f satisfies the condition of Lagrange's mean value theorem on [0, 1]

f satisfies the conditions of Rolle's theorem on [-1, 1]

f satisfies the condition of lagrange's mean value theorem on [-1, 1]

f satisfies the condition of Rolle's theorem on [0, 1]

g is increasing on (1,and decreasing on (2, ∞)

g is decreasing on (1,and increasing on (2, ∞)

there exists some c ϵ (0, ∞) such that f '(c) = 0

odd and is strictly decreasing in (-∞, ∞)

even and strictly increasing in (0, ∞)

odd and is strictly decreasing in (- ∞, ∞)

neither even nor odd but is strictly increasing in (-∞, ∞)

If f ( x ) = 3 x 4 + 4 x 3 - 12 x 2 + 12, then f ( x ) is

increasing in (-∞ -and in (0, 1)

decreasing in (-∞, -in (1, ∞)

Both A and R are correct are correct and R is not the correct reason for A

Both A and R are correct and R is the correct reason for A

A is correct but r is incorrect

A is incorrect but R is correct

JEE Questions for Maths Applications Of Derivatives Quiz 5

The set {x³ - 12x : -3 ≤ x ≤ 3} is equal to

0%
{x : -16 ≤ x < 16}
0%
{x : -12 ≤ x < 12}
0%
{x : -9 ≤ x < 9}
0%
{x : 0 ≤ x < 10}

The normal to the curve x = a(cos θ + sin θ), y = a(sin θ - θ cos θ) at any point θ is such that

0%
it is a constant distances from the origin
0%
it passes through (aπ/2, -a)
0%
it makes π/2 - θ with the X - axis
0%
it passes through the origin

The largest value of 2x³ - 3x² - 12x + 5 for -2 ≤ x ≤ 4 occurs at x is equal to

0%
-4
0%
0
0%
1
0%
4

If x + y = 8, then the maximum value of x² y is

0%
2048/9
0%
2048/81
0%
2048/3
0%
2048/27

The maximum value of xy when x + 2y = 8 is

0%
20
0%
16
0%
24
0%
8
0%
4

If f and g are differentiable functions in (0,satisfying f(= 2 = g(1), g(= 0 and f(= 6, then for some c ϵ ] 0, 1[

0%
2f '(c) = g'(c)
0%
2f '(c) = 3g'(c)
0%
f '(c) = g'(c)
0%
f '(c) = 2g'(c)

The length of the longest interval, in which f(x) = 3 sin x - 4 sin³ x is increasing, is

0%
π/3
0%
π/2
0%
3π/2
0%
π

Applying lagrange's mean value theorem for a suitable function f(x) in [0, h], we have f(h) = f(+ hf '(θh), 0 < θ < 1. Then, for f(x) = cos x, the value of
Maths-Applications of Derivatives-9315.png

Maths-Applications of Derivatives-9315.png

0%
1
0%
0
0%
1/2
0%
1/3

Maths-Applications of Derivatives-9317.png

0%
f satisfies the conditions of Rolle's theorem on [-1, 1]
0%
f satisfies the condition of lagrange's mean value theorem on [-1, 1]
0%
f satisfies the condition of Rolle's theorem on [0, 1]
0%
f satisfies the condition of Lagrange's mean value theorem on [0, 1]

If f : [0, 1] → R (the set of all real number) be a function. Suppose the function f is twice differentiable, f(= f(= 0 and satisfies f ''( x) - 2 f ''( x) + f( x) ≥ e^x, x ϵ [0, 1]. If the function e^-x f(x) assumes its minimum in the interval [0, 1] at x = 1/4, which of the following is correct ?

0%
f '(x) < f(x), 1/4 < x < 3/4
0%
f ' (x) > f (x), 0 < x < 1/4
0%
f ' (x) < f(x), 0 < f < 1/4
0%
f '(x) < f (x), 3/4 < x < 1

The value of c in the Lagrange's mean value theorem for
Maths-Applications of Derivatives-9320.png

0%
9/2
0%
5/2
0%
3
0%
4

Maths-Applications of Derivatives-9322.png

0%
g is increasing on (1, ∞)
0%
g is decreasing on (1, ∞)
0%
g is increasing on (1,and decreasing on (2, ∞)
0%
g is decreasing on (1,and increasing on (2, ∞)

Maths-Applications of Derivatives-9324.png

0%
f has a local maximum at x = 2
0%
f is decreasing on (2, 3)
0%
there exists some c ϵ (0, ∞) such that f '(c) = 0
0%
f has local minimum at x= 3

If f(x) = xe^x(1-x), then f(x) is

0%
increasing on R
0%
decreasing on [- (1/2), 1]
0%
increasing on [-(1/2), 1]
0%
decreasing on R

If f(x) = 2x² - |x| + 4 , x ϵ [-1, 2], then for some c ϵ (-1, 2), f ' (c) is equal to

0%
0%
2)
0%
0%
None of these

If f , g and h be real valued functions defined on thr interval [0, 1] by f(x) = x² e^x² + e^-x² . If a, b and c denote respectively, the absolute maximum of f, g and h on [0, 1], then

0%
a = b and c ≠ b
0%
a = c and a ≠ b
0%
a ≠ b and c ≠ b
0%
a = b = c

Maths-Applications of Derivatives-9332.png

0%
0%
2)
0%
0%

For what values of x, function f(x) = x⁴ + 4x³ + 4x² + 40 is monotonic decreasing ?

0%
0 < x < 1
0%
1 < x < 2
0%
2 < x < 3
0%
4 < x < 5

The Rolle's theorem is applicable in the interval - 1≤ x ≤ 1 for the function

0%
0%
2)
0%
0%

If f(x) = x³ - 36x + 2 is decreasing function, then x ϵ

0%
(6, ∞)
0%
(-∞, -2)
0%
(-2, 6)
0%
None of these

Maths-Applications of Derivatives-9345.png

0%
-1
0%
-0.5
0%
0.5
0%
1

Maths-Applications of Derivatives-9347.png

0%
even and strictly increasing in (0, ∞)
0%
odd and is strictly decreasing in (- ∞, ∞)
0%
odd and is strictly decreasing in (-∞, ∞)
0%
neither even nor odd but is strictly increasing in (-∞, ∞)

How many real solutions does the equation x⁷ + 14x⁵ + 16x³ + 30x - 560 = 0 have ?

0%
5
0%
7
0%
1
0%
3

Maths-Applications of Derivatives-9350.png

0%
an increasing
0%
a decreasing
0%
an even
0%
None of these

If the function f(x) ax³ + bx² + 11x - 6 satisfies the condition of Rolle's theorem in [1, 3] and
Maths-Applications of Derivatives-9352.png

0%
-1, 6
0%
-2, 1
0%
1, -6
0%
-1, 1/2

Select the correct statement from (a), (b), (c) and (d). The function f(x) = xe^{1 - x}

0%
strictly increases in the interval (1/2, 2)
0%
increases in the interval (0, ∞)
0%
decreases in the interval (0, 2)
0%
strictly decreases in the interval (1,∞)

Rolle's theorem is not applicable to the function f (x) = |x| for -2 ≤ x ≤ 2 because

0%
f is continuous for -2 ≤ x ≤ 2
0%
f is not derivable for x = 0
0%
f(-= f(2)
0%
f is not a constant function

If f(x) = 3x⁴ + 4x³ - 12x² + 12, then f(x) is

0%
increasing in (-∞ -and in (0, 1)
0%
increasing in (-2,and (1, ∞)
0%
decreasing in (-2,and (0, 1)
0%
decreasing in (-∞, -in (1, ∞)

The function f(x) = 2x³ + 13x² - 12x + 1 decreases in the interval

0%
(2, 3)
0%
(1, 2)
0%
(-2, -1)
0%
(-3, -2)

A value of c for which the conclusion of mean value theorem holds for the function f(x) = log_e x on the interval [1, 3] is

0%
2 log3 e
0%
1/2 loge 3
0%
log3 e
0%
loge 3

The value of b for which function f(x) = sin x - bx + c is decreasing in the interval (-∞, ∞) is given by

0%
b < 1
0%
b ≥ 1
0%
b > 1
0%
b ≤ 1

If f(x) = sin x/e^x in [0, π], then f(x)

0%
0%
2)
0%
0%

The function f defined by f(x) = 4x⁴ - 2x + 1 in increasing for

0%
x < 1
0%
x > 0
0%
x < 1/2
0%
x > 1/2

A function f is defined by f(x) = 2 + (x - 1)^2/3 in [0, 2]. Which of the following is not correct?

0%
f is not derivable in (0, 2)
0%
f is continuous in [0, 2]
0%
f(= f(2)
0%
Rolle's theorem is correct in [0, 2]

The function f(x) = cot^-1 x + x increases in the interval

0%
(1, ∞)
0%
(-1, ∞)
0%
(-∞, ∞)
0%
(0, ∞)

In the interval (-3,the function
Maths-Applications of Derivatives-9368.png

0%
increasing
0%
decreasing
0%
neither increasing nor decreasing
0%
partly increasing and partly decreasing

Maths-Applications of Derivatives-9370.png

0%
Both A and R are correct and R is the correct reason for A
0%
Both A and R are correct are correct and R is not the correct reason for A
0%
A is correct but r is incorrect
0%
A is incorrect but R is correct

Maths-Applications of Derivatives-9372.png

0%
0%
(-∞,∞) - {2}
0%
(-∞,∞) - {3}
0%
(-∞,∞) - {2,3}

A particle moves in a straight line so that it covered a distance at^{3 + bt + 5 metre in t seconds. If its acceleration after 4 seconds is 48 m/s², then a is equal to}

0%
1
0%
2
0%
3
0%
4

The approximate value of f(5.001), where
f(x) = x³ - 7 x² + 15 is

0%
-34.995
0%
-33.995
0%
-33.335
0%
-35.995

If there is an error of ± 0.04 cm in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is 10 cm, is

0%
± 1.2
0%
± 0.06
0%
± 0.006
0%
± 0.6

A spherical balloon is filled with 4500π cu m of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 π cu m/min, then the rate (in m/min) at which the radius of the balloon decreases 49 min after the leakage began is

0%
9/7
0%
7/9
0%
2/9
0%
9/2

If y = 2x³ - 2x² + 3x -5, then for x = 2 and ∆x = 0.1, value of ∆y is

0%
2.002
0%
1.9
0%
0
0%
0.9

The distance (in metres) travelled by a vehicle in time t (in seconds) is given by the equation s = 3t³ + 2t² + t +1. The difference in the acceleration between t = 2 and t = 4 is

0%
36 m/s2
0%
38 m/s2
0%
45 m/s2
0%
46 m/s2

The equation of motion of a particle moving along a straight line is s = 2t³ - 9t² + 12 t, where the units of s and t are cm and s. The acceleration of the particle will be zero after

0%
3/2 s
0%
2/3 s
0%
1/2 s
0%
1 s

A particle is moving in a straight line. At time t, the distance between the particle from its starting point is given by x = t - 6t² + t³. Its acceleration will be zero at

0%
t = 1 unit time
0%
t = 2 units time
0%
t = 3 units time
0%
t = 4 units time

The distance covered by a particle in t sec is given by x = 3 + 8t - 4t²

0%
0 unit
0%
3 units
0%
4 units
0%
7 units

If a particle moves such that the displacement is proportional to the square of the velocity acquired, then its acceleration is

0%
proportional to s2
0%
proportional to 1/s2
0%
proportional to 1/s
0%
a constant

Let f : [0,1] → R(the set of all real numbers) be a function. Suppose the function f is twice differentiable,
f(= f(= 0 and satisfies f \'\'(x) - 2f \'(x) + f(x) ≥ e^x, x ϵ [0,1].
Which of the following is correct for 0 < x < 1?