NDA 2024
Exam NDA
Year 2024
Subject Mathematics
Chapter Indefinite Integration NDA 2024 · Mathematics · Indefinite Integration
Let ∫ d x x+1 - x-1 =α(x+1)^ (3)/(2) +β(x-1)^ (3)/(2) +c What is...
Let
∫ d x x + 1 − x − 1 = α ( x + 1 ) 3 2 + β ( x − 1 ) 3 2 + c \int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}=\alpha(x+1)^{\frac{3}{2}}+\beta(x-1)^{\frac{3}{2}}+c ∫ x + 1 − x − 1 d x = α ( x + 1 ) 2 3 + β ( x − 1 ) 2 3 + c
What is the value of
α \alpha α ?
Options A 1 3 \frac{1}{3} 3 1 B 2 3 \frac{2}{3} 3 2 C 1 D 4 3 \frac{4}{3} 3 4 Correct answer A. 1 3 \frac{1}{3} 3 1
Step-by-step solution
Given,
∫ d x x + 1 − x − 1 = α ( x + 1 ) 3 2 + β ( x − 1 ) 3 2 + c . . . . ( i ) \int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}=\alpha(x+1)^{\frac{3}{2}}+\beta(x-1)^{\frac{3}{2}}+c\qquad....(i) ∫ x + 1 − x − 1 d x = α ( x + 1 ) 2 3 + β ( x − 1 ) 2 3 + c .... ( i )
Let,
\quad \mathrm{I}
= ∫ d x x + 1 − x − 1 =\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}} = ∫ x + 1 − x − 1 d x = ∫ ( x + 1 + x − 1 ) d x ( x + 1 − x − 1 ) ( x + 1 + x − 1 ) ⇒ I = 1 2 ∫ ( x + 1 ) d x + 1 2 ∫ x − 1 d x ⇒ I = 1 2 ⋅ ( x + 1 ) 3 2 ( 3 2 ) + 1 2 ( x − 1 ) 3 2 ( 3 2 ) + c ⇒ I = 1 3 ( x + 1 ) 3 2 + 1 3 ( x − 1 ) 3 2 + c ⇒ d x x + 1 − x − 1 = 1 3 ( x + 1 ) 3 2 + 1 3 ( x − 1 ) 3 2 + c … ( i i ) \begin{aligned}
& =\int \frac{(\sqrt{x+1}+\sqrt{x-1}) d x}{(\sqrt{x+1}-\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1})} \\
& \Rightarrow \quad \mathrm{I}=\frac{1}{2} \int \sqrt{(x+1)} d x+\frac{1}{2} \int \sqrt{x-1} d x \\
& \Rightarrow \quad \mathrm{I}=\frac{1}{2} \cdot \frac{(x+1)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+\frac{\frac{1}{2}(x-1)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c \\
& \Rightarrow \quad \mathrm{I}=\frac{1}{3}(x+1)^{\frac{3}{2}}+\frac{1}{3}(x-1)^{\frac{3}{2}}+c \\
& \Rightarrow \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}=\frac{1}{3}(x+1)^{\frac{3}{2}}+\frac{1}{3}(x-1)^{\frac{3}{2}}+c \ldots(ii)
\end{aligned} = ∫ ( x + 1 − x − 1 ) ( x + 1 + x − 1 ) ( x + 1 + x − 1 ) d x ⇒ I = 2 1 ∫ ( x + 1 ) d x + 2 1 ∫ x − 1 d x ⇒ I = 2 1 ⋅ ( 2 3 ) ( x + 1 ) 2 3 + ( 2 3 ) 2 1 ( x − 1 ) 2 3 + c ⇒ I = 3 1 ( x + 1 ) 2 3 + 3 1 ( x − 1 ) 2 3 + c ⇒ x + 1 − x − 1 d x = 3 1 ( x + 1 ) 2 3 + 3 1 ( x − 1 ) 2 3 + c … ( ii ) $
F r o m ( i ) a n d ( i i ) w e g e t From (i) and (ii) we get F ro m ( i ) an d ( ii ) w e g e t \alpha=\frac{1}{3}$
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