Q: Question 7: The normal to the curve x2 + 2xy – 3y2 = 0 at (1, 1) (a) meets the curve again in the second quadrant (b) does not meet the curve again (c) meets the curve again in the third quadr

(d) meets the curve again in fourth quadrant

Q: Question 8: The equation of the tangent to the curve y = 1 – at the point of intersection with Y-axis is (a) x + 2y = 0 (b) 2x + y = 0 (c) x – y = 2 (d) x + y = 2

(a) x + 2y = 0 Hint: The point of intersection of the curve with the Y-axis is the origin (0, 0).

Q: Question 9: If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, then P is (a) (4, 4) (b) (-1, 2) (c) (3, 6) (d)

(d) Hint: y2 = x(2 – x)2 = x(4 – 4x + x2) = x3 – 4x2 + 4x = slope of the tangent at (1, 1) ∴ equation of the tangent at (1, 1) is y – 1 = – (x – 1) ∴ 2y – 2 = -x + 1 ∴ x + 2y = 3 Only the coordinates satisfy both the equations y2 = x(2 – x)2 and x + 2y = 3 ∴ P is

Q: Question 1: If the curves ax2 + by2 = 1 and a’x2 + b’y2 = 1, intersect orthogonally, then prove that

Let P(x1, y1) be the point of intersection of the curves.

Q: Question 2: Determine the area of the triangle formed by the tangent to the graph of the function y = 3 – x2 drawn at the point (1, 2) and the coordinate axes.

y = 3 – x2 = slope of the tangent at (1, 2) ∴ equation of the tangent at (1, 2) is y – 2= -2(x – 1) ⇒ y – 2= -2x + 2 ⇒ 2x + y = 4 Let this tangent cuts the coordinate axes at A(a, 0) and B(0, b). ∴ 2a + 0 = 4 and 2(0) + b = 4 ∴ a = 2 and b = 4 ∴ area of required triangle = × l(OA) × l(OB) = ab = (2)

Q: Question 3: Find the equation of the tangent and normal drawn to the curve y4 – 4x4 – 6xy = 0 at the point M (1, 2).

y4 – 4x4 – 6xy = 0 Differentiating both sides w.r.t. x, we get = slope of the tangent at (1, 2) ∴ the equation of normal at M (1, 2) is y – 2 = (x – 1) ∴ 13y – 26 = 14x – 14 ∴ 14x – 13y + 12 = 0 The slope of normal at (1, 2) ∴ the equation of normal at M (1, 2) is y – 2 = (x – 1) 14y – 28 = -13x + 1

Q: Question 5: Find all points on the ellipse 9x2 + 16y2 = 400, at which the y-coordinate is decreasing and the x-coordinate is increasing at the same rate.

Let P(x1, y1) be the point on the ellipse 9x2 + 16y2 = 400 whose y-coordinate decreasing x-coordinate is increasing at the same rate.

Q: Question 6: Verify Rolle’s theorem for the function f(x) = on [-1, 1].

The functions ex, e-x, and 2 are continuous and differentiable in their respective domains. ∴ f(x) = is continuous on [-1, 1] and differentiable on (-1, 1), because ex + e-x ≠ 0 for all x ∈ [-1, 1]. Now, f(-1) = and f(1) = ∴ f(-1) = f(1) Thus, the function f satisfies all the conditions of the Rolle

Q: Question 7: The position of a particle is given by the function s(t) = 2t2 + 3t – 4. Find the time t = c in the interval 0 ≤ f ≤ 4 when the instantaneous velocity of the particle is equal to i

s(t) = 2t2 + 3t – 4 ∴ s(0) = 2(0)2 + 3(0) – 4 = -4 and s(4) = 2(4)2 + 3(4) – 4 = 32 + 12 – 4 = 40 ∴ average velocity = = = 11 Also, instantaneous velocity = = (2t2 + 3t – 4) = 2 × 2t + 3 × 1 – 0 = 4t + 3 ∴ instantaneous velocity at t = c is = 4c + 3 When instantaneous velocity at t = c equal to its

