Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3

Question 1.
Determine the maximum and minimum values of the following functions:
(i) f(x) = 2x3 – 21x2 + 36x – 20
Solution:
f(x) = 2x3 – 21x2 + 36x – 20
∴ f'(x) = \(\frac{d}{d x}\)(2x3 – 21x2 + 36x – 20)
= 2 × 3x2 – 21 × 2x + 36 × 1 – 0
= 6x2 – 42x + 36
and f”(x) = \(\frac{d}{d x}\)(6x2 – 42x + 36)
= 6 × 2x – 42 × 1 + 0
= 12x – 42
f'(x) = 0 gives 6x2 – 42x + 36 = 0.
∴ x2 – 7x + 6 = 0
∴ (x – 1)(x – 6) = 0
∴ the roots of f'(x) = 0 are x1 = 1 and x2 = 6.
For x = 1, f”(1) = 12(1) – 42 = -30 < 0
∴ by the second derivative test,
f has maximum at x = 1 and maximum value of f at x = 1
f(1) = 2(1)3 – 21(1)2 + 36(1) – 20
= 2 – 21 + 36 – 20
= -3
For x = 6, f”(6) = 12(6) – 42 = 30 > 0
∴ by the second derivative test,
f has minimum at x = 6 and minimum value of f at x = 6
f(6) = 2(6)3 – 21(6)2 + 36(6) – 20
= 432 – 756 + 216 – 20
= -128
Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3

(ii) f(x) = x . log x
Solution:
f(x) = x . log x
f'(x) = \(\frac{d}{d x}\)(x.log x)
= x.\(\frac{d}{d x}\)(log x) + log x.\(\frac{d}{d x}\)(x)
= x × \(\frac{1}{x}\) + (logx) × 1
= 1 + log x
and f”(x) = \(\frac{d}{d x}\)(1 + logx)
= 0 + \(\frac{1}{x}\)
= \(\frac{1}{x}\)
Now, f'(x) = 0, if 1 + log x = 0
i.e. if log x = -1 = -log e
i.e. if log x = log(e-1) = log \(\frac{1}{e}\)
i.e. if x = \(\frac{1}{e}\)
When x = \(\frac{1}{e}\), f”(x) = \(\frac{1}{(1 / e)}\) = e > 0
∴ by the second derivative test,
f is minimum at x = \(\frac{1}{e}\)
Minimum value of f at x = \(\frac{1}{e}\)
= \(\frac{1}{e}\) log(\(\frac{1}{e}\))
= \(\frac{1}{e}\) log(e-1)
= \(\frac{1}{e}\) (-1) log e
= \(\frac{-1}{e}\) ……..[∵ log e = 1]
Hence, the function f has minimum at x = \(\frac{1}{e}\) and minimum value is \(\frac{-1}{e}\).

(iii) f(x) = x2 + \(\frac{16}{x}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3 Q1(iii)
f'(x) = 0 gives 2x – \(\frac{16}{x^{2}}\) = 0
∴ 2x3 – 16 = 0
∴ x3 = 8
∴ x = 2
For x = 2, f”(2) = 2 + \(\frac{32}{(2)^{3}}\) = 6 > 0
∴ by the second derivative test, f has minimum at x = 2 and minimum value of f at x = 2
f(2) = (2)2 + \(\frac{16}{2}\)
= 4 + 8
= 12
Hence, the function f has a minimum at x = 2 and a minimum value is 12.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3

Question 2.
Divide the number 20 into two parts such that their product is maximum.
Solution:
Let the first part of 20 be x.
Then the second part is 20 – x.
∴ their product = x(20 – x) = 20x – x2 = f(x) …..(Say)
∴ f'(x) = \(\frac{d}{d x}\)(20x – x2)
= 20 × 1 – 2x
= 20 – 2x
and f”(x) = \(\frac{d}{d x}\)(20 – 2x)
= 0 – 2 × 1
= -2
The root of the equation f'(x) = 0
i.e. 20 – 2x = 0 is x = 10
and f”(10) = -2 < 0
∴ by the second derivative test, f is maximum at x = 10.
Hence, the required parts of 20 are 10 and 10.

Question 3.
A metal wire of 36 cm long is bent to form a rectangle. Find its dimensions where its area is maximum.
Solution:
Let x cm and y cm be the length and breadth of the rectangle.
Then its perimeter is 2(x + y) = 36
∴ x + y = 18
∴ y = 18 – x
Area of the rectangle = xy = x(18 – x)
Let f(x) = x(18 – x) = 18x – x2
Then f'(x) = \(\frac{d}{d x}\)(18x – x2)
= 18 × 1 – 2x
= 18 – 2x
and f”(x) = \(\frac{d}{d x}\)(18 – 2x)
= 0 – 2 × 1
= -2
Now, f(x) = 0, if 18 – 2x = 0
i.e. if x = 9
and f”(9) = -2 < 0
∴ by the second derivative test, f has maximum value at x = 9
When x = 9, y = 18 – 9 = 9
Hence, the rectangle is a square of side 9 cm.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3

Question 4.
The total cost of producing x units is ₹(x2 + 60x + 50) and the price is ₹(180 – x) per unit. For what units is the profit maximum?
Solution:
Let the number of units sold be x.
Then profit = S.P. – C.P.
∴ P(x) = (180 – x)x – (x2 + 60x + 50)
∴ P(x) = 180x – x2 – x2 – 60x – 50
∴ P(x) = 120x – 2x2 – 50
P'(x) = \(\frac{d}{d x}\)(120x – 2x2 – 50)
= 120 × 1 – 2 × 2x – 0
= 120 – 4x
and P”(x) = \(\frac{d}{d x}\)(120 – 4x)
= 0 – 4 × 1
= -4
P'(x) = 0 if 120 – 4x = 0
i.e. if x = 30 and P”(30) = -4 < 0
∴ by the second derivative test, P(x) is maximum when x = 30.
Hence, the number of units sold for maximum profit is 30.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

Question 1.
Test whether the following functions are increasing and decreasing:
(i) f(x) = x3 – 6x2 + 12x – 16, x ∈ R
Solution:
f(x) = x3 – 6x2 + 12x – 16
∴ f'(x) = \(\frac{d}{d x}\)(x3 – 6x2 + 12x – 16)
= 3x2 – 6 × 2x + 12 × 1 – 0
= 3x2 – 12x + 12
= 3(x2 – 4x + 4)
= 3(x – 2)2 > 0 for all x ∈ R, x ≠ 2
∴ f'(x) > 0 for all x ∈ R – {2}
∴ f is increasing for all x ∈ R – {2}.

(ii) f(x) = x – \(\frac{1}{x}\), x ∈ R, x ≠ 0
Solution:
f(x) = x – \(\frac{1}{x}\)
∴ f'(x) = \(\frac{d}{d x}\left(x-\frac{1}{x}\right)\)
= 1 – \(\left(-\frac{1}{x^{2}}\right)\)
= 1 + \(\frac{1}{x^{2}}\) > 0 for all x ∈ R, x ≠ 0
∴ f'(x) > 0 for all x ∈ R, where x ≠ 0
∴ f is increasing for all x > R, where x ≠ 0.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

(iii) f(x) = \(\frac{7}{x}\) – 3, x ∈ R, x ≠ 0
Solution:
f(x) = \(\frac{7}{x}\) – 3
∴ f'(x) = \(\frac{d}{d x}\left(\frac{7}{x}-3\right)=7\left(-\frac{1}{x^{2}}\right)-0\)
= \(-\frac{7}{x^{2}}\) < 0 for all x ∈ R, x ≠ 0
∴ f'(x) < 0 for all x ∈ R, where x ≠ 0.
∴ f is decreasing for all x ∈ R, where x ≠ 0.

Question 2.
Find the values of x, such that f(x) is increasing function:
(i) f(x) = 2x3 – 15x2 + 36x + 1
Solution:
f(x) = 2x3 – 15x2 + 36x + 1
∴ f'(x) = \(\frac{d}{d x}\)(2x3 – 15x2 + 36x + 1)
= 2 × 3x2 – 15 × 2x + 36 × 1 + 0
= 6x2 – 30x + 36
= 6(x2 – 5x + 6)
f is increasing, if f'(x) > 0
i.e. if 6(x2 – 5x + 6) > 0
i.e. if x2 – 5x + 6 > 0
i.e. if x2 – 5x > -6
i.e. if x – 5x + \(\frac{25}{4}\) > -6 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{1}{4}\)
i.e. if x – \(\frac{5}{2}\) > \(\frac{1}{2}\) or x – \(\frac{5}{2}\) < –\(\frac{1}{2}\)
i.e. if x > 3 or x < 2
i.e. if x ∈ (-∞, 2) ∪ (3, ∞)
∴ f is increasing, if x ∈ (-∞, 2) ∪ (3, ∞).

(ii) f(x) = x2 + 2x – 5
Solution:
f(x) = x2 + 2x – 5
∴ f'(x) = \(\frac{d}{d x}\)(x2 + 2x – 5)
= 2x + 2 × 1 – 0
= 2x + 2
f is increasing, if f'(x) > 0
i.e. if 2x + 2 > 0
i.e. if 2x > -2
i.e. if x > -1, i.e. x ∈ (-1, ∞)
∴ f is increasing, if x > -1, i.e. x ∈ (-1, ∞)

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

(iii) f(x) = 2x3 – 15x2 – 144x – 7
Solution:
f(x) = 2x3 – 15x2 – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\) (2x3 – 15x2 – 144x – 7)
= 2 × 3x2 – 15 × 2x – 144 × 1 – 0
= 6x2 – 30x – 144
= 6(x2 – 5x – 24)
f is increasing if, f'(x) > 0
i.e. if 6(x2 – 5x – 24) > 0
i.e. if x2 – 5x – 24 > 0
i.e. if x2 – 5x > 24
i.e. if x2 – 5x + \(\frac{25}{4}\) > 24 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{121}{4}\)
i.e. if \(x-\frac{5}{2}>\frac{11}{2} \text { or } x-\frac{5}{2}<-\frac{11}{2}\)
i.e. if x > 8 or x < -3
i.e. if x ∈ (-∞, -3) ∪ (8, ∞)
∴ f is increasing, if x ∈ (-∞, -3) ∪ (8, ∞).

Question 3.
Find the values of x such that f(x) is decreasing function:
(i) f(x) = 2x3 – 15x2 – 144x – 7
Solution:
f(x) = 2x3 – 15x2 – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x3 – 15x2 – 144x – 7)
= 2 × 3x2 – 15 × 2x – 144 × 1 – 0
= 6x2 – 30x – 144
= 6(x2 – 5x – 24)
f is decreasing, if f'(x) < 0
i.e. if 6(x2 – 5x – 24) < 0
i.e. if x2 – 5x – 24 < 0
i.e. if x2 – 5x < 24
i.e. if x2 – 5x + \(\frac{25}{4}\) < \(\frac{121}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{121}{4}\)
i.e. if \(-\frac{11}{2}<x-\frac{5}{2}<\frac{11}{2}\)
i.e. if \(-\frac{11}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{11}{2}+\frac{5}{2}\)
i.e. if -3 < x < 8
∴ f is decreasing, if -3 < x < 8.

(ii) f(x) = x4 – 2x3 + 1
Solution:
f(x) = x4 – 2x3 + 1
∴ f'(x) = \(\frac{d}{d x}\)(x4 – 2x3 + 1)
= 4x3 – 2 × 3x2 + 0
= 4x3 – 6x2
f is decreasing, if f'(x) < 0
i.e. if 4x3 – 6x2 < 0
i.e. if x2(4x – 6) < 0
i.e. if 4x – 6 < 0 …….[∵ x2 > 0]
i.e. if x < \(\frac{3}{2}\)
i.e. -∞ < x < \(\frac{3}{2}\)
∴ f is decreasing, if -∞ < x < \(\frac{3}{2}\).

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

(iii) f(x) = 2x3 – 15x2 – 84x – 7
Solution:
f(x) = 2x3 – 15x2 – 84x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x3 – 15x2 – 84x – 7)
= 2 × 3x2 – 15 × 2x – 84 × 1 – 0
= 6x2 – 30x – 84
= 6(x2 – 5x – 14)
f is decreasing, if f'(x) < 0
i.e. if 6(x2 – 5x – 14) < 0
i.e. if x2 – 5x – 14 < 0
i.e. if x2 – 5x < 14
i.e. if x – 5x + \(\frac{25}{4}\) < 14 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{81}{4}\)
i.e. if \(-\frac{9}{2}<x-\frac{5}{2}<\frac{9}{2}\)
i.e. if \(-\frac{9}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{9}{2}+\frac{5}{2}\)
i.e. if -2 < x < 7
∴ f is decreasing, if -2 < x < 7.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Question 1.
Find the equations of tangent and normal to the following curves at the given point on it:
(i) y = 3x2 – x + 1 at (1, 3)
Solution:
y = 3x2 – x + 1
∴ \(\frac{d y}{d x}=\frac{d}{d x}\) (3x2 – x + 1)
= 3 × 2x – 1 + 0
= 6x – 1
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}\) = 6(1) – 1
= 5
= slope of the tangent at (1, 3).
∴ the equation of the tangent at (1, 3) is
y – 3 = 5(x – 1)
∴ y – 3 = 5x – 5
∴ 5x – y – 2 = 0.
The slope of the normal at (1, 3) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}}=-\frac{1}{5}\)
∴ the equation of the normal at (1, 3) is
y – 3 = \(-\frac{1}{5}\)(x – 1)
∴ 5y – 15 = -x + 1
∴ x + 5y – 16 = 0
Hence, the equations of the tangent and normal are 5x – y – 2 = 0 and x + 5y – 16 = 0 respectively.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

(ii) 2x2 + 3y2 = 5 at (1, 1)
Solution:
2x2 + 3y2 = 5
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1 Q 1 (ii)
= slope of the tangent at (1, 1)
∴ the equation of the tangent at (1, 1) is
y – 1 = \(\frac{-2}{3}\)(x – 1)
∴ 3y – 3 = -2x + 2
∴ 2x + 3y – 5 = 0.
The slope of normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,1)}}=\frac{-1}{\left(\frac{-2}{3}\right)}=\frac{3}{2}\)
∴ the equation of the normal at (1, 1) is
y – 1 = \(\frac{3}{2}\)(x – 1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0
Hence, the equations of the tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively.

(iii) x2 + y2 + xy = 3 at (1, 1)
Solution:
x2 + y2 + xy = 3
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1 Q 1 (iii)
= slope of the tangent at (1, 1)
the equation of the tangent at (1, 1) is
y – 1= -1(x – 1)
∴ y – 1 = -x + 1
∴ x + y = 2
The slope of the normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{a t(1,1)}}\)
= \(\frac{-1}{-1}\)
= 1
∴ the equation of the normal at (1, 1) is y – 1 = 1(x – 1)
∴ y – 1 = x – 1
∴ x – y = 0
Hence, the equations of tangent and normal are x + y = 2 and x – y = 0 respectively.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Question 2.
Find the equations of the tangent and normal to the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Solution:
Let P(x1, y1) be the point on the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Differentiating y = x2 + 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(x2 + 5) = 2x + 0 = 2x
\(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=2 x_{1}\)
= slope of the tangent at (x1, y1)
Let m1 = 2x1
The slope of the line 4x – y + 1 = 0 is
m2 = \(\frac{-4}{-1}\) = 4
Since, the tangent at P(x1, y1) is parallel to the line 4x – y + 1 = 0,
m1 = m2
∴ 2x1 = 4
∴ x1 = 2
Since, (x1, y1) lies on the curve y = x2 + 5, y1 = \(x_{1}^{2}\) + 5
∴ y1 = (2)2 + 5 = 9 ……[x1 = 2]
∴ the coordinates of the point are (2, 9) and the slope of the tangent = m1 = m2 = 4.
∴ the equation of the tangent at (2, 9) is
y – 9 = 4(x – 2)
∴ y – 9 = 4x – 8
∴ 4x – y + 1 = 0
Slope of the normal = \(\frac{-1}{m_{1}}=-\frac{1}{4}\)
∴ the equation of the normal at (2, 9) is
y – 9 = \(-\frac{1}{4}\)(x – 2)
∴ 4y – 36 = -x + 2
∴ x + 4y – 38 = 0
Hence, the equations of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Question 3.
Find the equations of the tangent and normal to the curve y = 3x2 – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Solution:
Let P(x1, y1) be the point on the curve y = 3x2 – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Differentiating y = 3x2 – 3x – 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(3x2 – 3x – 5)
= 3 × 2x – 3 × 1 – 0
= 6x – 3
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=6 x_{1}-3\)
= slope of the tangent at (x1, y1)
Let m1 = 6x1 – 3
The slope of the line 3x – y + 1 = 0
m2 = \(\frac{-3}{-1}\) = 3
Since, the tangent at P(x1, y1) is parallel to the line 3x – y + 1 = 0,
m1 = m2
∴ 6x1 – 3 = 3
∴ 6x1 = 6
∴ x1 = 1
Since, (x1, y1) lies on the curve y = 3x2 – 3x – 5,
\(y_{1}=3 x_{1}{ }^{2}-3 x_{1}-5\), where x1 = 1
= 3(1)2 – 3(1) – 5
= 3 – 3 – 5
= -5
∴ the coordinates of the point are (1, -5) and the slope of the tangent = m1 = m2 = 3.
∴ the equation of the tangent at (1, -5) is
y – (-5) = 3(x – 1)
∴ y + 5 = 3x – 3
∴ 3x – y – 8 = 0
Slope of the normal = \(-\frac{1}{m_{1}}=-\frac{1}{3}\)
∴ the equation of the normal at (1, -5) is
y – (-5) = \(-\frac{1}{3}\)(x – 1)
∴ 3y + 15 = -x + 1
∴ x + 3y + 14 = 0
Hence, the equations of tangent and normal are 3x – y – 8 = 0 and x + 3y + 14 = 0 respectively.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

Question 1.
Obtain the differential equation by eliminating arbitrary constants from the following equations:
(i) y = Ae3x + Be-3x
Solution:
y = Ae3x + Be-3x ……(1)
Differentiating twice w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(i)
This is the required D.E.

(ii) y = \(c_{2}+\frac{c_{1}}{x}\)
Solution:
y = \(c_{2}+\frac{c_{1}}{x}\)
∴ xy = c2x + c1
Differentiating w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(ii)

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

(iii) y = (c1 + c2x) ex
Solution:
y = (c1 + c2x) ex
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(iii).1
This is the required D.E.

(iv) y = c1 e3x+ c2 e2x
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(iv)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(iv).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(iv).2
This is the required D.E.

(v) y2 = (x + c)3
Solution:
y2 = (x + c)3
Differentiating w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q1(v)
This is the required D.E.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

Question 2.
Find the differential equation by eliminating arbitrary constant from the relation x2 + y2 = 2ax.
Solution:
x2 + y2 = 2ax
Differentiating both sides w.r.t. x, we get
2x + 2y\(\frac{d y}{d x}\) = 2a
Substituting value of 2a in equation (1), we get
x2 + y2 = [2x + 2y \(\frac{d y}{d x}\)]x = 2x2 + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) = y2 – x2 is the required D.E.

Question 3.
Form the differential equation by eliminating arbitrary constants from the relation bx + ay = ab.
Solution:
bx + ay = ab
∴ ay = -bx + ab
∴ y = \(-\frac{b}{a} x+b\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{b}{a} \times 1+0=-\frac{b}{a}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 0 is the required D.E.

Question 4.
Find the differential equation whose general solution is x3 + y3 = 35ax.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2 Q4

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

Question 5.
Form the differential equation from the relation x2 + 4y2 = 4b2.
Sol ution:
x2 + 4y2 = 4b2
Differentiating w.r.t. x, we get
2x + 4(2y\(\frac{d y}{d x}\)) = 0
i.e. x + 4y\(\frac{d y}{d x}\) = 0 is the required D.E.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

Question 1.
Determine the order and degree of each of the following differential equations:
(i) \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
Solution:
The given D.E. is \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
This D.E. has highest order derivative \(\frac{d^{2} x}{d t^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=a^{x}\)
Solution:
The given D.E. is \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=a^{x}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 2.
∴ the given D.E. is of order 2 and degree 2.

(iii) \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\) with power 1.
∴ the given D.E. is of order 4 and degree 1.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

(iv) (y'”)2 + 2(y”)2 + 6y’ + 7y = 0
Solution:
The given D.E. is (y”‘)2 + 2(y”)2 + 6y’ + 7y = 0
This can be written as \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+2\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+6 \frac{d y}{d x}+7 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ the given D.E. is of order 3 and degree 2.

(v) \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
Solution:
The given D.E. is \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
On squaring both sides, we get
\(1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d y}{d x}\right)^{3}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}+1=\left(\frac{d y}{d x}\right)^{5}\)
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 5.
∴ the given D.E. is of order 1 and degree 5.

(vi) \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(vii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
i.e., \(\frac{d^{3} y}{d x^{3}}=9^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 1.
∴ the given D.E. is of order 3 and degree 1.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

Question 2.
In each of the following examples, verify that the given function is a solution of the corresponding differential equation:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1 Q2
Solution:
(i) xy = log y + k
Differentiating w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1 Q2(i)
Hence, xy = log y + k is a solution of the D.E. y'(1 – xy) = y2.

(ii) y = xn
Differentiating twice w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1 Q2(ii)
This shows that y = xn is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-n x \frac{d y}{d x}+n y=0\)

(iii) y = ex
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = ex = y
Hence, y = ex is a solution of the D.E. \(\frac{d y}{d x}\) = y.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

(iv) y = 1 – log x
Differentiating w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1 Q2(iv)
Hence, y = 1 – log x is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}=1\)

(v) y = aex + be-x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = a(ex) + b(-e-x) = aex – be-x
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = a(ex) – b(-e-x)
= aex + be-x
= y
Hence, y = aex + be-x is a solution of the D.E. \(\frac{d^{2} y}{d x^{2}}\) = y.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

(vi) ax2 + by2 = 5
Differentiating w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1 Q2(vi)
Hence, ax2 + by2 = 5 is a solution of the D.E.
\(x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}=y\left(\frac{d y}{d x}\right)\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

(I) Choose the correct alternative.

Question 1.
Regression analysis is the theory of
(a) Estimation
(b) Prediction
(c) Both a and b
(d) Calculation
Answer:
(c) Both a and b

Question 2.
We can estimate the value of one variable with the help of other known variable only if they are
(a) Correlated
(b) Positively correlated
(c) Negatively correlated
(d) Uncorrelated
Answer:
(a) Correlated

Question 3.
There are ________ types of regression equation
(a) 4
(b) 2
(c) 3
(d) 1
Answer:
(b) 2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 4.
In the regression equation of Y on X
(a) X is independent and Y is dependent
(b) Y is independent and X is dependent
(c) Both X and Y are independent
(d) Both X and Y are dependent.
Answer:
(a) X is independent and Y is dependent

Question 5.
In the regression equation of X on Y
(a) X is independent and Y is dependent
(b) Y is independent and X is dependent
(c) Both X and Y are independent
(d) Both X and Y are dependent
Answer:
(b) Y is independent and X is dependent

Question 6.
bxy is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(b) Regression coefficient of X on Y

Question 7.
byx is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(a) Regression coefficient of Y on X

Question 8.
‘r’ is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(d) Correlation coefficient between X and Y

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 9.
bxy . byx = _________
(a) v
(b) yx
(c) r2
(d) (yy)2
Answer:
(c) r2

Question 10.
If byx > 1 then bxy is ______
(a) > 1
(b) < 1
(c) > 0
(d) < 0
Answer:
(b) < 1

Question 11.
|bxy + byx| > ______
(a) |r|
(b) 2|r|
(c) r
(d) 2r
Answer:
(b) 2|r|

Question 12.
bxy and byx are ________
(a) Independent of change of origin and scale
(b) Independent of change of origin but not of the scale
(c) Independent of change of scale but not of origin
(d) Affected by change of origin and scale
Answer:
(b) Independent of change of origin but not of the scale

Question 13.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then byx = ________
(a) \(\frac{d}{c} b_{v u}\)
(b) \(\frac{c}{d} b_{v u}\)
(c) \(\frac{a}{b} b_{v u}\)
(d) \(\frac{b}{a} b_{v u}\)
Answer:
(a) \(\frac{d}{c} b_{v u}\)

