Test Index

KEAM 2017 Math Paper

© examsnet.com

Question : 39 of 120

Marks: +1, -0

The difference between the maximum and minimum value of the function f(x) =

\int\limits_{0}^{\pi}(t^2+t+1)dt

on [2,3] is

39/6
49/6
59/6
69/6
79/6

Solution:

Given

f(x)=\int\limits_{0}^{x}(t^{2}+t+1) dt

f'(x)=(x^{2}+x+1) \times 1-0

=x^{2}+x+1

For

x \in [2,3)

f'(x)>0

\therefore

Minimum is at

x=2

and maximum is at

x=3

.

Now, minimum value

=\int\limits_{0}^{2}(t^{2}+t+1) dt

=\left[\frac{t^{3}}{3}+\frac{t^{2}}{2}+t\right]_{b}^{2}

=\frac{8}{3}+\frac{4}{2}+2

=\frac{8}{3}+4=\frac{20}{3}

and maximum value

=\int\limits_{0}^{3}(t^{2}+t+1) dt

=\left[\frac{t^{3}}{3}+\frac{t^{2}}{2}+t\right]_{b}^{\beta}

=\frac{27}{3}+\frac{9}{2}+3

=\frac{9}{2}+12=\frac{33}{2}

\therefore

Difference between maximum and minimum value

=\frac{33}{2}-\frac{20}{3}

=\frac{99-40}{6}=\frac{59}{6}

© examsnet.com

Go to Question:

Prev Question

Next Question