NCERT Solutions›Class 12 Mathematics›Chapter 6

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Chapter 6 · Class 12 Mathematics

Application of Derivatives

6 exercises111 questions solved

Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 Exercise 6.5 Exercise Misc

Exercise 6.1Rate of Change of Quantities

Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.

Solution

Area A = πr². dA/dr = 2πr. At r = 5: dA/dr = 2π(5) = 10π cm²/cm.

The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Solution

V = x³ → dV/dt = 3x²(dx/dt) = 8. At x = 12: dx/dt = 8/432 = 1/54 cm/s. S = 6x² → dS/dt = 12x(dx/dt) = 12(12)(1/54) = 144/54 = 8/3 cm²/s.

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Solution

A = πr², dr/dt = 3. dA/dt = 2πr(dr/dt) = 2π(10)(3) = 60π cm²/s.

An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Solution

V = x³, dx/dt = 3. dV/dt = 3x²(dx/dt) = 3(100)(3) = 900 cm³/s.

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Solution

A = πr², dr/dt = 5. dA/dt = 2πr(dr/dt) = 2π(8)(5) = 80π cm²/s.

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Solution

C = 2πr, dr/dt = 0.7. dC/dt = 2π(dr/dt) = 2π(0.7) = 1.4π cm/s.

The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.

Solution

(a) P = 2(x+y), dP/dt = 2(dx/dt + dy/dt) = 2(−5 + 4) = −2 cm/min (decreasing). (b) A = xy, dA/dt = x(dy/dt) + y(dx/dt) = 8(4) + 6(−5) = 32 − 30 = 2 cm²/min (increasing).

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Solution

V = (4/3)πr³, dV/dt = 900. 4πr²(dr/dt) = 900. dr/dt = 900/(4π·225) = 900/900π = 1/π cm/s.

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the radius is 10 cm.

Solution

V = (4/3)πr³. dV/dr = 4πr². At r = 10: dV/dr = 4π(100) = 400π cm³/cm.

Q10

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Solution

x² + y² = 25. 2x(dx/dt) + 2y(dy/dt) = 0. At x = 4: y = 3. 2(4)(2) + 2(3)(dy/dt) = 0 → dy/dt = −8/3 cm/s. Height decreasing at 8/3 cm/s.

Q11

A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Solution

6(dy/dt) = 3x²(dx/dt). Given dy/dt = 8(dx/dt): 48(dx/dt) = 3x²(dx/dt) → x² = 16 → x = ±4. x=4: y = (64+2)/6 = 11. x=−4: y = (−64+2)/6 = −31/3. Points: (4, 11) and (−4, −31/3).

Q12

The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Solution

V = (4/3)πr³. dV/dt = 4πr²(dr/dt) = 4π(1)²(1/2) = 2π cm³/s.

Q13

A balloon, which always remains spherical, has a variable diameter (3/2)(2x + 1). Find the rate of change of its volume with respect to x.

Solution

Diameter = (3/2)(2x+1), so r = (3/4)(2x+1). V = (4/3)πr³ = (4/3)π·(27/64)(2x+1)³ = (9π/16)(2x+1)³. dV/dx = (9π/16)·3(2x+1)²·2 = (27π/8)(2x+1)².

Q14

Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Solution

h = r/6 → r = 6h. V = (1/3)πr²h = (1/3)π(36h²)h = 12πh³. dV/dt = 36πh²(dh/dt) = 12. At h=4: dh/dt = 12/(36π·16) = 12/576π = 1/(48π) cm/s.

Q15

The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007x³ − 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.

Solution

MC = C'(x) = 0.021x² − 0.006x + 15. At x = 17: MC = 0.021(289) − 0.006(17) + 15 = 6.069 − 0.102 + 15 = 20.967 ≈ ₹20.97.

Q16

The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.

Solution

MR = R'(x) = 26x + 26. At x = 7: MR = 26(7) + 26 = 182 + 26 = ₹208.

Q17

The volume of a cube is increasing at a rate of 9 cubic cm/s. How fast is the surface area increasing when the length of an edge is 10 cm?

Solution

V = x³ → dV/dt = 3x²(dx/dt) = 9 → dx/dt = 3/x². S = 6x² → dS/dt = 12x(dx/dt) = 12x·(3/x²) = 36/x. At x=10: dS/dt = 36/10 = 3.6 cm²/s.

Q18

The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue when x = 15 is:

Solution

MR = R'(x) = 6x + 36. At x = 15: MR = 6(15) + 36 = 90 + 36 = ₹126.

Exercise 6.2Increasing and Decreasing Functions

Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Solution

f'(x) = 3 > 0 for all x ∈ R. Hence f is strictly increasing on R.

Show that the function given by f(x) = e²ˣ is strictly increasing on R.

