Question

Let F: R → R be a thrice differentiable function. Suppose that F(1) = 0, F(3) = – 4 and F '(x) < 0 for all $x \in (\frac{1}{2}, 3)$ . Let f(x) $\equiv$ x F(x) for all $x \in R$ .

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Linked Question 1

If $\int_{1}^{3} x^{2} F' (x) dx = - 12 and \int_{1}^{3} x^{3} F " (x) dx = 40$ then the correct expression(s) is(are)

9f '(3) + f '(1) – 32 = 0

9f '(3) – f '(1) + 32 = 0

$\int_{1}^{3} f (x) dx = - 12$

$\int_{1}^{3} f (x) dx = 12$

$\int_{1}^{3} x^{2} F' (x) dx = - 12; \int_{1}^{3} x^{3} F " (x) dx = 40$

$\int_{1}^{3} x^{2} F' (x) dx = {x^{2} F (x) |}_{1}^{3} - 2 \int_{1}^{3} xF (x) dx = - 12$

$= - 36 - 2 \int_{1}^{3} f (x) dx = - 12 \Rightarrow \int_{1}^{3} f (x) dx = - 12$ ⇒ (D)

Now, $\int_{1}^{3} x^{3} F " (x) dx = {x^{3} F' (x) |}_{1}^{3} - 3 \int_{1}^{3} x^{2} F' (x) dx$

= 27 F'(3) – F'(1) – 3 $[{x^{2} F (x) |}_{1}^{3} - 2 \int_{1}^{3} xF (x) dx]$

= 27 F'(3) – F'(1) – 3 $[9 F (3) - F (1) - 2 \int_{1}^{3} f (x) dx]$

= 27F'(3) – 27F(3) – F'(1) + 3F(1) + 6 × (– 12)

= 27F'(3) + 108 – F'(1) – 72 = 9 (3 F'(3)) – F'(1) + 36

= 9 (f '(3) – F(3)) – (f '(1) – F(1)) + 36

= 9 f '(3) + 36 – f '(1) + 36 = 40

⇒ 9f '(3) – f '(1) + 32 = 0 ⇒ (C)

Linked Question 2

The correct statement(s) is(are)

f '(x) $\neq$ 0 for any $x \in (1, 3)$

f (2) < 0

f ' (x) = 0 for some $x \in (1, 3)$

f ' (1) < 0

$\underset{decreasing function}{\underset{⏟}{F' (x) < 0}}, x \in (\frac{1}{2}, 3)$

f(x) = x F(x)

F(1) = 0, F(3) = – 4

f '(x) = x F'(x) + F(x)

f '(1) = F'(1) + F(1) = F'(1) < 0 ⇒ (A)

f(2) = 2F(2)

Since, F is decreasing and F (1) = 0 ⇒ F (2) < 0

$∴$ f (2) = 2 F (2) < 0 ⇒ (B)

Again f '(x) = $\underset{< 0}{\underset{⏟}{x F' (x)}} + \underset{< 0}{\underset{⏟}{F (x)}} \forall x \in (1, 3)$

⇒ f '(x) < 0 " x $\in$ (1, 3) ⇒ f ' (x) ¹ 0 " x $\in$ (1, 3) $∴$ Ans. (C)