and 4x + 2y = 16 … (2)
Multiplying equation (1) by 2, we get 4x + 12y = 36
Subtracting equation (2) from this equation, we get
10y = 20
∴ y = 2
∴ from (1), 2x + 6(2) = 18
∴ 2x = 6
∴ x = 3
∴ P = (3, 2)
Q is the point of intersection of the lines
12x + 2y = 24 … (3)
and 4x + 2y = 16 … (2)
On subtracting, we get
8x = 8 ∴ x = 1
∴ from (2), 4(1) + 2y = 16
∴ 2y = 12 ∴ y = 6
∴ Q = (1, 6)
The values of the objective function z = 800x + 640y at these vertices are
z(E) = 800(9)+ 640(0) =7200 + 0 = 7200
z(P) = 800(3) + 640(2) = 2400 + 1280 = 3680
z(Q) = 800(1) + 640(6) =800 + 3840 =4640
z(D) = 800(0) + 640(12) = 0 + 7680 = 7680
∴ the minimum value of z is 3680 at the point (3, 2).
Hence, the company should buy 3 units of compound I and 2 units of compound II to have the minimum cost of ₹ 3680.

Question 12.
A person makes two types of gift items A and B requires the services of a cutter and a finisher. Gift item A requires 4 hours of cutter’s time and 2 hours of finisher’s time. B requires 2 hours of cutter’s time and 4 hours of finisher’s time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75/- and on gift item B is ₹ 125/-. Assuming that the person can sell all the gift items produced, determine how many gift items of each type should he make every month to obtain the best returns?
Solution:
Let x: number of gift item A
y: number of gift item B
As numbers of the items are never negative
x ≥ 0; y ≥ 0

Total time required for the cutter = 4x + 2y
Maximum available time 208 hours
∴ 4x+ 2y ≤ 208
Total time required for the finisher 2x +4y
Maximum available time 152 hours
2x + 4y ≤ 152
Total Profit is 75x + 125y
∴ L.P.P. of the above problem is
Minimize z = 75x + 125y
Subject to 4x+ 2y ≤ 208
2x + 4y ≤ 152
x ≥ 0; y ≥ 0
Graphical solution

Corner points
Now, Z at
x = (75x + 125y)
O(0, 0) = 75 × 0 + 125 × 0 = 0
A(52,0) = 75 × 52 + 125 × 0 = 3900
B(44, 16) = 75 × 44 + 125 × 16 = 5300
C(0, 38) = 75 × 0 + 125 × 38 = 4750
A person should make 44 items of type A and 16 Uems of type Band his returns are ₹ 5,300.

Question 13.
A firm manufactures two products A and B on which profit earned per unit ₹3/- and ₹4/- respectively. Each product is processed on two machines M₁ and M₂. The product A requires one minute of processing time on M₁ and two minute of processing time on M₂, B requires one minute of processing time on M₁ and one minute of processing time on M₂. Machine M₁ is available for use for 450 minutes while M₂ is available for 600 minutes during any working day. Find the number of units of product A and B to be manufactured to get the maximum profit.
Solution:
Let the firm manufactures x units of product
A and y units of product B.
The profit earned per unit of A is ₹3 and B is ₹ 4.
Hence, the total profit is z = ₹ (3x + 4y).
This is the linear function which is to be maximized.
Hence, it is the objective function.
The constraints are as per the following table :

From the table, the constraints are
x + y ≤ 450, 2x + y ≤ 600
Since, the number of gift items cannot be negative, x ≥ 0, y ≥ o.
∴ the mathematical formulation of LPP is,
Maximize z = 3x + 4y, subject to x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0.
Now, we draw the lines AB and CD whose equations are x + y = 450 and 2x + y — 600 respectively.


