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

Maths 35 Years JEE Mains & Advance Past year Papers Class 12

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

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 Q.1. Let f (x) be a polynomial of degree 5 such that x =±1 are its critical points. IfPrevious year Questions (2016-20) - Applications of Derivatives Notes | EduRevthen 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 = −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,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev    (1)
Let,
f(x) = 2x3 + ax4 + bx5 ⇒ f'(x) = 6x2 + 4ax3 + 5bx4
Since x = ±1 are the critical points of f (x), then
f'(1) = 6 + 4a + 5b = 0
⇒ 4a + 5b = -6...   (2)
Now, f'(-1) = 0 ⇒ 6 - 4a + 5b = 0
⇒ -4a + 5b = - 6....    (3)
On solving Eqs. (2) and (3), we get Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Therefore, Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
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 Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevthen which of the following is true?    (2020)
(1) f' is increasing in Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevand decreasing in Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

(2) f' (0) =Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(3) f is not differentiable at x = 0.
(4) f' is decreasing in Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevand increasing in Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (4)
We have,
f(x) = x[π - cos-1 Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Now,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Hence, f'(x) is decreasing in Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevand increasing in Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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) √2
(2) 4√2
(3) 2
(4) 2√2
Ans. (4)
The equation of the given curve is
x+ 2xy - 3y2 = 0
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Now, the slope of normal at (2,2) is
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
The equation of the normal is
y −2 = −1(x−2)⇒ x+ y - 4 = 0
Hence, the length of perpendicular from origin to the normal is
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.4. Let f(x) be a polynomial of degree 3 such that f(−1) = 10, f(1) = −6, f(x) has a critical point at x = −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
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Now, f '(-1) = 0 Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Therefore, Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Now, f(1) = - 6
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
⇒ −11k + 6d = −36...    (1)
Now, f(-1) = 10 ⇒ 5k + 6d = 60...    (2)
From Eqs. (1) and (2), we get
k=5,d=6
f(x) = x- 3x- 9x + 5 ⇒ f'(x) = 3(x + 1)(x-3)
⇒ f"(x) = 6x - 6 ⇒ 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) 5/6π
(2) 1/54π
(3) 1/36π
(4) 1/18π
Ans. (4)
Let the initial thickness of the ice is x cm, then
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
At x = 5 cm and Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev= 50 cm3/ min
We have,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.6. Let a function f :[0,5] →Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev be continuous f (1) = 3 and F be defined as Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevwhere Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevThen, 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,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev ....   (since f (1) = 0)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev    (since g (1) = 0)
⇒ F'(1) = 1.g (1) = 0....    (1)
⇒ F" (x) = 2x g (x) + x2 g' x
⇒ 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) 6π
(2) 3√3π
(3) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(4) 2√3π
Ans. (4)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev    ...(1)
Volume of cone
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev   ...(2)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
For maxima/minima,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev      Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev 
Then, h = √3 is point of maxima
Hence, the required maximum volume is,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.8. Let d ∈ R, and Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev θ ∈ [0, 2π]. If the minimum value of det (A) is 8, then a value of d is:     (2019)
(1) -5
(2) -7
(3) 2(√2 + 1)
(4) 2(√2 + 2)
Ans. (1)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Minimum value of det (A) is attained when sin2θ = 1
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
⇒ d = -5 or 1

Q.9.Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev If I is minimum then the ordered pair (a, b) is:     (2019)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (4)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Also,  Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
∴ I is minimum when (a, b) = (-√2, √2)

Q.10. The tangent to the curve,Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevpassing through the point (1, e) also passes through the point:     (2019)
(1) (2, 3e)
(2) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(3) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(4) (3, 6e)

