Jee Main 2024 27 01 2024 Shift 1 - Question4
Question 4
$\int _0^{1} \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}$, then
$2 a-3 b-4 c$ is equal to
(1) 10
(2) 0
(3) 12
(4) 20
Show Answer
Answer: (3)
Solution:
$I=\int _0^{1} \frac{1}{\sqrt{x+3}+\sqrt{x+1}} d x$
On rationalization
$I=\int _0^{1} \frac{\sqrt{x+3}-\sqrt{x+1}}{2} d x$ $I=\int _0^{1} \frac{(x+3)^{1 / 2}}{2} d x-\int _0^{1} \frac{(x+1)^{1 / 2}}{2} d x$
$I=\left.\frac{(x+3)^{3 / 2}}{3}\right| _0 ^{1}-\left.\frac{(x+1)^{3 / 2}}{3}\right| _0 ^{1}$
$=\frac{1}{3}\left[4^{3 / 2}-3^{3 / 2}\right]-\frac{1}{3}\left[2^{3 / 2}-1^{3 / 2}\right]$
$=\frac{8-3 \sqrt{3}-2 \sqrt{2}+1}{3}=3-\sqrt{3}-\frac{2}{3} \sqrt{2}$
$\Rightarrow 2 a=6$
$3 b=-2$
$4 c=-4$
$2 a-3 b-4 c=6+2+4=12$
BETA
