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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five Continuity and Differentiability

If m and n respectively are the number of local maximum and local minimum points of the function f(x) $= \int_{0}^{x^{2}} \frac{t^{2} - 5 t + 4}{2 + e^{t}} d t,$ then the ordered pair (m, n) is equal to....................

Option 1 -

(3, 2)

Option 2 -

(2, 3)

Option 3 -

(2, 2)

Option 4 -

(3, 4)

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Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$f (x) = \int_{0}^{x^{2}} \frac{t^{2} - 5 t + 4}{2 + e^{t}} d t$

$f' (x) = 2 x (\frac{x^{4} - 5 x^{2} + 4}{2 + e^{x^{2}}}) = 0$

$x = 0, o r (x^{2} - 4) (x^{2} - 1) = 0$

$x = 0, x = \pm 2, \pm 1$

Now,

$f' (x) = \frac{2 x (x + 1) (x - 1) (x + 2) (x - 2)}{e^{x^{2}} + 2}$

Changes sign from positive to negative at x = 1, 1 So, number of local maximum points = 2

Changes sign from negative to positive at

x = 2, 0, 2 So, number of local minimum points = 3

$∴ m = 2, n = 3$

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If m and n respectively are the number of local maximum and local minimum points of the function f(x) $= \int_{0}^{x^{2}} \frac{t^{2} - 5 t + 4}{2 + e^{t}} d t,$ then the ordered pair (m, n) is equal to....................

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If m and n respectively are the number of local maximum and local minimum points of the function f(x) $= \int_{0}^{x^{2}} \frac{t^{2} - 5 t + 4}{2 + e^{t}} d t,$ then the ordered pair (m, n) is equal to....................