If α, β are the roots of the equation a x^2+b x+c=0 then lim _x arrow β (1-cos (a x^2+b x+c ))/((x-β)^2) is

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MARKS Content Team · Accepted Answer

Correct answer: (D) (a^2)/2(α-β)^2. Hint: a x^2+b x+c=a(x-α)(x-β) lim _x arrow β (1-cos (a x^2+b x+c ))/((x-β)^2) =lim _x arrow β (1-cos (a(x-α)(x-β)))/((x-β)^2) =lim _x arrow β (2 sin^2 ((a(x-α)(x-β))/2 ))/((x-β)^2)=lim _x arrow β (2 sin^2 ((a(x-α)(x-β)/2 ))/((a^2(x-α)^2(x-β)^2)/4) × (a^2(x-α)^2)/4 =2 × 1 × (a^2(β-α)^2)/4=(a^2(α-β)^2)/2

If $\alpha, \beta$ are the roots of the equation $a x^2+b x+c=0$ then $\lim _{x \rightarrow \beta} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\beta)^2}$ is

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