Limit Continuity And Differentiability Ques 23
- If $e^y+x y=e$, the ordered pair $\left(\frac{d y}{d x}, \frac{d^2 y}{d x^2}\right)$ at $x=0$ is equal to
(2019 Main, 12 April I)
(a) $\left(\frac{1}{e},-\frac{1}{e^2}\right)$
(b) $\left(-\frac{1}{e}, \frac{1}{e^2}\right)$
(c) $\left(\frac{1}{e}, \frac{1}{e^2}\right)$
(d) $\left(-\frac{1}{e},-\frac{1}{e^2}\right)$
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Answer:
Correct Answer: 23.(b)
Solution: (b) Key Idea
Differentiating the given equation twice w.r.t. ’ $x$ ‘.
Given equation is
$ e^y+x y=e \quad ……..(i) $
On differentiating both sides w.r.t. $x$, we get
$ \begin{aligned} e^y \frac{d y}{d x}+x \frac{d y}{d x}+y =0 \quad ……..(ii) \\ \frac{d y}{d x} =-\left(\frac{y}{e^y+x}\right) \quad ……..(iii) \end{aligned} $
Again differentiating Eq. (ii) w.r.t. ’ $x$ ‘, we get
$ e^y \frac{d^2 y}{d x^2}+e^y\left(\frac{d y}{d x}\right)^2+x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=0 \quad ……..(iv) $
Now, on putting $x=0$ in Eq. (i), we get
$ \begin{gathered} e^y=e^1 \\ y=1 \end{gathered} $
On putting $x=0, y=1$ in Eq. (iii), we get
$ \frac{d y}{d x}=-\frac{1}{e+0}=-\frac{1}{e} $
Now, on putting $x=0, y=1$ and $\frac{d y}{d x}=-\frac{1}{e}$ in Eq. (iv), we get
$e^1 \frac{d^2 y}{d x^2}+e^1\left(-\frac{1}{e}\right)^2+0\left(\frac{d^2 y}{d x^2}\right)+\left(-\frac{1}{e}\right)+\left(-\frac{1}{e}\right)=0$
$\left.\Rightarrow \quad \frac{d^2 y}{d x^2}\right|_{(0,1)}=\frac{1}{e^2}$
So, $\quad\left(\frac{d y}{d x}, \frac{d^2 y}{d x^2}\right)$ at $(0,1)$ is $\left(-\frac{1}{e}, \frac{1}{e^2}\right)$
BETA
