Question
A function y =f(x) has a second order derivative f"=6(x-1). If its graph passes through the point(2,1) and at that point the tangent to the graph is y =3x -5, then the function is
Answer: Option B
:
B
Since f"(x)=6(x-1)
⇒f′(x)=3(x−1)2+c (integrating) ----(i)
Also, at the point (2,1), the tangent to the graph is y =3x-5and slope of thetangent = 3
⇒f′(2)=3
3(2−1)3+c=3 [from eq(i)
⇒3+c=3⇒c=0
From Eq (i) we have
f′(x)=3(x−1)2
⇒f(x)=(x−1)3+k (Integrating)----(ii)
∴1=(2−1)3+k⇒k=0
Hence the equation of the function is f(x)=(x−1)3.
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:
B
Since f"(x)=6(x-1)
⇒f′(x)=3(x−1)2+c (integrating) ----(i)
Also, at the point (2,1), the tangent to the graph is y =3x-5and slope of thetangent = 3
⇒f′(2)=3
3(2−1)3+c=3 [from eq(i)
⇒3+c=3⇒c=0
From Eq (i) we have
f′(x)=3(x−1)2
⇒f(x)=(x−1)3+k (Integrating)----(ii)
∴1=(2−1)3+k⇒k=0
Hence the equation of the function is f(x)=(x−1)3.
Was this answer helpful ?
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