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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

\text{If } \lim_{x \to 0} \frac{e^{ax} - \cos(bx) - \frac{cxe^{-cx}}{2}}{1 - \cos(2x)} = 17, \text{ then } 5a^2 + b^2 \text{ is equal to:}

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Explanation

Solution

Is problem ko solve karne ke liye hum Taylor Series Expansion ka use karenge kyunki denominator $x^2$ ki term mein convert ho jayega.

Step 1: Denominator ko simplify karna

Hum jaante hain ki:

$1 - \cos(2x) = 2\sin^2(x)$

Jab $x \to 0$ , tab $\sin(x) \approx x$ , isliye:

$1 - \cos(2x) \approx 2x^2$

Step 2: Numerator ka Expansion ( $x^2$ tak)

Ab numerator ki terms ko expand karte hain:

$e^{ax} = 1 + ax + \frac{a^2x^2}{2} + \dots$

$\cos(bx) = 1 - \frac{b^2x^2}{2} + \dots$

$\frac{cxe^{-cx}}{2} = \frac{cx}{2}(1 - cx + \dots) = \frac{cx}{2} - \frac{c^2x^2}{2}$

Step 3: Limit mein values put karna

Numerator $(N)$ ban jayega:

$N = \left(1 + ax + \frac{a^2x^2}{2}\right) - \left(1 - \frac{b^2x^2}{2}\right) - \left(\frac{cx}{2} - \frac{c^2x^2}{2}\right)$

$N = x\left(a - \frac{c}{2}\right) + x^2\left(\frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2}\right)$

Kyunki limit exist karti hai aur finite ( $17$ ) hai, isliye $x$ ka coefficient zero hona chahiye:

$a - \frac{c}{2} = 0 \implies c = 2a$

Step 4: Final calculation

Ab limit ki value nikaalte hain:

$\lim_{x \to 0} \frac{x^2\left(\frac{a^2 + b^2 + c^2}{2}\right)}{2x^2} = 17$

$\frac{a^2 + b^2 + c^2}{4} = 17 \implies a^2 + b^2 + c^2 = 68$

Kyunki $c = 2a$ , toh $c^2 = 4a^2$ . Isse equation mein rakhte hain:

$a^2 + b^2 + 4a^2 = 68$

$5a^2 + b^2 = 68$

Correct Option: (4) 68

Choose the correct answer:

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