Question

NCERT Solutions Class 12 Maths Chapter 6 Applications of Derivatives is designed as per the latest syllabus proposed by the CBSE. This will help students gain complete clarity of the intext questions of NCERT Solutions. Our study material covers easy to difficult concepts of NCERT Solutions Class 12 such as:

• Applications of Derivatives - there are various applications of derivatives in various fields apart from mathematics. The derivatives can be used to find out the rate of change of quantity, speed, distance, voltage, and current, to find approximate value, to find the equation of tangent which is normal to the curve, find out the value of maxima and minima of the algebraic expressions. It is also vastly used in the various fields of science, engineering, physics, etc.
• Rate of Change of Quantities – the constant change experienced by any given quantity with the respect to time is known as the rate of change of quantity. With the help of the derivative function at any given point, we can find small changes in the derivative functions at the given point with the help of the required equations.
• Increasing Functions – if the given function is an increasing function then an increment in y will simultaneously increase the value of x then the function is termed as the increasing function.
• Decreasing Functions - if the given function is an increasing function then a decrement in y will simultaneously increase the value of x then the function is termed the decreasing function.
• Tangents and Normal –tangent is a straight line that touches only one point of any curve line. There can be only one tangent at any given point on the curve. A line that is perpendicular to the tangent is called the normal to the tangent.

NCERT Solutions Class 12 Maths helps students gain complete clarity of all kinds of difficult concepts.

Answer 1

NCERT Solutions Class 12 Maths Chapter 6 Applications of Derivatives

1. Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm

Answer:

The area of a circle (A)with radius (r) is given by,

A = πr²

Now, the rate of change of the area with respect to its radius is given by,

\(\frac{dA}{dr} = \frac d {dr} {\pi r^2}= 2\pi r\)

(a) When r = 3 cm,

\(\frac{dA}{dr} = 2\pi(3) = 6\pi\)

Hence, the area of the circle is changing at the rate of 6π cm when its radius is 3 cm. 

(b) When r = 4 cm,

\(\frac{dA}{dr} = 2\pi(4) = 8\pi\)

Hence, the area of the circle is changing at the rate of 8π cm when its radius is 4 cm.

2. 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?

Answer:

Let x be the length of a side, V be the volume, and s be the surface area of the cube. Then, V = x³ and S = 6x² where x is a function of time t.

It is given that \(\frac{dV}{dt} = 8 cm^3 / s\)

Then, by using the chain rule, we have

\(\therefore 8 = \frac{dV}{dt} = \frac d{dt} (x^3) = \frac d{dx}(x^3).\frac{dx}{dt} = 3x^2. \frac{dx}{dt}\)

⇒ \(\frac{dx}{dt} = \frac{8}{3x^2}\)      ......(1)

Now,

\(\frac{dS}{dt} = \frac d {dt} (6x^2) = \frac{d}{dx}(6x^2).\frac{dx}{dt}\)    [By chain rule]

\(= 12 x.\frac{dx}{dt}= 12x.\left(\frac 8{3x^2}\right) = \frac {32}x\)

Thus, when x = 12 cm,

\(\frac{dS}{dt}= \frac{32}{12}\,cm^2/s= \frac83 cm^2/ s\)

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \(\frac 83\) cm²/s.

3. 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.

Answer:

The area of a circle (A) with radius (r) is given by, 

A = πr²

Now, the rate of change of area (A) with respect to time (t) is given by,

\(\frac{dA}{dt} = \frac{d}{dt}(\pi r^2). \frac{dr}{dt}= 2\pi r\frac{dr}{dt}\)      [By chain rule]

It is given that,

\(\frac{dr}{dt} = 3 cm /s\)

\(\therefore \frac{dA}{dt} = 2\pi r(3) = 6\pi r\)

Thus, when r = 10 cm,

\(\frac{dA}{dt} = 6 \pi(10)= 60 \pi\, cm^2 / s\)

Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm²/s.

4. 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?

Answer:

Let x be the length of a side and V be the volume of the cube. Then, 

V = x³

^{\(\therefore \frac{dV}{dt} = 3x^2.\frac{dx}{dt}\)}     [By chain rule]

It is given that,

\(\frac{dx}{dt} = 3 cm/s\)

\(\frac{dV}{dt} = 3x^2(3) = 9x^2\)

Thus, when x = 10 cm,

\(\frac{dV}{dt} = 9(10)^2 = 900 cm^3/s\)

Hence, the volume of the cube is increasing at the rate of 900 cm³/s when the edge is 10 cm long.

5. 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?

Answer:

The area of a circle (A) with radius (r) is given by A = πr².

Therefore, the rate of change of area (A) with respect to time (t) is given by

\(\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \frac{d}{dr}(\pi r^2) \frac{dr}{dt} = 2\pi r\frac{dr}{dt}\)   [By chain rule]

It is given that,

\(\frac{dr}{dt} = 5 cm/s\)

Thus, when r = 8 cm,

\(\frac{dA}{dt} = 2\pi(8)(5) = 80 \pi\)

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm²/s.

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

Answer:

The circumference of a circle (C) with radius (r) is given by C = 2πr. 

Therefore, the rate of change of circumference (C) with respect to time (t) is given by,

\(\frac{dC}{dt} = \frac{dC}{dr}.\frac{dr}{dt}\)     [By chain rule]

\(= \frac{d}{dr}(2\pi r) \frac{dr}{dt}\)

\(= 2\pi.\frac{dr}{dt}\)

It is given that

\(\frac{dr}{dt} = 0.7 cm/s\)

Hence, the rate of increase of the circumference is 2π(0.7) = 1.4π cm/s.

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

Answer:

Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:

\(\frac{dx}{dt} = -5 cm / min\) and \(\frac{dy}{dt} = 4 cm/ min\)

(a) The perimeter (P) of a rectangle is given by

P = 2(x + y)

\(\therefore \frac{dP}{dt} = 2 \left( \frac{dx}{dt} + \frac{dy}{dt}\right) = 2 (-5 +4) = -2 cm / min\)

Hence, the perimeter is decreasing at the rate of 2 cm/min. 

(b) The area (A) of a rectangle is given by, 

A = x⋅ y

\(\therefore \frac{dA}{dt} = \frac{dx}{dt}.y + x. \frac{dy}{dt} = -5y +4x\)

When x = 8 cm and y = 6 cm,

\(\frac{dA}{dt} = (-5 \times 6 +4 \times8)cm^2/ min = 2 \, cm^2/ min\)

Hence, the area of the rectangle is increasing at the rate of 2 cm²/min.