The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O(0, 0), C(300, 0), P and B (0, 450).
P is the point of intersection of the lines
2x + y = 600 … (1)
and x + y = 450 … (2)
On subtracting, we get
∴ x = 150
Substituting x = 150 in equation (2), we get
150 + y = 450
∴ y = 300
∴ P = (150, 300)
The values of the objective function z = 3x + 4y at these vertices are
z(O) = 3(0) + 4(0) = 0 + 0 = 0
z(C) = 3(300) + 4(0) = 900 + 0 = 900
z(P) = 3(150) + 4(300) = 450 + 1200 = 1650
z(B) = 3(0) + 4(450) = 0 + 1800 = 1800
∴ z has the maximum value 1800 when x = 0 and y = 450 Hence, the firm gets maximum profit of ₹ 1800 if it manufactures 450 units of product B and no unit product A.

Question 14.
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20/- per unit of A and ₹ 30/- per unit of B. Both A and B make use of two essential components a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each units of B requires 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should the manufacture per month to maximize profit? How much is the maximum profit?
Solution:
Let the firm manufactures x units of item A and y units of item B.
Firm can make profit of ₹ 20 per unit of A and ₹ 30 per unit of B.
Hence, the total profit is z = ₹ (20x + 30y).
This is the objective function which is to be maximized. The constraints are as per the following table :

From the table, the constraints are
3x + 2y ≤ 210, 2x + 4y ≤ 300
Since, number of items cannot be negative, x ≥ 0, y ≥ 0.
Hence, the mathematical formulation of given LPP is :
Maximize z = 20x + 30y, subject to 3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0.
We draw the lines AB and CD whose equations are 3x + 2y = 210 and 2x + 4y = 300 respectively.

The feasible region is OAPDO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), A (70, 0), P and D (0, 75).
P is the point of intersection of the lines
2x + 4y = 300 … (1)
and 3x + 2y = 210 … (2)
Multiplying equation (2) by 2, we get
6x + 4y = 420
Subtracting equation (1) from this equation, we get
∴ 4x = 120 ∴ x = 30
Substituting x = 30 in (1), we get
2(30) + 4y = 300
∴ 4y = 240 ∴ y = 60
∴ P is (30, 60)
The values of the objective function z = 20x + 30y at these vertices are
z(O) = 20(0) + 30(0) = 0 + 0 = 0
z(A) = 20(70) + 30(0) = 1400 + 0 = 1400
z(P) = 20(30) + 30(60) = 600 + 1800 = 2400
z(D) = 20(0) + 30(75) = 0 + 2250 = 2250
∴ z has the maximum value 2400 when x = 30 and y = 60. Hence, the firm should manufactured 30 units of item A and 60 units of item B to get the maximum profit of ₹ 2400.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.7

Question 1.
Find A^T, if
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
2 & -6 & 1 \\
-4 & 0 & 5
\end{array}\right]\)
Solution:
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rr}
1 & -4 \\
3 & 5
\end{array}\right]\)

ii. A = \(\left[\begin{array}{ccc}
-4 & 0 & 5
2 & -6 & 1 \\
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\)

[Note: Answer given in the textbook is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
6 & 0 \\
1 & 5
\end{array}\right]\). However, as per our calculation it is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\). ]

Question 2.
If [a_ij]_3×3 where a_ij = 2(i – j), find A and
A^T. State whether A and A^T are symmetric or skew-symmetric matrices?
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given a_ij = 2 (i — j)
∴ a₁₁ = 2(1-1) = 0,
a₁₂ = 2(1-2) = -2,
a₁₃ = 2(1-3) = -4,
a₂₁ = 2(2-1) = 2,
a₂₂ = 2(2-2) = 0,
a₂₃=2(2-3) = -2,
a₃₁ = 2(3-1) = 4,
a₃₂ = 2(3-2) = 2,
a₃₃=2(3-3) = 0

∴ A^T = -A and A = -A^T
∴ A and A^T both are skew-symmetric matrices.

Questionn 3.
If A = \(\left[\begin{array}{cc}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right]\), prove that (2A)^T = 2A^T.
Solution:

From (i) and (ii), we get
(2A)^T = 2A^T

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\), prove that (3A)^T = 3A^T.
Solution:

From (i) and (ii), we get
(3A)^T = 3A^T

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 1+2 i & 1-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\),
where i = \(\sqrt{-1}\), prove that A^T = – A.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\) , B = \(\left[\begin{array}{cc}
2 & 1 \\
4 & -1 \\
-3 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 4 \\
-2 & 3
\end{array}\right]\) then show that
i. (A + B)^T = AT + B^T
ii. (A – C)^T = A^T – C^T
Solution:

From (i) and (ii), we get
(A + B)^T = AT + B^T
[Note: The question has been modified.]

From (i) and (ii), we get
(A – C)^T = A^T – C^T</sup

Question 7.
If A = \(\left[\begin{array}{cc}
5 & 4 \\
-2 & 3
\end{array}\right]\) and \(\left[\begin{array}{cc}
-1 & 3 \\
4 & -1
\end{array}\right]\) then find C^T, such that 3A – 2B + C = I, where I is the unit matrix of order 2.
Solution:
3A – 2B + C = I
∴ C = I + 2B – 3A

Question 8.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 0 \\
0 & 4 & -2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
0 & -2 & 3 \\
2 & 1 & -4
\end{array}\right]\), then find
i. A^T + 4B^T
ii. 5A^T – 5B^T
Solution:

ii. ii. 5A^T – 5B^T = 5(A^T – B^T)

Question 9.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
3 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
2 & 1 & -4 \\
3 & 5 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
0 & 2 & 3 \\
-1 & -1 & 0
\end{array}\right]\), verify that (A + 2B + 3C)^T = A^T + 2B^T + 3C^T.
Solution:
A + 2B + 3C

∴ A^T + 2B^T + 3C^T = \(\left[\begin{array}{cc}
5 & 6 \\
8 & 8 \\
2 & -2
\end{array}\right]\)
From (i) and (ii), we get
(A + 2B + 3C)^T = A^T + 2B^T + 3C^T

Question 10.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right]\), prove that (A + B^T)^T = A^T + B.
prove that (A + B^T)^T = A^T + B
Solution:


From (i) and (ii), we get
(A + B^T)^T = A^T + B

Question 11.
Prove that A + A^T is a symmetric and A – A^T is a skew symmetric matrix, where
i. A = \(\left[\begin{array}{ccc}
1 & 2 & 4 \\
3 & 2 & 1 \\
-2 & -3 & 2
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
5 & 2 & -4 \\
3 & -7 & 2 \\
4 & -5 & -3
\end{array}\right]\)
Solution:

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.


∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.

∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

Question 12.
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
i. \(\left[\begin{array}{cc}
4 & -2 \\
3 & -5
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
Solution:
A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as
A = \(\frac{1}{2}\) (A + A^T) + \(\frac{1}{2}\) (A – A^T)

P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
A = P + Q
A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

∴ P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
∴ A = P + Q
∴ A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

Question 13.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
3 & -2 \\
4 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
0 & 3 & -4 \\
2 & -1 & 1
\end{array}\right]\), verify that
i. (AB)^T = B^TA^T
ii. (BA)^T = A^TB^T
Solution:


From (i) and (ii), we get
(AB)^T = B^TA^T

From (i) and (ii) we get
(BA)^T = A^TB^T

Question 14.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A^TA = I, where I is the unit matrix of order 2.
Solution:


∴ A^TA = I, where I is the unit matrix of order 2.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 1.
Evaluate:
i. \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
{[2} & -4 & 3
\end{array}\right]\)
ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
Solution:
i. \(\begin{aligned}
\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right] &=\left[\begin{array}{lll}
3(2) & 3(-4) & 3(3) \\
2(2) & 2(-4) & 2(3) \\
1(2) & 1(-4) & 1(3)
\end{array}\right] \\
&=\left[\begin{array}{ccc}
6 & -12 & 9 \\
4 & -8 & 6 \\
2 & -4 & 3
\end{array}\right]
\end{aligned}\)

ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
= [2(4)-1(3)+ 3(1)]
= [8 – 3 + 3] = [8]

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -3 \\
4 & 2
\end{array}\right]\) B = \(\left[\begin{array}{cc}
4 & 1 \\
3 & -2
\end{array}\right]\), = show that AB ≠ BA.
Solution:

From (i) and (ii), we get
AB ≠ BA

Question 3.
If A = \(\left[\begin{array}{ccc}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right]\) ,B = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right]\) state whether AB = BA? Justify your answer.
Solution:


From (i) and (ii), we get
AB ≠ BA

Question 4.
Show that AB = BA, where
i. A = \(\left[\begin{array}{rrr}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\) , B = \(\left[\begin{array}{rrr}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)
Solution:

From (i) and (ii), we get
AB = BA

From (i) and (ii), we get
AB = BA
[Note: The question has been modified.]

Question 5.
If A = \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\), prove that A² = 0.
Solution:
A² = A.A
= \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
16-16 & 32-32 \\
-8+8 & -16+16
\end{array}\right] \)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = 0

Question 6.
Verify A(BC) = (AB)C in each of the following cases:
i. A = \(=\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:



From (i) and (ii), we get
A(BC) = (AB)C.

Question 7.
Verify that A(B + C) = AB + AC in each of the following matrices:
i. A = \(\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(=\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:

From (i) and (ii), we get
A(B + C) = AB + AC.
[Note: The question has been modified.]

Question 8.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 6
\end{array}\right]\), B = \(\left[\begin{array}{cc}
3 & -1 \\
3 & 7
\end{array}\right]\), find AB – 2I, where I is unit matrix of order 2.
Solution:

Question 9.
If A = \(\left[\begin{array}{ccc}
4 & 3 & 2 \\
-1 & 2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 0 \\
1 & -2
\end{array}\right]\), show that matrix AB is non singular.
Solution:
im
∴ AB is non-singular matrix.

Question 10.
If A = \(\), find the product (A + I)(A – I).
Solution:

[Note : Answer given in the textbook is \(\left[\begin{array}{ccc}
9 & 6 & 4 \\
15 & 32 & -2 \\
35 & -7 & 29
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ccc}
10 & 10 & 4 \\
25 & 39 & 2 \\
35 & 7 & 22
\end{array}\right]\). ]

Question 11.
If A = \(\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\), find α, if A² = B.
Solution:
A² = B

∴ By equality of matrices, we get
α² = 1 and α + 1 = 2
∴ α = ± 1 and α = 1
∴ α = 1

Question 12.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), show that A² – 4A is scalar matrix.
Solution:
A² – 4A = A.A – 4A

Question 13.
If A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that A² – 8A – kI = O, where I is a unit matrix and O is a null matrix of order 2.
Solution:
A² – 8A – kI = O
∴ A.A – 8A – kI = O

∴ by equality of matrices, we get
1 – 8 – k = 0
∴ k = -7

Question 14.
If A = \(\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\), show that (A+B)² = A² + AB + B².
Solution:
We have to prove that (A + B)² = A² + AB + B²,
i.e., to prove A² + AB + BA + B² = A² + AB + B²,
i.e., to prove BA = 0.
BA = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\)
\(\left[\begin{array}{cc}
40-40 & 20-20 \\
80-80 & 40-40
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 15.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), prove that A² – 5A + 7I = 0, where I is unit matrix of order 2.
Solution:
A² – 5A + 7I = 0 = A.A – 5A + 7I = 0

Question 16.
If A = \(\left[\begin{array}{cc}
3 & 4 \\
-4 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-1 & 2
\end{array}\right]\), show that (A + B)(A – B) = A² – B².
Solution:
We have to prove that (A + B)(A – B) = A² – B²,
i.e., to prove A² – AB + BA – B² = A² – B²,
i.e., to prove – AB + BA = 0,
i.e., to prove AB – BA.

From (i) and (ii), we get AB = BA

Question 17.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & a \\
-1 & b
\end{array}\right]\) and (A + B)² = A² + B², find the values of a and b.
Solution:
Given, (A + B)² = A² + B²
∴ A² + AB + BA + B² = A² + B²
∴ AB + BA = 0
∴ AB = -BA

∴ by equality of matrices, we get
– 2 + a = 0 and 1 + b = 0
a = 2 and b = -1
[Note: The question has been modified.]

Question 18.
Find matrix X such that AX = B,
where A = \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{c}
-3 \\
-1
\end{array}\right]\)
Solution:
Let X = \(\left[\begin{array}{c}
a \\
b
\end{array}\right]\)
But AX = B
∴ \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{a} \\
\mathrm{b}
\end{array}\right]=\left[\begin{array}{r}
-3 \\
-1
\end{array}\right]\)
∴ \(\left[\begin{array}{c}
a-2 b \\
-2 a+b
\end{array}\right]=\left[\begin{array}{l}
-3 \\
-1
\end{array}\right]\)
By equality of matrices, we get
a – 2b = -3 …(i)
-2a + b = -l …(ii)
By (i) x 2 + (ii), we get
-3b =-7
∴ b = \(\frac{7}{3}\)
Substituting b = \(\frac{7}{3}\) in (i), we get
a – 2 (\(\frac{7}{3}\)) = -3
∴ a = -3 + \(\frac{14}{3}=\frac{5}{3}\)
∴ X = \(\left[\begin{array}{l}
\frac{5}{3} \\
\frac{7}{3}
\end{array}\right]\)

Question 19.
Find k, if A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and A² = KA – 2I
Solution:
A² = kA – 2I
∴ AA + 2I = kA

∴ \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]\)
∴ By equality of matrices, we get
3k = 3
∴ k = 1

Question 20.
Find x, if \(\left[\begin{array}{lll}
1 & x & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
3 & 2 & 5
\end{array}\right]\left[\begin{array}{c}
1 \\
-2 \\
3
\end{array}\right]\) = 0
Solution:

∴ [6 + 12x + 14] =[0]
∴ By equality of matrices, we get
∴ 12x + 20 = 0
∴ 12x =-20
∴ x = \(\frac{-5}{3}\)

Question 21.
Find x and y, if \(\left\{4\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 0 & 2
\end{array}\right]-\left[\begin{array}{ccc}
3 & -3 & 4 \\
2 & 1 & 1
\end{array}\right]\right\}\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
x = 19 andy = 12

Question 22.
Find x, y, z if
\(\left\{3\left[\begin{array}{ll}
2 & 0 \\
0 & 2 \\
2 & 2
\end{array}\right]-4\left[\begin{array}{cc}
1 & 1 \\
-1 & 2 \\
3 & 1
\end{array}\right]\right\}\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{c}
x-3 \\
y-1 \\
2 z
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
x – 3 = -6,y – 1 = 0, 2z = -2
∴ x = – 3, y = 1, z = – 1

Question 23.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\) show that A² = \(=\left[\begin{array}{cc}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\)
Solution:

Question 24.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)
show that AB ≠ BA, but |AB| = |A| . |B|.
Solution:
AB = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)

Now, |AB| = \(\left|\begin{array}{cc}
4 & 2 \\
10 & 7
\end{array}\right|\) = 28 – 20 = 8
|A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right|\) = 5 – 6 = -1
|B| = \(\left|\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right|\) = 0 – 8 = -8
∴ |A| . |B| = (-1).(-8) = 8 = |AB|
∴ AB ≠ BA, but |AB| = |A|.|B|

Question 25.
Jay and Ram are two friends in a class. Jay wanted to buy 4 pens and 8 notebooks, Ram wanted to buy 5 pens and 12 notebooks. Both of them went to a shop. The price of a pen and a notebook which they have selected was 6 and ₹ 10. Using matrix multiplication, find the amount required from each one of them.
Solution:
Let A be the matrix of pens and notebooks and B be the matrix of prices of one pen and one notebook.

The total amount required for each one of them is obtained by matrix AB.

∴ Jay needs ₹ 104 and Ram needs ₹ 150.