A line tangent to the graph of the function y=f(x) at the point x=a forms an angle π/3 with y-axis and at x=b and angle π/4 with x-axis then ∫baf′′(x)dx is
A
1/√3−1
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B
−π/12
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C
π/12
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D
√3−1
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Solution
The correct option is D1/√3−1 f′(a)=tan(π2−π3)=cotπ3=1√3 and f′(b)=tanπ4=1 So, ∫baf′′(x)dx=f′(b)−f′(a)=1√3−1