# Mathematics Questions and Answers – Derivatives Application – Approximations

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives Application – Approximations”.

1. Find the approximate value of $$\sqrt{64.3}$$.
a) 8.0675
b) 8.03465
c) 8.01875
d) 8.0665

Explanation: Let y=$$\sqrt{x}$$. Let x=64 and Δx=0.3
Then, Δy=$$\sqrt{x+Δx}-\sqrt{x}$$
Δy=$$\sqrt{64.3}-\sqrt{64}$$
$$\sqrt{64.3}$$=Δy+8
dy is approximately equal to Δy is equal to:
dy=$$\frac{dy}{dx}$$Δx
dy=$$\frac{1}{2\sqrt{x}}$$.Δx
dy=$$\frac{1}{2\sqrt{64}}$$ (0.3)
dy=0.3/16=0.01875
∴ The approximate value of $$\sqrt{64.3}$$ is 8+0.01875=8.01875

2. Find the approximate value of $$\sqrt{49.1}$$.
a) 7.0142
b) 7.087942
c) 7.022
d) 7.00714

Explanation: Let y=$$\sqrt{49.1}$$. Let x=49 and Δx=0.1
Then, Δy=$$\sqrt{x+Δx}-\sqrt{x}$$
Δy=$$\sqrt{49.1}-\sqrt{49}$$
$$\sqrt{49.1}$$=Δy+7
dy is approximately equal to Δy is equal to
dy=$$\frac{dy}{dx}$$Δx
dy=$$\frac{1}{(2\sqrt{x})}$$.Δx
dy=$$\frac{1}{(2\sqrt{49})}$$ (0.1)
dy=0.1/14=0.00714
∴ The approximate value of $$\sqrt{49.1}$$ is 7+0.00714=7.00714

3. Find the approximate value of f(5.03), where f(x)=4x2-7x+2.
a) 67.99
b) 56.99
c) 67.66
d) 78.09

Explanation: Let x=5 and Δx=0.03
Then, f(x+Δx)=4(x+Δx)2-7(x+Δx)+2
Δy=f(x+Δx)-f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f’ (x)Δx
⇒f(x+Δx)=f(x)+f’ (x)Δx
f(5.03)=(4(5)2-7(5)+2)+(8(5)-7)(0.03) (∵f’ (x)=8x-7)
f(5.03)=(100-35+2)+(40-7)(0.03)
f(5.03)=67+33(0.03)
f(5.03)=67+0.99=67.99

4. Find the approximate value of $$\sqrt{11}$$.
a) 3.34
b) 3.934
c) 3.0034
d) 3.544

Explanation: Let y=$$\sqrt{x}$$. Let x=9 and Δx=2
Then, Δy=$$\sqrt{x+Δx}-\sqrt{x}$$
Δy=$$\sqrt{11}-\sqrt{9}$$
$$\sqrt{11}$$=Δy+3
dy is approximately equal to Δy is equal to
dy=$$\frac{dy}{dx}$$Δx
dy=$$\frac{1}{(2\sqrt{x})}$$.Δx
dy=$$\frac{1}{(2\sqrt{9})}(2)$$
dy=2/6=0.34
∴ The approximate value of $$\sqrt{11}$$ is 3+0.34=3.34.

5. What will be the approximate change in the surface area of a cube of side xm caused by increasing the side by 2%.
a) 0.24x
b) 2.4x2
c) 0.4x2
d) 0.24x2

Explanation: Let the edge of the cube be x. Given that dx or Δx is equal to 0.02x(2%).
The surface area of the cube is A=6x2
Differentiating w.r.t x, we get
$$\frac{dA}{dx}$$=12x
dA=($$\frac{dA}{dx}$$)Δx=12x(0.02x)=0.24x2
Hence, the approximate change in volume is 0.24x2.

6. Find the approximate value of f(4.04), where f(x)=7x3+6x2-4x+3.
a) 346.2
b) 544.345
c) 546.2
d) 534.2

Explanation: Let x=4 and Δx=0.04
Then, f(x+Δx)=7(x+Δx)3+7(x+Δx)2-4(x+Δx)+3
Δy=f(x+Δx)-f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f'(x)Δx
⇒f(x+Δx)=f(x)+f’ (x)Δx
Here, f'(x)=21x2+12x-4
f(4.04)=(7(4)3+6(4)2-4(4)+3)+(21(4)2+12(4)-4)(0.04)
f(4.04)=(448+96-16+3)+(336+48-4)(0.04)
f(4.04)=531+380(0.04)=546.2

7. Find the approximate value of (127)1/3.
a) 5.0267
b) 2.0267
c) 8.0267
d) 5.04

Explanation: Let y=(x)1/3. Let x=125 and Δx=2
Then, Δy=(x+Δx)1/3-x1/3
Δy=(127)1/3-(125)1/3
(127)1/3=Δy+5
dy is approximately equal to Δy is equal to
dy=$$\frac{dy}{dx}$$Δx
dy=$$\frac{1}{3x^{2/3}}$$.Δx
dy=$$\frac{1}{3×125^{2/3}} (2)$$
dy=2/75=0.0267
∴ The approximate value of (127)1/3 is 5+0.0267=5.0267

8. Find the approximate change in the volume of cube of side xm caused by increasing the side by 6%.
a) 0.18x
b) 0.18x3
c) 0.18x2
d) 1.8x3

Explanation: We know that the volume V of a cube is given by
V=x3
Differentiating w.r.t x, we get
$$\frac{dV}{dx}$$=3x2
dV=($$\frac{dV}{dx}$$)Δx=3x2 Δx
dV=3x2 (6x/100)=0.18x3
Therefore, the approximate change in volume is 0.18x3.

9. Find the approximate value of (82)1/4.
a) 3.025
b) 3.05
c) 3.00925
d) 3.07825

Explanation: Let y=x1/4. Let x=81 and Δx=1
Then, Δy=(x+Δx)1/4-x1/4
Δy=821/4-811/4
821/4=Δy+3
dy is approximately equal to Δy is equal to
dy=$$\frac{dy}{dx}$$Δx
dy=$$\frac{1}{(4x^{3/4})}$$.Δx
dy=$$\frac{1}{(4×81^{3/4})} (1)$$
dy=$$\frac{1}{(4×27)}$$=0.00925
∴ The approximate value of 821/4 is 3+0.00925=3.00925

10. Find the approximate error in the volume of the sphere if the radius of the sphere is measured to be 6cm with an error of 0.07cm.
a) 10.08π cm3
b) 10.08cm3
c) 10.4πcm3
d) 9.08cm3

Explanation: Let x be the radius of the sphere.
Then, x=6cm and Δx=0.07cm
The volume of a sphere is given by V=$$\frac{4}{3}$$ πx3
∴$$\frac{dV}{dx}=\frac{4}{3}$$ π(3x2)=4πx2
dV=($$\frac{dV}{dx}$$)Δx=4πx2 Δx
dV=4×π×62×0.07
dV=10.08π cm3

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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