Let f:0,2→R be a function which is continuous on 0,2 and is differentiable on 0,2 f0=1. Let Fx=∫0x2f tdt for x∈0,2. If F'x=f'x for all x∈0,2, then F2 equals

JEE Advanced · Mathematics · 29. Differential Eqns

Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2) f (0) = 1$ . Let $F (x) = \int_{0}^{x^{2}} f (\sqrt{t}) d t$ for $x \in [0, 2] .$ If $F^{'} (x) = f^{'} (x)$ for all $x \in (0, 2),$ then $F (2)$ equals

A $e^{2} - 1$
B $e^{4} - 1$
C $e - 1$
D $e^{4}$

Verified Solution

Answer & Solution

Correct Answer

(B) $e^{4} - 1$

Step-by-step Solution

Detailed explanation

F (0) = 0

F^{'} (x) = 2 x f (x) = f^{'} (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

Consider the ellipse

\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1

. Let

H (α, 0), 0 < α < 2

, be a point. A straight line drawn through

H

parallel to the

y

-axis crosses the ellipse and its auxiliary circle at points

E

and

F

respectively, in the first quadrant. The tangent to the ellipse at the point

E

intersects the positive

x

-axis at a point

G

. Suppose the straight line joining

F

and the origin makes an angle

ϕ

with the positive

x

-axis.

	List- $I$		List- $II$
$(I)$	If $ϕ = \frac{π}{4}$ , then the area of the triangle $F G H$ is	$(P)$	$\frac{{(\sqrt{3} - 1)}^{4}}{8}$
$(II)$	If $ϕ = \frac{π}{3}$ , then the area of the triangle $F G H$ is	$(Q)$	$1$
$(III)$	If $ϕ = \frac{π}{6}$ , then the area of the triangle $F G H$ is	$(R)$	$\frac{3}{4}$
$(IV)$	If $ϕ = \frac{π}{12}$ , then the area of the triangle $F G H$ is	$(S)$	$\frac{1}{2 \sqrt{3}}$
		$(T)$	$\frac{3 \sqrt{3}}{2}$

The correct option is:

JEE Advanced 2022 Easy

Explore more questions on app

From JEE Advanced

Explore more questions on app