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# Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

## JEE : Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

The document Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev is a part of the JEE Course Maths 35 Years JEE Mains & Advance Past year Papers Class 12.
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Q.1. Let f (x) be a polynomial of degree 5 such that x =Â±1 are its critical points. Ifthen which one of the following is not true?    (2020)
(1)  f is an odd function.
(2) f(1) - 4f(-1) = 4.
(3) x = 1 is a point of maxima and x = âˆ’1 is a point of minimum of f.
(4) x = 1 is a point of minima and x = - 1 is a point of maxima of f.
Ans. (4)
Given,
(1)
Let,
f(x) = 2x3 + ax4 + bx5 â‡’ f'(x) = 6x2 + 4ax3 + 5bx4
Since x = Â±1 are the critical points of f (x), then
f'(1) = 6 + 4a + 5b = 0
â‡’ 4a + 5b = -6...   (2)
Now, f'(-1) = 0 â‡’ 6 - 4a + 5b = 0
â‡’ -4a + 5b = - 6....    (3)
On solving Eqs. (2) and (3), we get
Therefore,

Hence, x = 1 is a point of maxima and x = -1 is a point of minima of  f(x).

Q.2. Let f(x) = x cos-1 then which of the following is true?    (2020)
(1) f' is increasing in and decreasing in

(2) f' (0) =
(3) f is not differentiable at x = 0.
(4) f' is decreasing in and increasing in
Ans. (4)
We have,
f(x) = x[Ï€ - cos-1

Now,

Hence, f'(x) is decreasing in and increasing in

Q.3. The length of the perpendicular from the origin, on the normal to the curve, x2 + 2xy - 3yat the point (2, 2) is    (2020)
(1) âˆš2
(2) 4âˆš2
(3) 2
(4) 2âˆš2
Ans. (4)
The equation of the given curve is
x+ 2xy - 3y2 = 0

Now, the slope of normal at (2,2) is

The equation of the normal is
y âˆ’2 = âˆ’1(xâˆ’2)â‡’ x+ y - 4 = 0
Hence, the length of perpendicular from origin to the normal is

Q.4. Let f(x) be a polynomial of degree 3 such that f(âˆ’1) = 10, f(1) = âˆ’6, f(x) has a critical point at x = âˆ’1 and f'(x) has a critical point x = 1. Then f(x) has a local minima at x = _________.    (2020)
Ans.
(3.00)
Given, f'(x) has a critical point x = 1, let

Now, f '(-1) = 0
Therefore,

Now, f(1) = - 6

â‡’ âˆ’11k + 6d = âˆ’36...    (1)
Now, f(-1) = 10 â‡’ 5k + 6d = 60...    (2)
From Eqs. (1) and (2), we get
k=5,d=6
f(x) = x- 3x- 9x + 5 â‡’ f'(x) = 3(x + 1)(x-3)
â‡’ f"(x) = 6x - 6 â‡’ f"(3) = 6 x 3 - 6 = 12
Hence, f(x) has a local minima at x =3.

Q.5. A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which the thickness of ice decreases, is    (2020)
(1) 5/6Ï€
(2) 1/54Ï€
(3) 1/36Ï€
(4) 1/18Ï€
Ans. (4)
Let the initial thickness of the ice is x cm, then

At x = 5 cm and = 50 cm3/ min
We have,

Q.6. Let a function f :[0,5] â†’ be continuous f (1) = 3 and F be defined as where Then, for the function F, the point x = 1 is    (2020)
(1) a point of local minima.
(2) not a critical point.
(3) a point of local maxima.
(4) a point of inflection.

Ans. (1)
We have,
....   (since f (1) = 0)
(since g (1) = 0)
â‡’ F'(1) = 1.g (1) = 0....    (1)
â‡’ F" (x) = 2x g (x) + x2 g' x
â‡’ F" (1) = 2g (1) + 1 . g ' (1) = f(1) = 3....    (2)
From Eqs. (1) and (2), it is clear that x = 1 is the point of minima for function F.