Question 14.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then bxy = ________
(a) \(\frac{d}{c} b_{u v}\)
(b) \(\frac{c}{d} b_{u v}\)
(c) \(\frac{a}{b} b_{u v}\)
(d) \(\frac{b}{a} b_{u v}\)
Answer:
(b) \(\frac{c}{d} b_{u v}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 15.
Corr(x, x) = ________
(a) 0
(b) 1
(c) -1
(d) can’t be found
Answer:
(b) 1

Question 16.
Corr (x, y) = ________
(a) corr(x, x)
(b) corr(y, y)
(c) corr(y, x)
(d) cov(y, x)
Answer:
(c) corr(y, x)

Question 17.
Corr\(\left(\frac{x-a}{c}, \frac{y-b}{d}\right)\) = -corr(x, y) if,
(a) c and d are opposite in sign
(b) c and d are same in sign
(c) a and b are opposite in sign
(d) a and b are same in sign
Answer:
(a) c and d are opposite in sign

Question 18.
Regression equation of X and Y is
(a) y – \(\bar{y}\) = byx (x – \(\bar{x}\))
(b) x – \(\bar{x}\) = bxy (y – \(\bar{y}\))
(c) y – \(\bar{y}\) = bxy (x – \(\bar{x}\))
(d) x – \(\bar{x}\) = byx (y – \(\bar{y}\))
Answer:
(b) x – \(\bar{x}\) = bxy (y – \(\bar{y}\))

Question 19.
Regression equation of Y and X is
(a) y – \(\bar{y}\) = byx (x – \(\bar{x}\))
(b) x – \(\bar{x}\) = bxy (y – \(\bar{y}\))
(c) y – \(\bar{y}\) = bxy (x – \(\bar{x}\))
(d) x – \(\bar{x}\) = byx (y – \(\bar{y}\))
Solution:
(a) y – \(\bar{y}\) = byx (x – \(\bar{x}\))

Question 20.
byx = ________
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)
(c) \(\frac{1 \sigma_{y}}{r \sigma_{x}}\)
(d) \(\frac{1 \sigma_{y}}{r \sigma_{y}}\)
Answer:
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)

Question 21.
bxy = ________
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)
(c) \(\frac{1 \sigma_{y}}{r \sigma_{x}}\)
(d) \(\frac{1 \sigma_{y}}{r \sigma_{y}}\)
Answer:
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)

Question 22.
Cov (x, y) = ________
(a) Σ(x – \(\bar{x}\))(y – \(\bar{y}\))
(b) \(\frac{\sum(x-\bar{x})(y-\bar{y})}{n}\)
(c) \(\frac{\sum x y}{n}-\bar{x} \bar{y}\)
(d) b and c both
Answer:
(d) b and c both

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 23.
If bxy < 0 and byx < 0 then ‘r’ is ________
(a) > 0
(b) < 0
(c) > 1
(d) not found
Answer:
(b) < 0

Question 24.
If equation of regression lines are 3x + 2y – 26 = 0 and 6x + y – 31 = 0 then means of x and y are ________
(a) (7, 4)
(b) (4, 7)
(c) (2, 9)
(d) (-4, 7)
Answer:
(b) (4, 7)

(II) Fill in the blanks:

Question 1.
If bxy < 0 and byx < 0 then ‘r’ is ________
Answer:
negative

Question 2.
Regression equation of Y on X is ________
Answer:
(y – \(\bar{y}\)) = byx (x – \(\bar{x}\))

Question 3.
Regression equation of X on Y is ________
Answer:
(x – \(\bar{x}\)) = bxy (y – \(\bar{y}\))

Question 4.
There are ______ types of regression equations.
Answer:
2

Question 5.
Corr (x1 – x) = ______
Answer:
-1

Question 6.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then bxy = ______
Answer:
\(\frac{c}{d} b_{u v}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 7.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then byx = ______
Answer:
\(\frac{d}{c} b_{v u}\)

Question 8.
|bxy + byx| ≥ ______
Answer:
2|r|

Question 9.
If byx > 1 then bxy is ______
Answer:
< 1

Question 10.
bxy . byx = ______
Answer:
r2

(III) State whether each of the following is True or False.

Question 1.
Corr (x, x) = 1.
Answer:
True

Question 2.
Regression equation of X on Y is y – \(\bar{y}\) = bxy (x – \(\bar{x}\)).
Answer:
False

Question 3.
Regression equation of Y on X is y – \(\bar{y}\) = byx (x – \(\bar{x}\)).
Answer:
True

Question 4.
Corr (x, y) = Corr (y, x).
Answer:
True

Question 5.
bxy and byx are independent of change of origin and scale.
Answer:
False

Question 6.
‘r’ is the regression coefficient of Y on X.
Answer:
False

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 7.
byx is the correlation coefficient between X and Y.
Answer:
False

Question 8.
If u = x – a and v = y – b then bxy = buv.
Answer:
True

Question 9.
If u = x – a and v = y – b then rxy = ruv.
Answer:
True

Question 10.
In the regression equation of Y on X, byx represents the slope of the line.
Answer:
True

(IV) Solve the following problems.

Question 1.
The data obtained on X, the length of time in weeks that a promotional project has been in progress at a small business, and Y the percentage increase in weekly sales over the period just prior to the beginning of the campaign.
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q1
Find the equation of regression line to predict percentage increase in sales if the company has been in progress for 1.5 weeks.
Solution:
Let u = x – 3, v = y – 15
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q1.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q1.2
∴ Regression equation of Y on X is
(y – \(\bar{y}\)) = byx (x – \(\bar{x}\))
(y – 14.67) = 2.6(x – 2.5)
y – 14.67 = 2.6x – 6.5
y = 2.6x + 8.17
When x = 1.5
y = (2.6)(1.5) + 8.17
= 3.9 + 8.17
= 12.07

