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

(C) Statement -I is False; Statement -II is True

(A) Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I

(B) Statement -I is True; Statement -II is False.

(D) Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I

JEE Main · 2013 · 07 Apr 2013 · Offline · Definite Integrals

Statement - I : The value of the integral

\int_{π /6}^{π /3} \frac{d x}{1 + t a n x}

is equal to

\frac{π}{6}

.

Statement - II :

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

(A)Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
(B)Statement -I is True; Statement -II is False.
(C)Statement -I is False; Statement -II is True✓ Correct
(D)Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I

Answer: (C)

Statement II is the standard reflection property of definite integrals and is true. For Statement I, call the integral

I = \int_{π /6}^{π /3} \frac{d x}{1 + tan x}

and apply that property with

a + b - x = \frac{π}{2} - x

; since

tan (\frac{π}{2} - x) = cot x

, the integrand turns into

\frac{tan x}{1 + tan x}

. Adding the two forms,

2 I = \int_{π /6}^{π /3} d x = \frac{π}{6}

, so

I = \frac{π}{12}

. The value claimed in Statement I,

\frac{π}{6}

, is wrong, so Statement I is false while Statement II is true.

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option