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)
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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