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Test: 35 Year JEE Previous Year Questions: Applications of Derivatives


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33 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: 35 Year JEE Previous Year Questions: Applications of Derivatives

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Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 1

The maximum distance from origin of a point on the curve 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 1

Distance of origin from  

∴ Maximum distance from origin = a + b

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 2

If 2a + 3b + 6c = 0, (a, b, c ∈ R) then the quadratic equation ax2 + bx + c = 0 has

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 2

Also f (x) is continuous and differentiable in [0, 1] and [0, 1]. So by Rolle’s theorem, f ' (x) = 0.
i.e ax2 + bx + c = 0 has at least one root in [0, 1].

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 3

If th e function f ( x) = 2x3 - 9ax2 + 12a2x+1 , whe re a > 0 , attains its maximum and minimum at  p and  q respectively such that p 2 = q , then a equals

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 3

f (x) = 2x3 - 9ax2 + 12a2x +1
f '(x) = 6x2 - 18ax + 12a2; f"(x) = 12x- 18a
For max. or min.
6x2 - 18ax + 12a2 = 0 ⇒ x2 - 3ax + 2a2 = 0
⇒ x = aorx = 2a.At x = a max. and at x = 2a min

∴ p = a and q = 2a

As per question p2 = q 
∴ a = 2a ⇒ a = 2ora = 0
but a > 0, therefore, a = 2.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 4

A point on th e parabola y2 = 18x at which the ordinate increases at twice the rate of the abscissa is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 4

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 5

A function y = f (x) h as a second or der der iva tive f "( x) = 6(x- 1). If its graph passes through the point (2,1) and at that point the tangent to the graph is y = 3x – 5, then the function is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 5

f ''(x) = 6(x - 1). Inegrating, we get
f '( x) = 3x2 -6 x+c
Slope at (2, 1)  = f '(2) = c = 3
[∴ slope of tangent  at (2,1) is 3]
∴ f '( x) = 3x2 - 6 x + 3 = 3(x- 1)2
Inegrating again, we get  f (x) = (x -1)3+D
The curve passes through (2, 1)
⇒1= (2-1)3 +D ⇒ D = 0
∴ f (x) = (x – 1)3

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 6

The normal to the curve x = a(1 + cosθ), y = a sinθ at ‘θ’ always passes through the fixed point

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 6

∴ The slope of the normal  at θ = tan θ
∴ The equation of the normal at θ is
y -a sin θ = tan θ( x -a -a cosθ)

which always passes through (a, 0)

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 7

If 2a + 3b + 6c = 0, then at least one root of the equation ax2 + bx +c= 0 lies in the interval

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 7

Let us defin e a function

Being  polynomial, it is continuous and differentiable, also,

∴ f (x) satisfies all condition s of Rolle’s theor em therefore f ’(x) = 0 has a root in (0, 1)

i.e. ax2 + bx +c= 0 has at lease one root in (0, 1)

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 8

Area of the greatest rectangle that can be inscribed in the ellipse 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 8

Area of rectangle ABCD = 2a cosθ (2b sin θ) = 2ab sin2θ

⇒ Area of greatest rectangle is equal to 2ab When sin 2θ = 1 .

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 9

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: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 9

x =a (cos θ + θ sinθ

   .....(1)

From equations (1) and (2), we get

Equation of normal at ' θ ' is y – a (sin θ – θ cos θ)

Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 10

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm 3/min.
When the thickness of ice is 5 cm,then the rate at which the thickness of ice decreases is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 10

Given that

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 11

If the equation   has a positive root x = a , then the equation   has a positive root, which is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 11

Again f (x) has root a,  ⇒ f (a)=0
∴ f (0) = f (a)
∴ By Roll’s theorem f'(x) = 0 has root between (0, α)

Hence f '(x) has a positive root smaller than a.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 12

The function  has a local minimum at

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 12


 has local min at x = 2.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 13

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 13



Maximum value of sinq is 1 at 

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 14

A value of c for which conclusion of Mean Value Theorem holds for the function f (x) = loge x on the interval [1, 3] is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 14

Using Lagrange's Mean Value Theorem
Let f (x) be a function defined on [a, b]


Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 15

The function f (x) = tan–1(sin x  + cos x) is an increasing function in

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 15

Given f (x) = tan–1 (sin x + cos x)


if f ’ (x) > O then f (x) is increasing function.
Hence f (x) is increasing, if 

Hence, f (x) is increasing when  

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 16

If p and q are positive real numbers such that p2 + q2 = 1, then the maximum value of (p + q) is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 16

Given that p2 + q2 = 1
∴ p = cosθ and q = sinθ
Then p +q = cos θ + sinθ
We know that

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 17

Suppose the cubic x3 – px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 17



Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 18

How many real solutions does the equation

x7 + 14x5 + 16x3 + 30x – 560 = 0 have?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 18

Let f (x) = x7 + 14x5 + 16x3 + 30x –560


⇒The curve y = f (x) crosses x-axis only once.
∴ f (x) = 0 has exactly  one real root.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 19

Let f (x) = x | x | and g (x) = sin x.

Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point.
Statement-2 : gof is twice differentiable at x = 0.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 19

Given that f (x) = x | x | and g (x) = sin x
So that go f (x) = g (f (x)) = g (x | x |) = sin x | x |

Here we observe L (go f )' (0) =  0  =  R (go f)' (0)
⇒ go f is differentiable at x = 0 and (go f)' is continuous at x = 0

∴ L(go f)''  (0) ≠ R (go f )''  (0)
⇒ go f (x) is not twice differentiable at x = 0.
∴ Statement - 1 is true but statement -2 is false.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 20

Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P' (x) = 0. If P(–1) < P(1), then in the interval [ –1, 1] :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 20

We have P (x) = x4 + ax3 + bx2 + cx + d
⇒ P' (x) = 4x3 + 3ax2 + 2bx + c
But  P' (0) = 0 ⇒ c = 0
∴ P(x) = x4 + ax3 + bx2 + d
As given that P (– 1) < P (a)
⇒ 1 – a + b + d    <   1 + a + b + d ⇒ a > 0
Now P ' (x) = 4x3 + 3ax2 +2bx = x (4x2 + 3ax + 2b)
As P' (x) = 0, there is only one solution x = 0, therefore 4x2 + 3ax + 2b = 0 should not have any real roots i.e. D < 0


∴ P (x) is an increasing function on (0,1) ∴ P (0)  <  P (a)
Similarly we can prove P (x) is decreasing on (– 1, 0)
∴ P (– 1) >  P (0)
So we can conclude that
Max P (x) = P (1) and Min P (x) = P (0) ⇒ P(–1) is not minimum but P (1) is the maximum of  P.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 21

The equation of the tangent to the curve  that is parallel to the x-axis, is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 21

Since tangent is parallel to x-axis,

Equation of tangent is y – 3 = 0 (x – 2) ⇒ y = 3

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 22

If f has a local minimum at x = – 1 , then a possible value of k is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 22

This is true where k = – 1

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 23

Let f : R → R be a continuous function defined by

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 23


Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 24

The shortest distance between line y – x = 1 and curve x = y2 is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 24

Shortest distance between two curve occurred along the common normal Slope of normal to y2 = x at point P(t2, t) is – 2t and slope of line y – x = 1 is 1.
As they are perpendicular to each other

and shortest distance 

So shortest distance between them is

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 25

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 25


Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 26

A spherical balloon is filled with 4500π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 26

Volume of spherical balloon 

Differentiating both the sides, w.r.t 't' we get,


Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 27

Let a, b ∈ R be such that the function f given by f (x) = ln |x|+ bx2 + ax, x ≠ 0 has extreme values at  x = –1 and x = 2

Statement-1 : f has local maximum at x = –1 and at x = 2.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 27


          ...(ii)
On solving (i) and (ii) we get  

So maxima at x = –1,2
Hence both the statements are true and statement 2 is a correct explanation for 1.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 28

A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 28

Equation of a line passing through (x1,y1) having slope m is given by y – y1 = m (x–x1)
Since the line PQ is passing through (1,2) therefore its equation is
(y – 2) = m (x – 1)
where m is the slope of the line PQ.
Now, point P (x,0) will also satisfy the equation of PQ
∴ y –2 = m (x –1)
⇒ 0 – 2 = m (x – 1)




Similarly, point Q (0,y) will satisfy equation of PQ





Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 29

The intercepts on x-axis made by tangents to the curve,  which are parallel to the line y = 2x, are equal to :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 29

∴ equation of tangent is

y – 2 = 2(x – 2) or y + 2 = 2(x + 2)

⇒ x-intercept = ± 1.

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 30

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

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 30

Since, f and g both are continuous functions on [0, 1] and differentiable on (0, 1) then 

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 31

Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2.  then f(2) is equal to :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 31

Hence, f(x) = 2x2 + a2x3 + a3x4
f '(x) = 4x + 3a2x2 + 4a3x3
As given : f '(1) = 0 and f '(2) = 0

Hence, 4 + 3a2 + 4a3 = 0 ...(1)
and 8 + 12a2 + 32a3 = 0 ...(2)
By 4x (eq1) – eq (2), we get
16 + 12a2 + 16a3 – (8 + 12a2 + 32a3) = 0
⇒ 8 – 16a3 = 0 ⇒ a3 = 1/2
and by eqn. (1), 4 + 3a2 + 4/2 = 0 ⇒ a2 = –2

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 32

Consider :

A normal to y = f(x) at    also passes through the point :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 32







This equation is satisfied only by the point 

Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 33

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then: 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Applications of Derivatives - Question 33


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