This set of Mathematics Objective Questions and Answers for Class 12 focuses on “Application of Derivative for Error Determination”.

1. What will be the increment of the differentiable function f(x) = 2x^{2} – 3x + 2 when x changes from 3.02 to 3?

a) 0.18

b) 0.018

c) 0.16

d) 0.016

View Answer

Explanation: Let, y = f(x) = 2x

^{2}– 3x + 2

So, f’(x) = 4x – 3

Clearly, as x changes from 3 to 3.02, the increment of x is:

3.02 – 3 = 0.02

So, increment in f(x) = f(3.02) – f(3)

= 2(3.02)

^{2}– 3(3.02) + 2 – 2(3)

^{2}– 3(3) + 2

= 0.1808

Now, differential of y is dy

We get, f’(3) * 0.02 = (4.3 – 3)*0.02

= 0.18

2. If log_{10}3 = 0.4771 and log_{10}e = 0.4343, then what is the value of log_{10}30.5?

a) 1.43

b) 1.5

c) 1.484

d) 1.4

View Answer

Explanation: Let, y = log

_{10}x

Then f’(x) = d/dx[log

_{10}x]

= d/dx[log

_{e}x * log

_{10}e]

= 1/x(log

_{10}e)

Now, f(x + δx) = f(x) + f’(x) δx

Putting x = 30 and δx = 0.5 in the above equation we get,

f(30 + 0.5) = f(30) + f’(30) δx

=> f(30.5) = log

_{10}30 + (1/30) log

_{10}e * 0.5

Putting the values we get,

log

_{10}30.5 = 1.4843 = 1.484

3. If 1° = 0.01745 then, what is the value of cos62°?

a) 0.4588

b) 0.4788

c) 0.4688

d) 0.3688

View Answer

Explanation: Let, y = f(x) = cosx

And, x = 60° = π/3, δx = 2° = 2 * 0.01745 = 0.03490

Since f(x) = cosx, hence

f’(x) = -sinx

Now, we have f(x + δx) = f(x) + f’(x) δx

Or, f(60° + 2°) = f(60°) + f’(60°) * 0.0349

Or, f(62°) = cos(60°) – sin60° * 0.0349

Or, cos62° = 0.5 – 0.866 * 0.0349 = 0.4688

4. The length of a side of a cube is 10cm; if an error of 0.05cm is made in measuring the side, then what is the value of approximate error in calculating its volume?

a) 16 cu cm

b) 15 cu cm

c) 15.5 cu cm

d) 14 cu cm

View Answer

Explanation: Let, v cubic cm be the volume of the cube of the cube of side x cm.

Then, v = x

^{3}

Thus, dv/dx = d/dx(x

^{3}) = 3x

^{2}

Now, dv = dv/dx * δx = 3x

^{2}* δx

By problem, x = 10 and δx = 0.05

Therefore, approximate error = dv = 3 * 10

^{2}* 0.05 = 15 cu cm.

5. The length of a side of a cube is 10cm; if an error of 0.05cm is made in measuring the side, then what is the value of relative error in calculating its volume?

a) 0.0016

b) 0.014

c) 0.015

d) 0.0015

View Answer

Explanation: Let, v cubic cm be the volume of the cube of the cube of side x cm.

Then, v = x

^{3}

Thus, dv/dx = d/dx(x

^{3}) = 3x

^{2}

Now, dv = dv/dx * δx = 3x

^{2}* δx

By problem, x = 10 and δx = 0.05

Therefore, approximate error = dv = 3 * 10

^{2}* 0.05 = 15 cc cm.

Hence, relative error = dv/v = 15/10

^{3}= 0.015

6. The length of a side of a cube is 10cm; if an error of 0.05cm is made in measuring the side, then what is the value of percentage error in calculating its volume?

a) 1.65

b) 1.45

c) 0.015

d) 1.5

View Answer

Explanation: Let, v cubic cm be the volume of the cube of the cube of side x cm.

Then, v = x

^{3}

Thus, dv/dx = d/dx(x

^{3}) = 3x

^{2}

Now, dv = dv/dx * δx = 3x

^{2}* δx

By problem, x = 10 and δx = 0.05

Therefore, approximate error = dv = 3 * 10

^{2}* 0.05 = 15 cc cm.

Hence, relative error = dv/v = 15/10

^{3}= 0.015

Thus, percentage error = relative error * 100 = 0.015 * 100 = 1.5

7. What will be the estimate error made in calculating the area of the triangle ABC in which the sides a and b are measured accurately as 25 cm and 16 cm, while the angle C is measured as 60° but (1/2)° in error?

a) 55/63 sq cm

b) 53/63 sq cm

c) 55/67 sq cm

d) Data not sufficient

View Answer

Explanation: Let, y be the area of the triangle ABC.

Then, y = ½(ab)(sinC) where a and b are constants.

Now, dy = dy/dC * dC = ½(ab)(sinC) dC

Now, by problem, a = 25, b = 16, C = 60° and δC = (1/2)° = π/180 * ½ radian

Therefore, dy = 1/2 * 25 * 16 * cos60° * π/180 * ½ sq cm

=> dy = 25 * 8 * ½ * 22/7 * 1/180 * ½ sq cm

= 55/63 sq cm

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