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QJEEM23Ma337e: JEE Main 2023 Mathematics - Class 12 JEEM23Ma337e Solution: If are the roots of the equation , and lim x → 1 α ( 1 − cos ⁡…

If \alpha > \beta > 0 are the roots of the equation ax2+bx+1=0a x^{2}+b x+1=0 , and limx→1α(1−cos⁡(x2+bx+a)2(1−αx)2)12=1k(1β−1α), then k is equal to  :

Multiple Choice Options

  • A 2β2 \beta
  • B β\beta
  • C α\alpha
  • D 2α2 \alpha

Step-by-Step Verified Solution

Since, α,β\alpha, \beta are roots of ax2+bx+1=0a x^2+b x+1=0

Replace x→1xx \rightarrow \frac{1}{x}

ax2+bx+1=0⇒x2+bx+a=0\frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0

So, 1α,1β\frac{1}{\alpha}, \frac{1}{\beta} are the roots.

Now, lim⁡x→1α[1−cos⁡(x2+bx+a)2(1−αx)2]12\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}

=lim⁡x→1α[2sin⁡2(x2+bx+a2)2(1−αx)2]12=\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}

amp;=lim⁡x→1α[2sin⁡2(x−1α)(x−1β)24×2α2(x−1α)2(x−1β)24(x−1β)2]12amp;=lim⁡x→1α[±12sin⁡(x−1α)(x−1β)2α(x−1α)(x−1β)2(x−1β)]amp;=12α[−1α+1β)amp;⇒1k[1β−1α]=12α[1β−1α]amp;⇒k=2α\begin{aligned}& =\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2 \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{4 \times 2 \alpha^2 \frac{\left(x-\frac{1}{\alpha}\right)^2\left(x-\frac{1}{\beta}\right)^2}{4}\left(x-\frac{1}{\beta}\right)^2}\right]^{\frac{1}{2}} \\\\& =\lim _{x \rightarrow \frac{1}{\alpha}}\left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}\left(x-\frac{1}{\beta}\right)\right] \\\\& =\frac{1}{2 \alpha}\left[\frac{-1}{\alpha}+\frac{1}{\beta}\right) \\\\& \Rightarrow \frac{1}{k}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right]=\frac{1}{2 \alpha}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right] \\\\& \Rightarrow k=2 \alpha\end{aligned}

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