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Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced


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31 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced

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Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 1

If a + b + c = 0, then the quadratic equation 3ax2 + 2bx + c = 0 h as

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 1

Consider the function f (x) = ax3 + bx2 +  cx on [0, 1] then being a polynomial. It is continuous on  [0, 1], differentiable on (0, 1) and f (0) = f (1) = 0 [as given a + b + c = 0]
∴ By Rolle's theorem such that 2
f '(x) = 0 ⇒ 3ax2 + 2bx +c = 0
Thus equation 3ax2 + 2bx + c = 0 has at least one root in [0, 1].

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 2

AB is a diameter of a circle and C is any point on the circumference of the circle. Then

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 2


which is max. when sin 2α = 1
i.e. α = 45º
∴ ΔABC is an isosceles triangle.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 3

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

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 3

Dividing (2) by (1), we get

∴ Slope of normal = – cot θ
∴ Equation of normal is

⇒ y sin θ – a sin2 θ + a sin θ cos θ
= – x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
As q varies inclination is not constant.
∴ (a) is not correct.
Clearly does not pass through (0, 0).

which is constant

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 4

If y = a ln x + bx2 + x  has its extremum values at x = –1 and x = 2, then

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 4

y = a ln x + bx2 + x
has its extremum values at x = – 1 and 2

has – 1 and 2 as its roots.
∴ 2b – 1 + a = 0 … (1)
8b + 2 + a = 0 … (2)
Solving (1) and (2) we get a = 2, b = – 1/2.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 5

Which one of the following curves cut the parabola y2 = 4ax at right angles?

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 5

For y2 = 4ax, y-axis is tangent at (0, 0), while for x2 = 4ay, x-axis is tangent at (0, 0).
Thus the two curves cut each other at right angles. 

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 6

Th e fun ction defined by f(x) = (x + 2) e–x is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 6

f '( x) = -(x+ 2)e- x + e- x = -( x+ 1)e-x =0 ⇒ x = – 1
For x ∈ (-∞, -1), f '(x)>0 and for
x ∈ (-1, ∞), f '(x)< 0
∴ f (x) is increasing on (-∞,-1) and decr easing on (-1, -∞) 

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 7

The function 

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 7




Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 8

On the interval [0, 1] the function x25 (1 - x)75 takes its maximum value at the point

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 8



Also at x = 0, y = 0, at x = 1, y = 0, and at x = 1/4, y > 0

∴ Max. value of y occurs at x = 1/4

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 9

The slope of the tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. If the curve passes through the point (1, 2), then the area bounded by the curve, the x-axis and the line x = 1 is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 9

Slope of tangent at (x, f (x)) is 2x + 1
⇒ f ' (x) = 2x + 1 ⇒ f (x) = x2 + x + c
Also the curve passes through (1, 2)
∴ f (1) = 2
⇒ 2 = 1 + 1 + c ⇒ c = 0,  ∴ f (x) = x2 + x

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 10

then in this interval

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 10


where sin2 x is always +ve, when 0 < x < 1 . But to check Nr., we again let
h (x) = sin x – x cos x
⇒ h '(x) = x sin x > 0 for 0 < x<1  ⇒ h (x) is increasing
⇒ h (0) < h (x), when 0 < x<1
⇒ 0 <  sin x – x cos x, when 0 < x < 1
⇒ sin x – x cos x > 0, when 0 < x <1
⇒ f ' (x) > 0, x ∈ (0,1]

Here tan2 x > 0 But to check Nr. we consider
p (x) = tan x – x sec2 x
p'( x) = sec2 x - sec2 x- x.2 sec x.sec x tanx

⇒ p '(x) = -2 x sec2 x tanx<0 for 0 < x <1
⇒ p (x) is decreasing, when 0 < x<1
⇒ p (0) > p (x)  ⇒ 0 > tan x – x sec2 x
∴ g '(x) < 0

Hence g (x) is decreasing when 0 < x < 1.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 11

The function f(x) = sin4 x + cos4 x increases if

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 11

We are given f (x) = sin4 x + cos4 x
⇒ f '(x) = 4 sin3 x cos x- 4 cos3 x sinx
= – 4 sin x cos x (cos2 x – sin2 x)  
= – 2. sin 2x cos 2x = – sin 4x
Now for f (x) to be increasing function
f '(x) > 0 ⇒ -sin4x >0 ⇒ sin4x < 0

Since, If f (x) increasing on (π /4,π /2)

It will be increasing on (π / 4, 3π / 8) .

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 12

Consider the following statments in S and R

S : Both sin x and cos x are decreasing functions in the 

R: If a differentiable function decreases in an interval (a, b), then its derivative also decreases in (a, b).

Which of the following is true ?

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 12

From graph it is clear that both sin x and cos x in the interval (π / 2,π) are decreasing function.

∴ S is correct.
To disprove R let us consider the counter example :
f (x) = sin x on (0, π / 2) so that f '(x) = cosx
Again from graph it is clear that f (x) is increasing on (0, π / 2) but f '(x) is decreasing on (0, π / 2)
∴ R is wrong.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 13

Then f decreases in the interval

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 13

For decreasing function, f ' (x) < 0
⇒ ex (x – 1) (x – 2) < 0  ⇒ (x – 1) (x – 2) < 0
⇒ 1 < x < 2,

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 14

If the normal to the curve y = f(x) at the point (3,4) makes an angle 3π/4 with the positive x-axis, then f '(3) =

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 14


Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 15

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 15

It is clear from figur e that at x = 0, f (x) is n ot differentiable.

