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JEE Advanced 2007Paper 2
Exam
JEE Advanced
Year
2007
Paper
Paper 2
Subject
Mathematics
Chapter
Indefinite Integration

JEE Advanced 2007 · Paper 2 · Mathematics · Indefinite Integration

Let f(x)= x (1+x^n)^ 1 / n for n ≥ 2 and g(x)= (f f o o f) _ f...

Let f(x)=x(1+xn)1/nf(x)=\frac{x}{\left(1+x^n\right)^{1 / n}} for n≥2n \geq 2 and g(x)=(f∘fo…of)⏟f occurs n times (x)g(x)=\underbrace{(f \circ f o \ldots o f)}_{f \text{ occurs } n \text{ times }}(x). Then, ∫xn−2g(x)dx\int x^{n-2} g(x) d x equals

Options

  1. A1n(n−1)(1+nxn)1−1n+k\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k
  2. B1n−1(1+nxn)1−1n+k\frac{1}{n-1}\left(1+n x^n\right)^{1-\frac{1}{n}}+k
  3. C1n(n+1)(1+nxn)1+1n+k\frac{1}{n(n+1)}\left(1+n x^n\right)^{1+\frac{1}{n}}+k
  4. D1n+1(1+nxn)1+1n+k\frac{1}{n+1}\left(1+n x^n\right)^{1+\frac{1}{n}}+k

Correct answer

A. 1n(n−1)(1+nxn)1−1n+k\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k

Step-by-step solution

Here,  and ff(x)=f(x)[1+f(x)n]1/n=x(1+2xn)1/n∴fff(x)=x(1+3xn)1/n Let g(x)=(f∘fo…of)⏟n times (x)=x(1+nxn)1/nI=∫xn−2g(x)dx=∫xn−1dx(1+nxn)1/n=1n2∫n2xn−1dx(1+nxn)1/n=1n2∫ddx(1+nxn)(1+nxn)1/ndx\begin{array}{rlrl} & \text{ and } & f f(x) & =\frac{f(x)}{\left[1+f(x)^n\right]^{1 / n}}=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\ & \therefore \quad f f f(x) & =\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\ & \text{ Let } & g(x) & =\underbrace{(f \circ f o \ldots o f)}_{n \text{ times }}(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}} \\ I & =\int x^{n-2} g(x) d x=\int \frac{x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}} \\ & & =\frac{1}{n^2} \int \frac{n^2 x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}=\frac{1}{n^2} \int \frac{\frac{d}{d x}\left(1+n x^n\right)}{\left(1+n x^n\right)^{1 / n}} d x\end{array} ∴I=1n(n−1)(1+nxn)1−1n+k\therefore \quad I=\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k.
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