Let F : R→R be a thrice differentiable function. Suppose that F1=0, F3= -4 and F′x<0 all x∈12, 3. Let fx=xF(x) for all x∈ R . If ∫13x2F′xdx= -12 and ∫13x3 F′′xdx=40, then the correct expression(s) is(are)

JEE Advanced · Mathematics · 27. Definite Integration

Let $F : R \to R$ be a thrice differentiable function. Suppose that $F (1) = 0, F (3) = - 4$ and $F' (x) < 0$ all $x \in (\frac{1}{2}, 3) .$ Let $f (x) = x F (x)$ for all $x \in R$ .
If $\int_{1}^{3} x^{2} F^{'} (x) d x = - 12$ and $\int_{1}^{3} x^{3} F^{''} (x) d x = 40,$ then the correct expression(s) is(are)

A $9 f^{'} (3) + f^{'} (1) - 32 = 0$
B $\int_{1}^{3} f (x) d x = 12$
C $9 f^{'} (3) - f^{'} (1) + 32 = 0$
D $\int_{1}^{3} f (x) d x = - 12$

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\int_{1}^{3} x^{2} F^{'} (x) d x = - 12 a n d \int_{1}^{3} x^{3} F^{''} (x) d x = 40

\int_{1}^{3} x^{3} F^{''} (x) = x^{3} F^{'} (x)]_{1}^{3} - \int_{1}^{3} 3 x^{2} F' (x)

40 = 27 F^{'} (3) - F^{'} (1) + 36

4 = 9 (f (3) + 4) - f (1)

\Rightarrow 9 f^{'} (3) - f^{'} (1) + 32 = 0

Also

\int_{1}^{3} x^{2} F' (x) d x

= x^{2} F (x)]_{1}^{3} - \int_{1}^{3} 2 x F (x) d x = - 12

9 F (3) - F (1) - 2 \int_{1}^{3} x F (x) d x = - 12

- 36 - 0 - 2 \int_{1}^{3} f (x) d x = - 12

- 36 + 12 = 2 \int_{1}^{3} f (x) d x

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the following.

	List – I		List – II
(A)	Let $y (x) = \cos (3 \cos^{- 1} x), x ϵ [- 1, 1], x \neq \pm \frac{\sqrt{3}}{2} .$ Then $\frac{1}{y (x)} \{(x^{2} - 1) \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}\}$ equals	(P)	1
(B)	Let $A_{1}, A_{2}, \dots, A_{n} (n > 2)$ be the vertices of a regular polygon of n sides with its centre at the origin. Let $\vec{a_{k}}$ be the position vector of the point $A_{k}, k = 1, 2, \dots n .$ If $\|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \times \vec{a_{k + 1}})\| = \|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \cdot \vec{a_{k + 1}})\|$ , then the minimum value of n is	(Q)	2
(C)	If the normal from the point $P (h, 1)$ on the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ is perpendicular to the line $x + y = 8,$ then the value of h is	(R)	8
(D)	Number of positive solutions satisfying the equation $\tan^{- 1} (\frac{1}{2 x + 1}) + \tan^{- 1} (\frac{1}{4 x + 1}) = \tan^{- 1} (\frac{2}{x^{2}})$ is	(S)	9

JEE Advanced 2014 Hard

Explore more questions on app

From JEE Advanced

Let

p, q, r

be non-zero real numbers that are, respectively, the

10^{th}, 100^{th}

and

1000^{th}

terms of a harmonic progression. Consider the system of linear equations

x + y + z = 1

10 x + 100 y + 1000 z = 0

q r x + p r y + p q z = 0

	List- $I$		List- $II$
$(I)$	If $\frac{q}{r} = 10$ , then the system of linear equations has	$(P)$	$x = 0, y = \frac{10}{9}, z = - \frac{1}{9}$ as a solution
$(II)$	If $\frac{p}{r} \neq 100$ , then the system of linear equations has	$(Q)$	$x = \frac{10}{9}, y = - \frac{1}{9}, z = 0$ as a solution
$(III)$	If $\frac{p}{q} \neq 10$ , then the system of linear equations has	$(R)$	infinitely many solutions
$(IV)$	If $\frac{p}{q} = 10$ , then the system of linear equations has	$(S)$	no solution
		$(T)$	at least one solution

The correct option is:

JEE Advanced 2022 Medium

Explore more questions on app

Let F : R→R be a thrice differentiable function. Suppose that F1=0, F3= -4 and F′x<0 all x∈12, 3. Let fx=xF(x) for all x∈ R .If ∫13x2F′xdx= -12 and ∫13x3 F′′xdx=40, then the correct expression(s) is(are)

Step-by-step Solution