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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\cos\text{x}=\frac{1}{2}\Big(\text{a}+\frac{1}{\text{a}}\Big),$ and $3\text{x}=\lambda\Big(\text{a}^3+\frac{1}{\text{a}^3}\Big),$ then $\lambda=$
  • A
    $\frac{1}{4}$
  • ✓
    $\frac{1}{2}$
  • C
    $1$
  • D
    None of these
✓

Answer

Correct option: B.
$\frac{1}{2}$
Given:
$\cos\text{x}=\frac{1}{2}(\text{a}+\frac{1}{\text{a}})$
$\cos3\text{x}=\lambda\Big(\text{a}^2+\frac{1}{a^3}\Big)$
Now,
$\cos^3\text{x}=\frac{1}{8}\Big[\text{a}^3+\frac{1}{\text{a}^2}+3\text{a}\frac{1}{\text{a}^3}+\text{a}\frac{1}{\text{a}}\Big]$
$\cos^3\text{x}=\frac{1}{8}\Big(​a​^3+\frac{1}{​a​^3}+32\cos​​​\text{x}​\Big)$
$[\because\cos\text{x}=\frac{1}{2}(\text{a}+\frac{1}{\text{a}})]$
$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{\cos^3\text{x}}{\lambda}+6\cos\text{x}\Big)$
$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{4\cos^3\text{x}-3\cos\text{x}}{\lambda}+6\cos\text{x}\Big)$
$\Rightarrow\cos^3\text{x}=\frac{4\cos^3\text{x}}{8\lambda}\frac{3\cos^3\text{x}}{8\lambda}+\frac{6\cos\text{x}}{8}$
On comparing the powers of $\cos3x$ on both sides, we get
$1=\frac{4}{8\lambda}$
$1=\frac{4}{8\lambda}$
$\Rightarrow\lambda=\frac{1}{2}$

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