Q100 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

If $\lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$ then $\mathrm{m}(\mathrm{m}-1)=$

  • A
    0
  • B
    $\log _2 \mathrm{e}$
  • Correct answer
    C
    1
  • D
    $\frac{1+\sqrt{3}}{2}$

Solution

$\because \lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$ $\begin{aligned} & \Rightarrow m=\frac{1+\sqrt{1+4 \log _2 2}}{2+\left(4+\sin ^2 2+2 \cos 2\right)(4-4)} \\ & \Rightarrow m=\frac{1+\sqrt{1+4}}{2+0}=\frac{1+\sqrt{5}}{2}\end{aligned}$ Now, $m(m-1)=\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{1+\sqrt{5}}{2}-1\right)=1$

Hence, the correct answer is option: 'C'.

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