Ans. (2)
Solution. The equation of curve Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Since (1, e) lies on the curvePrevious year Questions (2016-20) - Applications of Derivatives Notes | EduRevthen equation of tangent at (1, e) is
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
So, equation of tangent to the curve passes through the pointPrevious year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.11. A helicopter is flying along the curve given by y - x3/2 = 7, (x ≥ 0). A soldier positioned at the point Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is:     (2019)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(4) 1/2
Ans. (3)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
⇒ f(x) is increasing function ∀ x > 0
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev   Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.12. The maximum value of the function f(x) = 3x3 -18x+ 27x - 40 on the set S = {x ∈ R: x2 + 30 ≤ 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 ∈ R : x2 + 30 ≤ 11x}
So x2 - 11x + 30 < 0
⇒ x ∈ [5, 6]
∴ 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 Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev is:      (2019)
(1) 1
(2) 1/2
(3) 1/4
(4) None of these
Ans. (3)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.14.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev 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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
⇒ 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) 20√2
(3) 32
(4) 18√3
Ans. 
(3)
Solution.

Given, the equation of parabola is,
x2 = 12 - y
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Area of the rectangle = (2t) (12 - t2)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
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 ΔPXQ 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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Parametric equations of the parabola y2 = 4x are, x = t2 and y = 2t.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
For maximum area t = 1/2
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.17. The maximum value of Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev for any real value of θ is:     (2019)
(1) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(2) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(3) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(4) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (1)
Solution. Let, the functions is,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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 Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev then the curve also passes through the point:     (2019)
(1) (3, 0)
(2) (√3, 0)
(3) (-1, 2)
(4)  (-√2, 1)

Ans. (2)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Solution of equation
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
∴ curve passes through point (1, -2)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
⇒ C = -9/4
Then, equation of curve
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Since, above curve satisfies the point.
Hence, the curve passes through (√3, 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)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (2)
Solution.
∵ Tangent to the given curve is parallel to line 2y = 4x + 1
∴ Slope of tangent (m) = 2
Then, the equation of tangent will be of the form
y = 2x + c    ...(1)
∵ Line (1) and curve y = x2 - 5x + 5 has only one point of intersection.
∴ 2x + c = x2 - 5x + 5
x2 - 7x + (5 - c) = 0
∴ D = 49 - 4(5 - c) = 0
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev 
Hence, the equation of tangent: Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.20. If the function/given by f(x) = x3 - 3(a - 2)x2 + 3ax + 7, for some a ∈ R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation,  Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev     (2019)
(1) -7
(2) 5
(3) 7
(4) 6

Ans. (3)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
⇒ a = 5
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Now,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.21. The shortest distance between the line y = x and the curve y2 = x - 2 is:     (2019)
(1) 2
(2) 7/8
(3) 7/4√2
(4) 11/4√2
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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Let equation of tangent of parabola having slope 1 is,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Distance between the line y - x = 0 and Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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, x∈R, 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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Here at -2 & 1, f'(x) changes from negative value to positive value.
⇒ -2 & 1 are local minimum points. At 0, f'(x) changes from positive value to negative value.
⇒ 0 is the local maximum point.
Hence, S1 = {-2, 1} and S2 = {0}

Q.23. Let f : [0 : 2] → R be a twice differentiable function such that f"(x) > 0, for all x∈(0, 2). If φ(x) = (x) + f(2 - x), then φ 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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev 
But f" (x) > 0 ⇒ f'(x) is an increasing function
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Hence, φ (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) isPrevious year Questions (2016-20) - Applications of Derivatives Notes | EduRev. 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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Equation (i) passes through the centre of the circle x2 + y2 - 2x - 2y = 0, i.e., (1,1)
∴ C = 2
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.25. The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:     (2019)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (3)
Solution.
Let radius of base and height of cylinder be r and h respectively.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev     ...(1)
Now, volume of cylinder, V = πr2h
Substitute the value of r2 from equation (i),
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Differentiating w.r.t. h,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
For maxima/minima,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
⇒ Volume is maximum when h = 2√3

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

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.
⇒ 1 + a - b = - 5
⇒ a - b = -6    ...(1)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Since, required line is perpendicular to y = x - 4, then slope of tangent at the point P (1, -5) = -1
∴ 3 + a = - 1
⇒ a = -4
⇒ b = 2
the equation of the curve is y = x3 - 4x - 2
⇒ (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)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (4)
Solution.

