Question

Statement-1 : The value of the integral \(\displaystyle\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\tan x}}\) is equal to \(\pi/6\)
Statement-2 : \(\displaystyle\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx\).

• A. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
• C. Statement- 1 is true; Statement-2 is False.
• D. Statement-1 is false; Statement-2 is true.

Hint:

Apply the property \(\int_a^b f(x)dx=\int_a^b f(a+b-x)dx\) and add the two forms of the integral.

Answer 1

The Correct Option is D

Statement-2 is the standard property of definite integrals, so it is true.
Let \(I=\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\tan x}}=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx\).
Here \(a+b=\pi/2\), so replacing \(x\) by \(\pi/2-x\): \(I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\).
Adding: \(2I=\int_{\pi/6}^{\pi/3}dx=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\), so \(I=\frac{\pi}{12}\neq\frac{\pi}{6}\).
Statement-1 is false, Statement-2 is true.
Answer: (4) Statement-1 is false; Statement-2 is true.