Q249 / 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 $p(x)=f(x)g(x)$, then which of the following statements is/are correct ?
I. $p(x)$ is continuous at $x=0$.
II. $p(x)$ 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 function $p(x)$ is given by $p(x) = f(x)g(x) = x|x|\tan(x^2)$.

We can redefine $p(x)$ as:
$p(x) = x^2 \tan(x^2)$ for $x \ge 0$
$p(x) = -x^2 \tan(x^2)$ for $x < 0$

Checking continuity at $x = 0$:
$\lim_{x \to 0^+} p(x) = \lim_{x \to 0^+} x^2 \tan(x^2) = 0$
$\lim_{x \to 0^-} p(x) = \lim_{x \to 0^-} -x^2 \tan(x^2) = 0$
$p(0) = 0$
Since $\lim_{x \to 0^+} p(x) = \lim_{x \to 0^-} p(x) = p(0)$, $p(x)$ is continuous at $x = 0$. Statement I is correct.

Checking differentiability at $x = 0$:
Right Hand Derivative (RHD) at $x = 0$:
$\lim_{h \to 0^+} \dfrac{p(h) - p(0)}{h} = \lim_{h \to 0^+} \dfrac{h^2 \tan(h^2) - 0}{h} = \lim_{h \to 0^+} h \tan(h^2) = 0$

Left Hand Derivative (LHD) at $x = 0$:
$\lim_{h \to 0^-} \dfrac{p(h) - p(0)}{h} = \lim_{h \to 0^-} \dfrac{-h^2 \tan(h^2) - 0}{h} = \lim_{h \to 0^-} -h \tan(h^2) = 0$

Since RHD = LHD = $0$, $p(x)$ 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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