Class 12

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

I. Evaluate the following:

Question 1.
∫x² log x dx
Solution:

Question 2.
∫x² sin 3x dx
Solution:

Question 3.
∫x tan^-1 x dx
Solution:

Question 4.
∫x² tan^-1 x dx
Solution:

Question 5.
∫x³ tan^-1 x dx
Solution:
Let I = ∫x³ tan^-1 x dx
= ∫(tan^-1 x) . x³ dx

Question 6.
∫(log x)² dx
Solution:
Let I = ∫(log x)² dx
Put log x = t

Question 7.
∫sec³ x dx
Solution:
Let I = ∫sec³ x dx
= ∫sec x sec² x dx
= sec x ∫sec² x dx – ∫[\(\frac{d}{d x}\)(sec x) ∫sec² x dx] dx
= sec x tan x – ∫(sec x tan x)(tan x) dx
= sec x tan x – ∫sec x tan² x dx
= sec x tan x – ∫sec x (sec² x – 1) dx
= sec x tan x – ∫sec³ x dx + ∫sec x dx
∴ I = sec x tan x – I + log|sec x + tan x|
∴ 2I = sec x tan x + log|sec x + tan x|
∴ I = \(\frac{1}{2}\) [sec x tan x + log|sec x + tan x|] + c.

Question 8.
∫x . sin² x dx
Solution:

Question 9.
∫x³ log x dx
Solution:

Question 10.
∫e^2x cos 3x dx
Solution:

Question 11.
∫x sin^-1 x dx
Solution:

Question 12.
∫x² cos^-1 x dx
Solution:

Question 13.
\(\int \frac{\log (\log x)}{x} d x\)
Solution:

= t(log t – 1) + c
= (log x) . [log(log x) – 1] + c.

Question 14.
\(\int \frac{t \cdot \sin ^{-1} t}{\sqrt{1-t^{2}}} d t\)
Solution:

Question 15.
∫cos√x dx
Solution:
Let I = ∫cos√x dx
Put √x = t
∴ x = t²
∴ dx = 2t dt
∴ I = ∫(cos t) 2t dt
= ∫2t cos t dt
= 2t ∫cos t dt – ∫[\(\frac{d}{d t}\)(2t) ∫cos t dt]dt
= 2t sin t – ∫2 sin t dt
= 2t sin t + 2 cos t + c
= 2[√x sin√x + cos√x] + c.

Question 16.
∫sin θ . log(cos θ) dθ
Solution:
Let I = ∫sin θ . log (cos θ) dθ
= ∫log(cos θ) . sin θ dθ
Put cos θ = t
∴ -sin θ dθ = dt
∴ sin θ dθ = -dt

= -t log t + t + c
= -cos θ . log(cos θ) + cos θ + c
= -cos θ [log(cos θ) – 1] + c.

Question 17.
∫x cos³ x dx
Solution:
cos 3x = 4 cos³ x – 3 cos x
∴ cos³ x + 3 cos x = 4cos3x
∴ cos³ x = \(\frac{1}{4}\) cos 3x + \(\frac{3}{4}\) cos x

Question 18.
\(\int \frac{\sin (\log x)^{2}}{x} \cdot \log x d x\)
Solution:

Question 19.
\(\int \frac{\log x}{x} d x\)
Solution:
Let I = \(\int \frac{\log x}{x} d x\)
Put log x = t
\(\frac{1}{x}\) dx = dt
∴ I = ∫t dt
= \(\frac{1}{2}\) t² + c
= \(\frac{1}{2}\) (log x)² + c

Question 20.
∫x sin 2x cos 5x dx.
Solution:
Let I = ∫x sin 2x cos 5x dx
sin 2x cos 5x = \(\frac{1}{2}\)[2 sin 2x cos 5x]
= \(\frac{1}{2}\) [sin(2x + 5x) + sin(2x – 5x)]
= \(\frac{1}{2}\) [sin 7x – sin 3x]
∴ ∫sin 2x cos 5x dx = \(\frac{1}{2}\) [∫sin 7x dx – ∫sin 3x dx]

Question 21.
\(\int \cos (\sqrt[3]{x}) d x\)
Solution:
Let I = \(\int \cos (\sqrt[3]{x}) d x\)

II. Integrate the following functions w.r.t. x:

Question 1.
e^2x sin 3x
Solution:

Question 2.
e^-x cos 2x
Solution:

Question 3.
sin(log x)
Solution:

Question 4.
\(\sqrt{5 x^{2}+3}\)
Solution:
Let I = \(\sqrt{5 x^{2}+3}\) dx

Question 5.
\(x^{2} \sqrt{a^{2}-x^{6}}\)
Solution:

Question 6.
\(\sqrt{(x-3)(7-x)}\)
Solution:

Question 7.
\(\sqrt{4^{x}\left(4^{x}+4\right)}\)
Solution:

Question 8.
(x + 1) \(\sqrt{2 x^{2}+3}\)
Solution:
Let I = ∫(x + 1) \(\sqrt{2 x^{2}+3}\) dx
Let x + 1 = A[\(\frac{d}{d x}\)(2x² + 3)] + B
= A(4x) + B
= 4Ax + B
Comparing the coefficients of x and constant term on both the sides, we get
4A = 1, B = 1
∴ A = \(\frac{1}{4}\), B = 1

Question 9.
\(x \sqrt{5-4 x-x^{2}}\)
Solution:
Let I = ∫\(x \sqrt{5-4 x-x^{2}}\) dx
Let x = A[\(\frac{d}{d x}\)(5 – 4x – x²)] + B
= A[-4 – 2x] + B
= -2Ax + (B – 4A)
Comparing the coefficients of x and the constant term on both sides, we get
-2A = 1, B – 4A = 0

Question 10.
\(\sec ^{2} x \sqrt{\tan ^{2} x+\tan x-7}\)
Solution:

Question 11.
\(\sqrt{x^{2}+2 x+5}\)
Solution:

Question 12.
\(\sqrt{2 x^{2}+3 x+4}\)
Solution:

III. Integrate the following functions w.r.t. x:

Question 1.
[2 + cot x – cosec² x] e^x
Solution:
Let I = ∫e^x [2 + cot x – cosec² x] dx
Put f(x) = 2 + cot x
∴ f'(x) = \(\frac{d}{d x}\)(2 + cot x)
= \(\frac{d}{d x}\)(2) + \(\frac{d}{d x}\)(cot x)
= 0 – cosec² x
= -cosec² x
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x (2 + cot x) + c.

Question 2.
\(\left(\frac{1+\sin x}{1+\cos x}\right) e^{x}\)
Solution:

Question 3.
\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Solution:
Let I = ∫\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Let f(x) = \(\frac{1}{x}\), f'(x) = \(-\frac{1}{x^{2}}\)
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x . \(\frac{1}{x}\) + c

Question 4.
\(\left[\frac{x}{(x+1)^{2}}\right] e^{x}\)
Solution:

Question 5.
\(\frac{e^{x}}{x}\) . [x(log x)² + 2 log x]
Solution:

Question 6.
\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)
Solution:
Let I = ∫\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)

Question 7.
\(e^{\sin ^{-1} x}\left[\frac{x+\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right]\)
Solution:

Question 8.
log(1 + x)^(1+x)
Solution :
Let I = ∫log(1 + x)^(1+x) dx

Question 9.
cosec (log x)[1 – cot(log x)]
Solution:
Let I = ∫cosec (log x)[1 – cot(log x)] dx
Put log x = t
x = e^t
dx = e^t dt
I = ∫cosec t (1 – cot t). e^t dt
= ∫e^t [cosec t – cosec t cot t] dt
= ∫e^t [cosec t + \(\frac{d}{d t}\) (cosec t)] dt
= e^t cosec t + c ….. [∵ e^t [f(t) +f'(t) ] dt = e^t f(t) + c ]
= x . cosec(log x) + c.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.4

Question 1.
Find \(\frac{d y}{d x}\) if
(i) x = at², y = 2at
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

(ii) x = a cot θ, y = b cosec θ
Solution:
x = a cot θ, y = b cosec θ
Differentiating x and y w.r.t. θ, we get

(iii) x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Solution:
x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Differentiating x and y w.r.t. m, we get
\(\frac{d x}{d m}=\frac{d}{d m}\left(\sqrt{a^{2}+m^{2}}\right)\)

(iv) x = sin θ, y = tan θ
Solution:
x = sin θ, y = tan θ
Differentiating x and y w.r.t. θ, we get

