Question 26
For the function f (x) = x + 1/π₯ , x β [1, 3], the value of c for mean value theorem is
1 (B) β3
(C) 2 (D) None of these
π(π₯)="x + " 1/π₯ in interval [1, 3]
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 if π(π₯)="x + " 1/π₯ is continuous in the interval [1, 3]
Let π(π)=π and π‘(π)=π/π
We know that,
π(π)=π₯ is continuous as it is a polynomial function
And,
π‘(π)=1/π₯ is continuous for all π₯ except for π=π
β΄ h(π₯)=1/π₯ is continuous in the interval [1, 3]
Hence,
π(π)=π(π)+π(π) is also continuous in the interval [1, 3]
Condition 2
We need to check if π(π₯)="x + " 1/π₯ is differentiable at ("1, 3")
A function is said to be differentiable if the derivative of the function exists.
Differentiating π(π₯) wrt π₯
π^β² (π)=πβπ/π^π
Since, derivative of the given function exists
Hence, π(π) is differentiable at ("1, 3")
Since both conditions are satisfied
From Mean Value Theorem,
There exists a c β (1, 3) such that,
π^β² (π) = (π(3) β π(1))/(3 β1)
πβπ/π^π = ((π + π/π) β (π + π/π))/π
1β1/π^2 = ((9 + 1)/3 β 2)/2
1β1/π^2 = (10/3 β 2)/2
1β1/π^2 = ((10 β 6)/3)/2
1β1/π^2 = 4/(3 Γ 2 )
πβπ/π^π = 2/3
πβπ/π= π/π^π
(π β π)/π= π/π^π
π/π= π/π^π
π^2=3
π=Β±β3
π=Β±π.ππ
Either,
c = β1.73
But,
β1.73 β(π, π)
Or,
c = 1.73
And,
1.73 β (π, π)
Therefore, π=βπ
So, the correct answer is (B)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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