JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

The value of the parameter '

a

' such that the area bounded by

y=a^2 x^2+a x+1

, coordinate axes and the line

x=1

attains its least value, is equal to

View answer and explanation

Correct option: C

-\frac{3}{4}

Option C (-

\frac{{3}}{{4}}

) is correct because it is the value of 'a' that minimizes the area under the given curve between the specified limits, found by completing the square and identifying the vertex.

More Questions from DEFINITE INTEGRATION

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

is equal to If

f\left( x \right)

is differentiable and

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

, then

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

equals The value of

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

, where

[*]

denotes the greatest integer function, is The integral

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

is equal to: The definite integral

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

equals If

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

,then the expression for

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

in terms of

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