If e^{y}+x y=e, then the ordered pair ≤ft(d y/d x, {d² y}{d x²}) at x=0 is equal to

Correct Answer is : (−1e,1e2)≤ft(-1/e, {1}{e²})(e−1,e21)

Correct Answer is : (−1e,1e2)≤ft(-1/e, {1}{e²})(e−1,e21)

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MHTCET 15 Apr 2026 Shift 2 Paper

Question : 123 of 150

Marks: +1, -0

e^{y}+x y=e

\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)

x=0

is equal to

Solution:

Concept:

Use implicit differentiation of the given equation to find

\frac{dy}{dx}

and

\frac{d^2y}{dx^2}

x=0

Explanation:

Given:

e^{y}+xy=e

x=0

e^{y}+0=e \Rightarrow e^{y}=e \Rightarrow y=1

Differentiate the equation with respect to

x

\frac{d}{dx}(e^{y})+\frac{d}{dx}(xy)=\frac{d}{dx}(e)

e^{y}\frac{dy}{dx}+x\frac{dy}{dx}+y=0

So,

(e^{y}+x)\frac{dy}{dx}+y=0

\frac{dy}{dx}=-\frac{y}{e^{y}+x}

x=0

y=1

\frac{dy}{dx}=-\frac{1}{e^{1}+0}=-\frac{1}{e}

Now differentiate

e^{y}\frac{dy}{dx}+x\frac{dy}{dx}+y=0

again:

e^{y}\left(\frac{dy}{dx}\right)^2+e^{y}\frac{d^2y}{dx^2}+x\frac{d^2y}{dx^2}+\frac{dy}{dx}+\frac{dy}{dx}=0

Thus,

(e^{y}+x)\frac{d^2y}{dx^2}+e^{y}\left(\frac{dy}{dx}\right)^2+2\frac{dy}{dx}=0

Substitute

x=0

y=1

\frac{dy}{dx}=-\frac{1}{e}

e\frac{d^2y}{dx^2}+e\left(\frac{1}{e^2}\right)+2\left(-\frac{1}{e}\right)=0

e\frac{d^2y}{dx^2}+\frac{1}{e}-\frac{2}{e}=0

e\frac{d^2y}{dx^2}-\frac{1}{e}=0

\frac{d^2y}{dx^2}=\frac{1}{e^2}

Answer:

\left(\frac{dy}{dx},\frac{d^2y}{dx^2}\right)=\left(-\frac{1}{e},\frac{1}{e^2}\right)

Therefore, the correct option is Option B.

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