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For \mathrm{k \in \mathbb{R},} let the solutions of the equation\mathrm{\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0<|x|<\frac{1}{\sqrt{2}}}  be \mathrm{\alpha \; and \; \beta,} where the inverse trigonometric functions take only principal values. If the solutions of the equation \mathrm{x^{2}-b x-5=0}  are  \mathrm{\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}} and   \mathrm{\frac{\alpha}{\beta},} then \mathrm{\frac{b}{k^{2}}}  is equal to__________.

Option: 1

12


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{\cos \left(\sin ^{-1} x\right)=\cos \left(\cos ^{-1} \sqrt{1-x^{2}}\right)=\sqrt{1-x^{2}}}\\ & \mathrm{\cos \left(\tan ^{-1} \sqrt{1-x^{2}}\right)=\cot ^{-x^{-1}} \mid\left(\sqrt{\frac{1}{\sqrt{1-x^{2}}}}\right)=\frac{1}{\sqrt{1-x^{2}}}}\\ & \mathrm{\Rightarrow \cos \left(\sin ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\right)=\frac{\sqrt{1-2 x^{2}}}{\sqrt{1-x^{2}}}}\\ \end{aligned}

\begin{aligned} &\mathrm{\Rightarrow \frac{\sqrt{1-2 x^{2}}}{\sqrt{1-x^{2}}}=k}\\ &\mathrm{\Rightarrow 1-2 x^{2}=k^{2}\left(1-x^{2}\right)}\\ &\mathrm{\Rightarrow\left(k^{2}-2\right) x^{2}=k^{2}-1}\\ &\mathrm{x^{2}=\frac{k^{2}-1}{k^{2}-2}}\\ \end{aligned}

\begin{aligned} & \mathrm{\alpha=\sqrt{\frac{k^{2}-1}{k^{2}-2}} \Rightarrow \alpha^{2}=\frac{k^{2}-1}{k^{2}-2}}\\ & \mathrm{\beta=\sqrt{\frac{k^{2}-1}{k^{2}-2}} \Rightarrow \beta^{2}=\frac{k^{2}-1}{k^{2}-2}}\\ & \mathrm{\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=2\left(\frac{k^{2}-2}{k^{2}-1}\right) \text { \& } \frac{\alpha}{\beta}=-1}\\ & \mathrm {\text { Sum of roots }=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{\alpha}{\beta}=b}\\ & \mathrm {\Rightarrow \frac{2\left(k^{2}-2\right)}{k^{2}+}-1=b \ldots \text { (1) }} \end{aligned}

\begin{aligned} & \text{Poduct of roots} =\left(\frac{1}{\alpha^{2}}+1-\frac{1}{\beta^{2}}\right) \frac{\alpha}{\beta}=-5\\ &\mathrm{\Rightarrow \frac{2\left(k^{2}-2\right)}{k^{2}-1}(-1)=-5} \\ &\mathrm{\Rightarrow 2 k^{2}-4=5 k^{2}-5} \\ &\mathrm{\Rightarrow 3 k^{2}=1 \Rightarrow k^{2}=\frac{1}{3} \cdots, \text { put in(1) } }\\ &\mathrm{\Rightarrow b=\frac{2\left(k^{2}-2\right)}{k^{2}-1}-1=5-1=4 }\\ &\mathrm{\frac{b}{k^{2}}=\frac{4}{\frac{1}{3}}=12 }\end{aligned}

 

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vishal kumar

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