8. 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.

Answer:

The volume of a sphere (V) with radius (r) is given by,

\(V = \frac43\pi r^3\)

∴ Rate of change of volume (V) with respect to time (t) is given by,

\(\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt}\)  [By chain rule]

\(= \frac{d}{dr}\left(\frac43\pi r^3\right).\frac{dr}{dt}\)

\(= 4\pi r^2. \frac{dr}{dt}\)

It is  given that 

\(\frac{dV}{dt} = 900 cm^3/s\)

\(\therefore 900 = 4\pi r^2.\frac{dr}{dt}\)

⇒ \(\frac{dr}{dt}=\frac{900}{4\pi r^2} = \frac{225}{\pi r^2}\)

Therefore, when radius = 15 cm,

\(\frac{dr}{dt} = \frac{225}{\pi(15)^2}= \frac 1{\pi}\)

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is \(\frac 1 {\pi} cm /s\).

Answer 2

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

Answer:

The volume of a sphere (V) with radius (r) is given by

\(V = \frac 43\pi r^3\)

Rate of change of volume (V) with respect to its radius (r) is given by,

\(\frac{dV}{dr}= \frac d{dr}\left(\frac43\pi r^3\right) = \frac{4}3{\pi}(3r^2) = 4\pi r^2\)

Therefore, when radius = 10 cm,

\(\frac{dV}{dr} = 4\pi(10)^2 = 400\pi\)

Hence, the volume of the balloon is increasing at the rate of 400π cm².

10. 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?

Answer:

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall. 

Then, by Pythagoras theorem, we have: 

x² + y² = 25 [Length of the ladder = 5 m]

⇒ y = \(\sqrt{25- x^2}{}\)

Then, the rate of change of height (y) with respect to time (t) is given by,

\(\frac{dy}{dt} = \frac{-x}{\sqrt{25 -x^2}}.\frac{dx}{dt}\)

It is given that \(\frac{dx}{dt} = 2 cm / s\)

Now, when x = 4 m, we have:

\(\frac{dy}{dt} = \frac{-2 \times 4}{\sqrt{25 -4^2}} = - \frac 83\)

Hence, the height of the ladder on the wall is decreasing at the rate of \(\frac 83 cm/s\).

11. 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.

Answer:

The equation of the curve is given as:

6y = x³ + 2

The rate of change of the position of the particle with respect to time (t) is given by,

\(6\frac{dy}{dt} = 3x^2\frac{dx}{dt} +0\)

⇒ \(2\frac{dy}{dt} = x^2 \frac{dx}{dt}\)

When the y-coordinate of the particle changes 8 times as fast as the x-coordinate i.e., \(\left(\frac{dy}{dt}=8\frac{dx}{dt}\right)\), we have:

\(2\left(8\frac{dx}{dt}\right) = x^2\frac{dx}{dt}\)

⇒ \(16\frac{dx}{dt} = x^2\frac{dx}{dt}\)

⇒ \((x^2 - 16) \frac{dx}{dt} = 0\)

⇒ \(x^2 = 16\)

⇒ \(x =\pm 4\)

When

\(x =4, y = \frac{4^3 + 2}{6} = \frac{66}{6}=11\)

When

\(x =-4, y = \frac{(-4)^3 + 2}{6} =- \frac{62}{6}=-\frac{31}{3}\)

Hence, the points required on the curve are (4, 11) and \(\left(-4, \frac{-31}{3}\right)\).

12. 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?

Answer:

The air bubble is in the shape of a sphere. 

Now, the volume of an air bubble (V) with radius (r) is given by,

\(V = \frac43\pi r^3\)

The rate of change of volume (V) with respect to time (t) is given by,

\(\frac{dV}{dt} = \frac43\pi\frac d{dr}(r^3).\frac{dr}{dt}\)   [By chain rule]

\(= \frac43\pi (3r^2)\frac{dr}{dt}\)

\(= 4\pi r^2 \frac{dr}{dt}\)

It is given that \(\frac{dr}{dt} = \frac12 cm/s\)

Therefore, when r = 1 cm,

\(\frac{dV}{dt} = 4\pi(1)^2\left(\frac12\right) = 2\pi \,cm^3/s\)

Hence, the rate at which the volume of the bubble increases is 2π cm³/s.

13. A balloon, which always remains spherical, has a variable diameter \(\frac32(2x +1)\). Find the rate of change of its volume with respect to x.

Answer:

The volume of a sphere (V) with radius (r) is given by,

\(V = \frac43\pi r^3\)

It is given that:

Diameter = \(\frac32(2x +1)\)

⇒ r = \(\frac34\) \((2x +1)\)

\(\therefore V = \frac43\pi(\frac34)^3 (2x +1)^3 =\frac 9{16}\pi (2x +1)^3\)

Hence, the rate of change of volume with respect to x is as

\(\frac{dV}{dx} = \frac9{16}\pi\frac d {dx}(2x +1)^3= \frac {9}{16}\pi \times3(2x +1)^2 \times 2 = \frac{27}{8}\pi(2x +1)^2\)

14. 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?

Answer:

The volume of a cone (V) with radius (r) and height (h) is given by,

\(V = \frac13\pi r^2 h\)

It is given that,

\(h = \frac16 r\)

⇒ \(r = 6h\)

\(\therefore V = \frac13\pi(6h)^2h = 12 \pi h^3\)

The rate of change of volume with respect to time (t) is given by,

\(\frac{dV}{dt} = 12 \pi\frac{d}{dh}(h^3). \frac{dh}{dt}\)     [By chain rule]

\(= 12 \pi(3h^2)\frac{dh}{dt}\)

\(= 36 \pi h^2\frac{dh}{dt}\)

It is also given that \(\frac{dV}{dt} = 12 cm^3 / s\)

Therefore, when h = 4 cm, we have:

\(12 = 36\pi (4)^2\frac{dh}{dt}\)

⇒ \(\frac{dh}{dt} = \frac{12}{36\pi(16)} = \frac1{48\pi}\)

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of \(\frac{1}{48\pi} cm/s\).

15. 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.

Answer:

Marginal cost is the rate of change of total cost with respect to output.

