New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

Definite Integration question

2020 · 9 Jan · Shift 2 · Q28

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
Definite Integration
2020 · 9 Jan · Shift 2 · Q28

Definite Integration question

2020 · 9 Jan · Shift 2 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

Let a function ƒ : [0, 5]

\to

R be continuous, ƒ(1) = 3 and F be defined as :

F(x) = \int\limits_1^x {{t^2}g(t)dt}

, where

g(t) = \int\limits_1^t {f(u)du}

Then for the function F, the point x = 1 is :

A
a point of inflection.
B
a point of local maxima.
C
a point of local minima.
D
not a critical point.

More from Definite Integration

If $f (x) = \frac{2 - x c o s x}{2 + x c o s x}$ and g(x) = logex, (x > 0) then the value of integral $- \frac{π}{4} \int \frac{π}{4} g (f (x)) d x$ is2019 · MCQ
Let $f (x) = 0 \int x g (t) d t$ where g is a non-zero even function. If ƒ(x + 5) = g(x), then $0 \int x f (t) d t$ equals-2019 · MCQ
The value of $0 \int π /2 \frac{s i n ^{3} x}{s i n x + c o s x} d x$ is2019 · MCQ
The value of the integral $0 \int 1 x cot^{- 1} (1 - x^{2} + x^{4}) d x$ is :-2019 · MCQ
If f : R $\to$ R is a differentiable function and f(2) = 6, then $x \to 2 lim \frac{6 \int f ( x ) 2 t d t}{( x - 2 )}$ is :-2019 · MCQ
The value of $0 \int π ∣ cos x ∣^{3} d x$ is :2019 · MCQ
If $0 \int \frac{π}{3} \frac{t a n θ}{2 k s e c θ} d θ = 1 - \frac{1}{2}, (k > 0),$ then value of k is :2019 · MCQ
Let f be a differentiable function from R to R such that $∣ f (x) - f (y) ∣ \leq 2 ∣ x - y ∣^{\frac{3}{2}},$ for all $x, y \in$ R. If $f (0) = 1$ then $0 \int 1 f^{2} (x) d x$ …2019 · MCQ