(v) x = a(1 – cos θ), y = b(θ – sin θ)
Solution:
x = a(1 – cos θ), y = b(θ – sin θ)
Differentiating x and y w.r.t. θ, we get

(vi) x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\), where a > 0, a ≠ 1 and t ≠ 0
Solution:
x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\) ………(1)
Differentiating x and y w.r.t. t, we get

(vii) x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Solution:
x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Put t = tan θ Then θ = tan^-1t

(viii) x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Solution:
x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Put t = cos θ. Then θ = cos^-1t
x = cos^-1(4cos³θ – 3cos θ)

Question 2.
Find \(\frac{d y}{d x}\), if
(i) x = cosec²θ, y = cot³θ at θ = \(\frac{\pi}{6}\)
Solution:
x = cosec²θ, y = cot³θ
Differentiating x and y w.r.t. θ, we get

(ii) x = a cos³θ, y = a sin³θ at θ = \(\frac{\pi}{3}\)
Solution:
x = a cos³θ, y = a sin³θ
Differentiating x and y w.r.t. θ, we get

(iii) x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\)) at t = 1
Solution:
x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\))
Differentiating x and y w.r.t. t, we get

(iv) x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = \(\frac{\pi}{4}\)
Solution:
x = 2 cos t + cos 2t, y = 2 sin t – sin 2t
Differentiating x and y w.r.t. t, we get

(v) x = t + 2 sin(πt), y = 3t – cos(πt) at t = \(\frac{1}{2}\)
Solution:
x = t + 2 sin(πt), y = 3t – cos(πt)
Differentiating x and y w.r.t. t, we get

Question 3.
(i) If x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\), then show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Solution:
x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\)

(ii) If x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\), then show that \(\frac{d y}{d x}=-\frac{y \log x}{x \log y}\)
Solution:
x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\)
log x = log \(e^{\sin 3 t}\), log y = log \(e^{\cos 3 t}\)
log x = (sin 3t)(log e), log y = (cos 3t)(log e)
log x = sin 3t, log y = cos 3t ….. (1) [∵ log e = 1]
Differentiating both sides w.r.t. t, we get

(iii) If x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\), then show that y2 – \(\frac{d y}{d x}\) = 0.
Solution:
x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\)

(iv) If x = a cos³t, y = a sin³t, then show that \(\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}\)
Solution:
x = a cos³t, y = a sin³t
Differentiating x and y w.r.t. t, we get

(v) If x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3), show that \(\frac{d y}{d x}=-\sqrt{\frac{3 y}{2 x}}\)
Solution:
x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3)

(vi) If x = log (1 + t²), y = t – tan^-1t, show that \(\frac{d y}{d x}=\frac{\sqrt{e^{x}-1}}{2}\)
Solution:
x = log (1 + t²), y = t – tan^-1t
Differentiating x and y w.r.t. t, we get

(vii) If x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\), show that sin x + \(\frac{d y}{d x}\) = 0
Solution:
x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\)
Differentiating x and y w.r.t. t, we get

(viii) If x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\), show that \(\frac{d x}{d y}=-\frac{b^{2} y}{a^{2} x}\)
Solution:
x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\)

Question 4.
(i) Differentiate x sin x w.r.t tan x.
Solution:
Let u = x sinx and v = tan x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

(ii) Differentiate \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) w.r.t \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Let u = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) and v = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)

(iii) Differentiate \(\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\) w.r.t \(\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)\)
Solution:

(iv) Differentiate \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) w.r.t. tan^-1x
Solution:
Let u = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) and v = tan^-1x
Then we want to find \(\frac{d u}{d v}\)
Put x = tan θ. Then θ = tan^-1x.

(v) Differentiate 3x w.r.t. log_x3.
Solution:
Let u = 3x and v = log_x3.
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(3^{x}\right)=3^{x} \cdot \log 3\)

(vi) Differentiate \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) w.r.t. sec^-1x.
Solution:
Let u = \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) and v = sec^-1x
Then we want to find \(\frac{d u}{d v}\).
Differentiating u and v w.r.t. x, we get

(vii) Differentiate x^x w.r.t. x^{sin x}.
Solution:
Let u = x^x and v = x^{sin x}
Then we want to find \(\frac{d u}{d x}\).
Take, u = x^x
log u = log x^x = x log x
Differentiating both sides w.r.t. x, we get

(viii) Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) w.r.t. \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Solution:
Let u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and v = \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)
u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\)
Put x = tan θ. Then θ = tan^-1x and

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.2(C) Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C)

I. Evaluate:

Question 1.
\(\int \frac{3 x+4}{x^{2}+6 x+5} d x\)
Solution:
Let I = \(\int \frac{3 x+4}{x^{2}+6 x+5} d x\)
Let 3x + 4 = A[\(\frac{d}{d x}\)(x² + 6x + 5)] + B
= A(2x + B) + B
∴ 3x + 4 = 2Ax + (6A + B)
Comparing the coefficient of x and constant on both sides, we get
2A = 3 and 6A + B = 4
∴ A = \(\frac{3}{2}\) and 6(\(\frac{3}{2}\)) + B = 4
∴ B = -5
3x + 4 = \(\frac{3}{2}\) (2x + 6) – 5

Question 2.
\(\int \frac{2 x+1}{x^{2}+4 x-5} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{x^{2}+4 x-5} d x\)
Let 2x + 1 = A[\(\frac{d}{d x}\)(x² + 4x – 5)] + B
2x + 1 = A(2x + 1) + B
∴ 2x + 1 = 2Ax + (4A + B)
Comparing the coefficient of x and constant on both sides, we get
4A = 2 and 4A + B = 4
∴ A = \(\frac{3}{2}\) and 6(\(\frac{3}{2}\)) + B = 4
∴ B = -5
∴ 2x + 1 = \(\frac{3}{2}\)(2x + 1) – 5

Question 3.
\(\int \frac{2 x+3}{2 x^{2}+3 x-1} d x\)
Solution:
Let I = \(\int \frac{2 x+3}{2 x^{2}+3 x-1} d x\)
Let 2x+ 3 = A[\(\frac{d}{d x}\)(2x² + 3x – 1)] + B
2x + 1 = A(4x + 3) + B
∴ 2x + 1 = 4Ax + (3A + B)
Comparing the coefficient of x and constant on both sides, we get
4A = 2 and 3A + B = 3
∴ A = \(\frac{1}{2}\) and 3(\(\frac{1}{2}\)) + B = 3
∴ B = \(\frac{3}{2}\)
∴ 2x + 3 = \(\frac{1}{2}\)(4x + 3) + \(\frac{3}{2}\)

Question 4.
\(\int \frac{3 x+4}{\sqrt{2 x^{2}+2 x+1}} d x\)
Solution:
Let I = \(\int \frac{3 x+4}{\sqrt{2 x^{2}+2 x+1}} d x\)
Let 3x + 4 = A[\(\frac{d}{d x}\)(2x² + 2x + 1)] + B
∴ 3x + 4 = A (4x + 2) + B
∴ 3x + 4 = 4Ax + (2A + B)
Comparing the coefficient of x and the constant on both the sides, we get
4A = 3 and 2A + B = 4
∴ A = \(\frac{3}{4}\) and 2(\(\frac{3}{4}\)) + B = 4
∴ B = \(\frac{5}{2}\)
∴ 3x + 4 = \(\frac{3}{4}\) (4x + 2) + \(\frac{5}{2}\)

Question 5.
\(\int \frac{7 x+3}{\sqrt{3+2 x-x^{2}}} d x\)
Solution:
Let I = \(\int \frac{7 x+3}{\sqrt{3+2 x-x^{2}}} d x\)
Let 7x + 3 = A[\(\frac{d}{d x}\)(3 + 2x – x²)] + B
7x + 3 = A(2 – 2x) + B
∴ 7x + 3 = -2Ax + (2A + B)
Comparing the coefficient of x and constant on both the sides, we get
-2A = 7 and 2A + B = 3

Question 6.
\(\int \sqrt{\frac{x-7}{x-9}} d x\)
Solution:

Comparing the coefficients of x and constant term on both sides, we get

Question 7.
\(\int \sqrt{\frac{9-x}{x}} d x\)
Solution:

Question 8.
\(\int \frac{3 \cos x}{4 \sin ^{2} x+4 \sin x-1} d x\)
Solution:

Question 9.
\(\int \sqrt{\frac{e^{3 x}-e^{2 x}}{e^{x}+1}} d x\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.2(B) Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B)

I. Evaluate the following:

Question 1.
\(\int \frac{1}{4 x^{2}-3} \cdot d x\)
Solution:

Question 2.
\(\int \frac{1}{25-9 x^{2}} \cdot d x\)
Solution:

Question 3.
\(\int \frac{1}{7+2 x^{2}} \cdot d x\)
Solution:

Question 4.
\(\int \frac{1}{\sqrt{3 x^{2}+8}} \cdot d x\)
Solution:

Question 5.
\(\int \frac{1}{\sqrt{11-4 x^{2}}} \cdot d x\)
Solution:

Question 6.
\(\int \frac{1}{\sqrt{2 x^{2}-5}} \cdot d x\)
Solution:

Question 7.
\(\int \sqrt{\frac{9+x}{9-x}} \cdot d x\)
Solution:

Question 8.
\(\int \sqrt{\frac{2+x}{2-x}} \cdot d x\)
Solution:

Question 9.
\(\int \sqrt{\frac{10+x}{10-x}} \cdot d x\)
Solution:

Question 10.
\(\int \frac{1}{x^{2}+8 x+12} \cdot d x\)
Solution:

Question 11.
\(\int \frac{1}{1+x-x^{2}} \cdot d x\)
Solution:

Question 12.
\(\int \frac{1}{4 x^{2}-20 x+17} \cdot d x\)
Solution:

Question 13.
\(\int \frac{1}{5-4 x-3 x^{2}} \cdot d x\)
Solution:

Question 14.
\(\int \frac{1}{\sqrt{3 x^{2}+5 x+7}} \cdot d x\)
Solution:

Question 15.
\(\int \frac{1}{\sqrt{x^{2}+8 x-20}} \cdot d x\)
Solution:

Question 16.
\(\int \frac{1}{\sqrt{8-3 x+2 x^{2}}} \cdot d x\)
Solution:

Question 17.
\(\int \frac{1}{\sqrt{(x-3)(x+2)}} \cdot d x\)
Solution:

Question 18.
\(\int \frac{1}{4+3 \cos ^{2} x} \cdot d x\)
Solution:

Question 19.
\(\int \frac{1}{\cos 2 x+3 \sin ^{2} x} \cdot d x\)
Solution:

Question 20.
\(\int \frac{\sin x}{\sin 3 x} \cdot d x\)
Solution:

II. Integrate the following functions w. r. t. x:

Question 1.
\(\int \frac{1}{3+2 \sin x} \cdot d x\)
Solution:

Question 2.
\(\int \frac{1}{4-5 \cos x} \cdot d x\)
Solution:

Question 3.
\(\int \frac{1}{2+\cos x-\sin x} \cdot d x\)
Solution:

Question 4.
\(\int \frac{1}{3+2 \sin x-\cos x} \cdot d x\)
Solution:

Question 5.
\(\int \frac{1}{3-2 \cos 2 x} \cdot d x\)
Solution:

Question 6.
\(\int \frac{1}{2 \sin 2 x-3} \cdot d x\)
Solution:

Question 7.
\(\int \frac{1}{3+2 \sin 2 x+4 \cos 2 x} \cdot d x\)
Solution:

Question 8.
\(\int \frac{1}{\cos x-\sin x} \cdot d x\)
Solution:

Question 9.
\(\int \frac{1}{\cos x-\sqrt{3} \sin x} \cdot d x\)
Solution:

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 6 Line and Plane Miscellaneous Exercise 6B Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6B

Question 1.
If the line \(\frac{x}{3}=\frac{y}{4}\) = z is perpendicular to the line \(\frac{x-1}{k}=\frac{y+2}{3}=\frac{z-3}{k-1}\) then the value of k is:

Solution:
(b) \(-\frac{11}{4}\)

Question 2.
The vector equation of line 2x – 1 = 3y + 2 = z – 2 is

Solution:
(a) \(\bar{r}=\left(\frac{1}{2} \hat{i}-\frac{2}{3} \hat{j}+2 \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})\)

Question 3.
The direction ratios of the line which is perpendicular to the two lines \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}\) are
(A) 4, 5, 7
(B) 4, -5, 7
(C) 4, -5, -7
(D) -4, 5, 8
Solution:
(A) 4, 5, 7

Question 4.
The length of the perpendicular from (1, 6, 3) to the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\)
(A) 3
(B) \(\sqrt {11}\)
(C) \(\sqrt {13}\)
(D) 5
Solution:
(C ) \(\sqrt {13}\)

Question 5.
The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is
Question is modified.
The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is

Solution:
(c) \(\frac{3}{\sqrt{2}}\)

Question 6.
The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\). and coplanar if
(A) k = 1 or -1
(B) k = 0 or -3
(C) k = + 3
(D) k = 0 or -1
Solution:
(B ) k = 0 or -3

Question 7.
The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{6}\) and are
(A) perpendicular
(B) inrersecting
(C) skew
(D) coincident
Solution:
(B) inrersecting

Question 8.
Equation of X-axis is
(A) x = y = z
(B) y = z
(C) y = 0, z = 0
(D) x = 0, y = 0
Solution:
(C) y = 0, z = 0

Question 9.
The angle between the lines 2x = 3y = -z and 6x = -y = -4z is
(A ) 45º
(B ) 30º
(C ) 0º
(D ) 90º
Solution:
(D ) 90º

Question 10.
The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are
(A ) 2, 1, 6
(B ) 2, 1, -6
(C ) 2, -1, 6
(D ) -2, 1, 6
Solution:
(B ) 2, 1, -6

Question 11.
The perpendicular distance of the plane 2x + 3y – z = k from the origin is \(\sqrt {14}\) units, the value
of k is
(A ) 14
(B ) 196
(C ) \(2\sqrt {14}\)
(D ) \(\frac{\sqrt{14}}{2}\)
Solution:
(A ) 14

Question 12.
The angle between the planes and \(\bar{r} \cdot(\bar{i}-2 \bar{j}+3 \bar{k})+4=0\) and \(\bar{r} \cdot(2 \bar{i}+\bar{j}-3 \bar{k})+7=0\) is

Solution:
(d) cos^-1\(\left(\frac{9}{14}\right)\)

Question 13.
If the planes \(\bar{r} \cdot(2 \bar{i}-\lambda \bar{j}+\bar{k})=3\) and \(\bar{r} \cdot(4 \bar{i}-\bar{j}+\mu \bar{k})=5\) are parallel, then the values of λ and μ are respectively.

Solution:
(d) \(\frac{1}{2}\), 2

Question 14.
The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is
(A ) x + y + z =1
(B ) x + y + z = 2
(C ) x + y + z = 3
(D ) x + y + z = 4
Solution:
(D ) x + y + z = 4

Question 15.
Measure of angle between the planes 5x – 2y + 3z – 7 = 0 and 15x – 6y + 9z + 5 = 0 is
(A ) 0º
(B ) 30º
(C ) 45º
(D ) 90º
Solution:
(A ) 0º

Question 16.
The direction cosines of the normal to the plane 2x – y + 2z = 3 are

Solution:
(a) \(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\)

Question 17.
The equation of the plane passing through the points (1, -1, 1), (3, 2, 4) and parallel to Y-axis is :
(A ) 3x + 2z – 1 = 0
(B ) 3x – 2z = 1
(C ) 3x + 2z + 1 = 0
(D ) 3x + 2z = 2
Solution:
(B ) 3x – 2z = 1

Question 18.
The equation of the plane in which the line \(\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\) and \(\frac{x-8}{7}=\frac{y-4}{1}=\frac{z+5}{3}\) lie, is
(A ) 17x – 47y – 24z + 172 = 0
(B ) 17x + 47y – 24z + 172 = 0
(C ) 17x + 47y + 24z +172 = 0
(D ) 17x – 47y + 24z + 172 = 0
Solution:
(A ) 17x – 47y – 24z + 172 = 0

Question 19.
If the line \(\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}\) lies in the plane 3x – 14y + 6z + 49 = 0, then the value of m is:
(A ) 5
(B ) 3
(C ) 2
(D ) -5
Solution:
(A ) 5

Question 20.
The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is
(A ) 4x + y + 5z = 14
(B ) 4x – 2y – 5z = 45
(C ) x – 2y – 5z = 10
(D ) 4x + y + 6z = 11
Solution:
(B ) 4x – 2y – 5z = 45

II. Solve the following :
Question 1.
Find the vector equation of the plane which is at a distance of 5 unit from the origin and which is normal to the vector \(2 \hat{i}+\hat{j}+2 \hat{k}\)
Solution:
If \(\hat{n}\) is a unit vector along the normal and p i the length of the perpendicular from origin to the plane, then the vector equation of the plane \(\bar{r} \cdot \hat{n}\) = p
Here, \(\overline{\mathrm{n}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and p = 5

Question 2.
Find the perpendicular distance of the origin from the plane 6x + 2y + 3z – 7 = 0
Solution:
The distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d is \(\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)
∴ the distance of the point (1, 1, -1) from the plane 6x + 2y + 3z – 7 = 0 is

= 1units.