∴ Marginal cost (MC) = \(\frac{dC}{dx} = 0.007(3x^2) - 0.003(2x +15)\)

\(= 0.021 x^2 - 0.006 x + 15\)

When x = 17, MC = 0.021 (172) − 0.006 (17) + 15 

= 0.021(289) − 0.006(17) + 15 

= 6.069 − 0.102 + 15 

= 20.967 

Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

16. 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.

Answer:

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

∴ Marginal Revenue (MR) = \(\frac {dR}{dx}\) = 13(2x) + 26 = 26x + 26

When x = 7, 

MR = 26(7) + 26 = 182 + 26 = 208 

Hence, the required marginal revenue is Rs 208.

Answer 3

17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is 

(A) 10π 

(B) 12π 

(C) 8π 

(D) 11π

Answer:

The area of a circle (A) with radius (r) is given by,

A = πr²

Therefore, the rate of change of the area with respect to its radius r is

\(\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r\)

∴ When r = 6 cm,

\(\frac{dA}{dr} = 2 \pi \times 6 = 12 \pi cm^2/s\)

Hence, the required rate of change of the area of a circle is 12π cm²/s. 

The correct answer is B.

18. 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 is 

(A) 116 

(B) 96 

(C) 90 

(D) 126

Answer:

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

∴ Marginal Revenue (MR) = \(\frac {dR}{dx}\) = 3(2x) + 36 = 6x + 36

∴ When x = 15, 

MR = 6(15) + 36 = 90 + 36 = 126

Hence, the required marginal revenue is Rs 126. 

The correct answer is D.

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

Answer:

Let x₁ and x₂ be any two numbers in R. 

Then, we have:

Hence, f is strictly increasing on R.

20. Show that the function given by f(x) = e^2x is strictly increasing on R.

Answer:

Let x₁ and y₂ be any two numbers in R.

Then, we have:

Hence, f is strictly increasing on R.

21. Show that the function given by f(x) = sin x is 

(a) strictly increasing in \(\left(0, \frac{\pi}{2}\right)\)

(b) strictly decreasing in \(\left(\frac{\pi}{2},\pi\right)\)

(c) neither increasing nor decreasing in (0, π)

Answer:

The given function is f(x) = sin x.

\(\therefore f'(x) = cos\, x\)

(a) Since for each \(x\in\left(0, \frac{\pi}{2}\right), cos\, x> 0\), we have \(f'(x)> 0\).

Hence, f is strictly increasing in \(\left(0, \frac{\pi}{2}\right)\).

(b) Since for each \(x\in\left(\frac{\pi}{2},\pi\right), cos \, x < 0\), we have \(f'(x)> 0\).

Hence, f is strictly decreasing in \(\left(\frac{\pi}{2},\pi\right)\).

(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).

22. Find the intervals in which the function f given by f(x) = 2x² − 3x is 

(a) strictly increasing 

(b) strictly decreasing

Answer:

The given function is f(x) = 2x² - 3x.

Now, the point \(\frac34\) divides the real line into two disjoint intervals i.e., \(\left(-\infty, \frac34\right)\) and \(\left(\frac34,\infty\right)\).

In interval \(\left(-\infty, \frac34\right), f'(x) = 4x - 3<0.\)

Hence, the given function (f) is strictly decreasing in interval \(\left(-\infty, \frac34\right)\).

In interval \(\left(\frac34,\infty\right)\), \( f'(x) = 4x - 3<0.\)

Hence, the given function (f) is strictly increasing in interval \(\left(\frac34,\infty\right)\).

23. Find the intervals in which the function f given by f(x) = 2x³ − 3x² − 36x + 7 is 

(a) strictly increasing 

(b) strictly decreasing

Answer:

The given function is f(x) = 2x³ − 3x² − 36x + 7

The points x = −2 and x = 3 divide the real line into three disjoint intervals i.e.,

\((-\infty, -2),(-2, 3)\) and \((3, \infty)\).

In intervals \((-\infty, -2) \) and \((3, \infty),f'(x)\) is positive while in interval (−2, 3), \(f'(x)\) is negative. 

Hence, the given function (f) is strictly increasing in intervals \((-\infty, -2) \) and \((3, \infty)\) , while function (f) is strictly decreasing in interval (−2, 3).

24. Find the intervals in which the following functions are strictly increasing or decreasing: 

(a) x² + 2x − 5 

(b) 10 − 6x − 2x² 

Answer:

(a) We have,

\(f(x) = x^2 + 2x - 5\)

\(\therefore f'(x) = 2x + 2\)

Now,

\(f'(x) = 0\)

⇒ x = −1 

Point x = −1 divides the real line into two disjoint intervals i.e., \((-\infty, -1) \) and \((-1, \infty)\) 

In interval \((-\infty, -1) , f'(x)= 2x + 2<0\). 

∴ f is strictly decreasing in interval \((-\infty, -1) \)

Thus, f is strictly decreasing for x < −1. 

In interval \((-1,\infty) , f'(x)= 2x + 2>0\)

∴ f is strictly increasing in interval Thus, f is strictly increasing for x > −1.

(b) We have, 

f(x) = 10 − 6x − 2x²

\(\therefore f'(x) = -6 -4x\)

Now,

\(f'(x)= 0 ⇒ x = -\frac32\)

The point \(x = -\frac32\) divides the real line into two disjoint intervals i.e., \(\left(-\infty,-\frac32\right)\) and \((-\frac32, \infty)\).

In interval \(\left(-\infty,-\frac32\right)\) i.e., when \(x< -\frac32\), f'(x) = −6 − 4x > 0.

∴ f is strictly increasing for \(x< -\frac32\)

In interval \((-\frac32, \infty)\) i.e., when , \(x> -\frac32\), f'(x) = −6 − 4x < 0.

∴ f is strictly decreasing for \(x> -\frac32\).

Answer 4

25. Show that \(y = log(1 +x) - \frac{2x}{2+x},x>-1\), is an increasing function of x throughout its domain.

Answer:

We have,

\(y = log(1 +x) - \frac{2x}{2+x},x>-1\)

Since x > −1, point x = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x < 0 and x > 0. 

When −1 < x < 0, we have:

Also, when x > 0:

Hence, function f is increasing throughout this domain.

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

Answer:

We have,

The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e., \((-\infty,0),(0, 1)(1,2) \) and \((2, \infty)\).

