If f(x)= [π²] x+ [-π²] x, then

Correct Answer is : f(π2)=−1f≤ft(π/2)=-1f(2π)=−1

Correct Answer is : f(π2)=−1f≤ft(π/2)=-1f(2π)=−1

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Manipal Entrance Test 2019 Math Paper

Question : 47 of 70

Marks: +1, -0

f(x)=\cos [\pi^{2}] x+\cos [-\pi^{2}] x,

then

Solution:

We have,

f(x)=\cos [\pi^{2}] x+\cos [-\pi^{2}] x

Put the value of π = 3.141

\Rightarrow \pi^{2}=

app. 9.6 in (i)

\Rightarrow f(x)=\cos 9x+\cos 10x

[\because \pi=3.141]

Put

x=\frac{\pi}{4}

f\left(\frac{\pi}{4}\right)=\cos\frac{9\pi}{4}+\cos\frac{10\pi}{4}

=\frac{1}{\sqrt{2}}+0

=\frac{1}{\sqrt{2}}

Now,

f(-\pi)=\cos 9\pi+\cos 10\pi

=-1+1=0

and

f(\pi)=\cos 9\pi+\cos 10\pi=0

\therefore f\left(\frac{\pi}{2}\right)=\cos\frac{9\pi}{2}+\cos\frac{10\pi}{2}

=\cos\frac{9\pi}{2}+\cos 5\pi

=0+(-1)=-1

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