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CAT 2020 Slot 2 — QA Question 19

Solving Quadratic EquationsMedium

The number of integers that satisfy the equality (x² - 5x + 7)x+1 = 1 is

Answer & solution

  • A

    5

  • B

    2

  • C

    4

  • 3

Solution

Medium

A power equals 11 in three ways: the exponent is 00 (base nonzero); the base is 11; or the base is −1-1 with an even exponent. Check each case for integer solutions and count the distinct ones.

1

Case A — exponent zero. x+1=0x+1=0, and the base x2−5x+7x^2-5x+7 at x=−1x=-1 is 1+5+7=13≠01+5+7=13\neq 0, so this is valid.

x+1=0⇒x=−1\begin{aligned} &x+1=0 \Rightarrow x=-1 \end{aligned}
2

Case B — base equal to 11.

x2−5x+7=1⇒ x2−5x+6=0⇒ (x−2)(x−3)=0⇒x=2, 3\begin{aligned} &x^2-5x+7 = 1\\ &\Rightarrow\ x^2-5x+6 = 0\\ &\Rightarrow\ (x-2)(x-3)=0 \Rightarrow x=2,\ 3 \end{aligned}
3

Case C — base equal to −1-1 with even exponent.

x2−5x+7=−1⇒ x2−5x+8=0⇒ discriminant=25−32=−7<0\begin{aligned} &x^2-5x+7 = -1\\ &\Rightarrow\ x^2-5x+8 = 0\\ &\Rightarrow\ \text{discriminant} = 25-32 = -7 < 0 \end{aligned}

No real (hence no integer) solution, so this case contributes nothing.

4

Collect solutions. The integers are x=−1, 2, 3x=-1,\,2,\,3 — three of them.

{−1, 2, 3}⇒3 integers\begin{aligned} &\{-1,\ 2,\ 3\} \Rightarrow 3 \text{ integers} \end{aligned}
Number of integers=3\text{Number of integers} = 3
CAT 2020 Slot 2 QA Q19: The number of integers that satisfy the equality (x&sup2; - 5x + 7) x+1 = 1 is — Solution | TheCATExam