Question 2.
The regression equation of y on x is given by 3x + 2y – 26 = 0. Find byx.
Solution:
Given, regression equation of Y on X is
3x + 2y – 26 = 0
∴ 2y = -3x + 26
∴ y = \(\frac{-3}{2}\)x + 13
∴ byx = \(\frac{-3}{2}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 3.
If for a bivariate data \(\bar{x}\) = 10, \(\bar{y}\) = 12, v(x) = 9, σy = 4 and r = 0.6. Estimate y when x = 5.
Solution:
Given, V(x) = 9
∴ σx = 3
byx = \(\frac{r \cdot \sigma_{y}}{\sigma_{x}}\)
= 0.6 × \(\frac{4}{3}\)
= 0.8
∴ Regression equation of Y on X is
(y – \(\bar{y}\)) = byx (x – \(\bar{x}\))
(y – 12) = 0.8(5 – 10)
y – 12 = 0.8(-5)
y – 12 = -4
y = 8

Question 4.
The equation of the line of regression of y on x is v = \(\frac{2}{9} x\) and x on y is x = \(\frac{y}{2}+\frac{7}{6}\). Find (i) r (ii) \(\sigma_{y}^{2} \text { if } \sigma_{x}^{2}=4\).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q4

Question 5.
Identify the regression equations of x on y and y on x from the following equations.
2x + 3y = 6 and 5x + 7y – 12 = 0
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q5
∴ Our assumption is correct
∴ Regression equation of Y on X is 2x + 3y = 6
∴ Regression equation of X on Y is 5x + 7y – 12 = 0

Question 6.
(i) If for a bivariate data byx = -1.2 and bxy = -0.3 then find r.
(ii) From the two regression equations y = 4x – 5 and 3x = 2y + 5, find \(\bar{x}\) and \(\bar{y}\).
Solution:
r2 = byx . bxy
r2 = (-1.2) × (-0.3)
r2 = 0.36
r = ±0.6
Since, byx . bxy are negative, r = -0.6
Also,(\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
y = 4x – 5, 3x = 2y + 5
8x – 2y = 10
3x – 2y = 5
on subtracting,
5x = 5
x = 1
Substituting x = 1 in y = 4x – 5
y = 4(1) – 5
y = -1
∴ \(\bar{x}\) = 1, \(\bar{y}\) = -1

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 7.
The equation of the two lines of regression are 3x + 2y – 26 = 0 and 6x + y – 31 = 0. Find
(i) Means of X and Y
(ii) Correlation coefficient between X on Y
(iii) Estimate of Y for X = 2
(iv) var (X) if var (Y) = 36
Solution:
(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression lines
3x + 2y = 26
6x + y = 31
3x + 2y = 26 …….(i)
12x + 2y = 62 ……..(ii)
on subtracting,
-9x = -36
x = 4
Substituting x = 4 in equation (i)
3(4) + 2y = 26
2y = 14
y = 7
∴ \(\bar{x}\) = 4, \(\bar{y}\) = 7
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q7
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q7.1

Question 8.
Find the line of regression of X on Y for the following data:
n = 8, Σ(xi – x)2 = 36, Σ(yi – y)2 = 44, Σ(xi – x)(yi – y) = 24
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q8

Question 9.
Find the equation of line regression of Y on X for the following data:
n = 8, Σ(xi – \(\bar{x}\))(yi – \(\bar{y}\)) = 120, \(\bar{x}\) = 20, \(\bar{y}\) = 36, σx = 2, σy = 3.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q9
Regression equation of Y on X is
(y – \(\bar{y}\)) = byx (x – \(\bar{x}\))
(y – 36) = 3.75(x – 20)
(y – 36) = 3.75x – 75
y = 3.75x – 39

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 10.
The following result was obtained from records of age (X) and systolic blood pressure (Y) of a group of 10 men.
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q10
and Σ(xi – \(\bar{x}\))(yi – \(\bar{x}\)) = 1120. Find the Prediction of blood pressure of a man of age 40 years.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q10.1
Regression equation of Y on X is
(y – \(\bar{y}\)) = byx (x – \(\bar{x}\))
(y – 140) = 0.7(40 – 50)
y – 140 = 0.7(-10)
y – 140 = -7
∴ y = 133

Question 11.
The equations of two regression lines are 10x – 4y = 80 and 10y – 9x = -40 Find:
(i) \(\bar{x}\) and \(\bar{y}\)
(ii) byx and bxy
(iii) If var(Y) = 36, obtain var(X)
(iv) r
Solution:
(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression
10x – 4y = 80 ……(i)
-9x + 10y = -40 ……..(ii)
50x – 20y = 400
-18x + 20y = -80
32x = 320
x = 10
x = 10 in equation (i)
10(10) – 4y = 80
4y = 20
y = 5
∴ \(\bar{x}\) = 10, \(\bar{y}\) = 5
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q11
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3 IV Q11.1
(iv) r2 = byx . bxy = 0.36
r = ±0.6
Since byx and bxy are positive
∴ r = 0.6

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

Question 12.
If byx = -0.6 and bxy = -0.216 then find correlation coefficient between X and Y comment on it.
Solution:
r2 = byx . bxy
r2 = -0.6 × -0.216
r2 = 0.1296
r = ±√0.1296
r = ± 0.36
Since byx and bxy are negative
r = -0.36

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

(I) Choose the correct alternatives:

Question 1.
Area of the region bounded by the curve x2 = y, the X-axis and the lines x = 1 and x = 3 is
(a) \(\frac{26}{3}\) sq units
(b) \(\frac{3}{26}\) sq units
(c) 26 sq units
(d) 3 sq units
Answer:
(a) \(\frac{26}{3}\) sq units

Question 2.
The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is
(a) 28 sq units
(b) 3 sq unit
(c) \(\frac{28}{3}\) sq units
(d) \(\frac{3}{28}\) sq units
Answer:
(c) \(\frac{28}{3}\) sq units

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

Question 3.
Area of the region bounded by x2 = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is
(a) 63 sq units
(b) \(\frac{3}{56}\) sq units
(c) \(\frac{56}{3}\) sq units
(d) \(\frac{63}{7}\) sq units
Answer:
(c) \(\frac{56}{3}\) sq units

Question 4.
Area of the region bounded by y = x4, x = 1, x = 5 and the X-axis is
(a) \(\frac{3142}{5}\) sq units
(b) \(\frac{3124}{5}\) sq units
(c) \(\frac{3142}{3}\) sq units
(d) \(\frac{3124}{3}\) sq units
Answer:
(b) \(\frac{3124}{5}\) sq units

