CUET PG 2026 — Mathematics PYQ

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Question

CUET PG 2026 — Mathematics PYQ

CUET PG | Mathematics | 2026

Let $f$ be function $f(x) = \begin{cases} \frac{\tan x}{x} & x \neq 0 \\ 1 & x = 0 \end{cases}$

Assertion: $x = 0$ is point of minima of $f$

Reason: $f'(0) = 0$

Choose the correct answer:

Explanation

Solution

Check Reason:

f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}

f'(0) = \lim_{h \to 0} \frac{\frac{\tan h}{h} - 1}{h} = \lim_{h \to 0} \frac{\tan h - h}{h^2}

Using expansion $\tan h = h + \frac{h^3}{3} + \dots$ :

f'(0) = \lim_{h \to 0} \frac{(h + \frac{h^3}{3}) - h}{h^2} = \lim_{h \to 0} \frac{h^3}{3h^2} = 0

Reason is true.

Check Assertion:

Expansion of $f(x)$ for $x \neq 0$ :

f(x) = \frac{\tan x}{x} = \frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots}{x} = 1 + \frac{x^2}{3} + \frac{2x^4}{15} + \dots

Since $x^2 > 0$ for all $x \neq 0$ (near $0$ ):

f(x) > 1 \implies f(x) > f(0)

Assertion is true.

Conclusion:

Both are true. $f'(0) = 0$ is the necessary condition for a local extremum at $x=0$ , and combined with the behavior $f(x) > f(0)$ , it confirms the minima.

Correct Option: (a)