Class 12

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.3 Answers Maharashtra Board

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

Std 12 Maths 1 Exercise 4.3 Solutions Commerce Maths

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.

(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:

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.

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.

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.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 4.1 Solutions
• 12th Commerce Maths Exercise 4.2 Solutions
• 12th Commerce Maths Exercise 4.3 Solutions
• 12th Commerce Maths Exercise 4.4 Solutions
• 12th Commerce Maths Miscellaneous Exercise 4 Solutions

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.2 Answers Maharashtra Board

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

Std 12 Maths 1 Exercise 4.2 Solutions Commerce Maths

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.

(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, ∞)

(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 < \(\frac{3}{2}\).

(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.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 4.1 Solutions
• 12th Commerce Maths Exercise 4.2 Solutions
• 12th Commerce Maths Exercise 4.3 Solutions
• 12th Commerce Maths Exercise 4.4 Solutions
• 12th Commerce Maths Miscellaneous Exercise 4 Solutions

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.1 Answers Maharashtra Board

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

Std 12 Maths 1 Exercise 4.1 Solutions Commerce Maths

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.

(ii) 2x² + 3y² = 5 at (1, 1)
Solution:
2x² + 3y² = 5
Differentiating both sides w.r.t. x, we get

= 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

= 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.

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.

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.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 4.1 Solutions
• 12th Commerce Maths Exercise 4.2 Solutions
• 12th Commerce Maths Exercise 4.3 Solutions
• 12th Commerce Maths Exercise 4.4 Solutions
• 12th Commerce Maths Miscellaneous Exercise 4 Solutions

Differentiation Class 12 Commerce Maths 1 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Miscellaneous Exercise 3 Questions and Answers.

Std 12 Maths 1 Miscellaneous Exercise 3 Solutions Commerce Maths

(I) Choose the correct alternative:

Question 1.
If y = (5x³ – 4x² – 8x)⁹, then \(\frac{d y}{d x}\) = ___________
(a) 9(5x³ – 4x² – 8x)⁸ (15x² – 8x – 8)
(b) 9(5x³ – 4x² – 8x)⁹ (15x² – 8x – 8)
(c) 9(5x³ – 4x² – 8x)⁸ (5x² – 8x – 8)
(d) 9(5x³ – 4x² – 8x)⁹ (5x² – 8x – 8)
Answer:
(a) 9(5x³ – 4x² – 8x)⁸ (15x² – 8x – 8)

Question 2.
If y = \(\sqrt{x+\frac{1}{x}}\), then \(\frac{d y}{d x}\) = ?
(a) \(\frac{x^{2}-1}{2 x^{2} \sqrt{x^{2}+1}}\)
(b) \(\frac{1-x^{2}}{2 x^{2} \sqrt{x^{2}+1}}\)
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
(d) \(\frac{1-x^{2}}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Answer:
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Hint:

Question 3.
If y = \(e^{\log x}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{e^{\log x}}{x}\)
(b) \(\frac{1}{x}\)
(c) 0
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{e^{\log x}}{x}\)

Question 4.
If y = 2x² + 2² + a², then \(\frac{d y}{d x}\) = ?
(a) x
(b) 4x
(c) 2x
(d) -2x
Answer:
(b) 4x

Question 5.
If y = 5^x . x⁵, then \(\frac{d y}{d x}\) = ?
(a) 5^x . x⁴(5 + log 5)
(b) 5^x . x⁵(5 + log 5)
(c) 5^x . x⁴(5 + x log 5)
(d) 5^x . x⁵(5 + x log 5)
Answer:
(c) 5^x . x⁴(5 + x log 5)

Question 6.
If y = \(\log \left(\frac{e^{x}}{x^{2}}\right)\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{2-x}{x}\)
(b) \(\frac{x-2}{x}\)
(c) \(\frac{e-x}{ex}\)
(d) \(\frac{x-e}{ex}\)
Answer:
(b) \(\frac{x-2}{x}\)
Hint:

Question 7.
If ax² + 2hxy + by² = 0, then \(\frac{d y}{d x}\) = ?
(a) \(\frac{(a x+h y)}{(h x+b y)}\)
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)
(c) \(\frac{(a x-h y)}{(h x+b y)}\)
(d) \(\frac{(2 a x+h y)}{(h x+3 b y)}\)
Answer:
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)

Question 8.
If x⁴ . y⁵ = (x + y)^(m+1) and \(\frac{d y}{d x}=\frac{y}{x}\) then m = ?
(a) 8
(b) 4
(c) 5
(d) 20
Answer:
(a) 8
Hint:
If x^p . y^q = (x + y)^p+q, then \(\frac{d y}{d x}=\frac{y}{x}\)
∴ m + 1 = 4 + 5 = 9
∴ m = 8.

Question 9.
If x = \(\frac{e^{t}+e^{-t}}{2}\), y = \(\frac{e^{t}-e^{-t}}{2}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{-y}{x}\)
(b) \(\frac{y}{x}\)
(c) \(\frac{-x}{y}\)
(d) \(\frac{x}{y}\)
Answer:
(d) \(\frac{x}{y}\)
Hint:

Question 10.
If x = 2at², y = 4at, then \(\frac{d y}{d x}\) = ?
(a) \(-\frac{1}{2 a t^{2}}\)
(b) \(\frac{1}{2 a t^{3}}\)
(c) \(\frac{1}{t}\)
(d) \(\frac{1}{4 a t^{3}}\)
Answer:
(c) \(\frac{1}{t}\)

(II) Fill in the blanks:

Question 1.
If 3x²y + 3xy² = 0 then \(\frac{d y}{d x}\) = …………
Answer:
-1
Hint:
3x²y + 3xy² = 0
∴ 3xy(x + y) = 0
∴ x + y = 0
∴ y = -x
∴ \(\frac{d y}{d x}\) = -1

Question 2.
If x^m . yⁿ = (x+y)^(m+n) then \(\frac{d y}{d x}=\frac{\ldots \ldots}{x}\)
Answer:
y

Question 3.
If 0 = log(xy) + a then \(\frac{d y}{d x}=\frac{-y}{\ldots . .}\)
Answer:
x

Question 4.
If x = t log t and y = t^t then \(\frac{d y}{d x}\) = …………
Answer:
y
Hint:
x = t log t = log t^t = log y
∴ 1 = \(\frac{1}{y} \cdot \frac{d y}{d x}\)
∴ \(\frac{d y}{d x}\) = y

Question 5.
If y = x . log x then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{1}{x}\)

Question 6.
If y = [log(x)]² then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{2(1-\log x)}{x^{2}}\)
Hint:

Question 7.
If x = y + \(\frac{1}{y}\) then \(\frac{d y}{d x}\) = …………
Answer:
\(\frac{y^{2}}{y^{2}-1}\)
Hint:

Question 8.
If y = e^ax, then x.\(\frac{d y}{d x}\) = …………
Answer:
axy

Question 9.
If x = t . log t, y = t^t then \(\frac{d y}{d x}\) = …………
Answer:
y

Question 10.
If y = \(\left(x+\sqrt{x^{2}-1}\right)^{m}\) then \(\sqrt{\left(x^{2}-1\right)} \frac{d y}{d x}\) = ………
Answer:
my
Hint:

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

Question 1.
If f’ is the derivative of f, then the derivative of the inverse of f is the inverse of f’.
Answer:
False

Question 2.
The derivative of log_a x, where a is constant is \(\frac{1}{x \cdot \log a}\).
Answer:
True

Question 3.
The derivative of f(x) = a^x, where a is constant is x . a^x-1
Answer:
False

Question 4.
The derivative of a polynomial is polynomial.
Answer:
True

Question 5.
\(\frac{d}{d x}\left(10^{x}\right)=x \cdot 10^{x-1}\)
Answer:
False

Question 6.
If y = log x, then \(\frac{d y}{d x}=\frac{1}{x}\).
Answer:
True

Question 7.
If y = e², then \(\frac{d y}{d x}\) = 2e.
Answer:
False

Question 8.
The derivative of a^x is a^x. log a.
Answer:
True

Question 9.
The derivative of x^m . yⁿ = (x + y)^(m+n) is \(\frac{x}{y}\)
Answer:
False