Question 5.
Using definite integration area of circle x2 + y2 = 25 is
(a) 5π sq units
(b) 4π sq units
(c) 25π sq units
(d) 25 sq units
Answer:
(c) 25π sq units

(II) Fill in the blanks:

Question 1.
Area of the region bounded by y = x4, x = 1, x = 5 and the X-axis is _________
Answer:
\(\frac{3124}{5}\) sq units

Question 2.
Using definite integration area of the circle x2 + y2 = 49 is ___________
Answer:
49π sq units

Question 3.
Area of the region bounded by x2 = 16y, y = 1, y = 4 and the Y-axis lying in the first quadrant is _________
Answer:
\(\frac{56}{3}\) sq units

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

Question 4.
The area of the region bounded by the curve x2 = y, the X-axis and the lines x = 3 and x = 9 is _________
Answer:
234 sq units

Question 5.
The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is __________
Answer:
\(\frac{28}{3}\) sq units

(III) State whether each of the following is True or False.

Question 1.
The area bounded by the curve x = g(y), Y-axis and bounded between the lines y = c and y = d is given by \(\int_{c}^{d} x d y=\int_{y=c}^{y=d} g(y) d y\)
Answer:
True

Question 2.
The area bounded by two curves y = f(x), y = g(x) and X-axis is \(\left|\int_{a}^{b} f(x) d x-\int_{b}^{a} g(x) d x\right|\)
Answer:
False

Question 3.
The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is \(\left|\int_{a}^{b} f(x) d x\right|\)
Answer:
True

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

Question 4.
If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis, and lines x = a, x = b is positive.
Answer:
False

Question 5.
The area of the portion lying above the X-axis is positive.
Answer:
True

(IV) Solve the following:

Question 1.
Find the area of the region bounded by the curve xy = c2, the X-axis, and the lines x = c, x = 2c.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q1
= c2 log(\(\frac{2 c}{c}\))
= c2 . log 2 sq units.

Question 2.
Find the area between the parabolas y2 = 7x and x2 = 7y.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q2
For finding the points of intersection of the two parabolas,
we equate the values of y2 from their equations.
From the equation x2 = 7y, y2 = \(\frac{x^{4}}{49}\)
∴ \(\frac{x^{4}}{49}\) = 7x
∴ x4 = 343x
∴ x4 – 343x = 0
∴ x(x3 – 343) = 0
∴ x = 0 or x3 = 343, i.e. x = 7
When x = 0, y = 0
When x = 7, 7y = 49
∴ y = 7
∴ the points of intersection are O(0, 0) and A(7, 7)
Required area = area of the region OBACO
= (area of the region ODACO) – (area of the region ODABO)
Now, area of the region ODACO = area under the parabola y2 = 7x
i.e. y = √7 √x
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q2.1
Area of the region ODABO = Area under the parabola
x2 = 7y
i.e. y = \(\frac{x^{2}}{7}\)
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q2.2
∴ required area = \(\frac{98}{3}-\frac{49}{3}=\frac{49}{3}\) sq units.

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

Question 3.
Find the area of the region bounded by the curve y = x2 and the line y = 10.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q3
By the symmetry of the parabola,
the required area is twice the area of the region OABCO
Now, the area of the region OABCO
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q3.1

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1.
Solution:
By the symmetry of the ellipse, the required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q4
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q4.1

Question 5.
Find the area of the region bounded by y = x2, the X-axis and x = 1, x = 4.
Solution:
Required area = \(\int_{1}^{4} y d x\), where y = x2
= \(\int_{1}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{1}^{4}=\frac{4^{3}}{3}-\frac{1}{3}=\frac{64-1}{3}\)
= 21 sq units.

Question 6.
Find the area of the region bounded by the curve x2 = 25y, y = 1, y = 4, and the Y-axis.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q6

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

Question 7.
Find the area of the region bounded by the parabola y2 = 25x and the line x = 5.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q7
Given the equation of the parabola is y2 = 25x
∴ y = 5√x …… [∵ IIn first quadrant, y > 0]
Required area = area of the region OQRPO
= 2(area of the region ORPO)
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 IV Q7.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1

Question 1.
Find the area of the region bounded by the following curves, the X-axis, and the given lines:
(i) y = x4, x = 1, x = 5
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q1(i)

(ii) y = \(\sqrt{6 x+4}\), x = 0, x = 2
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q1(ii)

(iii) \(\sqrt{16-x^{2}}\), x = 0, x = 4
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q1(iii)

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1

(iv) 2y = 5x + 7, x = 2, x = 8
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q1(iv)

(v) 2y + x = 8, x = 2, x = 4
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q1(v)

(vi) y = x2 + 1, x = 0, x = 3
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q1(vi)

(vii) y = 2 – x2, x = -1, x = 1
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q1(vii)

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1

Question 2.
Find the area of the region bounded by the parabola y2 = 4x and the line x = 3.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q2
Required area = area of the region OABO
= 2(area of the region OACO)
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q2.1

Question 3.
Find the area of the circle x2 + y2 = 25.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q3
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the circle, y2 = 25 – x2.
In the first quadrant y > 0
∴ y = \(\sqrt{25-x^{2}}\)
∴ area of the circle = 4(area of region OABO)
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q3.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q4
By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 2.
From the equation of the ellipse,
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q4.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1 Q4.2

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

(I) Choose the correct alternative:

Question 1.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\) = ________
(a) 0
(b) 3
(c) 9
(d) -9
Answer:
(a) 0

Question 2.
\(\int_{-2}^{3} \frac{d x}{x+5}\) = _________
(a) -log(\(\frac{8}{3}\))
(b) log(\(\frac{8}{3}\))
(c) log(\(\frac{3}{8}\))
(d) -log(\(\frac{3}{8}\))
Answer:
(b) log(\(\frac{8}{3}\))

Question 3.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\) = _________
(a) log(\(\frac{8}{3}\))
(b) -log(\(\frac{8}{3}\))
(c) \(\frac{1}{2}\) log(\(\frac{8}{3}\))
(d) \(\frac{-1}{2}\) log(\(\frac{8}{3}\))
Answer:
(c) \(\frac{1}{2}\) log(\(\frac{8}{3}\))

