A) \[1/4\]
B) \[1/2\]
C) \[1\]
D) \[2\]
Correct Answer: B
Solution :
[b] \[y=cosec\text{ }x-cot\text{ }x\] \[\Rightarrow \,\,\,\,\,\,\frac{dy}{dx}=-\cos ec\,\,x\,\,\cot x+\cos e{{c}^{2}}x\] \[\Rightarrow \,\,\,\,\,y'=\cos ec\,\,x(\cos ec\,x-\cot x)\] \[\Rightarrow \,\,\,\,\,y'=y\,\cos ec\,x\] \[\Rightarrow \,\,\,\,\,y'sin\,\,x=y\] \[\Rightarrow \,\,\,\,\,y''sin\,\,x+y'cos\,x=y'\] \[\Rightarrow \,\,\,\,\,y''sin\,\,x=y'(1-cos\,x)\] \[\Rightarrow \,\,\,\,\,y''=\frac{y}{{{\sin }^{2}}x}(1-\cos x)\] \[\Rightarrow \,\,\,\,\,y''=\frac{y\left( 2{{\sin }^{2}}\frac{x}{2} \right)}{4{{\sin }^{2}}\frac{x}{2}{{\cos }^{2}}\frac{x}{2}}\] \[\Rightarrow \,\,\,\,\,y''=\frac{1}{2}\,y\,{{\sec }^{2}}\frac{x}{2}\] \[\Rightarrow \,\,\,\,\,\frac{y''}{y}=\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\] ?..(1) Given that \[\frac{y''}{y}=\lambda {{\sec }^{2}}\frac{x}{2}\] ?..(2) On comparing (1) and (2), we get \[\lambda =\frac{1}{2}\]
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