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TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 65 of 160

Marks: +1, -0

If

f(x)=\frac{x-1}{e^x}

f'(0)+f^n(0)=

0
1
-1
2

Solution:

Given that,

f(x)=\frac{x-1}{e^{x}}

....(i)

Differentiating Eq. (i) w.r.t.

x

on both sides, we get

f'(x)=\frac{e^{x}\frac{d}{dx}(x-1)-(x-1)\frac{d}{dx}(e^{x})}{(e^{x})^{2}}

\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}u - u\frac{dv}{dx}}{v^{2}}\right]

=\frac{e^{x} \cdot 1-(x-1)e^{x}}{e^{2x}}

f'(x)=\frac{2-x}{e^{x}}

....(ii)

Putting

x=0

in Eq. (ii), we get

f'(0)=\frac{2-0}{e^{0}}=\frac{2}{1} \Rightarrow f'(0)=2

.....(iii)

Now, differentiating Eq. (ii) w.r.t.

x

on both sides, we get

f''(x)=\frac{e^{x}\frac{d}{dx}(2-x)-(2-x)\frac{d}{dx}(e^{x})}{(e^{x})^{2}}

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

f''(x)=\frac{x-3}{e^{x}}

....(iv)

Putting

x=0

, in Eq. (iv), we get

f'(0)=\frac{0-3}{e^{0}}=\frac{-3}{1} \Rightarrow f'(0)=-3

....(v)

Adding Eq. (iii) and Eq. (v), we get

f'(0)+f''(0)=2+(-3) \Rightarrow f'(0)+f''(0)=-1

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