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CBSE Class 12 Maths Revision Notes Chapter 6

Class 12 Mathematics Chapter 6 Notes

The Mathematics subject is very demanding. It requires a strong foundation and regular practice to excel in this subject. The knowledge of basic concepts introduced in previous classes is essential to understanding the complex topics introduced in Class 12 Mathematics.

The Class 12 Mathematics Chapter 6 notes- Application of Derivatives is a continuation of Chapter 5. In the notes, students can find definitions of derivatives, increasing and decreasing functions, maxima and minima, tangents and normals, and some more concepts related to differentiation. Students will be asked to solve various problems using the first and second derivative tests. With the help of the Class 12 Mathematics Chapter 6 notes, students will learn to illustrate basic principles involved in the application of derivatives.

Extramarks, an online learning platform, provides a helping guide to gain an in-depth understanding of the concepts included in this chapter. It also aids in the preparation of class 12 boards and other national-level competitive examinations. The Class 12 Mathematics Chapter 6 notes include detailed and apt information, which ensures deep knowledge and sufficient practice.

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Key Topics Covered In Class 12 Mathematics Chapter 6 Notes

The key topics covered under Extramarks class 12 Mathematics chapter 6 notes include the following.

Rate of change of quantity:

A function y=f(x), then ddxy = f’(x) gives the rate of change of function f(x).
Let the function x= f(t) and y= f(t) be two functions such that variable x and y are varying to parameter t, then using Chain Rule, we get ddxy= ddtyddtx, ddtx 0
The rate of change of volume for decreasing side is given ddxy.

Decreasing and increasing functions:

Let a function f(x) be continuous in [p,q] and differential on (p,q), then
If f’ (x) = 0 x ∈ (p, q), then F is a constant function

If f’ (x) < 0 x ∈ (p, q), then F is a decreasing function

If f’ (x) > 0 x ∈ (p, q), then F is an increasing function

Tangent and Normal to a Curve:

A normal is a perpendicular line to the tangent. The straight-line equation passing through a point with slope m is given as y – y1= m (x – x1). If y= f (x) is a tangent to the curve at point P(x1, y1), then the slope is given as ddxy = f’(x).
Therefore, the equation becomes y – y1= ddxy (x – x1)
Another way of writing the equation of normal is y – y1= -ddyx (x – x1)

Length of tangent = y cosec = y 1+ (ddxy)2(ddxy)
Length of normal = y 1+ (ddxy)2

Slope of the curve:

Slope of normal at point P= -1Slope of tangent at point P = -1(ddxy) = -ddyx
If the normal is perpendicular to x-axis and parallel to y-axis at (x,y), then f’(x) = 0
If the normal is parallel to x-axis and perpendicular to y-axis at (x,y), then f’(x) = ∞

The angle of intersection of two curves:

If y= f1(x) and y = f2(x) are two curves that meet at point P(x1 , y1), then the angle of intersection of two curves is equal to the angle between the tangent of two curves at the point P.
The angle is given as tan = m1 – m21 + m1 m2
Slope m1 = ddxf1 and m2 = ddxf2

Maxima and Minima:

Differentiation is used to find the minima and maxima of a function. Minima is the lowest point, and Maxima is the highest point of a graph.
If f (x) ≤ f (a), when x=a, x in the domain, f(x) has absolute maximum at point a.
If f (x) ≥ f (a), when x=a, x (p, q), f (x) has a relative minimum.
If f (x) ≥ f (a), when x=a, x in the domain, f(x) has an absolute minimum at point a.
If f (x) ≤ f (a), when x=a, x (p, q), f (x) has a relative maximum.

Monotonicity:

Monotonic function: If the function f(x) is either decreasing or increasing in the domain.

Strictly increasing function: In function f(x), for every x1, x2 D. If x1> x2, then f(x1)> f(x2). i.e., when the value of x increases there is an increase in the function f(x) value.

Strictly decreasing function: In function f(x), for every x1, x2 D. If x1 x2, then f(x1) f(x2). i.e., when the value of x decreases there is a decrease in the function f(x) value.

Non-increasing function: In function f(x), for every x1, x2 D. If x1> x2, then f(x1) f(x2). i.e., when the value of x increases, the value of function f(x) would never increase.

Non-decreasing function: In function f(x), for every x1, x2 D. If x1> x2, then f(x1) f(x2). i.e., when the value of x decreases, the value of function f(x) would never decrease.