(IV) Solve the following:

Question 1.
If y = (6x³ – 3x² – 9x)¹⁰, find \(\frac{d y}{d x}\)
Solution:
Given y = (6x³ – 3x² – 9x)¹⁰

Question 2.
If y = \(\sqrt[5]{\left(3 x^{2}+8 x+5\right)^{4}}\), find \(\frac{d y}{d x}\).
Solution:

Question 3.
If y = [log(log(log x))]², find \(\frac{d y}{d x}\).
Solution:

Question 4.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = 25 + 30x – x².
Solution:

Question 5.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = \(\frac{5 x+7}{2 x-13}\)
Solution:

Question 6.
Find \(\frac{d y}{d x}\) if y = x^x.
Solution:
y = x^x
∴ log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

Question 7.
Find \(\frac{d y}{d x}\) if y = \(2^{x^{x}}\).
Solution:

Question 8.
Find \(\frac{d y}{d x}\), if y = \(\sqrt{\frac{(3 x-4)^{3}}{(x+1)^{4}(x+2)}}\)
Solution:

Question 9.
Find \(\frac{d y}{d x}\) if y = x^x + (7x – 1)^x
Solution:

Question 10.
If y = x³ + 3xy² + 3x²y, find \(\frac{d y}{d x}\).
Solution:
y = x³ + 3xy² + 3x²y
Differentiating both sides w.r.t. x, we get

Question 11.
If x³ + y² + xy = 7, find \(\frac{d y}{d x}\).
Solution:
x³ + y² + xy = 7
Differentiating both sides w.r.t. x, we get

Question 12.
If x³y³ = x² – y², find \(\frac{d y}{d x}\).
Solution:
x³y³ = x² – y²
Differentiating both sides w.r.t. x, we get

Question 13.
If x⁷ . y⁹ = (x + y)¹⁶, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:

Question 14.
If x^a . y^b = (x + y)^a+b, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:

Question 15.
Find \(\frac{d y}{d x}\) if x = 5t², y = 10t.
Solution:
x = 5t², y = 10t
Differentiating x and y w.r.t. t, we get

Question 16.
Find \(\frac{d y}{d x}\) if x = e^3t, y = \(e^{\sqrt{t}}\).
Solution:
x = e^3t, y = \(e^{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get

Question 17.
Differentiate log(1 + x²) with respect to a^x.
Solution:
Let u = log(1 + x²) and v = a^x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

Question 18.
Differentiate e^(4x+5) with resepct to 10^4x.
Solution:
Let u = e^(4x+5) and v = 10^4x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

Question 19.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = log x.
Solution:
y = log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}\)

Question 20.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = 2at, x = at².
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

Question 21.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = x² . e^x
Solution:
y = x² . e^x
Differentiating w.r.t. x, we get

= e^x (2x + 2 + x² + 2x)
= e^x (x² + 4x + 2).

Question 22.
If x² + 6xy + y² = 10, then show that \(\frac{d^{2} y}{d x^{2}}=\frac{80}{(3 x+y)^{3}}\).
Solution:
x² + 6xy + y² = 10 ……..(1)
Differentiating both sides w.r.t. a, we get

Question 23.
If ax² + 2hxy + by² = 0, then show that \(\frac{d^{2} y}{d x^{2}}\) = 0.
Solution:
ax² + 2hxy + by² = 0 ……..(1)
∴ ax² + hxy + hxy + by² = 0
∴ x(ax + hy) + y(hx + by) = 0
∴ x(ax + hy) = -y(hx + by)
∴ \(\frac{a x+h y}{h x+b y}=-\frac{y}{x}\) …….(2)
Differentiating (1) w.r.t. x, we get

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Exercise 3.4 Solutions
• 12th Commerce Maths Exercise 3.5 Solutions
• 12th Commerce Maths Exercise 3.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Differentiation Class 12 Commerce Maths 1 Chapter 3 Exercise 3.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.6 Questions and Answers.