Q.7. The maximum volume (in cu.m) of the right circular cone having slant height 3 m is:    (2019)
(1) 6Ï€
(2) 3âˆš3Ï€
(3)
(4) 2âˆš3Ï€
Ans. (4)
Solution.

...(1)
Volume of cone
...(2)

For maxima/minima,

Then, h = âˆš3 is point of maxima
Hence, the required maximum volume is,

Q.8. Let d âˆˆ R, and  Î¸ âˆˆ [0, 2Ï€]. If the minimum value of det (A) is 8, then a value of d is:     (2019)
(1) -5
(2) -7
(3) 2(âˆš2 + 1)
(4) 2(âˆš2 + 2)
Ans. (1)
Solution.

Minimum value of det (A) is attained when sin2Î¸ = 1

â‡’ d = -5 or 1

Q.9. If I is minimum then the ordered pair (a, b) is:     (2019)

Ans. (4)
Solution.

Also,

âˆ´ I is minimum when (a, b) = (-âˆš2, âˆš2)

Q.10. The tangent to the curve,passing through the point (1, e) also passes through the point:     (2019)
(1) (2, 3e)
(2)
(3)
(4) (3, 6e)

Ans. (2)
Solution. The equation of curve

Since (1, e) lies on the curvethen equation of tangent at (1, e) is

So, equation of tangent to the curve passes through the point

Q.11. A helicopter is flying along the curve given by y - x3/2 = 7, (x â‰¥ 0). A soldier positioned at the point  wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is:     (2019)

(4) 1/2
Ans. (3)
Solution.

â‡’ f(x) is increasing function âˆ€ x > 0

Q.12. The maximum value of the function f(x) = 3x3 -18x+ 27x - 40 on the set S = {x âˆˆ R: x2 + 30 â‰¤ 11x } is     (2019)
(1) -122
(2) -222
(3) 122
(4) 222

Ans. (3)
Solution. Consider the function,
f(x) = 3x(x - 3)2 - 40
Now S = {x âˆˆ R : x2 + 30 â‰¤ 11x}
So x2 - 11x + 30 < 0
â‡’ x âˆˆ [5, 6]
âˆ´ f(x) will have maximum value for x = 6
The maximum value of function is,
f(6) = 3 x 6 x 3 x 3 - 40 = 122.

Q.13. Let x,y be positive real numbers and m, n positive integers. The maximum value of the expression  is:      (2019)
(1) 1
(2) 1/2
(3) 1/4
(4) None of these
Ans. (3)
Solution.

Q.14.
where a, b and d are non-zero real constants. Then:     (2019)
(1) f is an increasing function of x
(2) f is a decreasing function of x
(3) f is not a continuous function of x
(4) f is neither increasing nor decreasing function of x

Ans. (1)
Solution.

â‡’ f(x) is increasing function.
Hence, f(x) is increasing function.

Q.15. The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 - x2 such that the rectangle lies inside the parabola, is:     (2019)
(1) 36
(2) 20âˆš2
(3) 32
(4) 18âˆš3
Ans.
(3)
Solution.

Given, the equation of parabola is,
x2 = 12 - y

Area of the rectangle = (2t) (12 - t2)

At t = 2, area is maximum = 24(2) - 2(2)3
= 48 - 16 = 32 sq. units

Q.16. Let P(4, -4) and Q(9, 6) be two points on the parabola, y2 = 4x and let this X be any point arc POQ of this parabola, where O is vertex of the parabola, such that the area of Î”PXQ is maximum. Then this minimum area (in sq. units) is:     (2019)
(1) 75/2
(2) 125/4
(3) 625/4
(4) 125/2

Ans. (2)
Solution.

Parametric equations of the parabola y2 = 4x are, x = t2 and y = 2t.

