The maximum value of the function f x -01 t sin x - t dt - x - R is -A- 1-2-2-4B- 1-2-4C- -2-4D- 1-2 -2-2-4

The correct option is A 1/π^2√(π^2+4)f(x)= [ t/π cos(x+t π) ]^1_0 + 1/π∫^1_0 cos(x+t π)dt =1/π cos x 2/π^2 sin x f(x)_max.=√(1/π^2+4/π^4)=√(π^2+4)/π^2 ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

The maximum v...

Question

The maximum value of the function $f (x) = \int_{0}^{1} t sin (x + π t) d t, x \in R$ is -

\frac{1}{π} \sqrt{π^{2} + 4}

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\frac{1}{π^{2}} \sqrt{π^{2} + 4}

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\sqrt{π^{2} + 4}

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\frac{1}{2 π^{2}} \sqrt{π^{2} + 4}

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Solution

The correct option is B $\frac{1}{π^{2}} \sqrt{π^{2} + 4}$
$f (x) = {[- \frac{t}{π} cos (x + t π)]}_{0}^{1} + \frac{1}{π} \int_{0}^{1} cos (x + t π) d t$

$= \frac{1}{π} cos x - \frac{2}{π^{2}} sin x$

$f (x)_{m a x .} = \sqrt{\frac{1}{π^{2}} + \frac{4}{π^{4}}} = \frac{\sqrt{π^{2} + 4}}{π^{2}}$

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The maximum value of the function f(x)=∫10t sin(x+πt)dt, x∈R is -

The correct option is B 1π2√π2+4f(x)=[−tπ cos(x+tπ)]10+1π∫10cos(x+tπ)dt =1π cos x−2π2 sin x f(x)max.=√1π2+4π4=√π2+4π2

The maximum value of the function $f (x) = \int_{0}^{1} t sin (x + π t) d t, x \in R$ is -