Q54 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

The sum of all the real values of $x$ satisfying the equation $\left(x^2-7 x+11\right)^{x^2-6 x-7}=1$ is

  • A
    14
  • B
    20
  • C
    13
  • Correct answer
    D
    16

Solution

$f(x)^{g(x)}=1$ is possible when Case 1: $f(x)=1$ $\begin{aligned} & x^2-7 x+11=1 \\ & x^2-7 x+10=0 \\ & x=2,5\end{aligned}$ Case 2: $g(x)=0$ $\begin{aligned} & x^2-6 x-7=0 \\ & x=-i, 7\end{aligned}$ Case 3: $f(x)=-1$ and $g(x)=$ even $\begin{aligned} & x^2-7 x+11=-1 \\ & x^2-7 x+12=0 \\ & x=3,4\end{aligned}$ $\begin{aligned} & g(3)=3^2-6(3)-7=-16(\text { even }) \\ & g(4)=4^2-6(4)-7=-15(\text { odd }) \\ & \therefore \text { possible values of } x=-1,2,3,5,7 \\ & \text { sum }=16\end{aligned}$

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

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