Approximations - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 6 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
View Answer

Answer: c
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
View Answer

Answer: d
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
View Answer

Answer: a
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
View Answer

Answer: a
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
View Answer

Answer: d
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.

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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
View Answer

Answer: c
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
View Answer

Answer: a
Explanation: Let y=(x)^1/3. Let x=125 and Δx=2
Then, Δy=(x+Δx)^1/3-x^1/3
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
View Answer

Answer: b
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
View Answer

Answer: c
Explanation: Let y=x^1/4. Let x=81 and Δx=1
Then, Δy=(x+Δx)^1/4-x^1/4
Δy=82^1/4-81^1/4
82^1/4=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 82^1/4 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
View Answer

Answer: a
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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