f(x) : [0, 5] → R, F(x) = ∫x2 g(x), x ∈ [0, x], f(1) = 3g(x)

1.	f(x) : [0, 5] → R, F(x) = ∫x2 g(x), x ∈ [0, x], f(1) = 3g(x) = ∫f(t) dt, t ∈ [1, x], then correct choice is(1) F(x) has local minimum at x = 1 (2) F(x) has local maximum at x = 1(3) F(x) has point of inflection at x = 1(4) F(x) has no critical point
Answer» Answer is (1) ⇒ F'(x) = x²g(x) ⇒ F'(1) = 1.g(1) = 0 .....(1) (∴ g(1) = 0) Now F''(x) = 2xg(x) + x²g'(x) ⇒ F''(x) = 2xg(x) + x²f(x) ⇒ F''(1) = 0 + 1 x 3 (∴ g'(x) = f(x)) ⇒ F''(1) = 3 .....(2) From (1) and (2) F(x) has local minimum at x = 1

f(x) : [0, 5] → R, F(x) = ∫x2 g(x), x ∈ [0, x], f(1) = 3g(x) = ∫f(t) dt, t ∈ [1, x], then correct choice is(1) F(x) has local minimum at x = 1 (2) F(x) has local maximum at x = 1(3) F(x) has point of inflection at x = 1(4) F(x) has no critical point

Answer»

Answer is (1)

⇒ F'(x) = x²g(x)

⇒ F'(1) = 1.g(1) = 0 .....(1) (∴ g(1) = 0)

Now F''(x) = 2xg(x) + x²g'(x)

⇒ F''(x) = 2xg(x) + x²f(x)

⇒ F''(1) = 0 + 1 x 3 (∴ g'(x) = f(x))

⇒ F''(1) = 3 .....(2)

From (1) and (2) F(x) has local minimum at x = 1

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f(x) : [0, 5] → R, F(x) = ∫x2 g(x), x ∈ [0, x], f(1) = 3g(x) = ∫f(t) dt, t ∈ [1, x], then correct choice is(1) F(x) has local minimum at x = 1 (2) F(x) has local maximum at x = 1(3) F(x) has point of inflection at x = 1(4) F(x) has no critical point

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