In intervals \((-\infty,0)\) and \((1, 2), \frac{dy}{dx}<0.\)

∴ y is strictly decreasing in intervals \((-\infty,0)\) and \((1,2) \).

However, in intervals (0, 1) and (2, ∞), \(\frac{dy}{dx}>0\).

∴ y is strictly increasing in intervals (0, 1) and (2, ∞). 

y is strictly increasing for 0 < x < 1 and x > 2.

27. Prove that \(y = \frac{4sin\theta}{(2 + cos\theta)} - \theta\) is an increasing function of θ in \(\left[0, \frac{\pi}{2}\right]\).

Answer:

We have,

Since cos θ ≠ 4, cos θ = 0.

\(cos\theta = 0\)

⇒ \(\theta = \frac{\pi}2\)

Now,

In interval \(\left(0, \frac{\pi}{2}\right)\), we have cos θ > 0. 

Also, 

4 > cos θ ⇒ 4 − cos θ > 0.

Therefore, y is strictly increasing in interval \(\left(0, \frac{\pi}{2}\right)\).

Also, the given function is continuous at x = 0 and x = \(\frac{\pi}2\).

Hence, y is increasing in interval \(\left[0, \frac{\pi}{2}\right].\)

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

Answer:

The given function is f(x) = log x.

\(\therefore f'(x) = \frac1x\)

It is clear that for x > 0, \( f'(x) = \frac1x> 0.\)

Hence, f(x) = log x is strictly increasing in interval (0, ∞).

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

Answer:

The given function is f(x) = x² − x + 1.

\(\therefore f'(x) = 2x - 1\)

Now,

\(f'(x) = 0\)

⇒ \(x = \frac12\).

The point \(\frac12\) divides the interval (−1, 1) into two disjoint intervals

i.e., \(\left(-1, \frac12\right)\) and \(\left(\frac12, 1\right)\).

Now, in interval \(\left(-1, \frac12\right)\), \( f'(x) = 2x - 1>0.\)

Therefore, f is strictly decreasing in interval \(\left(-1, \frac12\right)\). 

However, in interval \(\left(\frac12, 1\right)\), \( f'(x) = 2x - 1>0.\)

Therefore, f is strictly increasing in interval \(\left(\frac12, 1\right)\).

Hence, f is neither strictly increasing nor decreasing in interval (−1, 1).

30. On which of the following intervals is the function f given by \(f(x)= x^{100} + sin\,x - 1\) strictly decreasing?

(A) (0, 1)

(B) \(\left(\frac{\pi}2, \pi\right)\)

(C) \(\left(0, \frac{\pi}2\right)\)

(D) None of these

Answer:

We have,

\(f(x) = x^{100} + sin \, x- 1 \)

\(\therefore f'(x) = 100x^{99} + cosx\)

In interval \((0, 1), cos\, x > 0 \) and \(100x^{99}> 0\).

\(\therefore f'(x)> 0\).

Thus, function f is strictly increasing in interval (0, 1). 

In interval \(\left(\frac{\pi}2, \pi\right)\), \(cos\, x > 0 \) and \(100x^{99}> 0\).

Also,

\(100x^{99}> cos\,x\)

\(\therefore f'(x)> 0\) in \(\left(\frac{\pi}2, \pi\right)\).

Thus, function f is strictly increasing in interval \(\left(\frac{\pi}2, \pi\right)\).

In interval \(\left(0, \frac{\pi}2\right)\), \(cos\, x > 0 \) and \(100x^{99}> 0\).

\(\therefore100x^{99}+cos\,x> 0\)

⇒ \(f'(x)> 0\) on \(\left(0, \frac{\pi}2\right)\)

∴ f is strictly increasing in interval \(\left(0, \frac{\pi}2\right)\).

Hence, function f is strictly decreasing in none of the intervals. 

The correct answer is D.

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

Answer:

We have,

\(f(x) = x^2 + ax + 1\)

\(\therefore f'(x) = 2x + a\)

Now, function f will be increasing in (1, 2), if f'(x) > 0 in (1, 2).

f(x) > 0

⇒ 2x + a > 0 

⇒ 2x > −a

⇒ \(x > \frac{-a}2\)

Therefore, we have to find the least value of a such that

\(x > \frac{-a}2\), when \(x \in(1, 2).\) 

⇒ \(x > \frac{-a}2\)  (when 1 < x < 2)

Thus, the least value of a for f to be increasing on (1, 2) is given by,

\(\frac{-a}{2} = 1\)

\(\frac{-a}{2} = 1\)

⇒ a = -2

Hence, the required value of a is −2.

Answer 5

32. Let I be any interval disjoint from (−1, 1). Prove that the function f given by \(f(x) = x + \frac1x\) is strictly increasing on I.

Answer:

We have,

\(f(x) = x + \frac1x\)

\(\therefore f'(x) = x + \frac1{x^2}\) 

Now,

\(f'(x) = 0\)

⇒ \(\frac1{x^2} = 1\)

⇒ \(x = \pm1\)

The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e., \((-\infty,-1),(-1, 1)\) and \((1, \infty)\).

In interval (−1, 1), it is observed that: 

f is strictly decreasing on (-1, 1) ~ {0}. 

In intervals \((-\infty,-1)\) and \((1, \infty)\), it is observed that:

\(\therefore\)f is strictly increasing on \((-\infty,-1)\) and \((1, \infty)\). 

Hence, function f is strictly increasing in interval I disjoint from (−1, 1). 

Hence, the given result is proved.

33. Prove that the function f given by f(x) = log sin x is strictly increasing on \(\left(0, \frac{\pi}2\right)\) and strictly decreasing on \(\left(\frac{\pi}2,\pi\right)\).

Answer:

We have,

f(x) = log sin x

\(\therefore f'(x) = \frac 1{sin\,x}cos\,x = cot\,x\)

In interval \(\left(0, \frac{\pi}2\right)\), f'(x) = cot x > 0.

∴ f is strictly increasing in \(\left(0, \frac{\pi}2\right)\).

In interval \(\left(\frac{\pi}2,\pi\right)\), f'(x) = cot x < 0.

∴ f is strictly decreasing in \(\left(\frac{\pi}2,\pi\right)\).

34. Prove that the function f given by f(x) = log cos x is strictly decreasing on \(\left(0, \frac{\pi}2\right)\) and strictly increasing on \(\left(\frac{\pi}2,\pi\right)\).

