JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Hard

1 of 1

Application +4 | -1

f:[-1,1] \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right], f(x)=\frac{x}{1+x^2}

. The area bounded by

y=f^{-1}(x), X

-axis,

x=\frac{1}{2}, x=-\frac{1}{2}

View answer and explanation

Correct option: B

\log \left(\frac{e}{2}\right)

Option B (

\log(\frac{e}{2})

) is correct because it represents the area bounded by the inverse function, X-axis, and the given limits.

More Questions from DEFINITE INTEGRATION

Practice All Chapter MCQs →

\int \nolimits_0^\infty \,\,\log \,\,\left( {x + \frac{1}{x}} \right)\,\,\frac{{dx}}{{1 + {x^2}}}

f\left( x \right)

\int \nolimits_0^{{t^2}} \,\,x\;\,f\left( x \right)\,\,dx = \frac{2}{5}\,\,{t^5}

f\left( {\frac{4}{{25}}} \right)

\int\limits_{{0}}^{11}[x]^{3} . \mathrm{dx}

[*]

\int\limits_{2}^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x

\int\limits_{{-2}}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x

I\,\,\left( {m,\,\,n} \right) = \int \nolimits_0^1 \,\,{t^m}\,\,{\left( {1 + t} \right)^n}\,\,dt

I\left( {m,\,\,n} \right)

I\,\,\left( {m + 1,\,\,n + 1} \right)