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Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsFunctions2 Marks
MCQ
If $f(x)=\cos \left[\pi^2\right] x+\cos \left[-\pi^2\right] x$, then
  • A
    $f\left(\frac{\pi}{4}\right)=2$
  • B
    $f(-\pi)=2$
  • C
    $f(\pi)=1$
  • ✓
    $f\left(\frac{\pi}{2}\right)=-1$
✓

Answer

Correct option: D.
$f\left(\frac{\pi}{2}\right)=-1$
(D)
$f (x)=\cos \left[\pi^2\right] x+\cos \left[-\pi^2\right] x$
$f(x)=\cos (9 x)+\cos (-10 x)$
$\cdots\left[\begin{array}{c}\because \pi=3.14 \Rightarrow[9.85]=9 \\ \text { and }[-9.85]=-10\end{array}\right]$
$=\cos (9 x)+\cos (10 x)$
$=2 \cos \left(\frac{19 x}{2}\right) \cos \left(\frac{x}{2}\right)$
$\therefore \quad f\left(\frac{\pi}{2}\right)=2 \cos \left(\frac{19 \pi}{4}\right) \cos \left(\frac{\pi}{4}\right) ;$
$\therefore \quad f\left(\frac{\pi}{2}\right)=2 \times \frac{-1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=-1$

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