Calculus

Application of Derivatives

Easy

common

If $x$ is real, the minimum value of $x^2 - 8x + 20$ is

-1

✅ Correct Option: 1

To find the minimum value of $x^2 - 8x + 20$ , complete the square.

Starting with:

$x^2 - 8x + 20$

Take the coefficient of $x$ , which is $-8$ , divide by 2 to get $-4$ , then square to get $16$ .

Add and subtract 16:

$x^2 - 8x + 20$

$= x^2 - 8x + 16 - 16 + 20$

$= (x - 4)^2 - 16 + 20$

$= (x - 4)^2 + 4$

Since $(x - 4)^2 \geq 0$ for all real $x$ , the minimum value of $(x - 4)^2$ is $0$ .

This occurs when $x = 4$ .

The minimum value of the expression is:

$(x - 4)^2 + 4$

$= 0 + 4$

$= 4$

Therefore, the minimum value is $4$ .

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Choose the correct answer from the options given below:

If xxx is real, the minimum value of x2−8x+20x^2 - 8x + 20x2−8x+20 is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $x$ is real, the minimum value of $x^2 - 8x + 20$ is