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For $x\in \left ( 0,\frac{5\pi }{2} \right ),$ define $f(x)=\int_{0}^{x}\sqrt{t}\sin t\, dt\; \; Then\; f\; has$

Option 1)
local minimum at $\pi$ and local maximum at $2\pi$

Option 2)
local maximum at $\pi$ and local minimum at $2\pi$

Option 3)
local maximum at $\pi$ and $2\pi$

Option 4)
local minimum at $\pi$ and $2\pi$

Answers (1)

As we learnt in

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

$\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}$

$\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)$

$\Rightarrow \frac{dV}{dt}(Volume)$

$\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)$

- wherein

Where dR / dt means Rate of change of radius.

$f(x)=\int ^{x}_{0}\sqrt{t}\:sint\:dt$

$f'(x)= \sqrt{x}\:sinx=0\\\therefore \sqrt{x}=0$

$x=0\:\:sinx=0$

$x=n \pi \\x=\pi$ & $2\pi$

x=0

$\therefore \pi$

$f"(x)=\frac{1}{2\sqrt{x}}\ sinx+\sqrt{x}\ cosx$

at $\pi \:\therefore \frac{1\times0}{2\sqrt{\pi }}$ $+\:\sqrt{\pi }\times\:cos\pi \:=\:-\sqrt{\pi }< 0$

at $2\pi$ It is $\sqrt{\pi }> 0$

Option 1)

local minimum at $\pi$ and local maximum at $2\pi$

This option is incorrect.

Option 2)

local maximum at $\pi$ and local minimum at $2\pi$

This option is correct.

Option 3)

local maximum at $\pi$ and $2\pi$

This option is incorrect.

Option 4)

local minimum at $\pi$ and $2\pi$

This option is incorrect.

Posted by

Sabhrant Ambastha

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