⇒ f (x) has neither maximum nor minimum at x = 0.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 16

For all x ∈ (0,1)

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 16

Let f ( x) = ex -1-x then f '(x) = ex - 1>0 for x ∈ (0,1)
∴ f (x) is an increasing function.


∴ g (x) is decreasing on (0, 1) ∴ x > 0
⇒ g (x) < g (0)
⇒ log (1 + x) – x < 0 ⇒ log (1 + x) < x
∴ (b) holds. Similarly it can be shown that (c) and (d) do not hold.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 17

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

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 17

f (x) = xex (1–x)


∴ f (x) is increasing on [–1/2, 1]

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 18

The triangle formed by the tangent to the curve f(x) = x2 + bx - b at the point (1, 1) and the coordinate axes, lies in the first quadrant. If its area is 2, then the value of b is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 18

Tangent to y = x2 + bx – b at (1, 1) is
y – 1 = (2 + b) (x – 1) ⇒ (b + 2) x – y = b + 1


⇒ b2 + 2b + 1 = – 4 (b + 2) ⇒ b2 + 6b + 9 = 0
⇒ (b + 3)2 = 0 ⇒ b = – 3

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 19

Let f(x) = (1 + b2)x2 + 2bx + 1 and let m(b) be the minimum value of f(x). As b varies, the range of m(b) is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 19

f (x) = (1 + b2) x2 + 2bx + 1
It is a quadratic expression with coeff. of x2 = 1 + b2 > 0.
∴ f (x) represents an upward parabola whose min value is   being the discreminant.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 20

The length of a longest interval in which the function 3 sin x – 4 sin3x is increasing, is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 20

3 sin x – 4 sin3 x = sin 3x which increases for

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 21

The point(s) on the curve y3 + 3x2 = 12y where the tangent is vertical, is (are)

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 21

The given curve is y3 + 3x2 = 12y

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 22

In [0,1] Lagranges Mean Value theorem is NOT applicable to

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 22

There is only one function in option (a) whose critical point    for the rest of the parts critical point 0 ∉ (0, 1). It can be easily seen that functions in options (b), (c) and (d) are continuous on [0, 1] and differentiable in (0, 1).



∴ f is not differentiable at 1 / 2 ∈ (0,1)
∴ LMV is not applicable for this function in [0, 1]

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 23

Tangent is drawn to ellipse

Then the value of θ such that sum of intercepts on axes made by this tangent is minimum, is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 23

Equation of tan gen t to the ellipse 


Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 24

If f (x) = x3 + bx2 + cx + d and 0 < b2 < c, then in (–∞, ∞)

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 24


Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 25

If f (x) = xα log x and f (0) = 0, then the value of α for which Rolle’s theorem can be applied in [0, 1] is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 25

For Rolle's theorem in [a, b]
f (a) = f (b), ln [0, 1] ⇒ f (0) = f (1) = 0
∴ The function has to be continuous in [0, 1]


Applying L' Hospital's Rule

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 26

If P(x) is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials so that P(0) = 0,

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 26

Let the polynominal be P (x) = ax2 + bx + c
Given P (0) = 0 and P (1) = 1 ⇒ c = 0 and a +  b = 1
⇒ a = 1 – b
∴ P (x) = (1 – b) x2 + bx
⇒ P' (x)  = 2 (1 – b) x + b

⇒ 2 (1 – b) x + b > 0
⇒ When x = 0, b > 0 and when x = 1, b < 2
⇒ 0 < b < 2
∴ S = {(1 - a) x2 + ax,a∈ (0, 2)}

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 27

The tangent to the curve y = ex drawn at the point (c, ec) intersects the line joining the points (c – 1, ec–1) and (c + 1, ec + 1)

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 27

The equation of tangent to the curve y = ex at (c, ec) is

y - ec = ec (x-c) … (1)

and equation of line joining (c –1, ec–1) and (c +1, ec+1) is


Subtracting equation (1) from (2), we get


∴ The two lines meet on the left of line x = c.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 28

Consider the two curves C1 : y2 = 4x, C2 : x2 + y2 – 6x + 1 = 0.

Then,

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 28

The given curves are
C1 : y2 = 4x ...(1)  and C2 : x2 + y2 – 6 x + 1 = 0...(2)
Solving (1) and (2) we get
x2 + 4x – 6x + 1 = 0 ⇒ x = 1 and ⇒ y = 2 or –2
∴ Points of intersection of the two curves are (1, 2) and  (1, –2).


∴ C1 and C2 touch each other at two points.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 29

The total number of local maxima and local minima of the function 

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 29


The graph of y = f (x) is as shown in the figure. From graph, clearly, there is one local maximum (at  x = –1) and one local minima (at x = 0)
∴ total number of local maxima or minima = 2.

Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 30

Let the function   be given by 

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 30



Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 31

The least value of  is

Detailed Solution for Test: Single Correct MCQs: Applications of Derivatives | JEE Advanced - Question 31

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