If 0 1 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: [2024]

Question

$If \int_{0}^{1} \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} d x = a + b \sqrt{2} + c \sqrt{3}, where a, b, c are rational numbers, then 2 a + 3 b - 4 c is equal to:$ [2024]

Smart Achievers · Accepted Answer

(4)

$\int_{0}^{1} \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} d x$

$= \int_{0}^{1} \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} \times \frac{\sqrt{3 + x} - \sqrt{1 + x}}{\sqrt{3 + x} - \sqrt{1 + x}} d x$ [ $∵$ Rationalising the denominator]

$= \int_{0}^{1} \frac{\sqrt{3 + x} - \sqrt{1 + x}}{(3 + x) - (1 + x)} d x = \frac{1}{2} \int_{0}^{1} (\sqrt{3 + x} - \sqrt{1 + x}) d x$

$= \frac{1}{2} [$ $\frac{2 {(3 + x)}^{3 / 2}}{3}$ $|_{0}^{1}$ $-$ $\frac{2}{3} {(1 + x)}^{3 / 2}$ $|_{0}^{1}$ $]$

$= 3 - \frac{2 \sqrt{2}}{3} - \sqrt{3} = a + b \sqrt{2} + c \sqrt{3} \Rightarrow a = 3, b = \frac{- 2}{3}, c = - 1$

$∴ 2 a + 3 b - 4 c = 2 \times 3 + 3 \times (\frac{- 2}{3}) - 4 (- 1) = 8$

Solution

Q.
$If \int_{0}^{1} \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} d x = a + b \sqrt{2} + c \sqrt{3}, where a, b, c are rational numbers, then 2 a + 3 b - 4 c is equal to:$ [2024]

1 10

2 7

3 4

4 8

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Solution

Q. If ∫0113+x+1+x dx=a+b2+c3, where a,b,c are rational numbers, then 2a+3b-4c is equal to: [2024]

1 10

2 7

3 4

4 8

Ans. (4) ∫0113+x+1+x dx =∫0113+x+1+x×3+x-1+x3+x-1+x dx [∵ Rationalising the denominator] =∫013+x-1+x(3+x)-(1+x) dx=12∫01(3+x-1+x) dx =12[2(3+x)3/23|01-23(1+x)3/2|01] =3-223-3=a+b2+c3⇒a=3, b=-23, c=-1 ∴ 2a+3b-4c=2×3+3×(-23)-4(-1)=8

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Q.
$If \int_{0}^{1} \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} d x = a + b \sqrt{2} + c \sqrt{3}, where a, b, c are rational numbers, then 2 a + 3 b - 4 c is equal to:$ [2024]