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CAT 2020 Slot 2 — QA Question 2

Number TheoryEasy

For real x, the maximum possible value of x1+x4 is

Answer & solution

  • A

    1/2

  • B

    1

  • 1/√2

  • D

    1/√3

Solution

Easy

Invert the expression and bring everything under one square root. The maximum of the original fraction occurs where the inside is minimum; the AM–GM inequality on x2x^2 and 1/x21/x^2 gives that minimum.

1

Rewrite by dividing by xx. Divide numerator and denominator inside the root by x2x^2.

x1+x4=11+x4x2⇒ =11x2+x2\begin{aligned} &\frac{x}{\sqrt{1+x^4}} = \frac{1}{\sqrt{\dfrac{1+x^4}{x^2}}}\\ &\Rightarrow\ = \frac{1}{\sqrt{\dfrac{1}{x^2}+x^2}} \end{aligned}
2

Apply AM–GM. For positive x2x^2, a number plus its reciprocal is at least 22.

x2+1x2≥2x2⋅1x2=2(AM–GM)⇒ minimum of the denominator is 2\begin{aligned} &x^2 + \frac{1}{x^2} \ge 2\sqrt{x^2\cdot\tfrac{1}{x^2}} = 2 \quad\text{(AM–GM)}\\ &\Rightarrow\ \text{minimum of the denominator is } \sqrt{2} \end{aligned}
3

Convert to a maximum. The fraction is largest when its denominator is smallest, i.e. at the bound from step 2.

max⁡x1+x4=12(at x=1)\begin{aligned} &\max \frac{x}{\sqrt{1+x^4}} = \frac{1}{\sqrt{2}} \quad\text{(at }x=1\text{)} \end{aligned}
12\frac{1}{\sqrt{2}}
CAT 2020 Slot 2 QA Q2: For real x, the maximum possible value of x 1 + x 4 is — Solution | TheCATExam