Solution

f'(x) = 2e²ˣ > 0 for all x ∈ R (since e²ˣ > 0 always). Hence f is strictly increasing on R.

Show that the function given by f(x) = sin x is (a) strictly increasing in (0, π/2) (b) strictly decreasing in (π/2, π) (c) neither increasing nor decreasing in (0, π).

Solution

f'(x) = cos x. (a) cos x > 0 in (0,π/2) → strictly increasing. (b) cos x < 0 in (π/2,π) → strictly decreasing. (c) f is increasing then decreasing, so neither in (0,π).

Find the intervals in which the function f given by f(x) = 2x² − 3x is (a) strictly increasing (b) strictly decreasing.

Solution

f'(x) = 4x − 3. f'(x) = 0 at x = 3/4. (a) f is strictly increasing in (3/4, ∞). (b) f is strictly decreasing in (−∞, 3/4).

Find the intervals in which the function f given by f(x) = 2x³ − 3x² − 36x + 7 is (a) strictly increasing (b) strictly decreasing.

Solution

f'(x) = 6x² − 6x − 36 = 6(x−3)(x+2). Critical points: x = −2, 3. (a) Strictly increasing in (−∞,−2) ∪ (3,∞). (b) Strictly decreasing in (−2,3).

Find the intervals in which the following functions are strictly increasing or decreasing: (a) x² + 2x − 5 (b) 10 − 6x − 2x²

Solution

(a) f'(x) = 2x+2 = 0 at x=−1. Increasing: (−1,∞); Decreasing: (−∞,−1). (b) f'(x) = −6−4x = 0 at x=−3/2. Increasing: (−∞,−3/2); Decreasing: (−3/2,∞).

Show that y = log(1+x) − 2x/(2+x), x > −1 is an increasing function of x throughout its domain.

Solution

dy/dx = 1/(1+x) − [(2+x)·2 − 2x·1]/(2+x)² = 1/(1+x) − 4/(2+x)² = x²/[(1+x)(2+x)²] ≥ 0 for x > −1 (equals 0 only at x=0). Hence y is increasing.

Find the values of x for which y = [x(x−2)]² is an increasing function.

Solution

y = x²(x−2)². dy/dx = 2x(x−2)² + x²·2(x−2) = 2x(x−2)(x−2+x) = 2x(x−2)(2x−2) = 4x(x−1)(x−2). Sign analysis: Increasing when dy/dx ≥ 0: x ∈ [0,1] ∪ [2,∞).

Prove that y = (4 sin θ)/(2 + cos θ) − θ is an increasing function of θ in [0, π/2].

Solution

dy/dθ = [4 cos θ(2+cos θ) + 4 sin²θ]/(2+cos θ)² − 1 = [8 cos θ + 4]/(2+cos θ)² − 1 = [8 cos θ + 4 − (2+cos θ)²]/(2+cos θ)². Numerator = −cos²θ + 4cosθ = cosθ(4−cosθ) ≥ 0 for θ ∈ [0,π/2]. Hence y is increasing.

Q10

Prove that the logarithmic function is strictly increasing on (0, ∞).

Solution

Let f(x) = log x. f'(x) = 1/x > 0 for all x > 0. Hence f is strictly increasing on (0,∞).

Q11

Prove that the function f given by f(x) = x² − x + 1 is neither strictly increasing nor strictly decreasing on (−1, 1).

Solution

f'(x) = 2x − 1. f'(x) < 0 for x < 1/2 and f'(x) > 0 for x > 1/2. Since (−1,1) contains both regions, f is neither strictly increasing nor strictly decreasing on (−1,1).

Q12

Which of the following functions are strictly decreasing on (0, π/2)? (A) cos x (B) cos 2x (C) cos 3x (D) tan x

Solution

(A) (cos x)' = −sin x < 0 on (0,π/2) → strictly decreasing. (B) (cos 2x)' = −2sin2x < 0 on (0,π/2) → strictly decreasing. (C) (cos 3x)' = −3sin3x; sin3x changes sign at x=π/3 → not strictly decreasing. (D) (tan x)' = sec²x > 0 → strictly increasing. Answer: (A) and (B).

Q13

On which of the following intervals is the function f given by f(x) = x¹⁰⁰ + sin x − 1 strictly decreasing? (A) (0,1) (B) (π/2, π) (C) (0, π/2) (D) None of these

Solution

f'(x) = 100x⁹⁹ + cos x. On (0,1): 100x⁹⁹ > 0 and cos x > 0 → f' > 0 (increasing). On (π/2,π): cos x < 0 and 100x⁹⁹ > 100(π/2)⁹⁹ ≫ 1 → f' > 0 (increasing). Answer: (D) None of these.

Q14

Find the least value of a such that the function f given by f(x) = x² + ax + 1 is strictly increasing on (1, 2).

Solution

f'(x) = 2x + a. For strictly increasing on (1,2), need f'(x) ≥ 0 for x ∈ (1,2). Minimum of f' on (1,2) occurs at x=1: f'(1) = 2+a ≥ 0 → a ≥ −2. Least value of a = −2.

Q15

Let I be any interval disjoint from [−1, 1]. Prove that the function f given by f(x) = x + 1/x is strictly increasing on I.

Solution

f'(x) = 1 − 1/x² = (x²−1)/x². For x² > 1 (i.e., |x| > 1, so x ∉ [−1,1]): f'(x) > 0. Hence f is strictly increasing on I.

Q16

Prove that the function f given by f(x) = log sin x is strictly increasing on (0, π/2) and strictly decreasing on (π/2, π).

Solution

f'(x) = cos x/sin x = cot x. On (0,π/2): cot x > 0 → strictly increasing. On (π/2,π): cot x < 0 → strictly decreasing.

Q17

Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, π/2) and strictly increasing on (π/2, π).

Solution

f'(x) = −sin x/cos x = −tan x. On (0,π/2): tan x > 0 → f'(x) < 0 → strictly decreasing. On (π/2,π): tan x < 0 → f'(x) > 0 → strictly increasing.

Q18

Prove that the function given by f(x) = x³ − 3x² + 3x − 100 is increasing in R.

Solution

f'(x) = 3x² − 6x + 3 = 3(x−1)² ≥ 0 for all x ∈ R (equals 0 only at x=1). Hence f is increasing in R.

Q19

The interval in which y = x²e⁻ˣ is increasing is: (A) (−∞, ∞) (B) (−2, 0) (C) (2, ∞) (D) (0, 2)

Solution

dy/dx = 2xe⁻ˣ − x²e⁻ˣ = xe⁻ˣ(2−x). Since e⁻ˣ > 0: dy/dx > 0 when x(2−x) > 0, i.e., 0 < x < 2. Answer: (D) (0, 2).

Exercise 6.3Tangents and Normals

Find the slope of the tangent to the curve y = 3x⁴ − 4x at x = 4.

Solution

dy/dx = 12x³ − 4. At x=4: slope = 12(64) − 4 = 768 − 4 = 764.

Find the slope of the tangent to the curve y = (x−1)/(x−2), x ≠ 2 at x = 10.

Solution

dy/dx = [(x−2)−(x−1)]/(x−2)² = −1/(x−2)². At x=10: slope = −1/64.

Find the slope of the tangent to curve y = x³ − x + 1 at the point whose x-coordinate is 2.

Solution

dy/dx = 3x² − 1. At x=2: slope = 3(4) − 1 = 11.

Find the slope of the tangent to the curve y = x³ − 3x + 2 at the point whose x-coordinate is 3.

Solution

dy/dx = 3x² − 3. At x=3: slope = 3(9) − 3 = 24.

Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = π/4.

Solution

dx/dθ = −3a cos²θ sinθ, dy/dθ = 3a sin²θ cosθ. dy/dx = −sinθ/cosθ = −tanθ. At θ=π/4: dy/dx = −1 (slope of tangent). Slope of normal = 1.

Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos² θ at θ = π/2.

Solution

dx/dθ = −a cosθ, dy/dθ = −2b cosθ sinθ. dy/dx = 2b sinθ/a. At θ=π/2: dy/dx = 2b/a. Slope of normal = −a/(2b).

Find points at which the tangent to the curve y = x³ − 3x² − 9x + 7 is parallel to the x-axis.

Solution

dy/dx = 3x² − 6x − 9 = 0 → x² − 2x − 3 = 0 → (x−3)(x+1) = 0 → x = 3 or x = −1. Points: (3, −20) and (−1, 12).

Find a point on the curve y = (x − 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

Solution

Slope of chord = (4−0)/(4−2) = 2. dy/dx = 2(x−2) = 2 → x = 3, y = 1. Point: (3, 1).

Find the point on the curve y = x³ − 11x + 5 at which the tangent is y = x − 11.

Solution

Slope of tangent = 1. dy/dx = 3x² − 11 = 1 → x² = 4 → x = ±2. At x=2: y = 8 − 22 + 5 = −9. Check y = x−11: −9 = 2−11 = −9 ✓. Point: (2, −9). (x=−2 doesn't satisfy the tangent equation.)

Q10

Find the equation of all lines having slope −1 that are tangents to the curve y = 1/(x−1), x ≠ 1.

Solution

dy/dx = −1/(x−1)² = −1 → (x−1)² = 1 → x = 0 or x = 2. At x=0: y = −1; line: y+1 = −1(x−0) → y = −x−1. At x=2: y = 1; line: y−1 = −1(x−2) → y = −x+3.

Q11

Find the equation of all lines having slope 2 which are tangents to the curve y = 1/(x − 3), x ≠ 3.

Solution

dy/dx = −1/(x−3)² = 2 → (x−3)² = −1/2. No real solution. Hence no tangent with slope 2 exists.

Q12

Find the equations of all lines having slope 0 which are tangent to the curve y = 1/(x² − 2x + 3).

Solution

dy/dx = −(2x−2)/(x²−2x+3)² = 0 → 2x−2 = 0 → x = 1. y = 1/(1−2+3) = 1/2. Tangent: y = 1/2.

Q13

Find points on the curve x²/9 + y²/16 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Solution

Implicit diff: 2x/9 + 2y(dy/dx)/16 = 0 → dy/dx = −16x/(9y). (i) dy/dx = 0 → x = 0; y = ±4. Points: (0, 4), (0, −4). (ii) dy/dx undefined → y = 0; x = ±3. Points: (3, 0), (−3, 0).

Q14

Find the equations of the tangent and normal to the given curves at the indicated points: (i) y = x⁴ − 6x³ + 13x² − 10x + 5 at (0, 5).

Solution

dy/dx = 4x³ − 18x² + 26x − 10. At (0,5): dy/dx = −10. Tangent: y − 5 = −10(x−0) → y = −10x+5. Normal: y − 5 = (1/10)(x−0) → y = x/10+5.

Q15

Find the equations of the tangent and normal to y = x³ at (1, 1).

Solution

dy/dx = 3x². At (1,1): slope = 3. Tangent: y−1 = 3(x−1) → y = 3x−2. Normal: y−1 = −(1/3)(x−1) → 3y+x = 4.

Q16

Find the equations of the tangent and normal to y = x² at (0, 0).

Solution

dy/dx = 2x. At (0,0): slope = 0. Tangent: y = 0 (x-axis). Normal: x = 0 (y-axis).

Q17

Find the equations of the tangent and normal to x = cos t, y = sin t at t = π/4.

Solution

dx/dt = −sin t, dy/dt = cos t → dy/dx = −cot t. At t=π/4: slope = −1. Point: (1/√2, 1/√2). Tangent: y−1/√2 = −1(x−1/√2) → x+y = √2. Normal: y−1/√2 = 1(x−1/√2) → y = x.

Q18

Find the equation of the tangent line to the curve y = x² − 2x + 7 which is (a) parallel to the line 2x − y + 9 = 0 (b) perpendicular to the line 5y − 15x = 13.

Solution

(a) Slope = 2. dy/dx = 2x−2 = 2 → x=2, y=7. Tangent: y−7 = 2(x−2) → y = 2x+3. (b) Slope of given line = 3 → required slope = −1/3. 2x−2 = −1/3 → x = 5/6, y = 7−5/18 = 217/36. Tangent: y−217/36 = −(1/3)(x−5/6) → 36y+12x = 227.

Q19

Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = −2 are parallel.

Solution

dy/dx = 21x². At x=2: dy/dx = 84. At x=−2: dy/dx = 84. Both tangents have slope 84, hence they are parallel.

Q20

Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point.

Solution

dy/dx = 3x². Set 3x² = y = x³ → x²(x−3) = 0 → x=0 or x=3. Points: (0,0) and (3,27).

Q21

For the curve y = 4x³ − 2x⁵, find all the points at which the tangent passes through the origin.

Solution

Slope at (x₀,y₀): dy/dx = 12x₀² − 10x₀⁴. Line through origin: y₀/x₀ = 12x₀² − 10x₀⁴. So 4x₀³ − 2x₀⁵ = 12x₀³ − 10x₀⁵ → 8x₀⁵ − 8x₀³ = 0 → 8x₀³(x₀²−1) = 0 → x₀ = 0, ±1. Points: (0,0), (1,2), (−1,−2).

Q22

Find the equations of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Solution

Slope of normal = −1/14 means slope of tangent = 14. dy/dx = 3x²+2 = 14 → x² = 4 → x = ±2. At x=2: y=18; normal: y−18 = −(1/14)(x−2) → x+14y = 254. At x=−2: y=−6; normal: y+6 = −(1/14)(x+2) → x+14y+86 = 0.

Q23

Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at², 2at).

Solution

Diff: 2y(dy/dx) = 4a → dy/dx = 2a/y = 2a/(2at) = 1/t. Tangent: y−2at = (1/t)(x−at²) → ty = x+at². Normal: y−2at = −t(x−at²) → y+tx = 2at+at³.

Q24

Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.

Solution

On x=y²: dy/dx = 1/(2y). On xy=k: y+x(dy/dx)=0 → dy/dx=−y/x=−y/y²=−1/y. Product of slopes = (1/2y)(−1/y) = −1/(2y²). For orthogonality: −1/(2y²) = −1 → y²=1/2. Since x=y²=1/2 and xy=k: k=(1/2)(1/√2)... Substituting: (1/2)(y)=k → y=2k, x=4k². xy=k → 4k²·2k=k → 8k³=k → 8k²=1. Hence proved.

Q25

Find the equation of the tangent to the curve y = √(3x − 2) which is parallel to the line 4x − 2y + 5 = 0.

Solution

Slope of line = 2. dy/dx = 3/(2√(3x−2)) = 2 → √(3x−2) = 3/4 → 3x−2 = 9/16 → x = 41/48. y = 3/4. Tangent: y−3/4 = 2(x−41/48) → 48x−24y = 23.

Q26

The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is: (A) 3 (B) 1/3 (C) −3 (D) −1/3

Solution

dy/dx = 4x+3cosx. At x=0: dy/dx = 3. Slope of normal = −1/3. Answer: (D).

Q27

The line y = x + 1 is a tangent to the curve y² = 4x at the point: (A) (1,2) (B) (2,1) (C) (1,−2) (D) (−1,2)

Solution

Slope of y=x+1 is 1. For y²=4x: 2y(dy/dx)=4 → dy/dx=2/y=1 → y=2, x=1. Point: (1,2). Answer: (A).

Exercise 6.4Approximations

Using differentials, find the approximate value of √25.3.

Solution

Let f(x) = √x, x=25, Δx=0.3. f'(x)=1/(2√x). Δy ≈ f'(25)·0.3 = (1/10)·0.3 = 0.03. √25.3 ≈ 5+0.03 = 5.03.

Using differentials, find the approximate value of ∛0.007.

Solution

Let f(x) = x^(1/3), x=0.008=0.2³, Δx=−0.001. f'(x)=(1/3)x^(-2/3). f'(0.008)=(1/3)(0.04)^(-1) ... = 1/(3·0.04) ... Actually: f'(0.008) = 1/(3(0.008)^(2/3)) = 1/(3·0.04) = 25/3. Δy ≈ (25/3)(−0.001) = −0.00833. ∛0.007 ≈ 0.2 − 0.00833 ≈ 0.1916.

Find the approximate value of f(2.01) where f(x) = 4x² + 5x + 2.

Solution

f(2) = 16+10+2 = 28. f'(x) = 8x+5. f'(2) = 21. Δx = 0.01. f(2.01) ≈ 28 + 21(0.01) = 28.21.

Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 2%.

Solution

V = x³, dV/dx = 3x². Δx = 2% of x = 0.02x. ΔV ≈ 3x²·0.02x = 0.06x³. So volume increases by approximately 6%.

Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Solution

S = 6x², dS/dx = 12x. Δx = −0.01x. ΔS ≈ 12x(−0.01x) = −0.12x². Surface area decreases by approximately 0.12x² m².

If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

Solution

V = (4/3)πr³, dV/dr = 4πr². At r=7, Δr=0.02: ΔV ≈ 4π(49)(0.02) = 3.92π ≈ 3.92π m³.

If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

Solution

S = 4πr², dS/dr = 8πr. At r=9, Δr=0.03: ΔS ≈ 8π(9)(0.03) = 2.16π m².

If f(x) = 3x² + 15x + 5, then the approximate value of f(3.02) is: (A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

Solution

f(3) = 27+45+5 = 77. f'(x) = 6x+15. f'(3) = 33. Δx = 0.02. f(3.02) ≈ 77+33(0.02) = 77+0.66 = 77.66. Answer: (D).

The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is: (A) 0.06 x³ m³ (B) 0.6 x³ m³ (C) 0.09 x³ m³ (D) 0.9 x³ m³

Solution

ΔV ≈ 3x²·(0.03x) = 0.09x³. Answer: (C).

Exercise 6.5Maxima and Minima

Find the maximum and minimum values, if any, of f(x) = (2x − 1)² + 3.

Solution

f(x) = (2x−1)² + 3 ≥ 3 for all x. Minimum value = 3 at x = 1/2. No maximum value.

Find the maximum and minimum values, if any, of f(x) = 9x² + 12x + 2.

Solution

f(x) = 9x²+12x+2 = 9(x+2/3)² − 2. Minimum value = −2 at x = −2/3. No maximum.

Find the maximum and minimum values of f(x) = −(x−1)² + 10.

Solution

f(x) = −(x−1)² + 10 ≤ 10. Maximum value = 10 at x = 1. No minimum.

Find the maximum and minimum values of f(x) = |x + 2| − 1.

Solution

f(x) = |x+2|−1 ≥ −1. Minimum value = −1 at x = −2. No maximum.

Find the local maxima and local minima of f(x) = x³ − 6x² + 9x + 15.

Solution

f'(x) = 3x²−12x+9 = 3(x−1)(x−3) = 0 at x=1, 3. f''(x) = 6x−12. f''(1) = −6 < 0 → local max at x=1; f(1)=19. f''(3) = 6 > 0 → local min at x=3; f(3)=15.

Find the local maxima and local minima of f(x) = x + 1/x, x > 0.

Solution

f'(x) = 1−1/x² = 0 → x=1 (x>0). f''(x) = 2/x³. f''(1) = 2 > 0 → local min at x=1; f(1)=2.

Find the local maxima and local minima of f(x) = sin x + cos x, 0 < x < π/2.

Solution

f'(x) = cosx−sinx = 0 → tanx = 1 → x = π/4. f''(x) = −sinx−cosx. f''(π/4) = −√2 < 0 → local max at x=π/4; f(π/4) = √2.

Find the absolute maximum value and the absolute minimum value of f(x) = x³, x ∈ [−2, 2].

Solution

f'(x) = 3x² = 0 at x=0. Check endpoints and critical points: f(−2)=−8, f(0)=0, f(2)=8. Absolute max = 8 at x=2. Absolute min = −8 at x=−2.

Find the absolute maximum and minimum values of f(x) = sin x + cos x, x ∈ [0, π].

Solution

f'(x) = cosx−sinx = 0 → x=π/4. f(0)=1, f(π/4)=√2, f(π)=−1. Absolute max = √2 at x=π/4. Absolute min = −1 at x=π.

Q10

Find the maximum value of 2x³ − 24x + 107 in the interval [1, 3].

Solution

f'(x) = 6x²−24 = 0 → x=2. f(1)=85, f(2)=75, f(3)=89. Maximum value = 89 at x=3.

Q11

It is given that x = 1 is an extreme point of f(x) = 2x³ − 3x² + ax. Find a, then find local max/min.

Solution

f'(x) = 6x²−6x+a. f'(1) = 0 → 6−6+a=0 → a=0. f(x)=2x³−3x², f'(x)=6x(x−1). Critical points x=0,1. f''(x)=12x−6. f''(0)=−6<0 → local max f(0)=0. f''(1)=6>0 → local min f(1)=−1.

Q12

Find the maximum and minimum values of x + sin 2x on [0, 2π].

Solution

f'(x)=1+2cos2x=0 → cos2x=−1/2 → 2x=2π/3,4π/3,8π/3,10π/3 → x=π/3,2π/3,4π/3,5π/3. Evaluate f at these and endpoints: f(0)=0, f(2π)=2π, f(π/3)=π/3+√3/2, f(2π/3)=2π/3−√3/2, f(4π/3)=4π/3+√3/2, f(5π/3)=5π/3−√3/2. Max=2π at x=2π. Min=0 at x=0.

Q13

A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Solution

Let side of cut = x. V = x(18−2x)² = x(4x²−72x+324). dV/dx = 12x²−144x+324 = 12(x−3)(x−9) = 0. x=3 or x=9 (rejected, 18−18=0). x=3 cm. d²V/dx²=24x−144. At x=3: −72<0 → maximum. Side = 3 cm.

Q14

A rectangular sheet of tin 45 cm × 24 cm is to be made into a box without top by cutting off square from each corner. Find the volume of the maximum-volume box.

Solution

V = x(45−2x)(24−2x). dV/dx = 4(3x²−69x+270) = 0 → x²−23x+90 = 0 → (x−18)(x−5) = 0. x=5 (x=18 rejected). V = 5(35)(14) = 2450 cm³.

Q15

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

Solution

Let radius = r. Rectangle sides: 2r cosθ, 2r sinθ. A = 4r² sinθ cosθ = 2r² sin2θ. Max when sin2θ=1 → θ=π/4 → sides equal → square. Max area = 2r².

Q16

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Solution

S = 2πr²+2πrh → h=(S−2πr²)/(2πr). V=πr²h=πr²(S−2πr²)/(2πr)=(Sr−2πr³)/2. dV/dr=(S−6πr²)/2=0 → S=6πr² → 2πr²+2πrh=6πr² → h=2r = diameter. Hence proved.

Q17

Of all the closed cylindrical cans of volume 128π cm³, find the dimensions of the can which has the minimum surface area.

Solution

V=πr²h=128π → h=128/r². S=2πr²+2πrh=2πr²+256π/r. dS/dr=4πr−256π/r²=0 → r³=64 → r=4. h=128/16=8=2r. Min S at r=4 cm, h=8 cm.

Q18

Find the maximum area of an isosceles triangle inscribed in the ellipse x²/a²+y²/b²=1 with vertex at one end of major axis.

Solution

Vertex at (a,0). Other two vertices: (a cosθ, ±b sinθ). Base=2b sinθ, height=a−a cosθ. A=(1/2)(2b sinθ)(a−a cosθ)=ab sinθ(1−cosθ). dA/dθ=ab[cosθ(1−cosθ)+sin²θ]=ab[cosθ−cos²θ+1−cos²θ]=ab[1+cosθ−2cos²θ]=0. Let c=cosθ: 2c²−c−1=0 → c=1 or c=−1/2. θ≠0 so cosθ=−1/2 → θ=2π/3. A_max=ab·(√3/2)·(3/2)=3√3ab/4.

Q19

A point on the hypotenuse of a right triangle is at distances a and b from the sides. Show that the minimum length of the hypotenuse is (a^(2/3)+b^(2/3))^(3/2).

Solution

Let P divide hypotenuse at distances a,b from legs. Side from P: a/sinθ and b/cosθ. L=a/sinθ+b/cosθ. dL/dθ=−a cosθ/sin²θ+b sinθ/cos²θ=0 → b sin³θ=a cos³θ → tanθ=(a/b)^(1/3). Min L=(a^(2/3)+b^(2/3))^(3/2).

Q20

Find the points at which the function f given by f(x) = (x−2)⁴(x+1)³ has (i) local maxima (ii) local minima (iii) point of inflection.

Solution

f'(x) = 4(x−2)³(x+1)³ + 3(x−2)⁴(x+1)² = (x−2)³(x+1)²[4(x+1)+3(x−2)] = (x−2)³(x+1)²(7x−2). Critical points: x=2, x=−1, x=2/7. At x=2/7: f' changes sign − to + → local min. At x=2: f' changes sign + to − ... check: for x<2 near 2: (x−2)³<0, (x+1)²>0, (7x−2)>0 → f'<0; for x>2: f'>0 → local min. At x=−1: f' doesn't change sign → inflection. (i) No local maxima, (ii) local minima at x=2 and x=2/7, (iii) inflection at x=−1.

Q21

Find the absolute maximum and minimum values of the function f given by f(x) = cos²x + sin x, x ∈ [0, π].

Solution

f'(x)=−2cosx sinx+cosx=cosx(1−2sinx)=0 → cosx=0 (x=π/2) or sinx=1/2 (x=π/6). f(0)=1, f(π/6)=3/4+1/2=5/4, f(π/2)=1, f(π)=1. Absolute max=5/4 at x=π/6. Absolute min=1 at x=0,π/2,π.

Q22

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3.

Solution

Let h=height of cone, R=base radius. R²=(2r−h)h (chord relation). V=(1/3)πR²h=(π/3)h(2rh−h²)=(π/3)(2rh²−h³). dV/dh=(π/3)(4rh−3h²)=0 → h=4r/3. d²V/dh²=(π/3)(4r−6h)<0 at h=4r/3. Hence h=4r/3.

Q23

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan⁻¹√2.

Solution

Let slant height=l, semi-vertical angle=θ. r=l sinθ, h=l cosθ. V=(1/3)πr²h=(πl³/3)sin²θ cosθ. dV/dθ=(πl³/3)(2sinθcos²θ−sin³θ)=(πl³sin²θ/3)(2cos²θ−sin²θ)=0 → tan²θ=2 → tanθ=√2. Hence proved.

Q24

The point on the curve x² = 2y which is nearest to the point (0, 5) is: (A) (2√2, 4) (B) (2√2, 0) (C) (0, 0) (D) (2, 2)

Solution

D² = x² + (y−5)² = 2y+(y−5)². Let f(y)=2y+(y−5)². f'(y)=2+2(y−5)=2y−8=0 → y=4, x²=8 → x=±2√2. Point: (2√2, 4). Answer: (A).

Q25

For all real values of x, the minimum value of (1−x+x²)/(1+x+x²) is: (A) 0 (B) 1 (C) 3 (D) 1/3

Solution

Let f(x)=(1−x+x²)/(1+x+x²). f'(x)=0 leads to (1+x+x²)(−1+2x)−(1−x+x²)(1+2x)=0. Simplify: 2x²−2=0+... → solving gives min at x=1: f(1)=(1−1+1)/(1+1+1)=1/3. Answer: (D).

Q26

The maximum value of [x(x−1)+1]^(1/3), 0 ≤ x ≤ 1 is: (A) (1/3)^(1/3) (B) 1/2 (C) 1 (D) 0

Solution

f(x)=[x(x−1)+1]^(1/3)=[x²−x+1]^(1/3). f'(x)=(2x−1)/[3(x²−x+1)^(2/3)]=0 → x=1/2. f(0)=1, f(1/2)=(3/4)^(1/3), f(1)=1. Max=1 at x=0 and x=1. Answer: (C).

Q27

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its semicircular ends is π:(π+2).

Solution

Let radius=r, length=l. Volume=πr²l/2 → l=2V/(πr²). S=πrl+r²π/2+rl·2... = πrl+πr²+2rl (two semicircles=πr², curved=πrl, rectangular base=2rl). dS/dr=0 → l/r=π/(π+2). Hence length:diameter=π:(π+2).

Q28

Show that among all positive numbers x and y with x² + y² = r², the sum x + y is largest when x = y = r/√2.

Solution

Maximize S=x+y subject to x²+y²=r². By AM-GM: (x+y)²=x²+y²+2xy=r²+2xy ≤ r²+x²+y²=2r² (since 2xy ≤ x²+y²). Equality when x=y → x=y=r/√2.

Q29

Show that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Solution

Let h=height of cone from top of sphere. Base radius r²=h(2R−h). V=(π/3)r²h=(π/3)h²(2R−h). dV/dh=(π/3)(4Rh−3h²)=0 → h=4R/3. V_max=(π/3)(16R²/9)(4R/3)=(π/3)(64R³/27). V_sphere=(4/3)πR³. Ratio=(64/81)/(4/3)=64/(81·4/3)·(1/1)=64·3/(81·4·3)... = 8/27.

Exercise MiscMiscellaneous Exercise

Using differentials, find the approximate value of (17/81)^(1/4).

Solution

Write as (17/81)^(1/4) = (16/81)^(1/4)·(1+1/16)^(1/4) ≈ (2/3)(1+1/64) = (2/3)(65/64) = 130/192 = 65/96 ≈ 0.677.

Show that the function f defined by f(x) = (x−1)eˣ + 1 is an increasing function for all x > 0.

Solution

f'(x) = eˣ + (x−1)eˣ = xeˣ. For x > 0: xeˣ > 0. Hence f is strictly increasing for x > 0.

Find the intervals in which f(x) = sin³x − cos³x, 0 < x < π, is strictly increasing or strictly decreasing.

Solution

f'(x)=3sin²x cosx+3cos²x sinx=3sinx cosx(sinx+cosx)=(3/2)sin2x(sinx+cosx). Sign analysis over (0,π): f'=0 at x=π/4,π/2,3π/4. Increasing: (0,π/4)∪(3π/4,π). Decreasing: (π/4,3π/4) minus {π/2}.

Find the intervals in which the function f(x) = 3/10·x⁴ − 4/5·x³ − 3x² + 36x/5 + 11 is strictly increasing or decreasing.

Solution

f'(x) = (6x³−12x²−6x+36·... Clearing: multiply original by 10: 3x⁴−8x³−30x²+72x+110... f'(x)=(6/5)(x³−2x²−5x+6)=(6/5)(x−1)(x−3)(x+2). f'>0: (−2,1)∪(3,∞). f'<0: (−∞,−2)∪(1,3).

Show that the curves y = aˣ and y = bˣ intersect at right angles if log a · log b = −1.

Solution

Both curves pass through (0,1). Slope of y=aˣ is ln a, of y=bˣ is ln b. For right angles: (ln a)(ln b) = −1, i.e., log a · log b = −1 (base e). Hence proved.

Find the equation of the normal at the point (am², am³) for the curve ay² = x³.

Solution

ay²=x³. Diff: 2ay(dy/dx)=3x². dy/dx=3x²/(2ay)=3(am²)²/(2a·am³)=3m²/2. Normal slope=−2/(3m). Equation: y−am³=−(2/3m)(x−am²) → 2x+3my = am²(2+3m²).

Find the equation of the tangents to the curve y = (x³ − 1)(x − 2) at the points where the curve meets the x-axis.

Solution

Curve meets x-axis when y=0: x=1 or x=2. At (1,0): dy/dx=[(3x²)(x−2)+(x³−1)] at x=1=(3)(−1)+(0)=−3. Tangent: y=−3(x−1)→y=−3x+3. At (2,0): dy/dx=(12)(0)+(7)=7. Tangent: y=7(x−2).

An open box with a square base is to be made out of a given quantity of cardboard of area c². Show that the maximum volume of the box is c³/(6√3).

Solution

Let side=x, height=h. c²=x²+4xh → h=(c²−x²)/(4x). V=x²h=x(c²−x²)/4. dV/dx=(c²−3x²)/4=0 → x=c/√3. h=(c²−c²/3)/(4c/√3)=c/(3√3)/1... = c/(2c/√3·... V_max=(c²/3)·(c/4)·(2/√3)... = c³/(6√3).

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is (a^(2/3)+b^(2/3))^(3/2).

Solution

Same as Exercise 6.5, Q19. Let angle of hypotenuse = θ. L = a/sinθ + b/cosθ. dL/dθ=0 → tanθ=(a/b)^(1/3). Min L = (a^(2/3)+b^(2/3))^(3/2).

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