Test Index

MHTCET 5 May 2025 Shift 2 Paper

Question : 115 of 150

Marks: +1, -0

\frac{1}{(2.002)^{2}}

Solution:

Concept:

Use the linear approximation formula

f(a+h) \approx f(a) + h f'(a)

to estimate the value of the function near a known point.

Explanation:

Let

f(x) = \frac{1}{x^2}

Then the derivative is

f'(x) = -\frac{2}{x^3}

Choose

a = 2

and

h = 0.002

, so that

a + h = 2.002

Now compute

f(2) = \frac{1}{2^2} = \frac{1}{4} = 0.25

Also compute

f'(2) = -\frac{2}{2^3} = -\frac{2}{8} = -\frac{1}{4} = -0.25

Applying linear approximation:

f(2.002) \approx f(2) + 0.002 \times f'(2)

Substitute the values:

\frac{1}{(2.002)^2} \approx 0.25 + 0.002 \times (-0.25)

Simplify:

0.25 - 0.0005 = 0.2495

Therefore, the approximate value of

\frac{1}{(2.002)^2}

0.2495

Answer:

Option A:

0.2495

Go to Question:

Prev Question

Next Question