Question 3.
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.
Solution:
The equation of the plane is 2x + 3y + 6z = 49
Dividing each term by
\(\sqrt{2^{2}+3^{2}+(-6)^{2}}\)
= \(\sqrt{49}\)
= 7
we get
\(\frac{2}{7}\)x + \(\frac{3}{7}\)y – \(\frac{6}{7}\)z = \(\frac{49}{7}\) = 7
This is the normal form of the equation of plane.
∴ the direction cosines of the perpendicular drawn from the origin to the plane are
l = \(\frac{2}{7}\), m = \(\frac{3}{7}\), n = \(\frac{6}{7}\)
and length of perpendicular from origin to the plane is p = 7.
the coordinates of the foot of the perpendicular from the origin to the plane are
(lp, ∓, np)i.e.(2, 3, 6)

Question 4.
Reduce the equation \(\bar{r} \cdot(\hat{i}+8 \hat{j}+24 \hat{k})=13\) to normal form and hence find
(i) the length of the perpendicular from the origin to the plane
(ii) direction cosines of the normal.
Solution:
The normal form of equation of a plane is \(\bar{r} \cdot \hat{n}\) = p where \(\hat{n}\) is unit vector along the normal and p is the length of perpendicular drawn from origin to the plane.
Given pane is \(\text { r. }(6 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+24 \hat{\mathrm{k}})=13\) …(1)

This is the normal form of the equation of plane.
Comparing with \(\bar{r} \cdot \hat{n}\) = p,
(i) the length of the perpendicular from the origin to plane is \(\frac{1}{2}\).
(ii) direction cosines of the normal are \(\frac{3}{13}, \frac{4}{13}, \frac{12}{13}\)

Question 5.
Find the vector equation of the plane passing through the points A(1, -2, 1), B (2, -1, -3) and C (0, 1, 5).
Solution:
The vector equation of the plane passing through three non-collinear points A(\(\bar{a}\)), B(\(\bar{b}\)) and C(\(\bar{c}\)) is \(\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) … (1)

Question 6.
Find the Cartesian equation of the plane passing through A(1, -2, 3) and the direction ratios of whose normal are 0, 2, 0.
Solution:
The Cartesian equation of the plane passing through (x₁, y₁, z₁), the direction ratios of whose normal are a, b, c, is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
∴ the cartesian equation of the required plane is
o(x + 1) + 2(y + 2) + 5(z – 3) = 0
i.e. 0 + 2y – 4 + 10z – 15 = 0
i.e. y + 2 = 0.

Question 7.
Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\)
Solution:
The cartesian equation of the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\) is 6x + 8y + 7z = 0 The required plane is parallel to it
∴ its cartesian equation is
6x + 8y + 7z = p …(1)
A (7, 8, 6) lies on it and hence satisfies its equation
∴ (6)(7) + (8)(8) + (7)(6) = p
i.e., p = 42 + 64 + 42 = 148.
∴ from (1), the cartesian equation of the required plane is 6x + 8y + 7z = 148.

Question 8.
The foot of the perpendicular drawn from the origin to a plane is M(1, 2,0). Find the vector equation of the plane.
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\).
M(1, 2, 0) is the foot of the perpendicular drawn from origin to the plane. Then the plane is passing through M and is
perpendicular to OM.
If \(\bar{m}\) is the position vector of M, then \(\bar{m}\) = \(\hat{\mathrm{i}}\).
Normal to the plane is
\(\bar{n}\) = \(\overline{\mathrm{OM}}\) = \(\hat{\mathrm{i}}\)
\(\overline{\mathrm{m}} \cdot \overline{\mathrm{n}}\) = \(\hat{\mathrm{i}}, \hat{i}\) = 5
∴ the vector equation of the required plane is
\(\bar{r} \cdot(\hat{i}+2 \hat{j})\) = 5

Question 9.
A plane makes non zero intercepts a, b, c on the co-ordinates axes. Show that the vector equation of the plane is \(\bar{r} \cdot(b c \hat{i}+c a \hat{j}+a b \hat{k})\) = abc
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)), B(\(\bar{b}\)).. C(\(\bar{c}\)), where A, B, C are non collinear is
\(\overline{\mathrm{r}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) …(1)
The required plane makes intercepts 1, 1, 1 on the coordinate axes.
∴ it passes through the three non collinear points A = (1, 0, 0), B = (0, 1, 0), C = (0, , 1)

Question 10.
Find the vector equation of the plane passing through the pointA(-2, 3, 5) and parallel to vectors \(4 \hat{i}+3 \hat{k}\) and \(\hat{i}+\hat{j}\)
Solution:
The vector equation of the plane passing through the point A(\(\bar{a}\)) and parallel to the vectors \(\bar{b}\) and \(\bar{c}\) is


= (-2)(-4) + (7)(-1) + (5)(4)
= 8 – 7 + 8
= 35
∴ From (1), the vector equation of the required plane is \(\overline{\mathrm{r}} \cdot(-3 \hat{\mathrm{i}}-3 a t \mathrm{j}+4 \hat{\mathrm{k}})\) = 35.

Question 11.
Find the Cartesian equation of the plane \(\bar{r}=\lambda(\hat{i}+\hat{j}-\hat{k})+\mu(\hat{i}+2 \hat{j}+3 \hat{k})\)
Solution:
The equation \(\bar{r}=\bar{a}+\lambda \bar{b}+\mu \bar{c}\) represents a plane passing through a point having position vector \(\overline{\mathrm{a}}\) and parallel to vectors \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\).
Here,

Question 12.
Find the vector equations of planes which pass through A(1, 2, 3), B (3, 2, 1) and make equal intercepts on the co-ordinates axes.
Question is modified
Find the cartesian equations of the planes which pass through A(1, 2, 3), B(3, 2, 1) and make equal intercepts on the coordinate axes.
Solution:
Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the cartesian equation of the plane is
ax + by + cz = 0 …..(1)
(1, 2, 3) and (3, 2, 1) lie on the plane.
∴ a + 2b + 3c = 0 and 3a + 2b + c = 0

∴ a, b, c are proportional to 1, -2, 1
∴ from (1), the required cartesian equation is x – 2y + z = 0.
Case 2 : Let the plane make non zero intercept p on each axis.
then its equation is \(\frac{x}{p}+\frac{y}{p}+\frac{z}{p}\) = 1
i.e. x + y + z = p …(2)
Since this plane pass through (1, 2, 3) and (3, 2, 1)
∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p
∴ p = 6
∴ from (2), the required cartesian equation is
x + y + z = 6
Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.

Question 13.
Find the vector equation of the plane which makes equal non-zero intercepts on the co-ordinates axes and passes through (1, 1, 1).
Solution:
Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the vector equation of the plane is ax + by + cz …(1)
(1, 1, 1) lie on the plane.
∴ 1a + 1b + 1c = 0

∴ from (1), the required cartesian equation is x – y + z = 0
Case 2 : Let he plane make non zero intercept p on each axis.
then its equation is \(\frac{\hat{\mathrm{i}}}{p}+\frac{\hat{\mathrm{j}}}{p}+\frac{\hat{\mathrm{k}}}{p}=1\) = 1
i.e. \(\hat{i}+\hat{j}+\hat{k}=p\) = p ….(2)
Since this plane pass through (1, 1, 1)
∴ 1 + 1 + 1 = p
∴ p = 3
∴ from (2), the required cartesian equation is \(\hat{i}+\hat{j}+\hat{k}\) = 3
Hence, the cartesian equations of required planes are \(\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=3\)

Question 14.
Find the angle between planes \(\bar{r} \cdot(-2 \hat{i}+\hat{j}+2 \hat{k})=17\) and \(\bar{r} \cdot(2 \hat{i}+2 \hat{j}+\hat{k})=71\).
Solution:
The acute angle between the planes

= (1)(2) + (1)(1) + (2)(1)
= 2 + 1 + 2
= 5
Also,

Question 15.
Find the acute angle between the line \(\bar{r}=\lambda(\hat{i}-\hat{j}+\hat{k})\) and the plane \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=23\)
Solution:
The acute angle θ between the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\) and the plane \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}\) = d is given by

= (2)(2) + (3)(-1) + (-6)(1)
= 4 – 3 – 6
= -5

Question 16.
Show that lines \(\bar{r}=(\hat{i}+4 \hat{j})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\bar{r}=(3 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})\)
Solution:

Question 17.
Find the distance of the point \(3 \hat{i}+3 \hat{j}+\hat{k}\) from the plane \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}+6 \hat{k})=21\)
Solution:
The distance of the point A(\(\bar{a}\)) from the plane \(\bar{r} \cdot \bar{n}\) = p is given by d = \(\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}\) ……(1)

= (3)(2) + (3)(3) + (1)(-6)
= 6 + 9 – 6
= 9

Question 18.
Find the distance of the point (13, 13, -13) from the plane 3x + 4y – 12z = 0.
Solution:

= 19units.

