If $\int_{0}^{1} \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}$, where $a, b, c$ are rational numbers, then $2a+3b-4c$ is equal to :

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Definite Integrals

\int_{0}^{1} \frac{1}{3 + x + 1 + x} d x = a + b 2 + c 3

, where

a, b, c

are rational numbers, then

2 a + 3 b - 4 c

is equal to :

Answer: (D)

Rationalize the integrand:

\frac{1}{3 + x + 1 + x} = \frac{3 + x - 1 + x}{2}

\int_{0}^{1} \frac{3 + x - 1 + x}{2} d x = \frac{1}{3} [(3 + x)^{3/2} - (1 + x)^{3/2}]_{0}^{1} = \frac{1}{3} [(8 - 22) - (33 - 1)]

= 3 - \frac{2}{3} 2 - 3

a = 3, b = - \frac{2}{3}, c = - 1

, giving

2 a + 3 b - 4 c = 6 - 2 + 4 = 8

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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