250+ TOP MCQs on Derivatives of Area and Answers

Differential and Integral Calculus Multiple Choice Questions on “Derivatives of Area”.

1. What is the maximum area of the rectangle with perimeter 620 mm?
a) 24,025 mm²
b) 22,725 mm²
c) 24,000 mm²
d) 24,075 mm²
View Answer

Answer: a
Explanation: Let x be the length of the rectangle and y be the width of the rectangle. Then, Area A is,
A=x*y …………………………………………………. (1)
Given: Perimeter of the rectangle is 620 mm. Therefore,
P=2(x+y)
620=2(x+y)
x+y=310
y=310-x
We can now substitute the value of y in (1)
A=x*(310-x)
A=310x-x²
To find maximum value we need derivative of A,
dA/dx=310-2x
To find maximum value, (frac{dA}{dx}=0 )
310-2x=0
2x=310
x=155 mm
Therefore, when the value of x=155 mm and the value of y=310-155=155 mm, the area of the rectangle is maximum, i.e., A=155*155=24,025 mm²

2. Power Rule of the derivative states that, (frac{d(x^n)}{dx}=nx^{n-1}.)
a) True
b) False
View Answer

Answer: b
Explanation: Power Rule is given by, (frac{d(x^n)}{dx}=nx^{n-1}.)
For example, (frac{d(x^4)}{dx}=4x^3. )

3. Which of the following trigonometric function derivatives is correct?
a) (frac{d(sinx)}{dx}=-cosx)
b) (frac{d(secx)}{dx}=secxtanx)
c) (frac{d(tanx)}{dx}=cot^2 x)
d) (frac{d(cosx)}{dx}=sinx)
View Answer

4. Product Rule of the derivative is given by ________
a) (frac{d(fg)}{dx}=frac{d(g)}{dx}*frac{d(f)}{dx})
b) (frac{df}{dx}*frac{dg}{dx}=frac{d(g)}{dx}+frac{d(f)}{dx})
c) (frac{d(fg)}{dx}=f frac{d(g)}{dx}*g frac{d(f)}{dx})
d) (frac{d(fg)}{dx}=f frac{d(g)}{dx}+g frac{d(f)}{dx})
View Answer

Answer: d
Explanation: Product Rule of derivative is given by,
( frac{d(fg)}{dx}=f frac{d(g)}{dx}+g frac{d(f)}{dx})

5. What is the derivative of (frac{3x^2}{(1-sinx)})?
a) (frac{6xsinx-3x^2 cosx}{(1-sinx)^2} )
b) (frac{6x}{1-sinx} )
c) (frac{6xsinx-3x^2 cosx-6x}{(1-sinx)^2} )
d) (frac{6xsinx-3cosx-6}{(1-sinx)^2} )
View Answer

Answer: c
Explanation: Given: y = (frac{3x^2}{(1-sinx)})
(frac{dy}{dx}=frac{3x^2(frac{d(1-sinx)}{dx})-(frac{d(3x^2)}{dx})(1-sinx)}{(1-sinx)^2})
(frac{dy}{dx}=frac{3x^2(-cosx)-6x(1-sinx)}{(1-sinx)^2})
(frac{dy}{dx}=frac{6xsinx-3x^2 cosx-6x}{(1-sinx)^2})

6. Which of the following is a type of Iterative method of solving non-linear equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods
View Answer

Answer: b
Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

7. The half-interval method in numerical analysis is also known as __________
a) Newton-Raphson method
b) Regula Falsi method
c) Taylor’s method
d) Bisection method
View Answer

Answer: d
Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.

8. Given (∫_0^8 x^{frac{1}{3}} dx,) find the error in approximating the integral using Simpson’s (frac{1}{3}) Rule with n=4.
a) 1.8
b) 2.9
c) 0.3
d) 0.35
View Answer

Answer: d
Explanation: Given: (∫_0^8 x^{frac{1}{3}} dx,), n = 3,
Let (f(x)= x^{frac{1}{3}},)
(∆x = frac{b-a}{2}=frac{8-0}{2}=4)………………since b=8, a=0 (limits of the given integral)
Hence endpoints x_i have coordinates {0, 2, 4, 6, 8}.
Calculating the function values at x_i, we get,
(f(0)= 0^{frac{1}{3}}=0)
(f(2)= 2^{frac{1}{3}})
(f(4)= 4^{frac{1}{3}})
(f(6)= 6^{frac{1}{3}})
(f(8)= 8^{frac{1}{3}} =2)
Substituting these values in the formula,
(∫_0^8x^{frac{1}{3}} dx ≈ frac{∆x}{3} [f(0)+4f(2)+2f(4)+4f(6)+f(8)])
(≈ frac{2}{3}[0+4(2^{frac{1}{3}})+ 2(4^{frac{1}{3}})+ 4(6^{frac{1}{3}})+2] ≈ 11.65)
Actual integral value,
(∫_0^8x^{frac{1}{3}}dx = left(frac{x^{frac{4}{3}}}{frac{4}{3}}right)_0^8=12)
Error in approximating the integral = 12 – 11.65 = 0.35

9. A sphere with the dimensions is shown in the figure. What is the error that can be incorporated in the radius such that the volume will not change more than 8%?
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a) 0.127%
b) 0.02546%
c) 0.002546%
d) 0.2546%
View Answer

Answer: b
Explanation: We know that volume of the sphere is,
(V = frac{4}{3} πR^3 )
Differentiating the above equation with respect to R we get,
(frac{dV}{dR} = frac{4}{3} π×3R^2=4πR^2)
Since the volume of the sphere should not exceed more than 8%,
(dR=frac{dV}{4πR^2}=frac{0.08}{4π(5)^2}=0.0002546)
Error in radius = 0.02546%

10. Half Range Fourier Series contains either sine or cosine terms.
a) True
b) False
View Answer

Answer: a
Explanation: A series which contains only sine or cosine terms is called Half Range Fourier Sine Series or Cosine Series respectively.

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