Definite IntegrationHard
Question
The tangent to the graph of the function y = f(x) at the point with abscissa x =1 form an angle of π/6 and at the point x = 2, an angle of π/3 and at the point x = 3, an angle of π/4 . The value of
(f″(x) is supposed to be continuous) is :
(f″(x) is supposed to be continuous) is :Options
A.
B.
C.
D.None of these
Solution
Given,
f′(1) =
= tan 
f′(2) =
= tan 
= √3
f′(3) =
= tan 
= 1
Let, I =
= I1 + I2
∴ I1 =
I1 =
dx
2I1 = {f′(3)}2 - {f′(1)}2
2I1 = 1 - 1/3
I1 = 1/3
and, I2 =
= f′(3) - f′(2)
= 1- √3
∴ I = I1 + I2 = 1/3 + 1 - √3
= 4/3 - √3
f′(1) =
= tan f′(2) =
= tan = √3f′(3) =
= tan = 1Let, I =
= I1 + I2∴ I1 =
I1 =
dx2I1 = {f′(3)}2 - {f′(1)}2
2I1 = 1 - 1/3
I1 = 1/3
and, I2 =
= f′(3) - f′(2)
= 1- √3
∴ I = I1 + I2 = 1/3 + 1 - √3
= 4/3 - √3
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