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CAT 2021 Slot 1 — QA Question 5

Discriminant and Roots of Quadratic EquationEasy

If r is a constant such that |x2 – 4x - 13| = r has exactly three distinct real roots, then the value of r is?

Answer & solution

Correct answer: 17

  • A

    21

  • B

    15

  • 17

  • D

    18

Solution

Easy

∣f(x)∣=r|f(x)|=r means f(x)=rf(x)=r or f(x)=−rf(x)=-r, i.e. two quadratics. Together they normally give four roots; exactly three distinct roots happens only when one of the two has a repeated root (discriminant 00). Find which value of rr does that consistently.

1

Split the modulus. Let f(x)=x2−4x−13f(x)=x^{2}-4x-13.

∣x2−4x−13∣=r⇒ x2−4x−13=rorx2−4x−13=−r\begin{aligned} &|x^{2}-4x-13|=r\\ &\Rightarrow\ x^{2}-4x-13=r \quad\text{or}\quad x^{2}-4x-13=-r \end{aligned}
2

Look at the minimum of ff. f(x)=(x−2)2−17f(x)=(x-2)^{2}-17, so its least value is −17-17 and the vertex is the only place a repeated root can occur. The graph of ∣f∣|f| has three roots exactly when the horizontal line y=ry=r passes through the vertex of the flipped-up portion, i.e. rr equals the depth 1717.

f(x)=(x−2)2−17⇒ min⁡f=−17 at x=2\begin{aligned} &f(x)=(x-2)^{2}-17\\ &\Rightarrow\ \min f=-17\ \text{at}\ x=2 \end{aligned}
3

Force a repeated root via the discriminant. Take x2−4x−13=−rx^{2}-4x-13=-r, i.e. x2−4x−(13−r)=0x^{2}-4x-(13-r)=0, and set D=0D=0.

D=(−4)2−4⋅1⋅(−(13−r))=0⇒ 16+4(13−r)=0⇒ 16+52−4r=0⇒ r=17\begin{aligned} &D=(-4)^{2}-4\cdot 1\cdot(-(13-r))=0\\ &\Rightarrow\ 16+4(13-r)=0\\ &\Rightarrow\ 16+52-4r=0\\ &\Rightarrow\ r=17 \end{aligned}
4

Confirm three distinct roots. With r=17r=17: the equation x2−4x−13=−17⇒(x−2)2=0x^{2}-4x-13=-17\Rightarrow(x-2)^{2}=0 gives the single root x=2x=2, while x2−4x−13=17⇒x2−4x−30=0x^{2}-4x-13=17\Rightarrow x^{2}-4x-30=0 has two distinct real roots. The other discriminant case would need r=−17r=-17, impossible since r=∣f∣≥0r=|f|\ge 0.

x=2(double, from −r branch)x=2±34(two more, from +r branch)⇒ 3 distinct roots\begin{aligned} &x=2 \quad\text{(double, from }-r\text{ branch)}\\ &x=2\pm\sqrt{34} \quad\text{(two more, from }+r\text{ branch)}\\ &\Rightarrow\ 3\ \text{distinct roots} \end{aligned}
r=17(option (c))r=17 \quad\text{(option (c))}

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