Std 12 Maths 1 Exercise 3.6 Solutions Commerce Maths

1. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = √x
Solution:

Question 2.
y = x⁵
Solution:

Question 3.
y = x^-7
Solution:

2. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = e^x
Solution:

Question 2.
y = e^(2x+1)
Solution:

Question 3.
y = e^{log x}
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Exercise 3.4 Solutions
• 12th Commerce Maths Exercise 3.5 Solutions
• 12th Commerce Maths Exercise 3.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Differentiation Class 12 Commerce Maths 1 Chapter 3 Exercise 3.5 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.5 Questions and Answers.

Std 12 Maths 1 Exercise 3.5 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if:

Question 1.
x = at², y = 2at
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

Question 2.
x = 2at², y = at⁴
Solution:

Question 3.
x = e^3t, y = e^(4t+5)
Solution:
x = e^3t, y = e^(4t+5)
Differentiating x and y w.r.t. t, we get

2. Find \(\frac{d y}{d x}\) if:

Question 1.
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\)
Solution:
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\) ……(1)
Differentiating x and y w.r.t. u, we get,

Question 2.
x = \(\sqrt{1+u^{2}}\), y = log(1 + u²)
Solution:
x = \(\sqrt{1+u^{2}}\), y = log(1 + u²) ……(1)
Differentiating x and y w.r.t. u, we get,

Question 3.
Differentiate 5^x with respect to log x.
Solution:
Let u = 5^x and v = log x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(5^{x}\right)=5^{x} \cdot \log 5\)

3. Solve the following:

Question 1.
If x = \(a\left(1-\frac{1}{t}\right)\), y = \(a\left(1+\frac{1}{t}\right)\), then show that \(\frac{d y}{d x}\) = -1
Solution:

Question 2.
If x = \(\frac{4 t}{1+t^{2}}\), y = \(3\left(\frac{1-t^{2}}{1+t^{2}}\right)\), then show that \(\frac{d y}{d x}=-\frac{9 x}{4 y}\)
Solution:

Question 3.
If x = t . log t, y = t^t, then show that \(\frac{d y}{d x}\) – y = 0.
Solution:
x = t log t
Differentiating w.r.t. t, we get

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Exercise 3.4 Solutions
• 12th Commerce Maths Exercise 3.5 Solutions
• 12th Commerce Maths Exercise 3.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

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Std 12 Maths 1 Exercise 3.4 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if:

Question 1.
√x + √y = √a
Solution:
√x + √y = √a
Differentiating both sides w.r.t. x, we get

Question 2.
x³ + y³ + 4x³y = 0
Solution:
x³ + y³ + 4x³y = 0
Differentiating both sides w.r.t. x, we get

Question 3.
x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiating both sides w.r.t. x, we get

2. Find \(\frac{d y}{d x}\) if:

Question 1.
y.e^x + x.e^y = 1
Solution:
y.e^x + x.e^y = 1
Differentiating both sides w.r.t. x, we get

Question 2.
x^y = e^(x-y)
Solution:
x^y = e^(x-y)
∴ log x^y = log e^(x-y)
∴ y log x = (x – y) log e
∴ y log x = x – y …..[∵ log e = 1]
∴ y + y log x = x
∴ y(1 + log x) = x
∴ y = \(\frac{x}{1+\log x}\)

Question 3.
xy = log(xy)
Solution:
xy = log (xy)
∴ xy = log x + log y
Differentiating both sides w.r.t. x, we get

3. Solve the following:

Question 1.
If x⁵ . y⁷ = (x + y)¹², then show that \(\frac{d y}{d x}=\frac{y}{x}\)
Solution:
x⁵ . y⁷ = (x + y)¹²
∴ log(x⁵ . y⁷) = log(x + y)¹²
∴ log x⁵ + log y⁷ = log(x + y)¹²
∴ 5 log x + 7 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get

Question 2.
If log(x + y) = log(xy) + a, then show that \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}}\)
Solution:
log (x + y) = log (xy) + a
∴ log(x + y) = log x + log y + a
Differentiating both sides w.r.t. x, we get

