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UPSEE Solved Model Paper 1

Question : 147 of 150

Marks: +1, -0

\lim\limits_{x\to 0}\frac{ae^{-x}-b\cos x-\frac{1}{2}cx}{x\cos x}

then the value of a + b + c is-

Solution:

\lim\limits_{x\to 0}\frac{ae^{-x}-b\cos x-\frac{1}{2}cx}{x\cos x}

= 2

At x = 0, denominator = 0, Numerator = a - b

For limit to be exist numerator should also be = 0

So,

a - b = 0 ⇒ a = b ...(i)

Now Apply L' Hospital rule

\lim\limits_{x\to 0}\frac{ae^{-x}-b\cos x-\frac{1}{2}cx}{x\cos x}

\lim\limits_{x\to 0}\frac{\frac{d}{dx}\left(ae^{-x}-b\cos x-\frac{1}{2}cx\right)}{\frac{d}{dx}(x\cos x)}

\lim\limits_{x\to 0}\frac{-ae^{-x}+b\sin x-\frac{1}{2}c}{-x\sin x+\cos x}

= - a -

\frac{1}{2}

⇒ - a -

\frac{1}{2}

c = 2

𝑐 = −2𝑎 − 4

𝑎 + 𝑏 + 𝑐 = 𝑎 + 𝑎 + (−2𝑎 − 4)

= −4

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