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 4.
\(\int_{4}^{9} \frac{d x}{\sqrt{x}}\) = ___________
(a) 9
(b) 4
(c) 2
(d) 0
Answer:
(c) 2

Question 5.
If \(\int_{0}^{a} 3 x^{2} d x=8\), then a = __________
(a) 2
(b) 0
(c) \(\frac{8}{3}\)
(d) a
Answer:
(a) 2

Question 6.
\(\int_{2}^{3} x^{4}\) dx = ________
(a) \(\frac{1}{2}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{5}{211}\)
(d) \(\frac{211}{5}\)
Answer:
(d) \(\frac{211}{5}\)

Question 7.
\(\int_{0}^{2} e^{x}\) dx = _______
(a) e – 1
(b) 1 – e
(c) 1 – e2
(d) e2 – 1
Answer:
(d) e2 – 1

Question 8.
\(\int_{a}^{b} f(x) d x\) = ________
(a) \(\int_{b}^{a} f(x) d x\)
(b) –\(\int_{a}^{b} f(x) d x\)
(c) –\(\int_{b}^{a} f(x) d x\)
(d) \(\int_{0}^{a} f(x) d x\)
Answer:
(c) –\(\int_{b}^{a} f(x) d x\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 9.
\(\int_{-7}^{7} \frac{x^{3}}{x^{2}+7} d x\) = _________
(a) 7
(b) 49
(c) 0
(d) \(\frac{7}{2}\)
Answer:
(c) 0

Question 10.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x\) = _________
(a) \(\frac{7}{2}\)
(b) \(\frac{5}{2}\)
(c) 7
(d) 2
Answer:
(b) \(\frac{5}{2}\)

(II) Fill in the blanks:

Question 1.
\(\int_{0}^{2} e^{x} d x\) = ________
Answer:
e2 – 1

Question 2.
\(\int_{2}^{3} x^{4} d x\) = __________
Answer:
\(\frac{211}{5}\)

Question 3.
\(\int_{0}^{1} \frac{d x}{2 x+5}\) = ____________
Answer:
\(\frac{1}{2} \log \left(\frac{7}{5}\right)\)

Question 4.
If \(\int_{0}^{a} 3 x^{2} d x\) = 8, then a = _________
Answer:
2

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 5.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\) = _________
Answer:
2

Question 6.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\) = _________
Answer:
\(\frac{1}{2} \log \left(\frac{8}{3}\right)\)

Question 7.
\(\int_{-2}^{3} \frac{d x}{x+5}\) = _________
Answer:
\(\log \left(\frac{8}{3}\right)\)

Question 8.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\) = _____________
Answer:
o

(III) State whether each of the following is True or False:

Question 1.
\(\int_{a}^{b} f(x) d x=\int_{-b}^{-a} f(x) d x\)
Answer:
True

Question 2.
\(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(t) d t\)
Answer:
True

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 3.
\(\int_{0}^{a} f(x) d x=\int_{a}^{0} f(a-x) d x\)
Answer:
False

Question 4.
\(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(x-a-b) d x\)
Answer:
False

Question 5.
\(\int_{-5}^{5} \frac{x^{3}}{x^{2}+7} d x=0\)
Answer:
True

Question 6.
\(\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x=\frac{1}{2}\)
Answer:
True

Question 7.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x=\frac{9}{2}\)
Answer:
False

Question 8.
\(\int_{4}^{7} \frac{(11-x)^{2}}{(11-x)^{2}+x^{2}} d x=\frac{3}{2}\)
Answer:
True

(IV) Solve the following:

Question 1.
\(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q1
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q1.1

Question 2.
\(\int_{1}^{2} \frac{x+3}{x(x+2)} d x\)
Solution:
Let I = \(\int_{1}^{2} \frac{x+3}{x(x+2)} d x\)
Let \(\frac{x+3}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}\)
∴ x + 3 = A(x + 2) + Bx
Put x = 0, we get
3 = A(2) + B(0)
∴ A = \(\frac{3}{2}\)
Put x + 2 = 0, i.e. x = -2, we get
-2 + 3 = A(0) + B(-2)
∴ 1 = -2B
∴ B = \(-\frac{1}{2}\)
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 3.
\(\int_{1}^{3} x^{2} \log x d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q3

Question 4.
\(\int_{0}^{1} e^{x^{2}} \cdot x^{3} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q4
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q4.1

Question 5.
\(\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q5

Question 6.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q6

Question 7.
\(\int_{-2}^{3} \frac{1}{x+5} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q7

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 8.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q8

Question 9.
\(\int_{0}^{1} \frac{x^{2}+3 x+2}{\sqrt{x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q9

Question 10.
\(\int_{3}^{5} \frac{d x}{\sqrt{x+4}+\sqrt{x-2}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q10

Question 11.
\(\int_{2}^{3} \frac{x}{x^{2}+1} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q11
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q11.1

Question 12.
\(\int_{1}^{2} x^{2} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q12

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 13.
\(\int_{-4}^{-1} \frac{1}{x} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q13

Question 14.
\(\int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q14
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q14.1

Question 15.
\(\int_{0}^{4} \frac{1}{\sqrt{x^{2}+2 x+3}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q15
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q15.1

Question 16.
\(\int_{2}^{4} \frac{x}{x^{2}+1} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q16

Question 17.
\(\int_{0}^{1} \frac{1}{2 x-3} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q17

Question 18.
\(\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q18
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q18.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

Question 19.
\(\int_{1}^{2} \frac{d x}{x(1+\log x)^{2}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q19

Question 20.
\(\int_{0}^{9} \frac{1}{1+\sqrt{x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q20
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6 IV Q20.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2

Evaluate the following integrals:

Question 1.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q1

Question 2.
\(\int_{0}^{a} x^{2}(a-x)^{3 / 2} d x\)
Solution:
We use the property
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2

Question 3.
\(\int_{1}^{3} \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q3

Question 4.
\(\int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q4

Question 5.
\(\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q5

Question 6.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q6

Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2

Question 7.
\(\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q7
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q7.1

Question 8.
\(\int_{0}^{1} x(1-x)^{5} d x\)
Solution:
We use the property
Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2 Q8