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CAT 2019 Slot 1 — QA Question 34

Trigonometric IdentitiesEasy

The number of the real roots of the equation 2cos(x(x + 1)) = 2x + 2–x is

Answer & solution

Correct answer: 1

  • A

    0

  • B

    2

  • C

    infinite

  • 1

Solution

Easy

Bound both sides. The left side 2cos⁡(⋅)2\cos(\cdot) never exceeds 22, while 2x+2−x≥22^{x}+2^{-x}\ge 2 by AM–GM. So both sides must equal 22 simultaneously — a very tight condition that pins down xx.

1

Bound the right side. Let a=2x>0a=2^{x}>0, so 2−x=1/a2^{-x}=1/a. By AM–GM:

2x+2−x=a+1a≥2(AM–GM, equality at a=1)\begin{aligned} &2^{x}+2^{-x}=a+\frac{1}{a}\ge 2\quad\text{(AM–GM, equality at }a=1) \end{aligned}
2

Bound the left side. Cosine is at most 11.

2cos⁡(x(x+1))≤2\begin{aligned} &2\cos\big(x(x+1)\big)\le 2 \end{aligned}
3

Force equality. LHS ≤2≤\le 2\le RHS, yet they are equal, so both equal 22.

2x+2−x=2 ⇒ a+1a=2 ⇒ (a−1)2=0 ⇒ a=1⇒ 2x=1 ⇒ x=0\begin{aligned} &2^{x}+2^{-x}=2\ \Rightarrow\ a+\tfrac1a=2\ \Rightarrow\ (a-1)^2=0\ \Rightarrow\ a=1\\ &\Rightarrow\ 2^{x}=1\ \Rightarrow\ x=0 \end{aligned}
4

Check x=0x=0. The cosine condition must also hold there.

2cos⁡(0⋅1)=2cos⁡0=2(satisfied)\begin{aligned} &2\cos\big(0\cdot 1\big)=2\cos 0=2\quad\text{(satisfied)} \end{aligned}

So exactly one value, x=0x=0, works.

Number of real roots=1\text{Number of real roots}=1

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CAT 2019 Slot 1 QA Q34 — Trigonometric Identities solution | TheCATExam