For a function f(x) differentiable on (a, b),
f’(x)>0, if the function is increasing and f’(x)<0, if the function is decreasing

Properties of Monotonicity:

If the function f(x) is strictly increasing, then f-1 exists on the domain and is strictly increasing.
If a continuous function f(x) is strictly increasing, then f-1 is also continuous on the domain [f(a), f(b)]
If the function f(x) and g(x) is strictly increasing (decreasing), then the composite function gof(x) exists on the domain and is strictly increasing (decreasing).
If either function f(x) or g(x) is strictly increasing and the other function is strictly decreasing, then the composite function gof(x) exists on the domain and is strictly decreasing.

Critical Points:

The points where the function f(x) is not differentiative, but its derivative is equal to zero are known as critical points. The point where the maxima and minima of a function occur are critical points; however, a critical point doesn’t necessarily imply that it is the maxima and minima of a function.

Points of inflection:

In a continuous function f(x), if the first derivative may or may not be zero but the second derivative at a point must be zero, then that point is known as the point of inflection. At this point, f’’(x) can change the sign.
Case 1: y=f(x) is concave downward when the f” (x) < 0, x ∈ (a, b) Case 2: y=f(x) is concave upward in interval (a, b) when the f” (x) > 0, x ∈ (a, b)

Applications of derivatives play a significant role in the branch of Physics. It helps in the study of seismology. With the help of graphs, loss and profit occurred can also be calculated. Also, derivatives have a range of applications like temperature, distance and speed.

Class 12 Mathematics Chapter 6 notes Exercises & Answer Solutions.

Chapter 6 mathematics class 12 notes provide a helping hand for students preparing for Class 12 board exams and other national level examinations such as the JEE Main, JEE Advanced, etc. The chapter- Application of derivatives holds relevance and is very important.
Students are advised to refer to the class 12 mathematics notes chapter 6 to resolve their doubts and clear the difficulties they face. These notes enable quick revision of formulas, definitions, and key points in a descriptive manner.

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Q.1 Find the maximum area of an isosceles triangle inscribed in the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

with its vertex at one end of the major axis.

Ans

$\begin{array}{l} A = area of isosceless Δ APP ‘ \\ AM = a - a cosθ \\ PP ‘ = \sqrt{{(bsinθ + bsinθ)}^{2}} \\ = \sqrt{{(2 bsinθ)}^{2}} = 2 bsinθ \\ A = \frac{1}{2} (a - acosθ) (2 bsinθ) \\ = ab (1 - cosθ) (sinθ) \\ Differentiating A w . r . t . θ \\ A ‘ = ab [(1 - cosθ) cosθ + sinθsinθ] \\ = ab [cosθ - \cos^{2} θ + \sin^{2} θ] \\ = ab [cosθ - \cos^{2} θ + 1 + \cos^{2} θ] \\ = ab [cosθ + 1 + 2 \cos^{2} θ] \\ A ‘ = 0 for maxima / minima \\ \Rightarrow ab [cosθ + 1 - 2 \cos^{2} θ] = 0 \\ \Rightarrow 2 \cos^{2} θ + cosθ - 1 = 0 \\ \Rightarrow 2 \cos^{2} θ - 2 cosθ + cosθ - 1 = 0 \\ \Rightarrow (2 cosθ + 1) (cosθ - 1) = 0 \\ \Rightarrow θ = \frac{2 π}{3} or θ = \frac{π}{2} (which is not possible) \\ A ‘ ‘ < 0 at θ = \frac{2 π}{3} \\ A = ab (1 - \cos \frac{2 π}{3}) (\sin \frac{2 π}{3}) or A = ab (1 + \frac{1}{2}) (\frac{\sqrt{3}}{2}) = \frac{3 \sqrt{3}}{4} ab \end{array}$

Q.2 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.

