The maximum value of f(x)=1/(3(x^2-x+1)) is : Option 1: 9/4 Option 2: 4/9 Option 3: 1/2 Option 4: -1/2

Answer: 4/9. Solution: f is largest when the denominator is smallest. Since x^2-x+1=(x-12)^2+34 has minimum 34 at x=12, the maximum of f is 1/(3x34)=49.

Calculus

Application of Derivatives

Medium

core

The maximum value of $f(x)=\dfrac{1}{3(x^2-x+1)}$ is :

$\dfrac{9}{4}$

$\dfrac{4}{9}$

$\dfrac{1}{2}$

$-\dfrac{1}{2}$

✅ Correct Option: 2

$f$ is largest when the denominator is smallest. Since $x^2-x+1=\left(x-\dfrac12\right)^2+\dfrac34$ has minimum $\dfrac34$ at $x=\dfrac12$ , the maximum of $f$ is $\dfrac{1}{3\times\frac34}=\dfrac49$ .

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Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The maximum value of $f(x)=\dfrac{1}{3(x^2-x+1)}$ is :