For maximum area t = 1/2

Q.17. The maximum value of  for any real value of Î¸ is:     (2019)
(1)
(2)
(3)
(4)
Ans. (1)
Solution. Let, the functions is,

Q.18. If a curve passes through the point (1, -2) and has slope of the tangent at any point (x, y) on it as  then the curve also passes through the point:     (2019)
(1) (3, 0)
(2) (âˆš3, 0)
(3) (-1, 2)
(4)  (-âˆš2, 1)

Ans. (2)
Solution.

âˆµ

Solution of equation

âˆ´ curve passes through point (1, -2)

â‡’ C = -9/4
Then, equation of curve

Since, above curve satisfies the point.
Hence, the curve passes through (âˆš3, o).

Q.19. The tangent to the curves y = x2 - 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point:     (2019)

Ans. (2)
Solution.
âˆµ Tangent to the given curve is parallel to line 2y = 4x + 1
âˆ´ Slope of tangent (m) = 2
Then, the equation of tangent will be of the form
y = 2x + c    ...(1)
âˆµ Line (1) and curve y = x2 - 5x + 5 has only one point of intersection.
âˆ´ 2x + c = x2 - 5x + 5
x2 - 7x + (5 - c) = 0
âˆ´ D = 49 - 4(5 - c) = 0
â‡’
Hence, the equation of tangent:

Q.20. If the function/given by f(x) = x3 - 3(a - 2)x2 + 3ax + 7, for some a âˆˆ R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation,       (2019)
(1) -7
(2) 5
(3) 7
(4) 6

Ans. (3)
Solution.

â‡’

â‡’
â‡’ a = 5

Now,

â‡’

Q.21. The shortest distance between the line y = x and the curve y2 = x - 2 is:     (2019)
(1) 2
(2) 7/8
(3) 7/4âˆš2
(4) 11/4âˆš2
Ans. (3)
Solution.
The shortest distance between line y = x and parabola = the distance LM between line y = x and tangent of parabola having slope 1.

Let equation of tangent of parabola having slope 1 is,

Distance between the line y - x = 0 and

Q.22. If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, f(x) = 9x4 + 12x3 - 36x2 + 25, xâˆˆR, then :     (2019)
(1) S1 = {-2}; S2 = {0, 1}
(2) S1 = {-2, 0}; S2 = {1}
(3) S1 = {-2, 1}; S2 = {0}
(4) S1 = {-1}; S2 = {0,2}
Ans. (3)
Solution.

Here at -2 & 1, f'(x) changes from negative value to positive value.
â‡’ -2 & 1 are local minimum points. At 0, f'(x) changes from positive value to negative value.
â‡’ 0 is the local maximum point.
Hence, S1 = {-2, 1} and S2 = {0}

Q.23. Let f : [0 : 2] â†’ R be a twice differentiable function such that f"(x) > 0, for all xâˆˆ(0, 2). If Ï†(x) = (x) + f(2 - x), then Ï† is:     (2019)
(1) increasing on (0, 1) and decreasing on (1, 2).
(2) decreasing on (0, 2)
(3) decreasing on (0, 1) and increasing on (1,2).
(4) increasing on (0, 2)

Ans. (3)
Solution.

But f" (x) > 0 â‡’ f'(x) is an increasing function

Hence, Ï† (x) is increasing on (1, 2) and decreasing on (0, 1).

Q.24. Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is. If the curve passes through the centre of the circle x2 + y2 - 2x - 2y = 0, then its equation is :     (2019)
(1) x loge|y| = 2(x - 1)
(2) x loge |y| = - 2(x - 1)
(3) x|loge|y| = -2(x - 1)
(4) x loge | y | = x - 1

Ans. (1)
Solution.

Equation (i) passes through the centre of the circle x2 + y2 - 2x - 2y = 0, i.e., (1,1)
âˆ´ C = 2

Q.25. The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:     (2019)

Ans. (3)
Solution.
Let radius of base and height of cylinder be r and h respectively.

...(1)
Now, volume of cylinder, V = Ï€r2h
Substitute the value of r2 from equation (i),

Differentiating w.r.t. h,

For maxima/minima,

â‡’ Volume is maximum when h = 2âˆš3

Q.26. If f(x) is a non-zero polynomial of degree four, having local extreme points at x = -1, 0, 1; then the set S = {x âˆˆ R : f(x) = f(0)} contains exactly:     (2019)
(1) four irrational numbers.
(2) four rational numbers.
(3) two irrational and two rational numbers.
(4) two irrational and one rational number.

Ans. (4)
Solution.
Since, function f(x) have local extreme points at x = -1,0, 1. Then
f(x) = K(x+ l)x(x - 1)
= K (x3 - x)

Q.27. If the tangent to the curve, y = x3 + ax - b at the point (1, -5) is perpendicular to the line, - x + y + 4 = 0, then which one of the following points lies on the curve?     (2019)
(1) (-2,1)
(2) (-2,2)
(3) (2,-1)
(4) (2,-2)

Ans. (4)
Solution.
y = x3 + ax - b
Since, the point (1, -5) lies on the curve.
â‡’ 1 + a - b = - 5
â‡’ a - b = -6    ...(1)

Since, required line is perpendicular to y = x - 4, then slope of tangent at the point P (1, -5) = -1
âˆ´ 3 + a = - 1
â‡’ a = -4
â‡’ b = 2
the equation of the curve is y = x3 - 4x - 2
â‡’ (2, -2) lies on the curve

Q.28. Let S be the set of all values of x for which the tangent to the curve y = f(x) = x3 - x2 - 2x at (x, y) is parallel to the line segment joining the points (1, f(1)) and (- 1, f(- 1)), then S is equal to:     (2019)

Ans. (4)
Solution.

y = f(x) = x3 - x2 - 2x

f(1) = 1 - 1 -2 = -2, f(-1) = -1 -1+2 = 0
Since the tangent to the curve is parallel to the line segment joining the points (1, -2) and (-1, 0)
And their slopes are equal.

Hence, the required set

Q.29. A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is  Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10m; is:     (2019)
(1) 1/15 Ï€
(2) 1/10 Ï€
(3) 2/Ï€
(4) 1/5 Ï€
Ans. (4)
Solution.

Given that water is poured into the tank at a constant rate of 5 m3/minute.

Volume of the tank is,
...(i)
where r is radius and h is height at any time. By the diagram,

...(ii)
Differentiate eq. (i) w.r.t. â€˜tâ€™, we get

Putting h = 10, r = 5 and dV/dt = 5 in the above equation.

Q.30. Let f(x) = ex - x and g(x) = x2 - x, âˆ€ x âˆˆ R. Then the set of all x âˆˆ R, where the function h(x) = (fog) (x) is increasing, is:     (2019)

Ans. (2)
Solution.
Given functions are, f(x) = ex - x and g(x) = x2 - x

Given f(g (x)) is increasing function.

are either both positive or negative

Q.31. If the tangent to the curve  at a point (Î±, Î²) is parallel to the line 2x + 6y - 11 = 0, then :     (2019)

Ans. (1)
Solution.

These values of Î± and Î² satisfies |6Î± + 2Î²| = 19

Q.32. A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is:     (2019)
(1) 1/18Ï€
(2) 1/36Ï€
(3) 5/6Ï€
(4) 1/9Ï€
Ans. (1)
Solution.
Given that ice melts at a rate of 50 cm3/min

Substitute r = 5,

Q.33. Let a1, a2, a3,.... be an A. P. with a6 = 2. Then the common difference of this A.P., which maximises the product a4 a4 a5 is:     (2019)
(1) 3/2
(2) 8/5
(3) 6/5
(4) 2/3

Ans. (2)
Solution.
a6 = a + 5d = 2
Here, a is first term of A.P and d is common difference

Q.34. If m is the minimum value of k for which the function is increasing in the interval [0, 3] and M is the maximum value off in [0,3] when k = m, then the ordered pair (m, M) is equal to:     (2019)
(1) (4, 3âˆš2)
(2) (4, 3âˆš3)
(3) (3, 3âˆš3)
(4) (5, 3âˆš6)
Ans. (2)
Solution.
Given function
Differentiating w. r. t. x,

[âˆ´ f(x) is increasing in [0, 3]]

Q.35. The equation of a common tangent to the curves, y2 = 16x and xy = - 4, is:     (2019)
(1) x - y + 4 = 0
(2) x + y + 4 = 0
(3) x - 2y + 16 = 0
(4) 2x - y + 2 = 0

Ans. (1)
Solution.
Given curves, y2 = 16x and xy = - 4
Equation of tangent to the given parabola;

âˆµ This is common tangent.

