Question
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 $\lambda=$
-
$\frac{1}{4}$
-
$\frac{1}{2}$
-
$1$
- None of these
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 $\lambda=$
$\frac{1}{4}$
$\frac{1}{2}$
$1$
Solution:
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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