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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives1 Mark
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})\ \text{and}\ \text{f}(1)=1, $ then $f1$ 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
    $\text{none of these}$
✓

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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