${(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}}$

Ans

$\begin{array}{l} Here AC = Hypotenuse of triangle ABC \\ And AC = AP + P \\ ∴ AC = \sqrt{b^{2} + y^{2}} + \sqrt{a^{2} + x^{2}} (with the help of figure) \\ We know that PD ∥ BC \\ Therefore, \\ \frac{AD}{AB} = \frac{PD}{BC} \Rightarrow \frac{y}{y + a} = \frac{b}{x + b} \\ \Rightarrow y (x + b) = b (y + a) \Rightarrow yx + yb = yb + ba \\ \Rightarrow y = \frac{ab}{x} \\ ∴ AC = \sqrt{b^{2} + {(\frac{ab}{x})}^{2}} + \sqrt{a^{2} + x^{2}} and \\ AC ‘ = \frac{{(ab)}^{2}}{2 \sqrt{b^{2} + {(\frac{ab}{x})}^{2}}} (\frac{- 2}{x^{3}}) + \frac{x}{\sqrt{a^{2} + x^{2}}} \Rightarrow AC ‘ = 0 for maxima / minima \\ \Rightarrow \frac{{(ab)}^{2}}{2 \sqrt{b^{2} + {(\frac{ab}{x})}^{2}}} (\frac{- 2}{x^{3}}) + \frac{x}{\sqrt{a^{2} + x^{2}}} = 0 \\ \Rightarrow \frac{{(ab)}^{2}}{2 \sqrt{b^{2} + {(\frac{ab}{x})}^{2}}} (\frac{1}{x^{3}}) = \frac{x}{\sqrt{a^{2} + x^{2}}} \\ Squaring both sides \\ \frac{{(ab)}^{2}}{(\frac{b^{2} x^{2} + a^{2} b^{2}}{x^{2}})} (\frac{1}{x^{6}}) = \frac{x^{2}}{a^{2} + x^{2}} \Rightarrow \frac{{(ab)}^{4}}{b^{2} (x^{2} + a^{2})} (\frac{1}{x^{6}}) = \frac{x^{2}}{a^{2} + x^{2}} \\ \frac{{(ab)}^{4}}{b^{2}} = x^{6} or x = a^{\frac{2}{3}} b^{\frac{1}{3}} \\ AC ‘ ‘ > 0 at x = a^{\frac{2}{3}} b^{\frac{1}{3}} \\ ∴ AC = \sqrt{b^{2} + \frac{a^{2} b^{2}}{a^{\frac{4}{3}} b^{\frac{2}{3}}}} + \sqrt{a^{2} + a^{\frac{4}{3}} b^{\frac{2}{3}}} \\ = \sqrt{b^{2} + a^{\frac{2}{3}} b^{\frac{4}{3}}} + \sqrt{a^{2} + a^{\frac{4}{3}} b^{\frac{2}{3}}} \\ = b^{\frac{2}{3}} \sqrt{b^{\frac{2}{3}} + a^{\frac{2}{3}}} + a^{\frac{2}{3}} \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}} \\ = \sqrt{b^{\frac{2}{3}} + a^{\frac{2}{3}}} (a^{\frac{2}{3}} + b^{\frac{2}{3}}) \\ = {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}} \end{array}$

Q.3 A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Ans

$\begin{array}{l} Step 1: \\ Perimeter of the window when the width of the window is x \\ and 2 r is the length . \\ \Rightarrow 2 x + 2 r + \frac{1}{2} \times 2 πr = 10 [Given] \\ 2 x + 2 r + πr = 10 \\ 2 x + r (2 + π) = 10 . \dots. (1) \\ For admitting the maximum light through the opening \\ the area of the window must be maximum . \\ Let A = Sum of areas of rectangle and semi circle . \\ Step 2 : \\ Area of circle = {πr}^{2} \\ Area of rectangle = l \times b = 2 \times r \times x \\ A = 2 rx + \frac{1}{2} {πr}^{2} \\ = r [10 - (π + 2) r] \frac{1}{2} {πr}^{2} \\ = 10 r - (\frac{1}{2} π + 2) r^{2} \\ For maximum aea \frac{dA}{dr} = 0 and \frac{d^{2} A}{{dr}^{2}} is negative . \\ \Rightarrow 10 - (π + 4) r = 0 \\ (π + 4) r = 10 \\ r = \frac{10}{π + 4} \\ Step 3: \\ \frac{d^{2} A}{{dr}^{2}} = - (π + 4) [Differentiating w . r . t . r] \\ i . e . \frac{d^{2} A}{{dr}^{2}} is negative for r = \frac{10}{π + 4} \\ \Rightarrow A ismaximum . \\ From (1) we have \\ 10 = (π + 2) r + 2 x \\ Put the value of r in above equation . \\ 10 = (π + 2) \times (\frac{10}{π + 4}) + 2 x \\ 10 = \frac{10 (π + 2)}{(π + 4)} + 2 x \\ \frac{10 (π + 2) + 2 x (π + 4)}{(π + 4)} \\ 10 (π + 4) = 10 (π + 2) + 2 x (π + 4) \\ 10 (π + 4) - 10 (π + 2) = 2 x (π + 4) \\ 10 π + 40 - 10 π - 20 = 2 x (π + 4) \\ x = \frac{10}{π + 4} \\ Step 4: \\ Length of rectangle = 2 r = 2 (\frac{10}{π + 4}) = \frac{20}{π + 4} \\ Breath = \frac{10}{π + 4} \end{array}$

Q.4 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is .

${sin}^{- 1} (\frac{1}{3})$

Ans

$\begin{array}{l} Here in ΔAOC, l^{2} = h^{2} + r^{2} \\ S = {πr}^{2} + πrl \Rightarrow \frac{S - {πr}^{2}}{πr} = l or l = \frac{S}{πr} - r \\ The total surface area of the cone is \\ Volume of the cone is given by \\ V = \frac{1}{3} {πr}^{2} h = \frac{1}{3} {πr}^{2} \sqrt{l^{2} - r^{2}} \\ V_{1} = V^{2} = \frac{1}{9} π^{2} r^{2} (l^{2} - r^{2}) \Rightarrow V_{1} = \frac{1}{9} π^{2} r^{4} ({[\frac{S}{πr} - r]}^{2} - r^{2}) \\ V_{1} = \frac{1}{9} π^{2} r^{4} (\frac{S^{2}}{π^{2} r^{2}} - \frac{2 S}{π}) = \frac{1}{9} (S^{2} r^{2} - 2 {Sπr}^{4}) \\ V_{1} ‘ = \frac{1}{9} (2 S^{2} r^{2} - 8 {Sπr}^{3}) \\ V_{1} ‘ = 0 for maxima / minima \\ \frac{1}{9} (2 S^{2} r - 8 {Sπr}^{3}) = 0 \Rightarrow 2 {Sπr}^{3} \Rightarrow \frac{S}{4 π} = r^{2} \\ V_{1} ‘ ‘ = \frac{1}{9} (2 S^{2} - 24 {Sπr}^{2}) < 0 \Rightarrow V is maximum at r^{2} = \frac{S}{4 π} \\ 4 {πr}^{2} = S \Rightarrow 4 {πr}^{2} = {πr}^{2} + πrl \Rightarrow 3 {πr}^{2} = πrl \\ or \frac{r}{l} = \frac{1}{3} \\ i . e \sin ∠ AOC = \frac{1}{3} \Rightarrow semi - vertical angle is \sin^{- 1} (\frac{1}{3}) \end{array}$

Q.5

$\begin{array}{l} Show that the right circular cone of least curved surface and given \\ volume has an altitude equal to \sqrt{2} times the radius of the base . \end{array}$

Ans

$\begin{array}{l} Here, volume of the cone v = \frac{1}{3} {πr}^{2} h \Rightarrow r^{2} = \frac{3 V}{πh} . \dots\dots\dots (1) \\ Surface area s = πrl = πr \sqrt{h^{2} + r^{2}} . \dots\dots\dots (2) \\ Whereh = height of the cone \\ r = radius of the cone \\ l = Slant height of the cone \\ S^{2} = π 2 r^{2} (h^{2} + r^{2}) by (2) \\ Let S_{1} = S^{2} then by (1) \\ S_{1} = \frac{3 πV}{h} (h^{2} + \frac{3 V}{πh}) = 3 πVh + \frac{9 V^{2}}{h^{2}} \\ \Rightarrow \frac{{dS}_{1}}{dh} = 3 πV + 9 V^{2} (\frac{- 2}{h^{3}}) \\ \frac{{dS}_{1}}{dh} = 0 for maxima / minima \\ \Rightarrow 3 πV + 9 V^{2} (\frac{- 2}{h^{3}}) = 0 \Rightarrow 3 πV + 9 V^{2} (\frac{- 2}{h^{3}}) = h^{3} = \frac{6 V}{π} \\ As surface area is least, we have \\ \frac{{dS}_{1}}{{dh}^{2}} > 0 when h^{3} = \frac{6 V}{π} \\ Therefore curved surface area is minimum when \frac{3 {πh}^{3}}{6} = V \\ Thus, \\ \frac{{πh}^{3}}{6} = \frac{1}{3} {πr}^{2} h \Rightarrow h^{2} = 2 r^{2} \\ h = \sqrt{2 r} \\ Hence for least curved surface the altitude is \sqrt{2} times radius . \end{array}$

Q.6 Find two numbers whose sum is 24 and whose product is as large as possible.