Question 19.
Find the vector equation of the plane passing through the origln and containing the line \(\bar{r}=(\hat{i}+4 \hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})\).
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) … (1)
We can take \(\bar{a}\) = \(\bar{0}\) since the plane passes through the origin.
The point M with position vector \(\bar{m}\) =\(\hat{i}+4 \hat{j}+\hat{k}\) lies on the line and hence it lies on the plane.
.’. \(\overline{\mathrm{OM}}=\bar{m}=\hat{i}+4 \hat{j}+\hat{k}\) lies on the plane.
The plane contains the given line which is parallel to \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
Let \(\bar{n}\) be normal to the plane. Then \(\bar{n}\) is perpendicular to \(\overline{\mathrm{OM}}\) as well as \(\bar{b}\)

Question 20.
Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B( 4, 3, -2) at right angle.
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) ….(1)
The position vectors \(\bar{a}\) and \(\bar{b}\) of the given points A and B are \(\bar{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}\) and \(\bar{b}=4 \hat{i}+3 \hat{j}-2 \hat{k}\)
If M is the midpoint of segment AB, the position vector \(\bar{m}\) of M is given by

Question 21.
Show thatlines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.
Solution:
Given lines are x = y, z = 0 and x + y = 0, z = 0.
It is clear that (0, 0, 0) satisfies both the equations.
∴ the lines intersect at O whose position vector is \(\overline{0}\)
Since z = 0 for both the lines, both the lines lie in XY- plane.
Hence, we have to find equation of XY-plane.
Z-axis is perpendicular to XY-plane.
∴ normal to XY plane is \(\hat{k}\).
0(\(\overline{0}\)) lies on the plane.
By using \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\), the vector equation of the required plane is \(\bar{r} \cdot \hat{k}=\overline{0} \cdot \bar{k}\)
i.e. \(\bar{r} \cdot \hat{k}=0\).
Hence, the given lines intersect each other and the vector equation of the plane determine by them is \(\bar{r} \cdot \hat{k}=0\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Miscellaneous Exercise 6A Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6A

Question 1.
Find the vector equation of the line passing through the point having position vector \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to \(6 \hat{i}-\hat{j}+\hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector
\(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to the vector \(6 \hat{i}-\hat{j}+\hat{k}\) is
\(\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+\lambda(\hat{6 \hat{\mathrm{i}}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\).

Question 2.
Find the vector equation of the line which passes through the point (3, 2, 1) and is parallel to the vector \(2 \hat{i}+2 \hat{j}-3 \hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector \(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and parallel to the vector
\(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \text { is } \overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})\)

Question 3.
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and parallel to the line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\)
Solution:
The line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\) has direction ratios 3, 5, 6. The required line has direction ratios 3, 5, 6 as it is parallel to the given line.
It passes through the point (-2, 4, -5).
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are

Question 4.
Obtain the vector equation of the line \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
Solution:
The cartesian equations of the line are \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
This line is passing through the point A(-5, -4, -5) and having direction ratios 3, 5, 6.
Let \(\bar{a}\) be the position vector of the point A w.r.t. the origin and \(\bar{b}\) be the vector parallel to the line.
Then \(\bar{a}=-5 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(\bar{b}=3 \hat{i}+5 \hat{j}+6 \hat{k}\).
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where λ is a scalar.
∴ the vector equation of the required line is \(\bar{r}=(-5 \hat{i}-4 \hat{j}-6 \hat{k})+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})\)

Question 5.
Find the vector equation of the line which passes through the origin and the point (5, -2, 3).
Solution:
Let \(\bar{b}\) be the position vector of the point B(5, -2, 3).
Then \(\bar{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}\)
Origin has position vector \(\overline{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\).
The vector equation the line passing through A(\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\) where λ is a scalar.
∴ the vector equation of the required line is \(\bar{r}=\overline{0}+\lambda(\bar{b}-\overline{0})=\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})\)

Question 6.
Find the Cartesian equations of the line which passes through points (3, -2, -5) and (3, -2, 6).
Solution:
Let A = (3, -2, -5), B = (3, -2, 6)
The direction ratios of the line AB are
3 – 3, -2 – (-2), 6 – (-5) i.e. 0, 0, 11.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, -2, -5) and having direction ratios are 0, 0, 11 are
x = 3 + (0)λ, y = -2 + 0(λ), z = -5 + 11λ
i.e. x = 3, y = -2, z = 11λ – 5
∴ the cartesian equations of the line are
x = 3, y = -2, z = 11λ – 5, λ is a scalar.

Question 7.
Find the Cartesian equations of the line passing through A(3, 2, 1) and B(1, 3, 1).
Solution:
The direction ratios of the line AB are 3 – 1, 2 – 3, 1 – 1 i.e. 2, -1, 0.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, 2, 1) and having direction ratios 2, -1, 0 are
x = 3 + 2λ, y = 2 – λ, z = 1 + 0(λ)
x – 3 = 2λ, y – 2 = -λ, z = 1
∴ \(\frac{x-3}{2}=\frac{y-2}{-1}\) = λ, z = 1
∴ the cartesian equations of the line are
\(\frac{x-3}{2}=\frac{y-2}{-1}\), z = 1.

Question 8.
Find the Cartesian equations of the line passing through the point A(1, 1, 2) and perpendicular to vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
Solution:
Let the required line have direction ratios p, q, r. ,
It is perpendicular to the vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3p + 2q – r = 0

∴ the required line has direction ratios -1, 1, -1.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equations of the line passing through the point (1, 1, 2) and having direction ratios -1, 1, -1 are
\(\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-2}{-1}\)

Question 9.
Find the Cartesian equations of the line which passes through the point (2, 1, 3) and perpendicular
to lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}\).
Solution:
Let the required line have direction ratios p, q, r.
It is perpendicular to the vector \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3 + 2q – r = 0

∴ the required line has direction ratios 2, -7, 4.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x=x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equation of the line passing through the point (2, -7, 4) and having directions ratios 2, -7, 4 are
\(\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-2}{4}\)

Question 10.
Find the vector equation of the line which passes through the origin and intersect the line x – 1 = y – 2 = z – 3 at right angle.
Solution:
The given line is \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}\) = λ … (Say)
∴ coordinates of any point on the line are
x = λ + 1, y = λ + 2, z = λ + 3
∴ position vector of any point on the line is
(λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … (1)
If \(\bar{b}\) is parallel to the given line whose direction ratios are 1, 1, 1, then \(\bar{b}=\hat{i}+\hat{j}+\hat{k}\).
Let the required line passing through O meet the given line at M.
∴ position vector of M
= \(\bar{m}\) = (λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … [By (1)]
The required line is perpendicular to given line

The vector equation of the line passing through A(\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\), λ is a scalar.
∴ the vector equation of the line passing through o(\(\bar{o}\)) and M(\(\bar{m}\)) is
\(\bar{r}=\overline{0}+\lambda(\bar{m}-\overline{0})=\lambda \bar{m}=\lambda(-\hat{i}+\hat{k})\) where λ is a scalar.
Hence, vector equation of the required line is \(\).

Question 11.
Find the value of λ so that lines \(\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angle.
Solution:
The equations of the given lines are

Question 12.
Find the acute angle between lines \(\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{2}\) and \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{1}\).
Solution:

Question 13.
Find the acute angle between lines x = y, z = 0 and x = 0, z = 0.
Solution:
The equations x = y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis J are 0, 1, 0.
∴ its directiton ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0.

If θ is the acute angle between the lines, then

Question 14.
Find the acute angle between lines x = -y, z = 0 and x = 0, z = 0.
Solution:
The equations x = -y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0.
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0.
∴ its direction ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0

Question 15.
Find the co-ordinates of the foot of the perpendicular drawn from the point (0, 2, 3) to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\).
Solution:
Let P = (0, 2, 3)
Let M be the foot of the perpendicular drawn from P to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\) = λ ……(Say)
The coordinates of any point on the line are given by
x = 5λ – 3, y = 2λ + 1, z = 3λ – 4
Let M = (5λ – 3, 2λ + 1, 3λ – 4) …(1)
The direction ratios of PM are
5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3 i.e. 5λ – 3, 2λ – 1, 3λ – 7
Since, PM is perpendicular to the line whose direcction ratios are 5, 2, 3,
5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0
25λ – 15 + 4λ – 2 + 9λ – 21 =0
38λ – 38 = 0 ∴ λ = 1
Substituting λ = 1 in (1), we get.
M = (5 – 3, 2 + 1, 3 – 4) = (2, 3, -1).
Hence, the coordinates of the foot of perpendicular are (2, 3, – 1).

Question 16.
By computing the shortest distance determine whether following lines intersect each other.
(i) \(\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}+2 \hat{j}-3 \hat{k})+\mu(\hat{i}+\hat{j}-2 \hat{k})\)
Solution:
The shortest distance between the lines


Shortest distance between the lines is 0.
∴ the lines intersect each other.

(ii) \(\frac{x-5}{4}=\frac{y-7}{5}=\frac{z+3}{5}\) and x – 6 = y – 8 = z + 2.
Solution:
The shortest distance between the lines

∴ x₁ = 5, y₁ = 7, z₁ = 3, x₂ = 6, y₂ = 8, z₂ = 2,
l₁ = 4, m₁ = 5, n₁ = 1, l₂ = 1, m₂ = -2, n₂ = 1

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64
= -116
and
(m₁n₂ – m₂n₁)² + (l₂n₁ – l₁n₂)² + (l₁m₂ – l₂m₁)²
= (-6 + 2)² + (1 – 7)² + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64
= 116
Hence, the required shortest distance between the given lines

or
Shortest distance between the lines is 0.
∴ the lines intersect each other.

Question 17.
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1}\) intersect each other then find m.
Solution:

Here, (x₁, y₁, z₁) ≡ (1, -1, 1),
(x₂, y₂, z₂) ≡ (2, -m, 2),
a₁ = 2, b₁ = 3, c₁ = 4,
a₂ = 1, b₂ = 2, c₂ = 1
Substituting these values in (1), we get

∴ 1(3 – 8) – (1 – m)(2 – 4) + 1 (4 – 3) = 0
∴ -5 + 2 – 2m + 1 = 0
∴ -2m = 2
∴ m = -1.

Question 18.
Find the vector and Cartesian equations of the line passing through the point (-1, -1, 2) and parallel to the line 2x – 2 = 3y + 1 = 6z – 2.
Solution:
Let \(\bar{a}\) be the position vector of the point A (-1, -1, 2) w.r.t. the origin.
Then \(\bar{a}=-\hat{i}-\hat{j}+2 \hat{k}\)
The equation of given line is
x – 2 = 3y + 1 = 6z – 2.

The direction ratios of this line are

Question 19.
Find the direction cosines of the line \(\bar{r}=\left(-2 \hat{i}+\frac{5}{2} \hat{j}-\hat{k}\right)+\lambda(2 \hat{i}+3 \hat{j})\).
Solution:

Question 20.
Find the Cartesian equation of the line passing through the origin which is perpendicular to x – 1 = y – 2 = z – 1 and intersects the \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\).
Solution:
Let the required line have direction ratios a, b, c
Since the line passes through the origin, its cartesian equations are
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) …(1)
This line is perpendicular to the line
x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1.

∴ 1(4b – 3c) + 1(4a – 2c) + 1(3a – 2b) = 0
∴ 4b – 3c + 4a – 2c + 3a – 2b = 0
∴ 7a + 2b – 5c = 0
From (2) and (3), we get

Question 21.
Write the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.
Solution:
4x – 3z + 5 = 0 can be written as

This line passes through the point A(0, 2, \(\frac{5}{3}\)) position vector is \(\bar{a}=2 \hat{j}+\frac{5}{3} \hat{k}\)
Also the line has direction ratio 3, 0, 4.
If \(\bar{b}\) is a vector parallel to the line, then \(\bar{b}=3 \hat{i}+4 \hat{k}\)
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where λ is \(\bar{a}\) scalar,
∴ the vector equation of the required line is
\(\bar{r}=\left(2 \hat{j}+\frac{5}{3} \hat{k}\right)+\lambda(3 \hat{i}+4 \hat{k})\).

Question 22.
Find the co-ordinates of points on the line \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) which are at the distance 3 unit from the base point A(1, 2, 3).
Solution:
The cartesian equations of the line are \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) = λ
The coordinates of any point on this line are given by
x = λ + 1, y = -2λ + 2, z = 2λ + 3
Let M(λ + 1, -2λ + 2, 2λ + 3) … (1)
be the point on the line whose distance from A(1, 2, 3) is 3 units.

When λ = 1, M = (1 + 1, -2 + 2, 2 + 3) … [By (1)]
i. e. M = (2, 0, 5)
When λ = -1, M = (1 – 1, 2 + 2, -2 + 3) … [By (1)]
i. e. M = (0, 4, 1)
Hence, the coordinates of the required points are (2, 0, 5) and (0, 4, 1).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.3

Question 1.
A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to machine shop for finishing. The number of man hours of labour required in each shop for production of A and B per unit and the number of man hours available for the firm are as follows :

Profit on the sale of A is ₹ 30 and B is ₹ 20 per units. Formulate the L.P.P. to have maximum profit.
Solution:
Let the number of gadgets A produced by the firm be x and the number of gadgets B produced by the firm be y.
The profit on the sale of A is ₹ 30 per unit and on the sale of B is ₹ 20 per unit.
∴ total profit is z = 30x + 20y.
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 total man hours of labour required for x units of gadget A and y units of gadget B in foundry is (10x + 6y) hours and total man hours of labour required in machine shop is (5x + 4y) hours.
Since, maximum time avilable in foundry and machine shops are 60 hours and 35 hours respectively.
Therefore, the constraints are 10x + 6y ≤ 60, 5x + 4y ≤ 35. Since, x and y cannot be negative, we have x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as :
Maximize z = 30x + 20y, subject to 10x + 6y ≤ 60, 5x + 4y ≤ 35, x ≥ 0, y ≥ 0.

Question 2.
In a cattle breading firm, it is prescribed that the food ration for one animal must contain 14, 22 and 1 units of nutrients A, B and C respectively. Two different kinds of fodder are available. Each unit of these two contains the following amounts of these three nutrients :

The cost of fodder 1 is ₹3 per unit and that of fodder ₹ 2, Formulate the L.P.P. to minimize the cost.
Solution:
Let x units of fodder 1 and y units of fodder 2 be prescribed.
The cost of fodder 1 is ₹ 3 per unit and cost of fodder 2 is ₹ 2 per unit.
∴ total cost is z = 3x + 2y
This is the linear function which is to be minimized. Hence it is the objective function. The constraints are as per the following table :

From table fodder contains (2x + y) units of nutrients A, (2x + 3y) units of nutrients B and (x + y) units of nutrients C. The minimum requirements of these nutrients are 14 units, 22 units and 1 unit respectively.
Therefore, the constraints are
2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1
Since, number of units (i.e. x and y) cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as
Minimize z = 3x + 2y, subject to
2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1, x ≥ 0, y ≥ 0.

Question 3.
A company manufactures two types of chemicals A and B. Each chemical requires two types of raw material P and Q. The table below shows number of units of P and Q required to manufacture one unit of A and one unit of B and the total availability of P and Q.

The company gets profits of ₹350 and ₹400 by selling one unit of A and one unit of B respectively. (Assume that the entire production of A and B can be sold). How many units of the chemicals A and B should be manufactured so that the company get maximum profit? Formulate the problem as L.P.P. to maximize the profit.
Solution:
Let the company manufactures x units of chemical A and y units of chemical B. Then the total profit f to the company is p = ₹ (350x + 400y).
This is a linear function which is to be maximized.
Hence, it is the objective function.
The constraints are as per the following table:

The raw material P required for x units of chemical A and y units of chemical B is 3x + 2y. Since, the maximum availability of P is 120, we have the first constraint as 3x + 2y ≤ 120.
Similarly, considering the raw material Q, we have : 2x + 5y ≤ 160.
Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as :
Maximize p = 350x + 400y, subject to
3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0.

Question 4.
A printing company prints two types of magazines A and B. The company earns ₹ 10 and ₹ 15 on magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on Machine I, 5 hours on Machine II and 2 hours on Machine III. Magazine B requires 3 hours on Machine I, 2 hours on Machine II and 6 hours on Machine III. Machines I, II, III are available for 36, 50, 60 hours per week respectively. Formulate the L.P.P. to determine weekly production of A and B, so that the total profit is maximum.
Solution:
Let the company prints x magazine of type A and y magazine of type B.
Profit on sale of magazine A is ₹ 10 per copy and magazine B is ₹ 15 per copy.
Therefore, the total earning z of the company is
z = ₹ (10x + 15y).
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 total time required for Machine I is (2x + 3y) hours, for Machine II is (5x + 2y) hours and for Machine III is (2x + 6y) hours. The machines I, II, III are available for 36,50 and 60 hours per week. Therefore, the constraints are 2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60.
Since x and y cannot be negative. We have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as :
Maximize z = 10x + 15y, subject to
2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.

Question 5.
A manufacture produces bulbs and tubes. Each of these must be processed through two machines M₁ and M₂. A package of bulbs require 1 hour of work on Machine M₁ and 3 hours of work on M₂. A package of tubes require 2 hours on Machine M₁ and 4 hours on Machine M₂. He earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes. Formulate the LLP to maximize the profit, if he operates the machine M₁, for atmost 10 hours a day and machine M₂ for atmost 12 hours a day.
Solution:
Let the number of packages of bulbs produced by manufacturer be x and packages of tubes be y. The manufacturer earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes.
Therefore, his total profit is p = ₹ (13.5x + 55y)
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 total time required for Machine M₁ is (x + 2y) hours and for Machine M₂ is (3x + 4y) hours.
Given Machine M₁ and M₂ are available for atmost 10 hours and 12 hours a day respectively.
Therefore, the constraints are x + 2y ≤ 10, 3x + 4y ≤ 12. Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as :
Maximize p = 13.5x + 55y, subject to x + 2y ≤ 10, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0.

Question 6.
A company manufactures two types of fertilizers F₁ and F₂. Each type of fertilizer requires
two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F₁ and F₂ and availability of the raw materials A and B per day are given in the
table below :

By selling one unit of F₁ and one unit of F₂, company gets a profit of ₹ 500 and ₹ 750
respectively. Formulate the problem as L.P.P. to maximize the profit.
Solution:
Let the company manufactures x units of fertilizers F₁ and y units of fertilizers F₁. Then the total profit to the company is
z = ₹(500x + 750y).
This is a linear function that is to be maximized. Hence, it is an objective function.

The raw material A required for x units of Fertilizers F₁ and y units of Fertilizers F₂ is 2x + Since the maximum availability of A is 40, we have the first constraint as 2x + 3y ≤ 40.
Similarly, considering the raw material B, we have x + 4y ≤ 70.
Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as:
Maximize z = 500x + 750y, subject to
2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.

Question 7.
A doctor has prescribed two different units of foods A and B to form a weekly diet for a sick person. The minimum requirements of fats, carbohydrates and proteins are 18, 28, 14 units respectively. One unit of food A has 4 units of fats. 14 units of carbohydrates and 8 units of protein. One unit of food B has 6 units of fat, 12 units of carbohydrates and 8 units of protein. The price of food A is ₹ 4.5 per unit and that of food B is ₹ 3.5 per unit. Form the L.P.P. so that the sick person’s diet meets the requirements at a minimum cost.
Solution:
Let the diet of sick person include x units of food A and y units of food B.
Then x ≥ 0, y ≥ 0.
The prices of food A and B are ₹ 4.5 and ₹ 3.5 per unit respectively.
Therefore, the total cost is z = ₹ (4.5x + 3.5y)
This is the linear function which is to be minimized.
Hence, it is objective function.
The constraints are as per the following table :

From the table, the sick person’s diet will include (4x + 6y) units of fats, (14x + 12y) units of carbohydrates and (8x + 8y) units of proteins. The minimum requirements of these ingredients are 18 units, 28 units and 14 units respectively.
Therefore, the constraints are
4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14.
Hence, the given LPP can be formulated as
Minimize z = 4.5x + 3.5y, subject to
4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14, x ≥ 0, y ≥ 0.

Question 8.
If John drives a car at a speed of 60 kms/hour he has to spend ₹ 5 per km on petrol. If he drives at a faster speed of 90 kms/hour, the cost of petrol increases to ₹ 8 per km. He has ₹ 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as L.P.P.
Solution:
Let John travel xl km at a speed of 60 km/ hour and x₁ km at a speed of 90 km/hour.
Therefore, time required to travel a distance of x₁ km is \(\frac{x_{1}}{60}\) hours and the time required to travel a distance of
x₂ km is \(\frac{x_{2}}{90}\) hours.
∴ total time required to travel is \(\left(\frac{x_{1}}{60}+\frac{x_{2}}{90}\right)\) hours.
Since he wishes to travel the maximum distance within an hour,
\(\frac{x_{1}}{60}+\frac{x_{2}}{90}\) ≤ 1
He has to spend ₹ 5 per km on petrol at a speed of 60 km/hour and ₹ 8 per km at a speed of 90 km/hour.
∴ the total cost of travelling is ₹ (5x₁ + 8x₂)
Since he has ₹ 600 to spend on petrol,
5x₁ + 8x₂ ≤ 600
Since distance is never negative, x₁ ≥ 0, x₂ ≥ 0.
Total distance travelled by John is z. = (x₁ + x₂) km.
This is the linear function which is to be maximized.
Hence, it is objective function.
Hence, the given LPP can be formulated as :
Maximize z = x₁ + x₂, subject to
\(\frac{x_{1}}{60}+\frac{x_{2}}{90}\) ≤ 1, 5x₁ + 8x₂ ≤ 600, x₁ ≥ 0, x₂ ≥ 0.

Question 9.
The company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be least 5 kg. Cement costs ₹ 20 per kg. and sand costs of ₹ 6 per kg. strength consideration dictate that a concrete brick should contain minimum 4 kg. of cement and not more than 2 kg. of sand. Form the L.P.P. for the cost to be minimum.
Solution:
Let the company use x₁ kg of cement and x₂ kg of sand to make concrete bricks.
Cement costs ₹ 20 per kg and sand costs ₹ 6 per kg.
∴ the total cost c = ₹ (20x₁ + 6x₂)
This is a linear function which is to be minimized.
Hence, it is the objective function.
Total weight of brick = (x₁ + x₂) kg
Since the weight of concrete brick has to be at least 5 kg,
∴ x₁ + x₂ ≥ 5.
Since concrete brick should contain minimum 4 kg of cement and not more than 2 kg of sand,
x₁ ≥ 4 and 0 ≤ x₂ ≤ 2
Hence, the given LPP can be formulated as :
Minimize c = 20x₁ + 6x₂, subject to
x₁ + x₂ ≥ 5, x₁ ≥ 4, 0 ≤ x₂ ≤ 2.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.2

I) Find the feasible solution of the following inequations graphically.
Question 1.
3x + 2y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + 2y = 18 and 2x + y = 10 respectively.


The feasible solution is OCPBO which is shaded in the graph.

Question 2.
2x + 3y ≤ 6, x + y ≥ 2, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CB whose equations are 2x + 3y = 6 and x + y = 2 respectively.


The feasible solution is ∆ABC which is shaded in the graph.

Question 3.
3x + 4y ≥ 12, 4x + 7y ≤ 28, y ≥ 1, x ≥ 0
Solution:
First we draw the lines AB, CD and EF whose equations are 3x + 4 y = 12, 4x + 7y = 28 and y = 1 respectively.


The feasible solution is PQDBP. which is shaded in the graph.

Question 4.
x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.


The feasible solution is OCPQBO. which is shaded in the graph.

Question 5.
0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≤ 5, 2x + y ≥ 4
Solution:
First we draw the lines AB, CD, EF and GH whose equations are x + y = 5, 2x + y = 4, x = 3 and y = 3 respectively.


The feasible solution is CEPQRC. which is shaded in the graph.

Question 6.
x – 2y ≤ 2, x + y ≥ 3, -2x + y ≤ 4, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB, CD and EF whose equations are x – 2y = 2, x + y = 3 and -2x + y = 4 respectively.


The feasible solution is shaded in the graph.

Question 7.
A company produces two types of articles A and B which requires silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find the feasible solution graphically.
Solution:
Let the company produces x units of article A and y units of article B.
The given data can be tabulated as:

Inequations are :
x + 2y ≤ 4 and 3x + 2y ≤ 6
x and y are number of items, x ≥ 0, y ≥ 0
First we draw the lines AB and CD whose equations are x + 2y = 4 and 3x + 2y = 6 respectively.


The feasible solution is OCPBO. which is shaded in the graph.

Question 8.
A furniture dealer deals in tables and chairs. He has Rs.1,50,000 to invest and a space to store at most 60 pieces. A table costs him Rs.1500 and a chair Rs.750. Construct the inequations and find the feasible solution.
Question is modified
A furniture dealer deals in tables and chairs. He has ₹ 15,000 to invest and a space to store at most 60 pieces. A table costs him ₹ 150 and a chair ₹ 750. Construct the inequations and find the feasible solution.
Solution:
Let x be the number of tables and y be the number of chairs. Then x ≥ 0, y ≥ 0.
The dealer has a space to store at most 60 pieces.
∴ x + y ≤ 60
Since, the cost of each table is ₹ 150 and that of each chair is ₹ 750, the total cost of x tables and y chairs is 150x + 750y. Since the dealer has ₹ 15,000 to invest, 150x + 750y ≤ 15,000
Hence the system of inequations are
x + y ≤ 60, 150x + 750y ≤ 15000, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are x + y = 60 and 150x + 750y = 15,000 respectively.


The feasible solution is OAPDO. which is shaded in the graph.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1

Question 1.
Solve graphically :
(i) x ≥ 0
Solution:
Consider the line whose equation is x = 0. This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1)
When x = 1, x ≥ 0
∴ (1, 1) lies in the required region
Therefore, the solution set is the Y-axis and the right
side of the Y-axis which is shaded in the graph.

(ii) x ≤ 0
Solution:
Consider the line whose equation is x = 0.
This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When x = 1, x ≰ 0
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the Y-axis and the left side of the Y-axis which is shaded in the graph.

(iii) y ≥ 0
Solution:
Consider the line whose equation is y = 0. This represents the X-axis. To find the solution set, we have to check any point other than origin. Let us check the point (1, 1).
When y = 1, y ≥ 0
∴ (1, 1) lies in the required region.
Therefore, the solution set is the X-axis and above the X-axis which is shaded in the graph.

(iv) y ≤ 0
Solution:
(iv) Consider the line whose equation is y = 0. This represents the X-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When y = 1, y ≰ 0.
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the X-axis and below the X-axis which is shaded in the graph.

Question 2.
Solve graphically :
(i) x ≥ 0 and y ≥ 0
Solution:
Consider the lines whose equations are x = 0, y = 0.
These represents the equations of Y-axis and X-axis respectively, which divide the plane into four parts.
(i) Since x ≥ 0, y ≥ 0, the solution set is in the first quadrant which is shaded in the graph.

(ii) x ≤ 0 and y ≥ 0
Solution:
Since x ≤ 0, y ≥ 0, the solution set is in the second quadrant which is shaded in the graph.

(iii) x ≤ 0 and y ≤ 0
Solution:
Since x ≤ 0, y ≤ 0, the solution set is in the third quadrant which is shaded in the graph.

(iv) x ≥ 0 and y ≤ 0
Solution:
Since x ≥ 0, y ≤ 0, the solution set is in the fourth ! quadrant which is shaded in the graph.

Question 3.
Solve graphically :
(i) 2x – 3 ≥ 0
Solution:
Consider the line whose equation is 2x – 3 = 0,
i.e. x = \(\frac{3}{2}\)
This represents a line parallel to Y-axis passing through the point (\(\frac{3}{2}\), 0)
Draw the line x =\(\frac{3}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 2x – 3 = 2 × 0 – 3 = -3 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(\frac{3}{2}\) and the non-origin side of the line which is shaded in the graph.

(ii) 2y – 5 ≥ 0
Solution:
Consider the line whose equation is 2y – 5 = 0, i.e. y = \(\frac{5}{2}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{5}{2}\)).
Draw the line y = \(\frac{5}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 2y – 5 = 2 × 0 – 5 = -5 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{5}{2}\) and the non-origin side of the line which is shaded in the graph.

(iii) 3x + 4 ≤ 0
Solution:
(iii) Consider the line whose equation is 3x + 4 = 0,
i.e. x = \(-\frac{4}{3}\)
This represents a line parallel to Y-axis passing through the point (\(-\frac{4}{3}\), 0).
Draw the line x = \(-\frac{4}{3}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 3x + 4 = 3 × 0 + 4= 4 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(-\frac{4}{3}\) and the non-origin side of the line which is shaded in the graph.

(iv) 5y + 3 ≤ 0
Solution:
(iv) Consider the line whose equation is 5y + 3 = 0,
i.e. y = \(\frac{-3}{5}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{-3}{5}\))
Draw the line y = \(\frac{-3}{5}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 5y + 3 = 5 × 0 + 3 = 3 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{-3}{5}\) and the non-origin side of the line which is shaded in the graph.

Question 4.
Solve graphically :
(i) x + 2y ≤ 6
Solution:
Consider the line whose equation is x + 2y = 6.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get x = 6.
∴ A = (6, 0) is a point on the line.
Put x = 0, we get 2y = 6, i.e. y = 3
∴ B = (0, 3) is another point on the line.

Draw the line AB joining these points. This line divide the line into two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line.
When x = 0, y = 0, then x + 2y = 0 which is less than 6.
∴ x + 2y ≤ 6 in this case.
Hence, origin lies in the required region. Therefore, the given inequality is the origin side which is
shaded in the graph.
This is the solution set of x + 2y ≤ 6.

(ii) 2x – 5y ≥ 10
Solution:
Consider the line whose equation is 2x – 5y = 10.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get 2x = 10, i.e. x = 5.
∴ A = (5, 0) is a point on the line.
Put x = 0, we get -5y = 10, i.e. y = -2
∴ B = (0, -2) is another point on the line.

Draw the line AB joining these points. This line J divide the plane in two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line. When x = 0, y = 0, then 2x – 5y = 0 which is neither greater nor equal to 10.
∴ 2x – 5y ≱ 10 in this case.
Hence (0, 0) will not lie in the required region.
Therefore, the given inequality is the non-origin side, which is shaded in the graph.
This is the solution set of 2x – 5y ≥ 10.

(iii) 3x + 2y ≥ 0
Solution:
Consider the line whose equation is 3x + 2y = 0.
The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is O ≡ (0, 0).
To find the another point, we can give any value of x and get the corresponding value of y.
Put x = 2, we get 6 + 2y = 0 i.e. y = – 3
∴ A = (2, -3) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check (0, 0) as it is already on the line.
We can check any other point which is not on the line.
Let us check the point (1, 1)

When x = 1, y = 1, then 3x + 2y = 3 + 2 = 5 which is greater than zero.
∴ 3x + 2y > 0 in this case.
Hence (1, 1) lies in the required region. Therefore, the required region is the upper side which is shaded in the graph.
This is the solution set of 3x + 2y ≥ 0.

(iv) 5x – 3y ≤ 0
Solution:
Consider the line whose equation is 5x – 3y = 0. The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is the origin O = (0, 0).
To find the other point, we can give any value of x and get the corresponding value of y.
Put x = 3, we get 15 – 3y = 0, i.e. y = 5
∴ A ≡ (3, 5) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check 0(0, 0), as it is already on the line. We can check any other point which is not on the line.
Let us check the point (1, -1).
When x = 1, y = -1 then 5x – 3y = 5 + 3 = 8
which is neither less nor equal to zero.
∴ 5x – 3y ≰ 0 in this case.
Hence (1, -1) will not lie in the required region. Therefore, the required region is the upper side which is shaded in the graph.

This is the solution set of 5x – 3y ≤ 0.

Question 5.
Solve graphically :
(i) 2x + y ≥ 2 and 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 – y ≤ 2 and x + 2y ≤ 8
Solution:
First we draw the lines AB and CD whose equations are x – y = 2 and x + 2y = 8 respectively.


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

(iii) x + y ≥ 6 and x + 2y ≤ 10
Solution:
First we draw the lines AB and CD whose equations are x + y = 6 and x + 2y = 10 respectively.


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

(iv) 2x + 3y ≤ 6 and x + 4y ≥ 4
Solution:
First we draw the lines AB and CD whose equations are 2x + 3y = 6 and x + 4y = 4 respectively.

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

(v) 2x + y ≥ 5 and x – y ≤ 1
Solution:
First we draw the lines AB and CD whose equations are 2x + y = 5 and x – y = 1 respectively.


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