Answer:

We have,

f(x) = log cos x

\(\therefore f'(x) = \frac 1{cos\,x}(-sin\,x) = -tan\,x\)

In interval \(\left(0, \frac{\pi}2\right)\), tan x > 0 ⇒ - tan x < 0.

\(\therefore f'(x) < 0 \) on \(\left(0, \frac{\pi}2\right)\)

∴ f is strictly decreasing on \(\left(0, \frac{\pi}2\right)\).

In interval \(\left(\frac{\pi}2,\pi\right)\), tan x < 0 ⇒ - tan x < 0.

\(\therefore f'(x) < 0 \) on \(\left(\frac{\pi}2,\pi\right)\)

∴ f is strictly increasing on \(\left(\frac{\pi}2,\pi\right)\).

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

Answer:

We have,

For any x∈R, (x − 1)² > 0. 

Thus, is always positive in R. 

Hence, the given function (f) is increasing in R.

36. The interval in which y = x² e^-x is increasing is

(A) \((-\infty, \infty)\)

(B) (−2, 0) 

(C) \((2, \infty)\)

(D) (0, 2)

Answer:

We have,

y = x² e^-x

\(\therefore \frac{dy}{dx} = 2xe^{-x} - x^2 e^{-x} = xe^{-x}(2 - x)\)

Now, 

\(\frac{dy}{dx} = 0\)

⇒ x = 0 and x = 2

The points x = 0 and x = 2 divide the real line into three disjoint intervals

i.e., \((-\infty, 0)\), (0, 2) and \((2, \infty)\).

In intervals \((-\infty, 0)\) and \((2, \infty)\), f'(x) < 0 as e^-x is always positive.

∴ f is decreasing on \((-\infty, 0)\) and \((2, \infty)\).

In interval (0, 2), f'(x) > 0.

∴ f is strictly increasing on (0, 2). 

Hence, f is strictly increasing in interval (0, 2). 

The correct answer is D.

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

Answer:

The given curve is y = 3x⁴ − 4x. 

Then, the slope of the tangent to the given curve at x = 4 is given by,

38. Find the slope of the tangent to the curve, \(y = \frac{x-1}{x -2}\) x ≠ 2 at x = 10.

Answer:

The given curve is \(y = \frac{x-1}{x -2}\)

Thus, the slope of the tangent at x = 10 is given by,

Hence, the slope of the tangent at x = 10 is \(\frac{-1}{64}\).

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

Answer:

The given curve is y = x³ − x + 1

\(\therefore \frac{dy}{dx} = 3x^2 - 1\)

The slope of the tangent to a curve at (x₀, y₀) is \(\left. \frac{dy}{dx}\right]_{(x_0,y_0)}\)

It is given that x₀ = 2. 

Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by,

Answer 6

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

Answer:

The given curve is y = x³ − 3x + 2

\(\therefore \frac{dy}{dx} = 3x^2 - 3\)

The slope of the tangent to a curve at (x₀, y₀) is \(\left. \frac{dy}{dx}\right]_{(x_0,y_0)}\)

Hence, the slope of the tangent at the point where the x-coordinate is 3 is given by,

41. Find the slope of the normal to the curve x = acos³θ, y = asin³θ at \(\theta = \frac{\pi}4\).

Answer:

It is given that x = acos³θ and y = asin³θ.

Therefore, the slope of the tangent at \(\theta = \frac{\pi}4\) is given by,

Hence, the slope of the normal at \(\theta = \frac{\pi}4\) is given by,

42. Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos²θ at \(\theta = \frac{\pi}2\).

Answer:

It is given that x = 1 − a sin θ and y = b cos²θ.

Therefore, the slope of the tangent at \(\theta = \frac{\pi}2\) is given by,

Hence, the slope of the normal at \(\theta = \frac{\pi}2\) is given by,

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

Answer:

The equation of the given curve is y = x³ − 3x² − 9x + 7

\(\therefore \frac{dy}{dx} = 3x^2 - 6x - 9\)

Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.

When x = 3, y = (3)³ − 3 (3)² − 9 (3) + 7 = 27 − 27 − 27 + 7 = −20.

When x = −1, y = (−1)³ − 3 (−1)² − 9 (−1) + 7 = −1 − 3 + 9 + 7 = 12. 

Hence, the points at which the tangent is parallel to the x-axis are (3, −20) and (−1, 12).

44. 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).

Answer:

If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord. 

The slope of the chord is \(\frac{4-0}{4-2}= \frac42 = 2\).

Now, the slope of the tangent to the given curve at a point (x, y) is given by, 

\(\frac{dx}{dy} = 2(x - 2)\)

Since the slope of the tangent = slope of the chord, we have:

2(x - 2) = 2

⇒ x - 2 = 1

⇒ x = 3

When x = 3, y = (3 - 2)² = 1

Hence, the required point is (3, 1).

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

Answer:

The equation of the given curve is y = x³ − 11x + 5. 

The equation of the tangent to the given curve is given as y = x − 11 (which is of the form y = mx + c). 

\(\therefore\) Slope of the tangent = 1

Now, the slope of the tangent to the given curve at the point (x, y) is given by,

\(\frac{dx}{dy} = 3x^2 - 11\)

Then, we have:

3x² - 11 = 1

⇒ 3x² = 12

⇒ x² = 4

⇒ x = \(\pm\)2

When x = 2, y = (2)³ − 11 (2) + 5 = 8 − 22 + 5 = −9. 

When x = −2, y = (−2)³ − 11 (−2) + 5 = −8 + 22 + 5 = 19. 

Hence, the required points are (2, −9) and (−2, 19).

46. Find the equation of all lines having slope −1 that are tangents to the curve

\(y = \frac 1 {x - 1} , x \ne 1 \) 

Answer:

The equation of the given curve is \(y = \frac 1 {x - 1} , x \ne 1 \)

The slope of the tangents to the given curve at any point (x, y) is given by,

\(\frac{dx}{dy} = \frac{-1}{(x - 1)^2}\)

If the slope of the tangent is −1, then we have:

\(\frac{-1}{(x - 1)^2} = -1\)

⇒ (x - 1)² = 1

⇒ x - 1 = \(\pm\)1

⇒ x = 2, 0

When x = 0, y = −1 and when x = 2, y = 1. 

Thus, there are two tangents to the given curve having slope −1. 

These are passing through the points (0, −1) and (2, 1).

