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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Limits

Let a>0 be a root of the equation 2 x^2+x-2=0 . If _ x arrow 1 a...

Let a>0\mathrm{a}>0 be a root of the equation 2x2+x−2=02 x^2+x-2=0. If lim⁡x→1a16(1−cos⁡(2+x−2x2))(1−ax)2=α+β17\lim _{x \rightarrow \frac{1}{\mathrm{a}}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-\mathrm{a} x)^2}=\alpha+\beta \sqrt{17}, where α,β∈Z\alpha, \beta \in Z, then α+β\alpha+\beta is equal to_______Let a>0 be a root of the equation 2 x^2+x-2=0 . If _ x arrow 1 a 16(1- (2+x-2 x^2)) (1-a x)^2 =α+β 17 , where α, β ∈ ZLet a>0 be a root of the equation 2 x^2+x-2=0 . If _ x arrow 1 a 16(1- (2+x-2 x^2)) (1-a x)^2 =α+β 17 , where α, β ∈ Z

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Step-by-step solution

Let a>0 be a root of the equation 2 x^2+x-2=0 . If _ x arrow 1 a 16(1- (2+x-2 x^2)) (1-a x)^2 =α+β 17 , where α, β ∈ Z
Let a>0 be a root of the equation 2 x^2+x-2=0 . If _ x arrow 1 a 16(1- (2+x-2 x^2)) (1-a x)^2 =α+β 17 , where α, β ∈ Z
lim⁡x→1a16⋅(1−cos⁡2(x−1a)(x−1b))4(x−1b)2×4(x−1b)2a2(x−1a)2=16×2a2(1a−1b)2=32a2(174)=17.8a2=17×8×16(−1+117)2=136.1618.27×18+2718+27=136256(18+27)⋅16=153+1717=α+β17α+β=153+17=170\begin{aligned} & \lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^2} \times \frac{4\left(x-\frac{1}{b}\right)^2}{a^2\left(x-\frac{1}{a}\right)^2} \\ & =16 \times \frac{2}{a^2}\left(\frac{1}{a}-\frac{1}{b}\right)^2 \\ & =\frac{32}{a^2}\left(\frac{17}{4}\right)=\frac{17.8}{a^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} \\ & =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} \\ & =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 \\ & =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} \\ & \alpha+\beta=153+17=170\end{aligned}
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