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  • For the curve  lying in first quadrant. If  <img alt="\lim _{x \

For the curve \sin x+\sin y=1 lying in first quadrant. If  \lim _{x \rightarrow 0} x^\alpha \frac{d^2 y}{d x^2} exists and non-zero then 2 \alpha=

 

Option: 1

3


Option: 2

4


Option: 3

5


Option: 4

1


Answers (1)

best_answer

\begin{aligned} & \sin x+\sin y=1 \\ & \Rightarrow \quad \cos x+\cos y y^{\prime}=0 \\ & \Rightarrow \quad-\sin x+\cos y y^{\prime \prime}+\left(y^{\prime}\right)(-\sin y)=0 \\ & \Rightarrow \quad y^{\prime \prime}=\frac{y^{\prime} \sin y+\sin x}{\cos y} \\ & =\frac{\left(-\frac{\cos x}{\cos y}\right)^2 \sin y+\sin x}{\cos y} \\ & =\frac{\sin x \cos ^2 y+\cos ^2 x \sin y}{\cos ^3 y} \\ & \end{aligned}

Putting \sin x=t, sin y=1-t, we get

\begin{aligned} & y^{\prime \prime}=\frac{t^{-3 / 2}\left(1-t+t^2\right)}{(2-t)^{3 / 2}} \\ \Rightarrow & \alpha=\frac{3}{2} \\ \Rightarrow & 2 \alpha=3 \end{aligned}

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himanshu.meshram

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