âˆ´ equation of common tangent is y = x + 4

Q.36. The tangents to the curve y = (x - 2)2 -1 at its points of intersection with the line x - y = 3, intersect at the point:     (2019)

Ans. (3)
Solution.

Tangent to the curve y = (x - 2)2 - 1 at any point (h, k) is,

Q.37.  then the local minimum value of value of h(x) is:     (2018)
(1) 3
(2) -3
(3) -2âˆš2
(4) 2âˆš2
Ans.
(4)
Solution.

Hence, local minimum value is 2âˆš2

Q.38. If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is:    (2018)
(1) 8âˆš2 Ï€
(2) 6âˆš2 Ï€
(3) 6âˆš3 Ï€
(4) 8âˆš3 Ï€
Ans.
(4)
Solution.

where r is radius and h is height of coin

Q.39. Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f (x) = 2x3 - 9x2 + 12x + 5 in the interval [0, 3]. Then M - m is equal to:     (2018)
(1) 1
(2) 9
(3) 5
(4) 4
Ans. (1)
Solution.

For maxima or minima put f'(x) = 0

Q.40. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is     (2017)
(1) 30
(2) 12.5
(3) 10
(4) 25
Ans.
(4)
Solution.

2r + Î¸r =20 ... (i)

A to be maximum

Hence for r = 5, A is maximum
Now, 10 + Î¸Â·5 = 20 â‡’ Î¸ = 2 (radian)

Q.41. The function f defined by f(x) = x3 â€“ 3x2 + 5x + 7, is:     (2017)
(1) decreasing in R
(2) increasing in R
(3) increasing in (0, âˆž) and decreasing in (â€“âˆž, 0)
(4) decreasing in (0, âˆž) and increasing in (â€“âˆž, 0)

Ans. (2)
Solution.
f(x) = x3 â€“ 3x2 + 5x + 7
f'(x) = 3x2 â€“ 6x + 5 > 0 ( âˆµD< 0, a>0)
â‡’x âˆˆ Ï•

Q.42. A tangent to the curve, y = f(x) at P(x,y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f(1)= 1, then the curve also passes through the point:    (2017)

Ans. (3)
Solution.

Let y = f(x) be a curve
slope of tangent = f'(x)
Equation of tangent (Y-y) = f'(x) (X- x)
Put Y = 0

y = 1/x3 is required curve (2, 1/8) passing through
y = 1/x3

Q.43. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If âˆ BPC = Î² then tan Î² is    (2016)
(1) 4/9
(2) 6/7
(3) 1/4
(4) 2/9
Ans.
(4)
Solution.

Q.44. If m and M are the minimum and the maximum values of   then M - m is equal to    (2016)
(1) 7/4
(2) 15/4
(3) 9/4
(4) 1/4
Ans.
(3)
Solution.

M = maximum value = 17/4
m = minimum value = 2
M-m = 17/4 - 2 = 9/4.

Q.45. Let f(x) = sin4x + cos4x. Then f is an increasing function in the interval    (2016)
(1)

(2)
(3)
(4)
Ans. (3)
Solution.
f(x) = sin4x + cos4x
f'(x) = 4sin3x cosx - 4cos3xsinx
= 4sinx cosx(sin2x - cos2x)
= - 2sin2x. cos2x
= - sin4x > 0
â‡’ sin4x < 0
â‡’ Ï€ < 4x < 2Ï€

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