\(\therefore \) The equation of the tangent through (0, −1) is given by,

y - (-1) = -1(x - 0)

⇒ y + 1 = -x

⇒ y + x + 1 = 0

∴ The equation of the tangent through (2, 1) is given by, 

y − 1 = −1 (x − 2) 

⇒ y − 1 = − x + 2 

⇒ y + x − 3 = 0 

Hence, the equations of the required lines are y + x + 1 = 0 and y + x − 3 = 0.

47. Find the equation of all lines having slope 2 which are tangents to the curve

\(y = \frac 1 {x - 3} , x \ne 3 \) 

Answer:

The equation of the given curve is \(y = \frac 1 {x - 3} , x \ne 3 \)

The slope of the tangent to the given curve at any point (x, y) is given by, 

\(\frac{dx}{dy} = \frac{-1}{(x - 3)^2}\)

If the slope of the tangent is 2, then we have:

\( \frac{-1}{(x - 3)^2} = 2\)

⇒ 2(x - 3)² = -1

⇒ (x - 3)² = \(\frac{-1}2\)

This is not possible since the L.H.S. is positive while the R.H.S. is negative.

Hence, there is no tangent to the given curve having slope 2.

Answer 7

48. Find the equations of all lines having slope 0 which are tangent to the curve

\(y = \frac 1 {x^2 - 2x + 3} \) 

Answer:

The equation of the given curve is \(y = \frac 1 {x^2 - 2x + 3} \).

The slope of the tangent to the given curve at any point (x, y) is given by,

If the slope of the tangent is 0, then we have:

When x = 1, 

\(y = \frac 1 {1 - 2 + 3} = \frac12 \) 

∴ The equation of the tangent through \((1, \frac12)\) is given by,

Hence, the equation of the required line is \(y = \frac12\).

49. 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.

Answer:

The equation of the given curve is y = x² − 2x + 7.

On differentiating with respect to x, we get:

\(\frac{dy}{dx} = 2x - 2\)

(a) The equation of the line is 2x − y + 9 = 0. 

2x − y + 9 = 0 

∴ y = 2x + 9

This is of the form y = mx + c. 

∴ Slope of the line = 2 

If a tangent is parallel to the line 2x − y + 9 = 0, then the slope of the tangent is equal to the slope of the line. 

Therefore, we have: 

2 = 2x − 2

⇒ 2x = 4

⇒ x = 2

Now, x = 2 

⇒ y = 4 − 4 + 7 = 7

Thus, the equation of the tangent passing through (2, 7) is given by,

y - 7 = 2(x - 2)

⇒ y - 2x - 3 = 0

Hence, the equation of the tangent line to the given curve (which is parallel to line 2x − y + 9 = 0) is y - 2x - 3 = 0.

(b) The equation of the line is 5y − 15x = 13. 

5y − 15x = 13 

∴ \(y = 3x + \frac{13}5\)

This is of the form y = mx + c. 

∴ Slope of the line = 3 

If a tangent is perpendicular to the line 5y − 15x = 13, then the slope of the tangent is

\(\frac{-1}{\text{slope of the line}} = \frac{-1}3\).

Thus, the equation of the tangent passing through \(\left(\frac56,\frac{217}{36}\right)\) is given by,

Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y − 15x = 13) is 36y + 12x - 227 = 0.

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

Answer:

The equation of the given curve is y = 7x³ + 11.

\(\therefore \frac{dy}{dx} = 21 x^2\)

The slope of the tangent to a curve at (x₀, y₀) is \(\left.\frac{dy}{dx}\right]_{(x_0,y_0)}\).

Therefore, the slope of the tangent at the point where x = 2 is given by,

\(\left.\frac{dy}{dx}\right]_{x =-2 }=21(2)^2 = 84 \)

It is observed that the slopes of the tangents at the points where x = 2 and x = −2 are equal. 

Hence, the two tangents are parallel.

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

Answer:

The equation of the given curve is y = x³.

\(\therefore \frac{dy}{dx} = 3x^2\)

The slope of the tangent at the point (x, y) is given by,

\(\left.\frac{dy}{dx}\right]_{(x,y)} = 3x^2\)

When the slope of the tangent is equal to the y-coordinate of the point, then y = 3x². 

Also, we have y = x³. 

∴ 3x² = x³ 

∴ x² (x − 3) = 0 

∴ x = 0, x = 3 

When x = 0, then y = 0 and when x = 3, then y = 3(3)² = 27. 

Hence, the required points are (0, 0) and (3, 27).

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

Answer:

The equation of the given curve is y = 4x³ − 2x⁵.

\(\therefore \frac{dy}{dx} = 12x^2 - 10x^4\)

Therefore, the slope of the tangent at a point (x, y) is 12x² − 10x⁴. 

The equation of the tangent at (x, y) is given by,

Y - y = (12x² - 10x⁴) (X - x)

When the tangent passes through the origin (0, 0), then X = Y = 0. 

Therefore, equation (1) reduces to:

When x = 0, y = 4(0)³ - 2(0)⁵ = 0

When x = 1, y = 4 (1)³ − 2 (1)⁵ = 2. 

When x = −1, y = 4 (−1)³ − 2 (−1)⁵ = −2. 

Hence, the required points are (0, 0), (1, 2), and (−1, −2).

53. Find the points on the curve x² + y² − 2x − 3 = 0 at which the tangents are parallel to the x-axis.

Answer:

The equation of the given curve is x² + y² − 2x − 3 = 0. 

On differentiating with respect to x, we have:

Now, the tangents are parallel to the x-axis if the slope of the tangent is 0.

But, x² + y² − 2x − 3 = 0 for x = 1. 

y² = 4 

⇒ y = ±2 

Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, −2).

Answer 8

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

Answer:

The equation of the given curve is ay² = x³. 

On differentiating with respect to x, we have:

The slope of a tangent to the curve at (x₀, y₀) is \(\left.\frac{dy}{dx}\right]_{(x_0 , y_0)}\).

The slope of the tangent to the given curve

∴ Slope of normal at (am², am³)

\(= \frac{-1}{\text{slope of the tangent at}(am^2, \,am^3)}= \frac{-2}{3m}\)

Hence, the equation of the normal at (am², am³) is given by,

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

Answer:

The equation of the given curve is y = x³ + 2x + 6. 

The slope of the tangent to the given curve at any point (x, y) is given by,

\(\frac{dy}{dx} = 3x^2 + 2\)

∴ Slope of the normal to the given curve at any point (x, y)

\(= \frac{-1}{\text{slope of the tangent at the point}(x,\,y)}\)

\(= \frac{-1}{3x^2 + 2}\)

The equation of the given line is x + 14y + 4 = 0. x + 14y + 4 = 0 

∴ (which is of the form y = mx + c)

\(y = -\frac1{14}x - \frac4{14}\)

∴ Slope of the given line = \(\frac{-1}{14}\)

If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.

When x = 2, y = 8 + 4 + 6 = 18. 

When x = −2, y = − 8 − 4 + 6 = −6.

Therefore, there are two normals to the given curve with slope \(\frac{-1}{14}\) and passing through the points (2, 18) and (−2, −6). 

Thus, the equation of the normal through (2, 18) is given by,

And, the equation of the normal through (−2, −6) is given by,

Hence, the equations of the normals to the given curve (which are parallel to the given line) are x + 14y - 254 = 0 and x + 14y + 86 = 0.

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

Answer:

The equation of the given parabola is y² = 4ax. 

On differentiating y² = 4ax with respect to x, we have:

∴ The slope of the tangent at (at², 2at) is 

Then, the equation of the tangent at (at², 2at) is given by

Now, the slope of the normal at (at², 2at)

\( \frac{-1}{\text{slope of the tangent at}(at^2, \,2at)}= -t\)

Thus, the equation of the normal at (at² , 2at) is given as:

57. Find the equations of the tangent and normal to the hyperbola \(\frac{x^2}{a^2}- \frac{y^2}{b^2} = 1\) at the point (x₀, y₀).

Answer:

Differentiating \(\frac{x^2}{a^2}- \frac{y^2}{b^2} = 1\)

Therefore, the slope of the tangent at 

Then, the equation of the tangent at (x₀, y₀)

Now, the slope of the normal at (x₀, y₀) is given by,

\( \frac{-1}{\text{slope of the tangent at}(x_0, y_0)}= \frac{-a^2y_0}{b^2x_0}\)

Hence, the equation of the normal at (x₀, y₀) is given by,

58. The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is 

(A) 3 

(B) \(\frac13\)

(C) −3 

(D) \(-\frac13\) 

Answer:

The equation of the given curve is y = 2x² + 3 sin x . 

Slope of the tangent to the given curve at x = 0 is given by,

Hence, the slope of the normal to the given curve at x = 0 is

\( \frac{-1}{\text{slope of the tangent at}\, x \,=\, 0}= \frac{-1}3\)

The correct answer is D.

59. 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)

Answer:

The equation of the given curve is y² = 4x. 

Differentiating with respect to x, we have:

\(2y \frac{dy}{dx}= 4\)

⇒ \(\frac{dy}{dx} = \frac2y\)

Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,

\(\frac{dy}{dx} = \frac2y\)

The given line is y = x + 1 (which is of the form y = mx + c) 

∴ Slope of the line = 1 The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve. 

Thus, we must have:

\( \frac2y= 1\)

⇒ y = 2

Now, 

y = x  + 1

⇒ x = y - 1

⇒ x = 2 - 1 = 1

Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2). 

The correct answer is A.

Answer 9

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

Answer:

Let x = 2 and ∆x = 0.01. 

Then, we have: 

f(2.01) = f(x + ∆x) = 4(x + ∆x)² + 5(x + ∆x) + 2 

Now, ∆y = f(x + ∆x) − f(x)

f(x + ∆x) = f(x) + ∆y

Hence, the approximate value of f (2.01) is 28.21.

61. Find the approximate value of f (5.001), where f (x) = x³ − 7x² + 15.

Answer:

Let x = 5 and ∆x = 0.001. Then, we have:

Hence, the approximate value of f (5.001) is −34.995.

62. Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.

Answer:

The volume of a cube (V) of side x is given by V = x³.

Hence, the approximate change in the volume of the cube is 0.03x³ m³.

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

Answer:

The surface area of a cube (S) of side x is given by S = 6x².

Hence, the approximate change in the surface area of the cube is 0.12x² m².

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

Answer:

Let r be the radius of the sphere and ∆r be the error in measuring the radius. 

Then, 

r = 7 m and ∆r = 0.02 m 

Now, the volume V of the sphere is given by,

Hence, the approximate error in calculating the volume is 3.92 π m³.

65. 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 in surface area.

Answer:

Let r be the radius of the sphere and ∆r be the error in measuring the radius. 

Then, r = 9 m and ∆r = 0.03 m 

Now, the surface area of the sphere (S) is given by, 

S = 4πr²

Hence, the approximate error in calculating the surface area is 2.16π m².

66. 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

Answer:

Let x = 3 and ∆x = 0.02. 

Then, we have:

Hence, the approximate value of f(3.02) is 77.66. 

The correct answer is D.

67. 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

Answer:

The volume of a cube (V) of side x is given by V = x³.

Hence, the approximate change in the volume of the cube is 0.09x³ m³ . 

The correct answer is C.

68. Find the maximum and minimum values, if any, of the following functions given by 

(i) f(x) = (2x − 1)² + 3 

(ii) f(x) = 9x² + 12x + 2 

(iii) f(x) = −(x − 1)² + 10 

(iv) g(x) = x³ + 1

Answer:

(i) The given function is f(x) = (2x − 1)² + 3. 

It can be observed that (2x − 1)² ≥ 0 for every x ∈ R. 

Therefore, f(x) = (2x − 1)² + 3 ≥ 3 for every x ∈ R. 

The minimum value of f is attained when 2x − 1 = 0.

2x − 1 = 0 

∴ \(x = \frac12\)

∴ Minimum value of f = \(f(\frac12) = (2 .\frac12 - 1) + 3 = 3\)

Hence, function f does not have a maximum value. 

(ii) The given function is f(x) = 9x² + 12x + 2 = (3x + 2)² − 2. 

It can be observed that (3x + 2)² ≥ 0 for every x ∈ R. 

Therefore, f(x) = (3x + 2)² − 2 ≥ −2 for every x ∈ R. 

The minimum value of f is attained when 3x + 2 = 0.

3x + 2 = 0 

∴ \(x = \frac{-2}3\)

∴ Minimum value of f = \(f(\frac23) = (3 (\frac{-2}3) + 2)^2 - 2 = -2\)

Hence, function f does not have a maximum value.