Ans

Let one number be x. Then, the other number is (24 − x).

Let P(x) denote the product of the two numbers. Thus,

P(x) = x(24 – x) = 24 – x²

∴ P'(x) = 24 – 2x

P”(x) = -2

Now, P'(x) = 0 ⇒ x = 12

Also, P”(12)= -2 < 0

∴ By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.

Q.7 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Ans

Let circle piece length = x m

Then square piece length = 28 – x m

For circle,
Perimeter of circle = 2

$π$

r =x ⇒ r = x/2

$π$

And for square,

Perimeter of square = 28 – x = 4a ⇒ a = (28 – x)/4

$\begin{array}{l} Now the sum of the areas ‘ A ‘ of circle and square \\ A = {πr}^{2} + a^{2} \\ = π {(\frac{x}{2 π})}^{2} + {(\frac{28 - x}{4})}^{2} \\ = \frac{x^{2}}{4 π} + {(\frac{28 - x}{4})}^{2} \\ \frac{dA}{dx} = \frac{x}{2 π} - \frac{28 - x}{8} \\ For maxima / minima \\ \frac{dA}{dx} = 0 \Rightarrow \frac{x}{2 π} - \frac{28 - x}{8} = 0 or x = \frac{28 π}{4 + π} \\ \frac{d^{2} A}{{dx}^{2}} = 0 or x = \frac{28 π}{4 + π} \\ Hence, A is minimum at circle piece = x = \frac{28 π}{4 + π} \\ Square piece = 28- x = \frac{112}{4 + x} . \end{array}$

Q.8 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 to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Ans

Let each side of the square to be cut off be x cm

Therefore, for the box:

Length = 18 – 2x

Breadth = 18 – 2x

And Height = x
Volume ‘V’ of the box: V = (18 – 2x)² x

$\begin{array}{l} \frac{dV}{dx} = {(18 - 2 x)}^{2} . 1 + \times . 2 (18 - 2 x) (- 2) \\ = (18 - 2 x) (18 - 6 x) \\ For maxima / minima, \frac{dV}{dx} = 0 \Rightarrow (18 - 2 x) (18 - 6 x) = 0 \\ \Rightarrow x = 3 and x = 9 \\ x = 9 is not possible as the length and breadth can not be zero . \\ now, \\ \frac{d^{2} y}{{dx}^{2}} = (18 - 2 x) (- 6) + (18 - 6 x) (- 2) \\ at x = 3 \frac{d^{2} y}{{dx}^{2}} < 0 \\ ∴ Volume of the box is maximum when x = 3. \end{array}$

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

Q.9 Find both the maximum and minimum value of 3x⁴ – 8x + 12x² – 48x + 25 on the interval [0, 3].

Ans

Let f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25, therefore,

f'(x) = 12x³ – 24x² + 24x – 48

f'(x) = 12(x³ – 2x² + 2x – 4)

f'(x) = 12(x² + 2) ( x – 2)

For maxima/minima f'(x) = 0

12(x² + 2) ( x – 2) = 0 x = 2.

Now, we have to find f(x) at x = 0, 2 and 3.

f(0) = 25, f(2) = – 39 and f(3) = 16

At x = 0, maximum value is 25

At x = 2, minimum value is – 39.

Q.10

$Use differential to approximate \sqrt{0 .037} .$

Ans

$\begin{array}{l} Let y = \sqrt{x}, x = 0.040 and x + Δx = 0.037. \\ Then Δx = 0.037-0.047 = -0.003 \\ for x = 0.040, y = \sqrt{0 .040} = 0 .2 \\ Let dx = Δx = -0.003 \\ Now, y = \sqrt{x} \\ i . e ., \frac{dy}{dx} = \frac{1}{2} \sqrt{x} = \frac{1}{0 .4} \\ Therefore, dy = \frac{dy}{dx} x dx \\ i . e ., dy = (\frac{1}{0 .4}) (- 0 .003) = \frac{- 3}{100} \\ \Rightarrow Δy = - \frac{3}{400} \\ Hence, \sqrt{0 .037} = y + Δy = 0.2 - (\frac{3}{400}) = 0 .1925. \end{array}$

Q.11 Use differential to approximate (25)^1/3.

