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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 65 of 160

Marks: +1, -0

If

(x^2-3x+2)^{\frac{y}{x^{2-1}}}=x+2

\left(\frac{dy}{dx}\right)_{x=0}=

2
-2
1
-1

Solution:

Given equation

(x^2-3x+2)e^{\frac{y}{x-1}}=x+2

Differentiate both sides w.r.t., we get

\frac{d}{dx}\left((x^2-3x+2)\cdot e^{\frac{y}{x-1}}\right)=\frac{d}{dx}(x+2)

\Rightarrow(2x-3)e^{\frac{y}{x-1}}+(x^2-3x+2)\cdot e^{\frac{y}{x-1}}

\Rightarrow(2x-3)e^{\frac{y}{x-1}}+(x^2-3x+2)\cdot e^{\frac{y}{x-1}}\left(\frac{y}{x-1}\right)=1

\left[\frac{\frac{dy}{dx}(x-1)-y}{(x-1)^2}\right]=1

Now, substitute

x=0

into the given equation

(0^2-3(0)+2)e^{\frac{y}{0-1}}=0+2

\Rightarrow 2e^{-y}=2\Rightarrow e^{-y}=1

\Rightarrow -y=\ln(1)\Rightarrow -y=0

\Rightarrow y=0

Substitute

x=0

and

y=0

in Eq. (i), we get

(2(0)-3)e^{\frac{0}{0-1}}+(0^2-3(0)+2)e^{\frac{0}{0-1}}

\Rightarrow-\frac{\frac{dy}{dx}(0-1)-0}{(0-1)^2}]=1

\Rightarrow-3e^0+2e^0\left(\frac{-dy}{dx}\right)=1

\Rightarrow-3+2\left(\frac{-dy}{dx}\right)=1

\Rightarrow-2\frac{dy}{dx}=1+3=4\Rightarrow\frac{dy}{dx}=\frac{4}{-2}=-2

\therefore\left(\frac{dy}{dx}\right)_{x=0}=-2

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