JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

Let

f(x)=x^2+p x+q

. The value of

(p, q)

so that

f(1)=3

is an extreme value of

f

[0,2]

View answer and explanation

Correct option: C

(-2,4)

Option C ((-2,4)) is correct because the given conditions f(1)=3 and f'(1)=0 lead to p=-2 and q=4.

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