Ans

$\begin{array}{l} Let y = x^{\frac{1}{3}} and x = 27 \\ Therefore, \frac{dy}{dx} = \frac{1}{2 {(x^{2})}^{\frac{1}{3}}} \\ Suppose, Δx = -2 \\ Now, y + Δy = {(x + Δx)}^{\frac{1}{3}} or {(25)}^{\frac{1}{3}} = x^{\frac{1}{3}} + Δy \\ As Δy \approx dy \\ Δy = \frac{dy}{dx} Δx = \frac{1}{3 {(x^{2})}^{\frac{1}{3}}} . (- 2) \\ = \frac{- 2}{27} = - 0 .074 \\ ∴ {(25)}^{\frac{1}{3}} = 3 + (- 0 .074) = 2 .926 \end{array}$

Q.12 If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.

Ans

Let r be the radius of the sphere and Δr be the error in measuring the radius.
Then r = 9 cm and Δr = 0.03 cm. Now, the volume V of the sphere is given by

V = (4/3)

$π r^{3}$

, therefore,

dV/dr =

$4 π r^{2}$

$4 π$

(9)2(0.03) = 9.72

$π$

cm³.

Thus, the approximate error in calculating the volume will be 9.72

$π$

cm³.

Q.13

$\begin{array}{l} Find the equation of the tangent to the curve y = \sqrt{(3 x - 2)} which is \\ parallel to the line 4 x -2 y +5 = 0. \end{array}$

Ans

$\begin{array}{l} Let the point of contact of the tangent line parallel to the given line be (x_{1}, y_{1}) . \\ Equation of curve is y = \sqrt{(3 x - 2)} . \dots\dots. (1) \\ and equation of given line is 4 x -2 y +5 = 0. \\ Differentiating equation (1), w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {(3 x - 2)}^{\frac{1}{2}} \\ = \frac{3}{2 {(3 x - 2)}^{\frac{1}{2}}} \\ {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = \frac{3}{2 \sqrt{(3 x_{1} - 2)}} \\ Differentiating equation (2), w . r . t . x, we get \\ 4-2 \frac{dy}{dx} + 0 = 0 \\ \Rightarrow {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = 2 \\ Sinceline is parallel to tangent to the curve \\ So, \frac{3}{2 \sqrt{(3 x_{1} - 2)}} = 2 \\ \Rightarrow 3 = 4 \sqrt{(3 x_{1} - 2)} \\ Squarring both sides, we get \\ 9 = 16 {(\sqrt{(3 x_{1} - 2)})}^{2} \\ 3 x_{1} - 2 = \frac{9}{16} \Rightarrow x_{1} = \frac{41}{48} \\ Since point (x_{1}, y_{1}) lies on curve (1), so \\ y_{1} = \sqrt{(3 x_{1} - 2)} \\ y_{1} = \sqrt{(3 \times \frac{41}{48}) - 2} \\ = \sqrt{(\frac{41}{16} - 2)} \\ = \frac{3}{4} \\ Thus, the equation of tangent at (\frac{41}{48}, \frac{3}{4}) is \\ \Rightarrow y - y_{1} = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} (x - x_{1}) \\ \Rightarrow y - \frac{3}{4} = 2 (x - \frac{41}{48}) \\ \Rightarrow \frac{4 y - 3}{4} = 2 (\frac{48 \times - 41}{48}) \\ \Rightarrow 6 (4 y - 3) = 48 \times - 41 \\ \Rightarrow 24 y - 18 = 48 \times - 41 \\ \Rightarrow 48 x - 24 y - 41 + 18 = 0 \\ \Rightarrow 48 x - 24 y - 23 = 0 \end{array}$

Q.14 Prove that the curves x = y³ and xy = k cut at right angles if 8k² = 1.

Ans

Here x = y³ ..(1)

and xy = k ..(2)

Let (x₁, y₁) be the intersecting point

By solving given equations we have (x₁, y₁) = (k^2/3, k^1/3)

Slope of the equation (1) at (k^2/3, k^1/3) is

dy/dx = 1/2y or dy/dx at (k^2/3, k^1/3) = 1/ 2k^1/3

And slope of the equation (2) at (k^2/3, k^1/3)is

dy/dx = -(y/x) or dy/dx at (k^2/3, k^1/3) = – 1/k^1/3

Now curve cuts at right angles
⇒ (1/2)k^2/3.(- 1/k^1/3)= – 1
⇒ (1)³ = 8k²

⇒ 8k² = 1.

