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MHTCET 10 May 2024 Shift 2 Paper

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Question : 104 of 150

Marks: +1, -0

If

y

is a function of

x

\log (x+y)=2 x y

, then the value of

y^{\prime}(0)

is

1
-1
2
0

Solution:

Concept:

Differentiate the given implicit equation with respect to

x

, then substitute

x=0

to find

y

and

y'(0)

.

Explanation:

Given:

\log(x+y)=2xy

Differentiate both sides w.r.t.

x

:

\frac{1}{x+y}\left(1+\frac{dy}{dx}\right)=2\left(x\frac{dy}{dx}+y\right)

Simplify and solve for

\frac{dy}{dx}

:

\frac{dy}{dx}=\frac{1-2xy-2y^2}{2x^2+2xy-1}

At

x=0

, substitute into

\log(x+y)=2xy

:

\log(0+y)=0 \Rightarrow \log y=0 \Rightarrow y=1

Now substitute

x=0

and

y=1

into the derivative expression:

y'(0)=\frac{1-0-2(1)^2}{0+0-1}=\frac{-1}{-1}=1

Answer:

The value of

y'(0)

is

1

, i.e. option A.

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