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CAT 2020 Slot 1 — QA Question 11

IndicesEasy

The number of real-valued solutions of the equation 2x + 2-x = 2 – (x – 2)² is

Answer & solution

  • 0

  • B

    Infinite

  • C

    2

  • D

    1

Solution

Easy

Bound each side. The left side 2x+2−x2^{x}+2^{-x} is always at least 22 (AM–GM), while the right side 2−(x−2)22-(x-2)^2 is always at most 22. They can be equal only if both hit 22 simultaneously — check whether that single candidate works.

1

Bound the left side from below. For y=2x>0y=2^{x}>0, y+1y≥2y+\tfrac{1}{y}\ge 2 by AM–GM, with equality only at y=1y=1 i.e. x=0x=0.

2x+2−x≥2(equality ⇔x=0)\begin{aligned} &2^{x}+2^{-x}\ge 2\quad\text{(equality }\Leftrightarrow x=0) \end{aligned}
2

Bound the right side from above. A square is non-negative, so subtracting it from 2 cannot exceed 2.

2−(x−2)2≤2(equality ⇔x=2)\begin{aligned} &2-(x-2)^{2}\le 2\quad\text{(equality }\Leftrightarrow x=2) \end{aligned}
3

Both extremes can't occur together. Equality on the left needs x=0x=0; equality on the right needs x=2x=2. Since LHS≥2≥RHS\text{LHS}\ge 2\ge\text{RHS}, any solution forces both sides =2=2, requiring x=0x=0 and x=2x=2 at once — impossible. Confirm at each candidate:

x=0: LHS=2, RHS=2−4=−2 ⇒ LHS≠RHSx=2: RHS=2, LHS=4+14=4.25 ⇒ LHS≠RHS\begin{aligned} &x=0:\ \text{LHS}=2,\ \text{RHS}=2-4=-2\ \Rightarrow\ \text{LHS}\ne\text{RHS}\\ &x=2:\ \text{RHS}=2,\ \text{LHS}=4+\tfrac14=4.25\ \Rightarrow\ \text{LHS}\ne\text{RHS} \end{aligned}
0 real solutions0\ \text{real solutions}
CAT 2020 Slot 1 QA Q11: The number of real-valued solutions of the equation 2 x + 2 -x = 2 – (x – 2)² is — Solution | TheCATExam