Q.15 For the curve y = 4x³ – 2x⁵, find all the points at which the tangent passes through the origin.

Ans

Let (x₁, y₁) be the required point on the given curve y = 4x³ – 2x⁵.
Slope of the curve dy/dx = 12x² – 10x⁴

Slope of the curve at (x₁, y₁) = 12x₁² – 10x₁⁴

Thus, the equation of tangent is

y – y₁ = (12x₁² – 10x₁⁴)(x – x₁)

As tangent passes through the origin

y₁ = (12x₁² – 10x₁⁴)(x₁)

⇒ 4x₁³ – 2x₁⁵= 12x₁³ – 10x₁⁵ or 8x₁⁵ – 8x₁³ =0

⇒ 8x₁³ ( x₁²– 1) =0

⇒ x₁ = 0, – 1, 1

at x₁ = 0 ⇒ y₁ = 0

at x₁ = – 1 ⇒ y₁ = – 2, and

at x₁ = 1 ⇒ y₁ = 2

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

Q.16 Find the equation of tangent to the curve given by x =asin³t, y = bcos³t at a point where t =

$\frac{π}{2}$

Ans

Here, x =asin³t, y = bcos³t …(1)
Differentiating (1) w.r.t.t, we get

dx/dt = 3asin²tcost and dydt = – 3bcos²t sint

Therefore, dy/dx = -(b/a) cot t

Hence, slope of the tangent at t =

$\frac{π}{2}$

is given by

dy/dx = – (b/a) cot

$\frac{π}{2}$

= 0.

Hence, the equation of tangent is given by

y – 0 = 0(x – a), i.e. y = 0.

Q.17 Let I be any interval disjoint from (-1, 1). Prove that the function

$\int$

given by f(x) = x + (1/x) is strictly increasing on I.

Ans

$\begin{array}{l} We have f (x) = x + \frac{1}{x} \\ f (x) = 1 - \frac{1}{x^{2}} = \frac{x^{2} - 1}{x^{2}} \\ Now x \in I \Rightarrow x \notin (- 1, 1) \\ \Rightarrow x \leq - 1 or x \geq 1 \Rightarrow x^{2} \geq 1 \\ \Rightarrow x^{2} - 1 \geq 0 \Rightarrow \frac{x^{2} - 1}{x^{2}} \geq 0 \\ \Rightarrow f ‘ (x) \geq 0 \\ Thus f ‘ (x) \geq 0 for all x \in I \\ Hence f (x) is strictly increasing on I . \end{array}$

Q.18

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

Ans

$\begin{array}{l} f (x) = \log (1 + x) - \frac{2 x}{2 + x}, x > - 1 \\ f ‘ (x) = \frac{1}{1 + x} - 2 [\frac{(x + 2) - x}{{(x + 2)}^{2}}] = \frac{1}{1 + x} - \frac{4}{{(x + 2)}^{2}} \\ = \frac{x^{2}}{(x + 1) {(x + 2)}^{2}} \\ for to be increasing f ‘ (x) > 0 \\ \Rightarrow \frac{x^{2}}{(x + 1) {(x + 2)}^{2}} > 0 \\ \Rightarrow \frac{1}{x + 1} > 0 \\ \Rightarrow x > - 1 \\ Hence, y = logx - \frac{2 x}{x + 2} is an increasing function of x for all values of x > -1. \end{array}$

Q.19 Find the intervals in which the function f given by f(x) = sinx + cosx, 0 ≤ x ≤ 2

$π$

is strictly increasing or strictly decreasing.

Ans

We have,

f(x) = sinx + cosx

or, f'(x) = cosx – sinx, for turning point f'(x) = 0 it gives sinx = cosx

$i.e., x= \frac{1}{4}, 5 \frac{π}{4}$

$\begin{array}{l} three disjoint intervals [0, \frac{π}{4}) (\frac{π}{4}, \frac{5 π}{4}) and (\frac{5 π}{4}, 2 π] \\ f ‘ (x) > 0 for all x \in [0, \frac{π}{4}) and (\frac{5 π}{4}, 2 π] \\ f ‘ (x) < 0 for all x \in (\frac{π}{4}, \frac{5 π}{4}) \\ f ‘ (x) is strictly increasing on x \in [0, \frac{π}{4}) and (\frac{5 π}{4}, 2 π] \\ and f (x) is strictly decreasing on (\frac{π}{4}, \frac{5 π}{4}) . \end{array}$

Q.20 Find the intervals in which the function f given by f(x) = 4x³ – 6x² – 72x + 30 is (a) strictly increasing (b) strictly decreasing.

