IIT-JEE 2007 Paper 2 Offline | Indefinite Integrals Question 4 | Mathematics

IIT-JEE 2007 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let $$f(x)=\frac{x}{\left(1+x^{n}\right)^{1 / n}}$$ for $$n \geq 2$$ and $$g(x)=\underbrace{(f o f o \ldots . o f)}_{f \text { occurs } n \text { times }}(x)$$. Then $$\int x^{n-2} g(x) d x$$ equals :

$$\frac{1}{n(n-1)}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+k$$

$$\frac{1}{n(n+1)}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+k$$

$$\frac{1}{n(n-1)}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+k$$

$$\frac{1}{(n+1)}\left(1+n x^{n}\right)^{1+\frac{1}{n}}+k$$

IIT-JEE 2006

MCQ (Single Correct Answer)

-1

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$ is equal to

$\frac{\sqrt{2 x^4-2 x^2+1}}{x^2}+\mathrm{C}$

$\frac{\sqrt{2 x^4-2 x^2+1}}{x^3}+\mathrm{C}$

$\frac{\sqrt{2 x^4-2 x^2+1}}{x}+\mathrm{C}$

$\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+C$

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-0.75

If $$\int\limits_{\sin x}^1 {{t^2}f\left( t \right)dt = 1 - \sin x,} $$ then f$$\left( {{1 \over {\sqrt 3 }}} \right)$$ is

$${1 \over 3}$$

$${{1 \over {\sqrt 3 }}}$$

$$3$$

$${\sqrt 3 }$$

IIT-JEE 1995 Screening

MCQ (Single Correct Answer)

-0.75

The value of the integral $$\int {{{{{\cos }^3}x + {{\cos }^5}x} \over {{{\sin }^2}x + {{\sin }^4}x}}} \,dx\,$$ is

$$\sin x - 6{\tan ^{ - 1}}\left( {\sin x} \right) + c$$

$$\sin x - 2{\left( {\sin x} \right)^{ - 1}} + c$$

$$\sin x - 2{\left( {\sin x} \right)^{ - 1}} - 6{\tan ^{ - 1}}\left( {\sin x} \right) + c$$

$$\,\sin x - 2{\left( {\sin x} \right)^{ - 1}} + 5{\tan ^{ - 1}}\left( {\sin x} \right) + c$$

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