MCQ
If $\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})(\cos3\text{x}+\text{i}\sin3\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})$ and $\text{f}(1)=1, $ then $f_1$ is equals to :
  • A
    $\frac{\text{n}(\text{n}+1)}{2}$
  • B
    $\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
  • $-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
  • D
    None of these

Answer

Correct option: C.
$-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
$\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos\text{x}+\text{i}\sin\text{x})^2...(\cos\text{x}+\text{i}\sin\text{x})^\text{n}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{1+2+3 ........ \text{n}}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{\frac{\text{n}(\text{n}+1)}{2}}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^\text{a}$
$\Rightarrow\text{f}(\text{ax})=(\cos\text{ax}+\text{i}\sin\text{ax})...1$
$\Rightarrow\text{f}(1)=(\cos\text{a}+\text{i}\sin\text{a})$
$\Rightarrow1=(\cos\text{a}+\text{i}\sin\text{a})...2\ [\because\text{f}(1)=1]$
Differentiating eqn. $1,$ we get
$\text{f}\ '(\text{x})=\text{a}(-\sin\text{ax}+\text{i}\cos\text{ax})$
$\Rightarrow\text{f}\ ''(\text{x})=\text-{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$
$\Rightarrow\text{f}\ ''\text{x}=\text{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$
$\Rightarrow\text{f}\ ''(\text{x})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{ax}+\text{i}\sin\text{ax})$
$\Rightarrow\text{f}\ ''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{a}+\text{i}\sin\text{a})$
$\Rightarrow\text{f}\ ''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}\ [\text{using }2]$

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