Question

Let $f\left(x\right)=\prod _{k=1}^n(\cos (2k-1)x+i\sin (2k-1)x),$ then ${\left(Ref\left(x\right)\right)}^{\prime \prime }+i{\left(Imf\left(x\right)\right)}^{\prime \prime }$ is equal to

• A. $n^{2}f\left(x\right)$
• B. $-n^{4}f\left(x\right)$
• C. $-n^{2}f\left(x\right)$
• D. $n^{4}f\left(x\right)$

Answer 1

The correct option is B $-n^{4}f\left(x\right)$

$f\left(x\right)=\prod _{k=1}^n\left(\cos (2k-1)x+i\sin (2k-1)x\right)$

$=\cos \sum _{k=1}^n(2k-1)x+i\sin \sum _{k=1}^n(2k-1)x$

$=\cos n^{2}x+i\sin n^{2}x$ $($Using De Moivre's Theorem$)$

$\therefore {\left(Ref\left(x\right)\right)}^{\prime \prime }=-n^4\cos n^{2}x$ and ${\left(Imf\left(x\right)\right)}^{\prime \prime }=-n^4\sin n^{2}x.$

Thus, ${\left(Ref\left(x\right)\right)}^{\prime \prime }+i{\left(Imf\left(x\right)\right)}^{\prime \prime }=-n^4[\cos n^{2}x+i\sin n^{2}x]$ $=-n^{4}f\left(x\right).$