NIMCET 2026 — Mathematics PYQ

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Question

NIMCET 2026 — Mathematics PYQ

NIMCET | Mathematics | 2026

Let $f : [0, \infty) \to R$ be a function defined by $f (x) = \frac{3 x ^{2} + 4 x + 1}{x ^{2} + 3 x + 2}$ . Then the value of $(f^{- 1})^{'} (2)$ is equal to:

Choose the correct answer:

Explanation

The derivative of the inverse function is given by the formula:

$(f^{- 1})^{'} (y) = \frac{1}{f ^{'} ( x )}, where f (x) = y$

1. Find $x$ such that $f (x) = 2$ :

$\frac{3 x ^{2} + 4 x + 1}{x ^{2} + 3 x + 2} = 2$

$3 x^{2} + 4 x + 1 = 2 (x^{2} + 3 x + 2)$

$3 x^{2} + 4 x + 1 = 2 x^{2} + 6 x + 4$

$x^{2} - 2 x - 3 = 0$

$(x - 3) (x + 1) = 0$

Since the domain is $[0, \infty)$ , we have $x = 3$ .

2. Find $f^{'} (x)$ :

Using the quotient rule $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ :

$u = 3 x^{2} + 4 x + 1 ⟹ u^{'} = 6 x + 4$

$v = x^{2} + 3 x + 2 ⟹ v^{'} = 2 x + 3$

$f^{'} (x) = \frac{( 6 x + 4 ) ( x ^{2} + 3 x + 2 ) - ( 3 x ^{2} + 4 x + 1 ) ( 2 x + 3 )}{( x ^{2} + 3 x + 2 ) ^{2}}$

3. Evaluate $f^{'} (3)$ :

At $x = 3$ :

$u = 3 (9) + 4 (3) + 1 = 40$

$u^{'} = 6 (3) + 4 = 22$

$v = 3^{2} + 3 (3) + 2 = 20$

$v^{'} = 2 (3) + 3 = 9$

$f^{'} (3) = \frac{( 22 ) ( 20 ) - ( 40 ) ( 9 )}{( 20 ) ^{2}} = \frac{440 - 360}{400} = \frac{80}{400} = \frac{1}{5}$

4. Calculate $(f^{- 1})^{'} (2)$ :

$(f^{- 1})^{'} (2) = \frac{1}{f ^{'} ( 3 )} = \frac{1}{1/5} = 5$

Correct Option: (b)