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Limit of an expression in a quadratic at one of its roots

This type of question, asked 3 times: 2021 Paper 20, 8 Apr 2023, 5 Apr 2024. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 20

    76
    If α,β\alpha,\beta are the distinct roots of x2+bx+c=0x^{2}+bx+c= 0, then lim⁡x→βe2(x2+bx+c)−1−2(x2+bx+c)(x−β)2\lim_{x\rightarrow\beta} \frac{e^{2\left( x^{2}+bx+c \right)}- 1 - 2\left( x^{2}+bx+c \right)}{(x-\beta)^{2}} is equal to:

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  2. Question 2

    8 Apr 2023

    167
    If α>β>0\alpha > \beta > 0 are the roots of the equation ax2+bx+1=0ax^{2}+ bx + 1 = 0, and lim⁡x→1α(1−cos⁡(x2+bx+a)2(1−αx)2)12=1k(1β−1α)\lim_{x \rightarrow\frac{1}{\alpha}} \left( \frac{1 - \cos\left( x^{2}+ bx + a \right)}{2(1 - \alpha x)^{2}} \right)^{\frac{1}{2}}=\frac{1}{k}\left( \frac{1}{\beta}-\frac{1}{\alpha} \right), then kk is equal to

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  3. Question 3

    5 Apr 2024

    176
    Let a>0a > 0 be a root of the equation 2x2+x−2=02x^{2}+ x - 2 = 0. If lim⁡x→1a16(1−cos⁡(2+x−2x2))(1−ax)2=α+β17\lim_{x \rightarrow\frac{1}{a}} \frac{16\left( 1 - \cos\left( 2 + x - 2x^{2} \right) \right)}{\left( 1 - ax \right)^{2}}= \alpha + \beta\sqrt{17}, where α,β∈Z\alpha,\beta \in Z then α+β\alpha + \beta is equal to