y = f(x) = x3 - x2 - 2x
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Hence, the required set Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.29. A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev 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 π
(2) 1/10 π
(3) 2/π
(4) 1/5 π
Ans. (4)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Given that water is poured into the tank at a constant rate of 5 m3/minute.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Volume of the tank is,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev   ...(i)
where r is radius and h is height at any time. By the diagram,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev   ...(ii)
Differentiate eq. (i) w.r.t. ‘t’, we get
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Putting h = 10, r = 5 and dV/dt = 5 in the above equation.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.30. Let f(x) = ex - x and g(x) = x2 - x,  x ∈ R. Then the set of all x ∈ R, where the function h(x) = (fog) (x) is increasing, is:     (2019)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (2)
Solution.
Given functions are, f(x) = ex - x and g(x) = x2 - x
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Given f(g (x)) is increasing function.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev are either both positive or negative
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.31. If the tangent to the curve Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev at a point (α, β) is parallel to the line 2x + 6y - 11 = 0, then :     (2019)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (1)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
These values of α and β satisfies |6α + 2β| = 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) 1/18π
(2) 1/36π
(3) 5/6π
(4) 1/9π
Ans. (1)
Solution.
Given that ice melts at a rate of 50 cm3/min
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Substitute r = 5,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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

Q.34. If m is the minimum value of k for which the functionPrevious year Questions (2016-20) - Applications of Derivatives Notes | EduRev 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, 3√2)
(2) (4, 3√3)
(3) (3, 3√3)
(4) (5, 3√6)
Ans. (2)
Solution.
Given function Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Differentiating w. r. t. x,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
[∴ f(x) is increasing in [0, 3]]
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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;
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
∵ This is common tangent.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
∴ 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)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (3)
Solution.

Tangent to the curve y = (x - 2)2 - 1 at any point (h, k) is,
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.37. Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev then the local minimum value of value of h(x) is:     (2018)
(1) 3
(2) -3
(3) -2√2
(4) 2√2
Ans. 
(4)
Solution.

Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev    Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Hence, local minimum value is 2√2

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) 8√2 π
(2) 6√2 π
(3) 6√3 π
(4) 8√3 π
Ans. 
(4)
Solution.

Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
where r is radius and h is height of coin
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
For maxima or minima put f'(x) = 0
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
2r + θr =20 ... (i)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRevPrevious year Questions (2016-20) - Applications of Derivatives Notes | EduRev
A to be maximum
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Hence for r = 5, A is maximum
Now, 10 + θ·5 = 20 ⇒ θ = 2 (radian)
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

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

Ans. (2)
Solution.
f(x) = x3 – 3x2 + 5x + 7
f'(x) = 3x2 – 6x + 5 > 0 ( ∵D< 0, a>0)
⇒x ∈ ϕ

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

Ans. (3)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Let y = f(x) be a curve
slope of tangent = f'(x)
Equation of tangent (Y-y) = f'(x) (X- x)
Put Y = 0
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
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 ∠BPC = β then tan β is    (2016)
(1) 4/9
(2) 6/7
(3) 1/4
(4) 2/9
Ans.
(4)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

Q.44. If m and M are the minimum and the maximum values of Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev  then M - m is equal to    (2016)
(1) 7/4
(2) 15/4
(3) 9/4
(4) 1/4
Ans.
(3)
Solution.
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
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)Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev

(2)Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(3)Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
(4) Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev
Ans. (3)
Solution.
f(x) = sin4x + cos4x
f'(x) = 4sin3x cosx - 4cos3xsinx
= 4sinx cosx(sin2x - cos2x)
= - 2sin2x. cos2x
= - sin4x > 0
⇒ sin4x < 0
⇒ π < 4x < 2π
Previous year Questions (2016-20) - Applications of Derivatives Notes | EduRev 


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