If \alpha \gt \beta \gt 0 are the roots of the equation a x^2+b x+1=0 , and \lim_x arrow \frac1\alpha(\frac1-\cos (x^2+b x+a)2(1-\alpha x)^2)^\frac12=\frac1k(\frac1\beta-\frac1\alpha), then k is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2023 · 8 Apr · Shift 2 · Q32 with Solution

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Correct answer: D

Use the fact that $\alpha,\beta$ are roots of $ax^2+bx+1=0$

Since $\alpha,\beta$ are roots of $ax^2+bx+1=0,$ we have by Vieta: $\alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{1}{a}.$ Hence $a=\frac{1}{\alpha\beta},\qquad b=-a(\alpha+\beta)=-\frac{\alpha+\beta}{\alpha\beta}.$

Factor $x^2+bx+a$ using the roots

The polynomial $ax^2+bx+1$ can be written as $a(x-\alpha)(x-\beta).$ Dividing by $a$ , $x^2+\frac{b}{a}x+\frac{1}{a}=(x-\alpha)(x-\beta).$ But $\frac{1}{a}=\alpha\beta$ and $\frac{b}{a}=-(\alpha+\beta)$ , so $x^2+bx+a$ should be handled directly by substituting $a,b$ : $x^2+bx+a=x^2-\frac{\alpha+\beta}{\alpha\beta}x+\frac{1}{\alpha\beta}.$ This factors as $x^2+bx+a=\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right).$

Rewrite the limit

We need $L=\lim_{x\to 1/\alpha}\left(\frac{1-\cos(x^2+bx+a)}{2(1-\alpha x)^2}\right)^{1/2}.$ Let $u=x^2+bx+a=\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right).$ As $x\to \frac1\alpha$ , we have $u\to 0$ . Using $1-\cos u \sim \frac{u^2}{2} \quad (u\to 0),$ we get $\frac{1-\cos u}{2(1-\alpha x)^2}\sim \frac{u^2/2}{2(1-\alpha x)^2}=\frac{u^2}{4(1-\alpha x)^2}.$ Therefore

=\lim_{x\to 1/\alpha} \frac{|u|}{2|1-\alpha x|}.$$ 4. **Simplify $u$ and denominator** Now $$u=\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right),$$ and $$1-\alpha x=-\alpha\left(x-\frac{1}{\alpha}\right).$$ So $$|1-\alpha x|=\alpha\left|x-\frac{1}{\alpha}\right|$$ (since $\alpha>0$). Thus $$L=\lim_{x\to 1/\alpha}\frac{\left|x-\frac1\alpha\right|\left|x-\frac1\beta\right|}{2\alpha\left|x-\frac1\alpha\right|} =\lim_{x\to 1/\alpha}\frac{\left|x-\frac1\beta\right|}{2\alpha}.$$ Hence $$L=\frac{1}{2\alpha}\left|\frac1\alpha-\frac1\beta\right|.$$ Since $\alpha>\beta>0$, we have $$\frac1\alpha<\frac1\beta,$$ so $$\left|\frac1\alpha-\frac1\beta\right|=\frac1\beta-\frac1\alpha.$$ Therefore $$L=\frac{1}{2\alpha}\left(\frac1\beta-\frac1\alpha\right).$$ 5. **Compare with the given form** Given $$L=\frac{1}{k}\left(\frac1\beta-\frac1\alpha\right).$$ So $$\frac1k=\frac{1}{2\alpha}\implies k=2\alpha.$$ 6. **Check options** - A: $2\beta$ - B: $\beta$ - C: $\alpha$ - D: $2\alpha$ ✅ Therefore, the correct option is **D**.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: D

Use the fact that $\alpha,\beta$ are roots of $ax^2+bx+1=0$

Factor $x^2+bx+a$ using the roots

Rewrite the limit

=\lim_{x\to 1/\alpha} \frac{|u|}{2|1-\alpha x|}.$$ 4. **Simplify $u$ and denominator** Now $$u=\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right),$$ and $$1-\alpha x=-\alpha\left(x-\frac{1}{\alpha}\right).$$ So $$|1-\alpha x|=\alpha\left|x-\frac{1}{\alpha}\right|$$ (since $\alpha>0$). Thus $$L=\lim_{x\to 1/\alpha}\frac{\left|x-\frac1\alpha\right|\left|x-\frac1\beta\right|}{2\alpha\left|x-\frac1\alpha\right|} =\lim_{x\to 1/\alpha}\frac{\left|x-\frac1\beta\right|}{2\alpha}.$$ Hence $$L=\frac{1}{2\alpha}\left|\frac1\alpha-\frac1\beta\right|.$$ Since $\alpha>\beta>0$, we have $$\frac1\alpha<\frac1\beta,$$ so $$\left|\frac1\alpha-\frac1\beta\right|=\frac1\beta-\frac1\alpha.$$ Therefore $$L=\frac{1}{2\alpha}\left(\frac1\beta-\frac1\alpha\right).$$ 5. **Compare with the given form** Given $$L=\frac{1}{k}\left(\frac1\beta-\frac1\alpha\right).$$ So $$\frac1k=\frac{1}{2\alpha}\implies k=2\alpha.$$ 6. **Check options** - A: $2\beta$ - B: $\beta$ - C: $\alpha$ - D: $2\alpha$ ✅ Therefore, the correct option is **D**.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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Limits Continuity and Differentiability question

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