JEE Advanced 2024 — Paper 1 — MATH-IV solution

JEE Advanced2024Paper 1MATH-IV

Let $f : R \to R$ and $g : R \to R$ be functions defined by

$f (x) = ⎩ ⎨ ⎧ x ∣ x ∣ sin (\frac{1}{x}), x \neq = 0 0, x = 0$ and $g (x) = ⎩ ⎨ ⎧ 1 - 2 x, 0 \leq x \leq \frac{1}{2} 0, otherwise$

Let $a, b, c, d \in R$ . Define the function $h : R \to R$ by

$h (x) = a f (x) + b (g (x) + g (\frac{1}{2} - x)) + c (x - g (x)) + d g (x)$ , $x \in R$ .

Match each entry in List-I to the correct entry in List-II.

The correct option is

Solution

Compute $g (\frac{1}{2} - x) = 2 x$ for $0 \leq x \leq \frac{1}{2}$ , and $0$ otherwise. Therefore

$g (x) + g (\frac{1}{2} - x) = ⎩ ⎨ ⎧ 1, 0 \leq x \leq \frac{1}{2} 0, otherwise$

(P) $b = 1$ : $h (x) = g (x) + g (\frac{1}{2} - x)$ takes only values ${0, 1}$ . (P) $\to$ (5).

(Q) at $a = 1$ , $h (x) = f (x)$

$f'(0^+)=\lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h} =\lim_{h\to 0}\dfrac{h^2\sin\left(\dfrac{1}{h}\right)}{h}$

$= lim_{h \to 0} h sin (\frac{1}{h}) = 0$

$f'(0^-)=\lim_{h\to 0}\dfrac{f(0-h)-f(0)}{-h} =\lim_{h\to 0}h\sin\left(\dfrac{1}{h}\right)=0$

$\Rightarrow f^{'} (0^{+}) = f^{'} (0^{-})$

As $f^{'} (0^{+}) = f^{'} (0^{-})$ , the function is differentiable at $x = 0$ .

Therefore, $h (x) = f (x)$ is differentiable at $x = 0$ .

At all other points, $f (x) = h (x)$ is differentiable as it is the product of two differentiable functions. (Q) $\to$ (3).

(R) $c = 1$ : $h (x) = x - g (x) = {3 x - 1, 0 \leq x \leq \frac{1}{2} x, otherwise$ .

From $h (x)$ graph: Range covers all of $R$ , and $h$ is onto. (R) $\to$ (2).

(S) $d = 1$ : $h (x) = g (x)$ , which takes values in $[0, 1]$ . Range is $[0, 1]$ . (S) $\to$ (4).

$∴ Option (C)$

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