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CAT 2020 Slot 1 — QA Question 10

Solving Quadratic EquationsMedium

How many distinct positive integer-valued solutions exist to the equation (x2-7x+11)(x2-13x+42) = 1?

Answer & solution

  • A

    2

  • B

    8

  • C

    4

  • 6

Solution

Medium

A power PQ=1P^{Q}=1 in exactly three ways: (i) base =1=1 (any exponent), (ii) exponent =0=0 (with non-zero base), or (iii) base =−1=-1 with an even exponent. Test each case as an equation and collect all distinct integer roots.

1

Case (i): base =1=1. Solve x2−7x+11=1x^{2}-7x+11=1.

x2−7x+10=0⇒ (x−2)(x−5)=0⇒ x=2, 5\begin{aligned} &x^{2}-7x+10=0\\ &\Rightarrow\ (x-2)(x-5)=0\\ &\Rightarrow\ x=2,\ 5 \end{aligned}
2

Case (ii): exponent =0=0. Solve x2−13x+42=0x^{2}-13x+42=0 (base is non-zero at these roots, so valid).

(x−6)(x−7)=0⇒ x=6, 7\begin{aligned} &(x-6)(x-7)=0\\ &\Rightarrow\ x=6,\ 7 \end{aligned}
3

Case (iii): base =−1=-1 with even exponent. Solve x2−7x+11=−1x^{2}-7x+11=-1, then check the exponent's parity.

x2−7x+12=0⇒ (x−3)(x−4)=0⇒ x=3, 4\begin{aligned} &x^{2}-7x+12=0\\ &\Rightarrow\ (x-3)(x-4)=0\\ &\Rightarrow\ x=3,\ 4 \end{aligned}

At both x=3x=3 and x=4x=4, the exponent x2−13x+42x^{2}-13x+42 equals 1212 and 66 respectively — both even — so (−1)even=1(-1)^{\text{even}}=1. Both are valid.

4

Collect distinct solutions. No overlaps among the cases.

x∈{2,5}∪{6,7}∪{3,4}={2,3,4,5,6,7}⇒ 6 distinct solutions\begin{aligned} &x\in\{2,5\}\cup\{6,7\}\cup\{3,4\}=\{2,3,4,5,6,7\}\\ &\Rightarrow\ 6\ \text{distinct solutions} \end{aligned}
66
CAT 2020 Slot 1 QA Q10: How many distinct positive integer-valued solutions exist to the equation x 2 - 7 x + 11 ( x 2 - 13 x + 42 ) = — Solution | TheCATExam