(iii) The given function is f(x) = − (x − 1)² + 10. 

It can be observed that (x − 1)² ≥ 0 for every x ∈ R. 

Therefore, f(x) = − (x − 1)² + 10 ≤ 10 for every x ∈ R. 

The maximum value of f is attained when (x − 1) = 0. 

(x − 1) = 0 

∴ x = 0 

∴ Maximum value of f = f(1) = − (1 − 1)² + 10 = 10 

Hence, function f does not have a minimum value. 

(iv) The given function is g(x) = x³ + 1. 

Hence, function g neither has a maximum value nor a minimum value.

Answer 10

69. Find the maximum and minimum values, if any, of the following functions given by 

(i) f(x) = |x + 2| − 1 

(ii) g(x) = − |x + 1| + 3 

(iii) h(x) = sin(2x) + 5 

(iv) f(x) = |sin 4x + 3|

(v) h(x) = x + 4, x \(\in\) (−1, 1)

Answer:

(i) f(x) = |x + 2|-1

We know that |x + 2| \(\ge\) 0 for every x ∈ R. 

Therefore, f(x) = |x + 2| -1 \(\ge\) -1 for every x ∈ R. 

The minimum value of f is attained when |x + 2| = 0.

|x + 2| = 0

⇒ x = -2

∴ Minimum value of f = f(−2) = |-2 + 2| -1 = -1

Hence, function f does not have a maximum value.

(ii) g(x) = -|x + 1| + 3

We know that -|x + 1| \(\le \) 0 for every x ∈ R. 

Therefore, g(x) = - |x + 1| + 3 \(\le \) 3 for every x ∈ R. 

The maximum value of g is attained when |x + 1| = 0

|x + 1| = 0

⇒ x = -1

∴ Maximum value of g = g(−1) = - |-1 + 1| + 3 = 3

Hence, function g does not have a minimum value.

(iii) h(x) = sin2x + 5 

We know that − 1 ≤ sin 2x ≤ 1. 

∴ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 

∴ 4 ≤ sin 2x + 5 ≤ 6 

Hence, the maximum and minimum values of h are 6 and 4 respectively.

(iv) f(x) = |sin 4x + 3| 

We know that −1 ≤ sin 4x ≤ 1. 

∴ 2 ≤ sin 4x + 3 ≤ 4 

∴ 2 ≤ |sin 4x + 3| ≤ 4 

Hence, the maximum and minimum values of f are 4 and 2 respectively.

(v) h(x) = x + 1, x \(\in\) (−1, 1)

Here, if a point x₀ is closest to −1, then we find \(\frac{x_0}2 + 1 < x_0 + 1\) for all x₀ ∈ (−1, 1). 

Also, if x₁ is closest to 1, then \(x_1 +1 < \frac{x_1 + 1}{2}\) for all x ∈ (−1, 1).

Hence, function h(x) has neither maximum nor minimum value in (−1, 1).

70. Prove that the following functions do not have maxima or minima: 

(i) f(x) = e^x 

(ii) g(x) = logx 

(iii) h(x) = x³ + x² + x + 1

Answer:

(i). We have, 

f(x) = e^x 

Now, if f'(x) = 0, then e^x. But, the exponential function can never assume 0 for any value of x. 

Therefore, there does not exist c

∴ R such that f'(c) = 0.

Hence, function f does not have maxima or minima.

(ii). We have, g(x) = log x 

\(\therefore g'(x) = \frac1x\)

Since log x is defined for a positive number x, g'(x) > 0 for any x.

Therefore, there does not exist c

∴ R such that \(g'(c) = 0\)

Hence, function g does not have maxima or minima.

(iii). We have, h(x) = x³ + x² + x + 1 

∴ h'(x) = 3x² + 2x + 1

Now, 

h(x) = 0 

∴ 3x² + 2x + 1 = 0 

∴ \(x = \frac{-2\pm2\sqrt2i}{6} = \frac{-1+\sqrt 2i}{3}\notin R\)

Therefore, there does not exist c

∴ R such that. \(h'(c) = 0\)

Hence, function h does not have maxima or minima.

71. Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 − 24x − 18x².

Answer:

The profit function is given as p(x) = 41 − 24x − 18x².

By second derivative test, \(x = -\frac23\) is the point of local maxima of p.

∴ Maximum profit = \(p\left(-\frac23\right)\)

Hence, the maximum profit that the company can make is 49 units.

72. At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

Answer:

Let f(x) = sin 2x.

∴ f'(x) = 2cos 2x

Now,

Then, we evaluate the values of f at critical points \(x = \frac{\pi}4,\frac{3\pi}4,\frac{5\pi}{4},\frac{7\pi}4\) and at the end points of the interval [0, 2π].

Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring at \(x = \frac{\pi}4\) and \(x = \frac{5\pi}4\).

73. What is the maximum value of the function sin x + cos x?

Answer:

Let f(x) = sin x + cos x.

Now, \(f^n(x)\) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive. Also, we know that sin x and cos x both are positive in the first quadrant. Then, \(f^n(x)\) will be negative when \(x \in\left(0, \frac{\pi}{2}\right)\)

Thus, we consider \(x = \frac{\pi}4\)

∴ By second derivative test, f will be the maximum at \(x = \frac{\pi}4\) and the maximum value of f is

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

Answer:

Let f(x) = 2x³ − 24x + 107.

∴ f'(x) = 6x² - 24 = 6 (x² - 4)

Now,

f'(x) = 0

⇒ 6(x² - 4) = 0

⇒ x² = 4

⇒ x = \(\pm\) 2

We first consider the interval [1, 3]. 

Then, we evaluate the value of f at the critical point x = 2 

∴ [1, 3] and at the end points of the interval [1, 3].

f(2) = 2(8) − 24(2) + 107 = 16 − 48 + 107 = 75 

f(1) = 2(1) − 24(1) + 107 = 2 − 24 + 107 = 85 

f(3) = 2(27) − 24(3) + 107 = 54 − 72 + 107 = 89 

Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3. 

Next, we consider the interval [−3, −1]. 

Evaluate the value of f at the critical point x = −2 

∴ [−3, −1] and at the end points of the interval [1, 3].

f(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125 

f(−1) = 2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129 

f(−2) = 2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139 

Hence, the absolute maximum value of f(x) in the interval [−3, −1] is 139 occurring at x = −2.