Question 3.
If e^x + e^y = e^(x+y), then show that \(\frac{d y}{d x}=-e^{y-x}\).
Solution:
e^x + e^y = e^(x+y) ……….(1)
Differentiating both sides w.r.t. x, we get

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Exercise 3.4 Solutions
• 12th Commerce Maths Exercise 3.5 Solutions
• 12th Commerce Maths Exercise 3.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

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Std 12 Maths 1 Exercise 3.3 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if:

Question 1.
y = \(x^{x^{2 x}}\)
Solution:

Question 2.
y = \(x^{e^{x}}\)
Solution:

Question 3.
y = \(e^{x^{x}}\)
Solution:

2. Find \(\frac{d y}{d x}\) if:

Question 1.
y = \(\left(1+\frac{1}{x}\right)^{x}\)
Solution:

Question 2.
y = (2x + 5)^x
Solution:

Question 3.
y = \(\sqrt[3]{\frac{(3 x-1)}{(2 x+3)(5-x)^{2}}}\)
Solution:

3. Find \(\frac{d y}{d x}\) if:

Question 1.
y = (log x)^x + x^{log x}
Solution:

Question 2.
y = x^x + a^x
Solution:

Question 3.
y = \(10^{x^{x}}+10^{x^{10}}+10^{10^{x}}\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Exercise 3.4 Solutions
• 12th Commerce Maths Exercise 3.5 Solutions
• 12th Commerce Maths Exercise 3.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Differentiation Class 12 Commerce Maths 1 Chapter 3 Exercise 3.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.2 Questions and Answers.

Std 12 Maths 1 Exercise 3.2 Solutions Commerce Maths

1. Find the rate of change of demand (x) of a commodity with respect to price (y) if:

Question 1.
y = 12 + 10x + 25x²
Solution:
Given y = 12 + 10x + 25x²

Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{1}{10+50 x}\)

Question 2.
y = 18x + log(x – 4)
Solution:
Given y = 18x + log (x – 4)

Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{x-4}{18 x-71}\)

Question 3.
y = 25x + log(1 + x²)
Solution:
Given y = 25x + log(1 + x²)

Hence, the rate of change of demand (x) with respect to price (y) \(\frac{d x}{d y}=\frac{1+x^{2}}{25 x^{2}+2 x+25}\)

2. Find the marginal demand of a commodity where demand is x and price is y.

Question 1.
y = xe^-x + 7
Solution:

Question 2.
y = \(\frac{x+2}{x^{2}+1}\)
Solution:

Question 3.
y = \(\frac{5 x+9}{2 x-10}\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Exercise 3.4 Solutions
• 12th Commerce Maths Exercise 3.5 Solutions
• 12th Commerce Maths Exercise 3.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Differentiation Class 12 Commerce Maths 1 Chapter 3 Exercise 3.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.1 Questions and Answers.

Std 12 Maths 1 Exercise 3.1 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if,

Question 1.
y = \(\sqrt{x+\frac{1}{x}}\)
Solution:

Question 2.
y = \(\sqrt[3]{a^{2}+x^{2}}\)
Solution:

Question 3.
y = (5x³ – 4x² – 8x)⁹
Solution:

2. Find \(\frac{d y}{d x}\) if:

Question 1.
y = log(log x)
Solution:
Given y = log(log x)
Let u = log x
Then y = log u

Question 2.
y = log(10x⁴ + 5x³ – 3x² + 2)
Solution:
Given y = log(10x⁴ + 5x³ – 3x² + 2)
Let u = 10x⁴ + 5x³ – 3x² + 2
Then y = log u

Question 3.
y = log(ax² + bx + c)
Solution:
Given y = log(ax² + bx + c)
Let u = ax² + bx + c
Then y = log u

3. Find \(\frac{d y}{d x}\) if:

Question 1.
y = \(e^{5 x^{2}-2 x+4}\)
Solution:

Question 2.
y = \(a^{(1+\log x)}\)
Solution:

Question 3.
y = \(5^{(x+\log x)}\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Exercise 3.4 Solutions
• 12th Commerce Maths Exercise 3.5 Solutions
• 12th Commerce Maths Exercise 3.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions