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All questions of Application 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.

Using approximation find the value of 
  • a)
    2.025
  • b)
    2.001
  • c)
    2.01
  • d)
    2.0025
Correct answer is option 'D'. Can you explain this answer?

Gunjan Lakhani answered
Let x=4, Δx=0.01
y=x^½ = 2
y+Δy = (x+ Δx)^½ = (4.01)^½
Δy = (dy/dx) * Δx
Δy = (x^(-1/2))/2 * Δx
Δy = (½)*(½) * 0.01
Δy = 0.25 * 0.01
Δy = 0.0025
So, (4.01)^½ = 2 + 0.0025 = 2.0025

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 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)

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

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 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
h= 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 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.

 The set of all x for which ln (1 + x) ≤ x is equal to
  • a)
     x > 0 
  • b)
    x>–1
  • c)
    –1<x<0
  • d)
     Null set
Correct answer is option 'B'. Can you explain this answer?

Praveen Kumar answered
f(x) = ln(1+x) - x ≤ 0
f(x) = ln(1+x) - x
f’(x) = 1/(1+x) - 1
= (1 - 1 - x)/(1+x) 
= -x/(1+x)
f’(x) ≤ 0
-x/(1+x) ≤ 0
0 ≤ x /(1+x)
for(x = 2) (-2)/(1-2)
= 2 > 0, therefore 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?

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 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

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

The maximum value of  is​
  • a)
    (1/e)1/e
  • b)
    (e)2/e
  • c)
    (e)-1/e
  • d)
    (e)1/e
Correct answer is option 'D'. Can you explain this answer?

For every real number (or) valued function f(x), the values of x which satisfies the equation f1(x)=0 are the point of it's local and global maxima or minima.
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x 
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y. 
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating  dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e

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

In case of strict decreasing functions, slope of tangent and hence derivative is
  • a)
    Negative
  • b)
    either negative or zero.
  • c)
    Zero
  • d)
    Positive
Correct answer is option 'A'. Can you explain this answer?

Avantika Joshi answered
In the case of strictly decreasing functions, the slope of the tangent line is always negative. This implies that the derivative of a strictly decreasing function is always negative. The derivative represents the rate of change of the function, and if the function is strictly decreasing, the rate of change is negative.

Let f (x) = x3−6x2+9x+8, then f (x) is decreasing in
  • a)
    (−∞,1) ∪ (3,∞)
  • b)
    [1, 3]
  • c)
    [3,∞]
  • d)
    (−∞,1)
Correct answer is option 'B'. Can you explain this answer?

Pranjal Saini answered
Understanding the Function f(x)
The function given is f(x) = x^3 - 6x^2 + 9x + 8. To determine where this function is decreasing, we need to analyze its first derivative.
Finding the First Derivative
- The first derivative f'(x) helps identify increasing and decreasing intervals.
- f'(x) = 3x^2 - 12x + 9.
Finding Critical Points
- Set the first derivative equal to zero to find critical points:
3x^2 - 12x + 9 = 0.
- Simplifying gives us:
x^2 - 4x + 3 = 0.
- Factoring results in:
(x - 1)(x - 3) = 0, leading to critical points at x = 1 and x = 3.
Test Intervals
Now we need to evaluate the sign of f'(x) in the intervals defined by these critical points:
- Interval (-∞, 1): Choose x = 0.
f'(0) = 3(0)^2 - 12(0) + 9 = 9 (positive).
- Interval (1, 3): Choose x = 2.
f'(2) = 3(2)^2 - 12(2) + 9 = -3 (negative).
- Interval (3, ∞): Choose x = 4.
f'(4) = 3(4)^2 - 12(4) + 9 = 9 (positive).
Conclusion
- f'(x) > 0 in (-∞, 1) and (3, ∞) (increasing).
- f'(x) < 0="" in="" (1,="" 3)="">
Thus, f(x) is decreasing in the interval [1, 3], which corresponds to option 'B'. 0="" in="" (1,="" 3)="" (decreasing).="" thus,="" f(x)="" is="" decreasing="" in="" the="" interval="" [1,="" 3],="" which="" corresponds="" to="" option="">
Thus, f(x) is decreasing in the interval [1, 3], which corresponds to option 'B'.>

Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2 < c < 3, f’ (c) = 0} then
  • a)
    S has exactly one point
  • b)
    S = { }
  • c)
    S has atleast one point
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Conditions of Rolle’s Theorem are satisfied by f(x) in [2,3].Hence there exist atleast one real c in (2, 3) s.t. f ‘(c) = 0 . Therefore , the set S contains atleast one element

Find the approximate change in total surface area of a cube of side x metre caused by increase in side by 1%.​
  • a)
    12 m2
  • b)
    0.12x2 m2
  • c)
    1.2x m2
  • d)
    12x m2
Correct answer is option 'B'. Can you explain this answer?

Ayush Joshi answered
Toolbox:
Let y=f(x)
Δx denote a small increment in x
Δy=f(x+Δx)−f(x)
dy=(dy/dx)Δx
Surface area of cube =6s^2

Step 1:
The side of the cube =x meters
Decrease in side =1%
= 0.01x
Increase in side =Δx
= −0.01x

Step 2:
Surface area of cube = 6s^2
= 6 * x^2
S = 6x^2
ds/dx = 12x [Differentiating with respect to x] 
Approximate change in surface area of cube = ds/dx * Δx
=12x *(0.01x)
= 0.12x^2 m^2

If the line y=x is a tangent to the parabola y=ax2+bx+c at the point (1,1) and the curve passes through (−1,0), then
  • a)
     
    a=b=−1, c=3
  • b)
     
    a=b=1/2, c=0
  • c)
     
    a=c=1/4, b=1/2
  • d)
     
    a=0, b=c=1/2
Correct answer is option 'C'. Can you explain this answer?

Explanation:

Given Conditions:
- The line y = x is a tangent to the parabola y = ax^2 + bx + c at the point (1,1).
- The curve passes through (-1,0).

Using the tangent condition:
- Since the line y = x is a tangent to the parabola y = ax^2 + bx + c at the point (1,1), the slope of the tangent at this point should be equal to 1 (slope of the line y = x).
- The derivative of y = ax^2 + bx + c is y' = 2ax + b.
- At x = 1, the derivative should be equal to 1. So, 2a + b = 1.

Using the point condition:
- Since the curve passes through (-1,0), substituting x = -1 and y = 0 in the equation of the parabola gives 0 = a(-1)^2 + b(-1) + c.
- This simplifies to a - b + c = 0.

Solving the equations:
- From the tangent condition, we have 2a + b = 1.
- From the point condition, we have a - b + c = 0.
- Substituting a = c/4 into the equations above, we get a = c = 1/4 and b = 1/2.

Final Answer:
- Therefore, the correct values for a, b, and c are a = c = 1/4 and b = 1/2, which corresponds to option 'C'.

Given that f (x) = x1/x , x>0, has the maximum value at x = e,then
  • a)
    eπe
  • b)
    eπe
  • c)
    eπe
  • d)
    eπ⩽πe
Correct answer is option 'C'. Can you explain this answer?

Anirudh Malik answered
Understanding the Function
The function f(x) = x^(1/x) is given, and we need to find its maximum value.
Finding the Maximum Value
To find the maximum, we take the natural logarithm of f(x):
- Let y = f(x) = x^(1/x)
- Then, ln(y) = (1/x) * ln(x)
Next, we differentiate ln(y) with respect to x and set the derivative to zero:
- d/dx[ln(y)] = d/dx[(1/x) * ln(x)]
- Using quotient rule and simplifying leads to critical points.
Setting the Derivative to Zero
After simplification, we find the critical points by solving:
- 1 - ln(x) = 0
- Hence, ln(x) = 1
- This gives us x = e.
Verifying Maximum at x = e
To confirm that x = e is a maximum, we can use the second derivative test or analyze the behavior of the function around e. The function f(x) increases until x = e and then decreases afterwards.
Conclusion
Since the maximum value of the function f(x) occurs at x = e, we conclude:
- The correct answer is c) e.
This shows that the maximum value of the function f(x) = x^(1/x) indeed occurs at x = e, making option 'C' the correct choice.

The equation of the tangent to the curve y=(4−x2)2/3 at x = 2 is
  • a)
    x = 2
  • b)
    x = – 2
  • c)
    y = – 1.
  • d)
    y = 2
Correct answer is option 'A'. Can you explain this answer?

Sushant Khanna answered
, which does not exist at x = 2 . However , we find that  , at x = 2 . Hence , there is a vertical tangent to the given curve at x = 2 .The point on the curve corresponding to x = 2 is (2 , 0). Hence , the equation of the tangent at x = 2 is x = 2

The true set of real values of x for which the function, f(x) = x ln x – x + 1 is positive is
  • a)
     (1, ∞)
  • b)
    (1/e, ∞)
  • c)
    [e, ∞)
  • d)
    (0, 1) and (1, ∞)
Correct answer is option 'D'. Can you explain this answer?

Anjali Sen answered
The true set of real values of x for which the function f(x) = xln x is defined is x > 0. This is because the natural logarithm function ln x is only defined for positive real numbers.

The normal at any point q to the curve x = a (cos q + q sin q), y = a (sin q – q cos q) is at distance from the origin that is equal to… .​
  • a)
    a
  • b)
    2a
  • c)
    4a
  • d)
    3a
Correct answer is option 'A'. Can you explain this answer?

Athul Yadav answered
Clearly,  dx/dy =tan q = slope of normal = −cotq
Equation of normal at θ' is y−a(sinq − qcosq) = −cotq(x−a(cosq + qsinq)
= ysinq − asin2q + aqcosqsinq 
= −xcosq + acos2q + aqsinqcosq
= xcosq + ysinq = a

The points on the curve 4 y = |x2−4| at which tangents are parallel to x – axis, are
  • a)
    (4, 3) and (– 4, – 3)
  • b)
    (0, 1) only
  • c)
    (2, 0) and (– 2, 0)
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Ameya Sengupta answered
Given curve: 4y = |x-24|

To find the points on the curve at which tangents are parallel to the x-axis, we need to find the points where the slope of the tangent is zero (since the slope of the x-axis is zero).

First, let's find the derivative of the curve:

Differentiating both sides of the equation with respect to x, we get:

4(dy/dx) = d/dx |x-24|

To find the derivative of the absolute value function, we consider two cases: x-24 > 0 and x-24 <>

Case 1: x-24 > 0
In this case, the absolute value function simplifies to x-24. Taking the derivative, we get:

4(dy/dx) = d/dx (x-24)
4(dy/dx) = 1

Case 2: x-24 <>
In this case, the absolute value function simplifies to -(x-24). Taking the derivative, we get:

4(dy/dx) = d/dx (-(x-24))
4(dy/dx) = -1

Simplifying both cases, we get:

Case 1: dy/dx = 1/4
Case 2: dy/dx = -1/4

Now, let's find the points on the curve where the slope of the tangent is zero (parallel to the x-axis).

Setting dy/dx = 0, we get:

Case 1: 1/4 = 0 (No solution)
Case 2: -1/4 = 0 (No solution)

Therefore, there are no points on the curve where the tangents are parallel to the x-axis. Hence, the correct answer is option 'D' (none of these).

Note: It is important to carefully analyze the given curve and consider all possible cases when finding the points where tangents are parallel to a particular line. In this case, since the slope of the x-axis is zero, we need to find the points where the derivative is zero. However, after considering all cases, we find that there are no such points on the given curve.

The function f (x) = x2, for all real x, is
  • a)
    neither decreasing nor increasing
  • b)
    Decreasing
  • c)
    Increasing
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Pragati Patel answered

Since f ‘(x) = 2x > 0 for x > 0,and f ‘ (x) = 2x < 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing. Infact , f is strict increasing on [0 , ∞) and strict decreasing on (- ∞,0].

​Find the points of local maxima or minima for the function f(x) = x3.ex.
  • a)
    x=-3 is a point of local maxima
  • b)
    x=-3 is a point of local minima
  • c)
    x=0 is a point of local maxima
  • d)
    x=0 is a point of local minima
Correct answer is option 'B'. Can you explain this answer?

Solution:

The given function is f(x) = x3.ex.

To find the points of local maxima or minima, we need to find the critical points of the function.

Critical points: The points where the derivative of the function is either zero or does not exist.

f'(x) = 3x2.ex + x3.ex

Let f'(x) = 0, then

3x2.ex + x3.ex = 0

x2(ex + x) = 0

x = 0 or x = -ex

Now, we need to check the nature of critical points using the second derivative test.

f''(x) = 6x.ex + 6x2.ex + 2x3.ex

At x = 0,

f''(0) = 0

Thus, x = 0 is not a point of local maxima or minima.

At x = -ex,

f''(-ex) = 6(-ex).ex + 6(-ex)2.ex + 2(-ex)3.ex

f''(-ex) = -2ex3 <>

Thus, x = -ex is a point of local maxima.

Hence, option B is the correct answer.

Note: The second derivative test is used to determine the nature of critical points. If f''(x) > 0, then the critical point is a point of local minima. If f''(x) < 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" inconclusive.="" 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="">

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