Question 12 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Nov. 17, 2021 by Teachoo

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. If Teachoo has been of any help to you in your Board exam preparation, then please support us by clicking on this link to make a donation

The value of c in Mean value theorem for the function
f
(x) = x (x β 2), x β [1, 2] is

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. If Teachoo has been of any help to you in your Board exam preparation, then please support us by clicking on this link to make a donation

Question 12
The value of c in Mean value theorem for the function f (x) = x (x β 2), x β [1, 2] is
(A) 3/2 (B) 2/3 (C) 1/2 (D) 3/2
π(π₯)=π₯" (" π₯" β 2)"
π(π₯) = π₯^2 β 2π₯ in interval [1, 2].
Checking conditions for
Mean value Theorem
Conditions of Mean value theorem
π(π₯) is continuous at {π, π}
π(π₯) is differentiable at (π , π)
If both conditions satisfied, then there exist some c in (π , π)
such that πβ²(π) = (π(π) β π(π))/(π β π)
Condition 1
We need to check ifπ(π₯) = π₯^2 β 2π₯ is continuous in interval [1, 2]
Since π(π₯) is polynomial .
it is continuous
β΄ π(π) is continuous at [1, 2]
Condition 2
We need to check if π(π₯) = π₯^2 β 2π₯ is differentiable in interval (1, 2)
Since π(π₯) is a polynomial .
it is Differentiable
β΄ π(π) is differentiable in (1, 2)
Since both conditions are satisfied
From Mean Value Theorem,
There exists a c β (1, 2) such that,
π^β² (π) = (π(π) β π(π))/(π β π)
2πβ2= ((2^2 β 2(2)) β (1^2 β 2(1)))/1
ππβπ = ((π β π) β (π β π))/π
ππβπ = (π β (βπ))/π
2πβ2 = 1
2π = 1+2
2π = 3
π=π/π
Since c = π/π β(π, π)
So, the correct answer is (A)

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.