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Linear Differential Equation and Integral Bound | JEE

JEE Maths question with a full step-by-step solution.

Question

A function

y = f(x)

satisfies

f'(x)\sin x + f(x)\cos x = 1

, with

f(x)

bounded as

x \to 0

. If

I = \displaystyle\int_0^{\pi/2} f(x)\,dx

, then:

\dfrac{\pi}{2} < I < \dfrac{\pi^2}{4}

correct

\dfrac{\pi}{4} < I < \dfrac{\pi^2}{2}

1 < I < \dfrac{\pi}{2}

0 < I < 1

Solution

Step 1: The left side is an exact derivative:

\frac{d}{dx}\big(f(x)\sin x\big) = 1 \Rightarrow f(x)\sin x = x + c \Rightarrow f(x) = \frac{x}{\sin x} + \frac{c}{\sin x}.

Step 2: For

f

to stay bounded as

x\to 0

, take

c = 0

, so

f(x) = \dfrac{x}{\sin x}

. Step 3: On

\left(0, \dfrac{\pi}{2}\right)

\dfrac{x}{\sin x}

increases from

1

\dfrac{\pi}{2}

, so

1 < \dfrac{x}{\sin x} < \dfrac{\pi}{2}

. Integrating over

\left[0, \dfrac{\pi}{2}\right]

\frac{\pi}{2}\cdot 1 < \int_0^{\pi/2}\frac{x}{\sin x}\,dx < \frac{\pi}{2}\cdot\frac{\pi}{2} \Rightarrow \frac{\pi}{2} < I < \frac{\pi^2}{4}.

Correct answer: (1)

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