Q51 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

If $\alpha$ and $\beta$ are the roots of the equation $2^{6 x}-3\left(2^{3 x+2}\right)+32$ $=0$ with $\beta < 1$, then $2 \alpha+3 \beta=$

  • A
    -3
  • B
    -4
  • C
    3
  • Correct answer
    D
    4

Solution

Since, $2^{6 x}-3\left(2^{3 x+2}\right)+32=0$ $\begin{aligned} & \Rightarrow\left(2^{3 x}\right)^2-12\left(2^{3 x}\right)+32=0 \\ & \text { Let } y=2^{3 x} \\ & \therefore y^2-12 y+32=0 \\ & (y-4)(y-8)=0 \\ & \Rightarrow y=4 \quad \text { or, } y=8 \\ & \Rightarrow 2^{3 x}=4 \quad \text { or, } 2^{3 x}=8 \\ & \Rightarrow 2^{3 x}=22 \quad \text { or, } 2^{3 x}=23\end{aligned}$ $\Rightarrow x=\frac{2}{3} \quad$ or, $x=1$ Hence, $\alpha=1 \& \beta=\frac{2}{3}$ Then, $2 \alpha+3 \beta=2 \times 1+3 \times \frac{2}{3}=4$

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

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