Question 1

Question 1: If the function f(x) = ax3 + bx2 + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] and f'(2 + ) = 0, then values of a and b are respectively. (a) 1, -6 (b) -2, 1 (c) -1,

Accepted Answer

(a) 1, -6

Question 2

Question 2: If f(x) = , for every real x, then the minimum value of f is (a) 1 (b) 0 (c) -1 (d) 2

Accepted Answer

(c) -1

Question 3

Question 3: A ladder 5 m in length is resting against a vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. The length of the higher

Accepted Answer

(b) 2

Question 4

Question 4: Let f(x) and g(x) be differentiable for 0 < x < 1 such that f(0) = 0, g(0) = 0, f(1) = 6. Let there exist a real number c in (0, 1) such that f'(c) = 2g'(c), then the value o

Accepted Answer

(b) 3

Question 5

Question 5: Let f(x) = x3 – 6x2 + 9x + 18, then f(x) is strictly decreasing in (a) (-∞, 1) (b) [3, ∞) (c) (-∞, 1] ∪ [3, ∞) (d) (1, 3)

Accepted Answer

(d) (1, 3)

Question 6

Question 6: If x = -1 and x = 2 are the extreme points of y = α log x + βx2 + x, then (a) α = -6, β = (b) α = -6, β = (c) α = 2, β = (d) α = 2, β =

Accepted Answer

(c) α = 2, β =

Question 7

Question 7: The normal to the curve x2 + 2xy – 3y2 = 0 at (1, 1) (a) meets the curve again in the second quadrant (b) does not meet the curve again (c) meets the curve again in the third quadr

Accepted Answer

(d) meets the curve again in fourth quadrant

Question 8

Question 8: The equation of the tangent to the curve y = 1 – at the point of intersection with Y-axis is (a) x + 2y = 0 (b) 2x + y = 0 (c) x – y = 2 (d) x + y = 2

Accepted Answer

(a) x + 2y = 0 Hint: The point of intersection of the curve with the Y-axis is the origin (0, 0).

Question 9

Question 9: If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, then P is (a) (4, 4) (b) (-1, 2) (c) (3, 6) (d)

Accepted Answer

(d) Hint: y2 = x(2 – x)2 = x(4 – 4x + x2) = x3 – 4x2 + 4x = slope of the tangent at (1, 1) ∴ equation of the tangent at (1, 1) is y – 1 = – (x – 1) ∴ 2y – 2 = -x + 1 ∴ x + 2y = 3 Only the coordinates satisfy both the equations y2 = x(2 – x)2 and x + 2y = 3 ∴ P is

Question 10

Question 10: The approximate value of tan (44° 30′), given that 1° = 0.0175, is (a) 0.8952 (b) 0.9528 (c) 0.9285 (d) 0.9825

Accepted Answer

(d) 0.9825

Question 11

Question 1: If the curves ax2 + by2 = 1 and a’x2 + b’y2 = 1, intersect orthogonally, then prove that

Accepted Answer

Let P(x1, y1) be the point of intersection of the curves.

Question 12

Question 2: Determine the area of the triangle formed by the tangent to the graph of the function y = 3 – x2 drawn at the point (1, 2) and the coordinate axes.

Accepted Answer

y = 3 – x2 = slope of the tangent at (1, 2) ∴ equation of the tangent at (1, 2) is y – 2= -2(x – 1) ⇒ y – 2= -2x + 2 ⇒ 2x + y = 4 Let this tangent cuts the coordinate axes at A(a, 0) and B(0, b). ∴ 2a + 0 = 4 and 2(0) + b = 4 ∴ a = 2 and b = 4 ∴ area of required triangle = × l(OA) × l(OB) = ab = (2)

Question 13

Question 3: Find the equation of the tangent and normal drawn to the curve y4 – 4x4 – 6xy = 0 at the point M (1, 2).

Accepted Answer

y4 – 4x4 – 6xy = 0 Differentiating both sides w.r.t. x, we get = slope of the tangent at (1, 2) ∴ the equation of normal at M (1, 2) is y – 2 = (x – 1) ∴ 13y – 26 = 14x – 14 ∴ 14x – 13y + 12 = 0 The slope of normal at (1, 2) ∴ the equation of normal at M (1, 2) is y – 2 = (x – 1) 14y – 28 = -13x + 1

Question 14

Question 4: A water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of ch

Accepted Answer

Let r be the radius of the base, h be the height and V be the volume of the water level at any time t. Since, the height of the cone is 8 feet and the radius is 4 feet, r = ……..(1) Hence, the rate of change of water level is ft/sec.

Question 15

Question 5: Find all points on the ellipse 9x2 + 16y2 = 400, at which the y-coordinate is decreasing and the x-coordinate is increasing at the same rate.

Accepted Answer

Let P(x1, y1) be the point on the ellipse 9x2 + 16y2 = 400 whose y-coordinate decreasing x-coordinate is increasing at the same rate.

Question 16

Question 6: Verify Rolle’s theorem for the function f(x) = on [-1, 1].

Accepted Answer

The functions ex, e-x, and 2 are continuous and differentiable in their respective domains. ∴ f(x) = is continuous on [-1, 1] and differentiable on (-1, 1), because ex + e-x ≠ 0 for all x ∈ [-1, 1]. Now, f(-1) = and f(1) = ∴ f(-1) = f(1) Thus, the function f satisfies all the conditions of the Rolle

Question 17

Question 7: The position of a particle is given by the function s(t) = 2t2 + 3t – 4. Find the time t = c in the interval 0 ≤ f ≤ 4 when the instantaneous velocity of the particle is equal to i

Accepted Answer

s(t) = 2t2 + 3t – 4 ∴ s(0) = 2(0)2 + 3(0) – 4 = -4 and s(4) = 2(4)2 + 3(4) – 4 = 32 + 12 – 4 = 40 ∴ average velocity = = = 11 Also, instantaneous velocity = = (2t2 + 3t – 4) = 2 × 2t + 3 × 1 – 0 = 4t + 3 ∴ instantaneous velocity at t = c is = 4c + 3 When instantaneous velocity at t = c equal to its

Question 18

Question 8: Find the approximate value of the function f(x) = at x = 1.02.

Accepted Answer

f(x) =

Question 19

Question 9: Find the approximate value of cos-1(0.51), given π = 3.1416, = 1.1547.

Accepted Answer

Let f(x) = cos-1 x The formula for approximation is f(a + h)= f(a) + h. f'(a) ∴ cos-1 (0.51) = f(0.51) = f(0.5 + 0.01) = f(0.5) + (0.01) f'(0.5) = + 0.01 × (-1.1547) = – 0.011547 = 1.0472 – 0.011547 = 1.035653 ∴ cos-1 (0.51) = 1.035653.

Question 20

Question 10: Find the intervals on which the function y = xx, (x > 0) is increasing and decreasing.

Accepted Answer

y = xx ∴ log y = log xx = x log x Differentiating both sides w.r.t. x, we get y is increasing if ≥ 0 i.e. if xx (1 + log x) ≥ 0 i.e. if 1 + log x ≥ 0 ……[∵ x > 0] i.e. if log x ≥ -1 i.e. if log x ≥ -log e …….[∵ log e = 1] i.e. if logx ≥ log i.e. if x ≥ ∴ y is increasing in y is decreasing if ≤ 0 i

Question 21

Question 11: Find the intervals on which the function f(x) = is increasing and decreasing.

Accepted Answer

f(x) = f is increasing if f'(x) ≥ 0 i.e. if ≥ 0 i.e. if log x – 1 ≥ 0 ……..[∵ (log x)2 > 0] i.e. if log x ≥ 1 i.e. if log x ≥ log e ………[∵ log e = 1] i.e. if x ≥ e ∴ f is increasing on [e, ∞) f is decreasing if f'(x) ≤ 0 i.e. if ≤ 0 i.e. if log x – 1 ≤ 0 ……..[∵ (log x)2 > 0] i.e. if log x ≤ 1 i.

Question 22

Question 12: An open box with a square base is to be made out of the given quantity of sheet of area a2. Show that the maximum volume of the box is .

Accepted Answer

Let x be the side of square base and h be the height of the box. Then x2 + 4xh = a2 ∴ h = …….(1) Let V be the volume of the box. Then V = x2h Hence, the maximum volume of the box is cu units.

Question 23

Question 13: Show that of all rectangles inscribed in a given circle, the square has the maximum area.

Accepted Answer

Let ABCD be a rectangle inscribed in a circle of radius r. Let AB = x and BC = y. Then x2 + y2 = 4r2 …….(1) Area of rectangle = xy = ……[By (1)] Let f(x) = x2(4r2 – x2) = 4r2x2 – x4 ∴ f'(x) = (4r2x2 – x4) = 4r2 × 2x – 4x3 = 8r2x – 4x3 and f”(x) = (8r2x – 4x3) = 8r2 × 1 – 4 × 3x2 = 8r2 – 12x2 For maxi

Question 24

Question 14: Show that a closed right circular cylinder of a given surface area has maximum volume if its height equals the diameter of its base.

Accepted Answer

Let r be the radius of the base, h be the height and V be the volume of the closed right circular cylinder, whose surface area is a2 sq units (which is given). 2πrh + 2πr2 = a2 ⇒ 2πr(h + r) = a2 ⇒ h = – r ……(1) Hence, the volume of the cylinder is maximum if its height is equal to the diameter of th

Question 25

Question 15: A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is 30 m, find the dimensions so that the greatest possible amount of light may be admitted.

Accepted Answer

Let x be the length, y be the breadth of the rectangle and r be the radius of the semicircle. Then perimeter of the window = x + 2y + πr, where x = 2r This is given to be 30 m ⇒ 2r + 2y + πr = 30 ⇒ 2y = 30 – (π + 2)r ⇒ y = 15 – ……..(1) The greatest possible amount of light may be admitted if the are

Question 26

Question 16: Show that the height of a right circular cylinder of greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.

Accepted Answer

Given the right circular cone of fixed height h and semi-vertical angle a. Let R be the radius of the base and H be the height of the right circular cylinder that can be inscribed in the right circular cone. In the figure, ∠GAO = α, OG = r, OA = h, OE = R, CE = H. We have, = tan α ∴ r = h tan α ……(1

Question 27

Question 17: A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least if th

Accepted Answer

Let r be the radius of the circle and x be the length of the side of the square. Then (circumference of the circle) + (perimeter of the square) = l ∴ 2πr + 4x = l ∴ r = A = (area of the circle) + (area of the square) = πr2 + x2 This shows that the sum of the areas of circle and square is least when

Question 28

Question 18: A rectangular sheet of paper of fixed perimeter with the sides having their lengths in the ratio 8 : 15 converted into an open rectangular box by folding after removing the squares

Accepted Answer

The sides of the rectangular sheet of paper are in the ratio 8 : 15. Let the sides of the rectangular sheet of paper be 8k and 15k respectively. Let x be the side of the square which is removed from the comers of the sheet of paper. The total area of removed squares is 4x2, which is given to be 100.

Question 29

Question 19: Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is .

Accepted Answer

Let x be the radius of the base and h be the height of the cone which is inscribed in a sphere of radius r. In the figure, AD = h and CD = x = BD Since, ΔABD and ΔBDE are similar, BD2 = AD. DE = AD (AE – AD) x2 = h(2r – h) …… (1) Let V be the volume of the cone. ∴ V is maximum when h = Hence, the al

Question 30

Question 20: Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is . Also, find the maximum Volume.

Accepted Answer

Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm. Then from the figure, ∴ R2 = r2 – ………(1) Let V be the volume of the cylinder. Then V = πR2h Hence, the volume of the largest cylinder inscribed in a sphere of radius ‘r’ cm = cu cm.

Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 Solutions