Q250 / 270 · NDA 2026 (Phase 1) · Mathematics

Passage: Let $f(x)=\tan(x^2)$ and $g(x)=x|x|$ for $|x|<\sqrt{\dfrac{\pi}{2}}$.

Question: If $q(x)=f\circ g(x)$, then which of the following statements is/are correct ?
I. $q(x)$ is continuous at $x=0$.
II. $q(x)$ is differentiable at $x=0$.
Select the answer using the code given below :

  • A
    I only
  • B
    II only
  • Correct answer
    C
    Both I and II
  • D
    Neither I nor II

Solution

Given $f(x) = \tan(x^2)$ and $g(x) = x|x|$.

The composite function $q(x)$ is given by:

$q(x) = f(g(x)) = f(x|x|) = \tan((x|x|)^2)$

Since $(x|x|)^2 = x^2|x|^2 = x^4$, we have:

$q(x) = \tan(x^4)$

Checking continuity at $x=0$:

$\lim_{x \to 0} q(x) = \lim_{x \to 0} \tan(x^4) = 0$

Also, $q(0) = \tan(0) = 0$. Since $\lim_{x \to 0} q(x) = q(0)$, $q(x)$ is continuous at $x=0$. Statement I is correct.

Checking differentiability at $x=0$:

$q'(0) = \lim_{x \to 0} \dfrac{q(x) - q(0)}{x - 0} = \lim_{x \to 0} \dfrac{\tan(x^4)}{x}$

$q'(0) = \lim_{x \to 0} \left( \dfrac{\tan(x^4)}{x^4} \times x^3 \right) = 1 \times 0 = 0$

Since the limit exists and is finite, $q(x)$ is differentiable at $x=0$. Statement II is correct.

Both statements I and II are correct.

Answer: Both I and II

Hence, the correct answer is option: 'C'.

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