All Courses
Get the App Signup / Login
Sign in Open App
All Exams  >   JEE  >   Online MCQ Tests for JEE  >   All Questions

All questions of Applications of Derivatives for JEE Exam

The function f(x) = ax, 0 < a < 1 is​
  • a)
    increasing
  • b)
    strictly decreasing on R
  • c)
    neither increasing or decreasing
  • d)
    decreasing
Correct answer is option 'D'. Can you explain this answer?

Krishna Iyer answered
 f(x) = ax
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga   {ax > 0 for all x implies R, 
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
Clear all your JEE doubts with EduRev

The maximum value of f (x) = sin x in the interval [π,2π] is​
a) 6
b) 0
c) -2
d) -4
Correct answer is option 'B'. Can you explain this answer?

Kiran Mehta answered
f(x) = sin x
f’(x) =cosx 
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0 
Hence, 0 is the maxima.

Find the approximate value of f(10.01) where f(x) = 5x2 +6x + 3​
  • a)
    564.06
  • b)
    564.01
  • c)
    563.00
  • d)
    563.01
Correct answer is option 'A'. Can you explain this answer?

Naina Sharma answered
f(x) = 5x2 +6x + 3
f(10.01) = 5*(10.01)2 + 6*(10.01) + 3
To find (10.01)2
Let p=10, Δp=0.01
y=p2 = 100
y+Δy = (p+ Δp)2 = (10.01)2
Δy = (dy/dp) * Δp
Δy = 2*p* Δx
Δy = 2*10* 0.01
Δy = 20 * 0.01
Δy = 0.2
So, (10.01)2 = y + Δy
= 100.2
So,
f(10.01) = 5*(100.2) + 6*(10.01) + 3
= 501 + 60.06 + 3
= 564.06

The abscissa of the point on the curve ay2 = x3, the normal at which cuts off equal intercepts from the coordinate axes is
  • a)
     
  • b)
     
  • c)
     –
  • d)
    –
Correct answer is option 'B'. Can you explain this answer?

Kavita Joshi answered
(That should be "axes" -- plural.) 
Find the normal; find the intercepts -- for a point (u, v) say. 
To get there, differentiate to find the gradient - of the tangent, hence of the normal. 
2 a y dy/dx = 3 x^2 
So dy/dx = 3 x^2 / (2 a y) . 
At the point (u, v), this is 3 u^2 / (2 a v) 
So the gradient of the normal there is (-2 a v) / (3 u^2) 
So the equation of the normal is 
y - v = (-2 a v) / (3 u^2) * (x - u) 
When x = 0, 
y = v - 2 a v / (3 u^2) (-u) 
= v (1 + 2 a / (3 u) ) 
When y = 0, 
x = u + 3 u^2 / (2 a v) * v 
= u (1 + 3 u / (2 a) ) 
Now, "equal intercepts" could mean just by size, or by size and sign. Take the latter: 
Then 
v (2a + 3u) / (3u) = u (2a + 3u) / (2a) 
So either u = -3 a / 2, 
or v / (3u) = u / (2a), 
2 a v = 3 u^2 -- 
each to be taken with a v^2 = u^3 
One possibility, which fits all three equations, is (u, v) = (0, 0).; and if either u = 0 or v = 0, the so is the other. 
If u ≠ 0, v ≠ 0 then 
first case: 
u = -3 a / 2 
so a v^2 = -27 a^3 / 8 
so v^2 = -27 a^2 / 8 -- not possible 
second case: 
2 a v = 3 u^2 
and a v^2 = u^3 divide 
So v / 2 = u /3 substitute 
so a 4u^2 / 9 = u^3, 
(u = 0 or) u = 4 a / 9 substitute, again: 
a v^2 = 64 a^3 / 729 
v = (-/)+ 8 a / 27 
The (only) point that cuts off equal intercepts on the axes is (4a/9, 8a/27). 
(if we allow equal distances, we get one other point -- y changes sign.)

Find slope of normal to the curve y=5x2-10x + 7 at x=1​
  • a)
    not defined
  • b)
    -1
  • c)
    1
  • d)
    zero
Correct answer is option 'A'. Can you explain this answer?

Neha Sharma answered
y = 5x2 - 10x + 7
dy/dx = 10x - 10
(At x = 1) 10(1) - 10 
m1 = 0
As we know that slope, m1m2 = -1 
=> 0(m2) = -1
m2 = -1/0 (which is not defined)

The radius of air bubble is increasing at the rate of 0. 25 cm/s. At what rate the volume of the bubble is increasing when the radius is 1 cm.​
  • a)
    4π cm3/s
  • b)
    22π cm3/s
  • c)
    2π cm3/s
  • d)
    π cm3/s
Correct answer is option 'D'. Can you explain this answer?

Rohan Yadav answered
Given, the rate of increase of radius of the air bubble = 0.25 cm/s

We need to find the rate of increase of volume of the bubble when the radius is 1 cm.

Formula used:

Volume of a sphere = (4/3)πr^3

Differentiating both sides with respect to time t, we get:

dV/dt = 4πr^2(dr/dt)

where dV/dt is the rate of change of volume of the sphere with respect to time t and dr/dt is the rate of change of radius of the sphere with respect to time t.

Substituting the given values, we get:

dV/dt = 4π(1)^2(0.25) = π cm^3/s

Therefore, the rate of increase of volume of the bubble when the radius is 1 cm is π cm^3/s, which is the correct answer.

Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24 x – 18x2​
  • a)
    56
  • b)
    49
  • c)
    23
  • d)
    89
Correct answer is option 'B'. Can you explain this answer?

Aryan Khanna answered
p’(x) = -24 - 36x
p”(x) = -36
Now, p’(x) = 0  ⇒ x = (-24)/36
x = -⅔
Also, p”(-⅔) = -36 < 0
By the second derivative test,  x = -⅔
Therefore, maximum profit = p(-⅔)
= 41 - 24(-⅔) - 18(-⅔)^2 
= 41 +16 - 8  
⇒ 49

The equation of the normal to the curve x2 = 4y which passes through the point (1, 2) is.​
  • a)
    x + y – 3 = 0
  • b)
    4x – y = 2
  • c)
    4x – 2y = 0
  • d)
    4x – 3y + 2= 0
Correct answer is option 'B'. Can you explain this answer?

Sushil Kumar answered
h2 = 4k 
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h2) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3

The length of the subtangent to the curve =3 at the point (4, 1) is
  • a)
    2
  • b)
    1/2
  • c)
    3
  • d)
    4
Correct answer is option 'A'. Can you explain this answer?

Anjali Sharma answered
Length of subtangent =y(dx/dy)
Now given √x+√y=3
√y=3−√x
⇒1/2√ydy/dx = −1/(2√x) (On differentiating)
⇒dx/dy=−√x/√y
⇒ydx/dy=−√x/√y
⇒dx/dy/(4,1) = −√1√4
=−2
But the length can never be negative.
So length = 2.

Find the equation of tangent to  which has slope 2.
  • a)
    2x – y = 1
  • b)
    No tangent
  • c)
    y – 2x = 0
  • d)
    y – 2x = 3
Correct answer is option 'B'. Can you explain this answer?

Raghava Rao answered
Y=1/(x-3)^2

dy/dx=(-1/(x-3)^2)

given slope=2

-1/(x-3)^2 =2

-1/2=(x-3)^2

negative number is not equal to square. so no tangent

 The minimum value of the function defined by f(x) = max (x, x + 1, 2 – x) is
  • a)
    0
  • b)
    1/2
  • c)
    1
  • d)
    3/2
Correct answer is option 'D'. Can you explain this answer?

Leelu Bhai answered
Bhai dekh iss func ka graph bana ....mtlb ki f(x) = x, f(x) = x+ 1 aur f(x) = 2 - x , ye teeno func ek hi graph pe.now see, jo f(x) = x aur f(x) = x+1 hai woh to kahi bhi intersect nhi krege qki dono ||ᵉˡ line hai..aur jab tu f(x) = 2 - x ka graph banaega to inn dono ko intersect karte hue jyega.ab jo top most intersection point wohi sbse tagda hoga aur wohi f(x) = max. {x, x + 1, 2 - x} ki value hogi...so, f(x) = x + 1 and f(x) = 2 - x ka intersection iss graph ka top point hai ... to isko solve krke x = 1/2 aa rha hai aur iss x ki value ko f(x) me put krde to uski value 3/2 aayegi...and according to ques the value of f(x) = max. {x, x + 1, 2 - x} is the minimum value of function.i hope u will understand....

A real number x when added to its reciprocal give minimum value to the sum when x is
  • a)
    1/2
  • b)
    -1
  • c)
    1
  • d)
    2
Correct answer is option 'C'. Can you explain this answer?

Krish Das answered
Finding the Real Number that Gives Minimum Value to the Sum

Solution:

Let x be the real number. Then, its reciprocal is 1/x.

The sum of x and its reciprocal is x + 1/x.

To find the minimum value of this sum, we can use the concept of the arithmetic mean and geometric mean inequality.

We know that for any two positive numbers a and b, the arithmetic mean is (a+b)/2 and the geometric mean is √(ab).

The arithmetic mean is always greater than or equal to the geometric mean, i.e., (a+b)/2 ≥ √(ab).

Let's apply this inequality to x and 1/x.

The arithmetic mean of x and 1/x is (x + 1/x)/2.

The geometric mean of x and 1/x is √(x * 1/x) = √1 = 1.

By the arithmetic mean and geometric mean inequality, we have:

(x + 1/x)/2 ≥ √(x * 1/x) = 1

Multiplying both sides by 2 gives:

x + 1/x ≥ 2

Therefore, the minimum value of x + 1/x is 2, which is attained when x=1.

Hence, the real number x that gives minimum value to the sum x + 1/x is 1.

A point c in the domain of a function f is called a critical point of f if​
  • a)
    f’ (x) = 0 at x = c
  • b)
    f is not differentiable at x = c
  • c)
    Either f’ (c) = 0 or f is not differentiable
  • d)
    f” (x) = 0, at x = c
Correct answer is option 'B'. Can you explain this answer?

Ishan Choudhury answered
A point C in the domain of a function f at which either f(c) = 0 or f is not differentiable.  
The point f  is called the critical point.
c is called the point of local maxima
If f ′(x) changes sign from positive to negative as x increases through c, that is, if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c.
c is called the point of local minima
If f ′(x) changes sign from negative to positive as x increases through c, that is, if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c.
c is called the point of inflexion
If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima.

The maximum and minimum values of f(x) =  are
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Aryan Khanna answered
f(x) = sinx + 1/2cos2x  
⇒ f'(x) = cos x – sin2x 
Now, f'(x) = 0 gives cosx – sin2x = 0 
⇒ cos x (1 – 2 sinx) = 0 
⇒ cos x = 0, (1 – 2 sinx) = 0 
⇒ cos x = 0, sinx = 1/2 
⇒ x = π/6 , π/2 
Now, f(0) = 1/2, 
f(π/6) = 1/2 + 1/4 = 3/4, 
f(π/2) = 1 – 1/2 = 1/2 
Therefore, the absolute max value = 3/4 and absolute min = 1/2

f(x) = x5 – 5x4 + 5x3 – 1. The local maxima of the function f(x) is at x =
  • a)
    1
  • b)
    5
  • c)
    0
  • d)
    3
Correct answer is option 'A'. Can you explain this answer?

Rajat Patel answered
f′(x)=5x4−20x3+15x2 
f′(x)=5x2(x2−4x+3)
when f′(x)=0
⇒5x2(x2−4x+3)=0
⇒5x2(x−3)(x−1)=0
⇒x=0,x=3,x=1

The function  increases in​
  • a)
    (0, 1)
  • b)
    (-∞,e)
  • c)
    (e,∞)
  • d)
    (1, e)
Correct answer is option 'C'. Can you explain this answer?

Vikas Kapoor answered
 f'(x) = [(logx).2−2x.1/x](logx)2
= 2(logx−1)/(logx)^2
∴f'(x)>0
⇔logx−1>0
⇔logx>1
⇔logx>loge
⇔x>e
∴f(x) is increasing in (e,∞)

Let f be a real valued function defined on (0, 1) ∪ (2, 4) such that f ‘ (x) = 0 for every x, then
  • a)
    f is constant function if f  1/2 = f (3)
  • b)
    f is a constant function
  • c)
    f is a constant function if f  1/2 = 0
  • d)
    f is not a constant function
Correct answer is option 'A'. Can you explain this answer?

Nandini Choudhury answered
f ‘ (x) = 0 ⇒ f (x)is constant in (0 , 1)and also in (2, 4). But this does not mean that f (x) has the same value in both the intervals . However , if f (c) = f (d) , where c ∈ (0 , 1) and d ∈ (2, 4) then f (x) assumes the same value at all x ∈ (0 ,1) U (2, 4) and hence f is a constant function.

The equation of the tangent line to the curve y =  which is parallel to the line 4x -2y + 3 = 0 is​
  • a)
    80x +40y – 193 = 0
  • b)
    4x – 2y – 3 = 0
  • c)
    80x – 40y + 193 = 0
  • d)
    80x – 40y – 103 = 0
Correct answer is option 'D'. Can you explain this answer?

Dabhi Bharat answered
Given curve y=√5x-3 -2
y+2=√5x-3
dy/dx=5/2×√5x-3
m1=5/2(y+2)
given that line 4x-2y+3=0is parallel to tangent of curve
slope of line m2=2
m1=m2
5/2(y+2)=2
5/4=(y+2)
y=-3/4
from y+2=√5x-3
-3/4+2=√5x-3
5/4=√5x-3
5x=25/16+3
X=73/80
points are p(73/80,-3/4)
equ. of tangent :y-y1=m(x-x1)
y+3/4=2(x-73/80)
80x-73=40y+30
80x-40y-103=0

Find two positive numbers x and y such that x + y = 60 and xy3 is maximum
  • a)
    x = 45, y = 15
  • b)
    x = 15, y = 45
  • c)
    x = 10, y = 50
  • d)
    x = 30, y = 30
Correct answer is option 'B'. Can you explain this answer?

Gaurav Kumar answered
two positive numbers x and y are such that x + y = 60.
 x + y = 60
⇒ x = 60 – y  ...(1)
Let P = xy3
∴ P =(60 – y)y3 = 60y3 – y4
Differentiating both sides with respect to y, we get

For maximum or minimum dP/dy = 0
⇒ 180y2 - 4y3 = 0
⇒ 4y2 (45 - y) = 0
⇒ y = 0 or 45 - y = 0
⇒ y = 0 or y = 45
⇒ y = 45 (∵ y = 0 is not possible)


Thus, the two positive numbers are 15 and 45.

Area of the greatest rectangle that can be inscribed in the ellipse
  • a)
    2ab
  • b)
    ab
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Anjana Sharma answered
Let (acosθ.bsinθ) be one of the vertices of the rectangle.
Hence the length = 2acosθ and the width = 2bsinθ
⇒ Area of the rectangle A = 2acosθx2bsinθ = 2ab.2sinθcosθ
⇒ A = 2absin2θ
⇒ A′ = 2ab.2cos2θ

For maximun or minimum values of A, →A′= 0
⇒ 4abcos2θ = 0
⇒ cos2θ = π2 →θ = π4
⇒ A = 2absin2π4
⇒ A = 2absin(π2) = 2ab
Now A′′ = −8absin2θ→Area is maximum when A = 2ab

Every continuous function is
  • a)
    not differentiable
  • b)
    not decreasing
  • c)
    decreasing
  • d)
    differentiable.
Correct answer is option 'A'. Can you explain this answer?

Siddharth Rane answered
Explanation:

Continuous functions are those functions which can be drawn without lifting a pen from the paper. In other words, a function is said to be continuous if it has no abrupt changes or jumps in its graph.

Not Necessory Differentiable:
A function is said to be differentiable at a point if its derivative exists at that point. However, a continuous function need not be differentiable at every point. There are many examples of continuous functions that are not differentiable at some points. One such example is the absolute value function, |x|. This function is continuous everywhere, but it is not differentiable at x = 0.

Not Decreasing:
A function is said to be decreasing if its value decreases as the input increases. However, a continuous function need not be decreasing. For example, the function f(x) = x^2 is continuous everywhere, but it is not decreasing.

Decreasing:
Some functions are decreasing. For example, the function f(x) = -x is a decreasing function. However, not all continuous functions are decreasing.

Differentiable:
A function is said to be differentiable if its derivative exists at every point in its domain. However, a continuous function need not be differentiable at every point.

Conclusion:
Therefore, the correct answer is option 'A' that every continuous function is not necessarily differentiable.

The value of ‘a’ for which x3 - 3x + a = 0 has two distinct roots in [0, 1] is given by
  • a)
    -1
  • b)
    1
  • c)
    3
  • d)
    does not exists
Correct answer is option 'D'. Can you explain this answer?

Rithika Mehta answered
Let α, β ∈[0,1].f (x) is continuous on [a,b] & differentiable on (a,b) and f (α) = f (β) = 0
∴ c ∈ (α, β) such that f' (c) = 0 ⇒ c = ±1∉ (0,1)

Chapter doubts & questions for Applications of Derivatives - Online MCQ Tests for JEE 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Applications of Derivatives - Online MCQ Tests for JEE in English & Hindi are available as part of JEE exam. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Related JEE Content

Signup to see your scores go up within 7 days!

Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
© EduRev
Education Revolution
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily