JEE Advanced 2021 Paper 2 Q03 Mathematics Integration & Differential Equations Definite Integrals Hard

JEE Advanced 2021 Paper 2 · Q03 · Definite Integrals

Let $f:\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\to\mathbb{R}$ be a continuous function such that $$f(0)=1\ \text{ and }\ \int_0^{\pi/3} f(t)\,dt = 0.$$ Then which of the following statements is (are) TRUE?

  1. A. The equation $f(x)-3\cos 3x = 0$ has at least one solution in $\left(0,\dfrac{\pi}{3}\right)$
  2. B. The equation $f(x)-3\sin 3x = -\dfrac{6}{\pi}$ has at least one solution in $\left(0,\dfrac{\pi}{3}\right)$
  3. C. $\displaystyle\lim_{x\to 0}\dfrac{x\int_0^x f(t)\,dt}{1-e^{x^2}} = -1$
  4. D. $\displaystyle\lim_{x\to 0}\dfrac{\sin x\int_0^x f(t)\,dt}{x^2} = -1$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

(A) Let $g(x)=f(x)-3\cos 3x$. Then $\int_0^{\pi/3} g(x)\,dx = 0 - 0 = 0$ (since $\int_0^{\pi/3} 3\cos 3x\,dx = [\sin 3x]_0^{\pi/3} = 0$). So $g$ vanishes somewhere in $(0,\pi/3)$. TRUE. (B) Let $h(x)=f(x)-3\sin 3x + \dfrac{6}{\pi}$. $\int_0^{\pi/3} h(x)\,dx = 0 - 2 + 2 = 0$, so $h$ vanishes in $(0,\pi/3)$. TRUE. (C) Using L'Hopital and $\dfrac{d}{dx}\int_0^x f(t)\,dt = f(x)$: $\dfrac{x\cdot f(x) + \int_0^x f(t)\,dt}{-2xe^{x^2}}$. As $x\to 0$, numerator $\to 0$. Apply L'Hopital again or note $\int_0^x f(t)\,dt \approx x f(0) = x$, so the ratio $\to \dfrac{2x}{-2x} = -1$. TRUE. (D) $\dfrac{\sin x \cdot \int_0^x f(t)\,dt}{x^2}\to \dfrac{1\cdot 1}{1} = 1 \ne -1$. FALSE.

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