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The value of \( x^{2} - 4x + 11 \) can never be less than

A) 7
B) 8
C) 11
D) 22

Correct Answer:

A) 7

Description for Correct answer:
The value of \( \Large x^{2} - 4x + 11 \) can never be less than

i.e. we have to find minimum value.

Differentiate the given equation and equate it to zero

\( \Large f(x) = x^{2} + 4x + 11 \)

\( \Large f'(x) = 2x - 4 = 0 \)

x = 2

and f''(x) = 2 > 0 hence at x = 2 function will have minimum value

\( \Large f(x) = x^{2} - 4x + 1 \)

=> \( \Large f(2) = (2)^{2} - 4(2) + 11 \)

= 4 - 8 + 11 = 7

Part of solved CDS Maths(2) questions and answers : Exams >> CDSE >> CDS Maths(2)

Comments

Similar Questions

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