Ans

We have f(x) = 4x³ – 6x² – 72x + 30 or f'(x) = 12x² – 12x – 72

= 12(x² – x -6) = 12(x – 3) (x + 2).

Therefore, f'(x) = 0 gives x = – 2 , 3.

There are three disjoint intervals, (– , – 2), (– 2, 3) and (3, ).

f'(x) > 0 for all x ∈ (– ∞, – 2) and (3, ∞),

f'(x) < 0 for all x ∈ (– 2, 3)

Q.21 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?

Ans

Let height of the cone = h = 4 cm, and radius of the cone = r

Given h = (1/6)r or 6h = r

Sand is pouring from a pipe at the rate of 12 cm/s

i.e., dv/dt = 12, where v = volume of the cone.

We know that the volume of cone

v = (1/3)

$π r^{2}$ h = (1/3)

$π$ (6h)²h = 12

$π$ h³

On differentiating w.r.t.t, we get

dv/dt = 12 x 3

$π$ h²dh/dt

⇒dh/dt = 1/(

$π$ x 3 x 4 x 4) = 1/48

$π$ cm/s

Q.22 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.

Ans

Here 6y = x³ + 2 ...(1)

Differentiating w.r.t t. we get

6 dy/dt = 3x² dx/dt …(2)

the y-coordinate is changing 8 times as fast as the x-coordinate.

i.e., dy/dt = 8 dx/dt, putting in (2), we get

48 dx/dt = 3x² dx/dt

x² = 16, or

x = ± 4

by (1), we get

y = 11 at x = 4 and

y = – 31/3 at x = – 4.

Q.23 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 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Ans

$\begin{array}{l} Let AC be the ladder = 5 m and BC is distance the distance between the \\ foot of the ladder and wall = 4 m . BC is increasing at the rate of 2 cm / s \\ \Rightarrow \frac{dy}{dx} = 2 m / s \\ Now in ΔABC \\ x^{2} + y^{2} = 25 . \dots. (1) \\ x = 3 (∵ y = 4 given) \\ diff . equation (1) w . r . t . t, we get \\ 2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 0 \\ 6 \frac{dx}{dt} + 8 \times 2 = 0 \\ \frac{dx}{dt} = - \frac{16}{6} = - \frac{8}{3} m / s \\ height is decreasing at the rate of \frac{8}{3} m / s . \end{array}$

Q.24 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cm³ of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Ans

$\begin{array}{l} As volume of the baloon with radius r which always remains \\ spherical is V = \frac{4}{3} {nrr}^{3} \\ Volume ‘ V ‘ of the baloon is increasing at the rate of 900 \frac{{cm}^{3}}{s} \\ \frac{dV}{dt} = 900 \frac{{cm}^{3}}{s} \\ \frac{dV}{dt} = 4 {πr}^{2} \frac{dr}{dt} \\ 900 = 4 π {(15)}^{2} \frac{dr}{dt} \\ \frac{dr}{dt} = \frac{900}{4 π {(15)}^{2}} = \frac{900}{4 \times 15 \times 15 π} = \frac{1}{π} cm / s \end{array}$

Q.25 The length

$π$

of a rectangle is decreasing at the rate of 3 cm/minute and the width yis increasing at the rate of 2 cm/minute. When x = 10 cm and y = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.

Ans

Since the length x is decreasing and the width y is increasing with respect to time, we have

dx/dt = – 3 cm/min and dy/dt = 2 cm/min

(a) the perimeter of rectangle is P = 2 (x + y)

Therefore, rate of change of perimeter and dy/dt = 2 cm/min

dP/dt = 2{dx/dt + dy/dt} = 2( – 3 + 2) = – 2 cm/min
Thus, perimeter is decreasing at the rate of 2 cm/min.

(b) The area A of the rectangle is given by A = xy.

Therefore, the rate of change of area A with respect to time t is

dA/dt = x dy/dt + ydx/dt = 10 x 2 + 6 x ( – 3) = 20 -18cm²/min

Thus, area is increasing at the rate of 2 cm/min.

Q.26 A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?

Ans

Area A of a circle with radius r is given by A =

${πr}^{2}$

Therefore, the rate of change of area A with respect to time t is

dA/dt = 2

$π$

r dr/dt.

It is given that dr/dt = 4 cm/s

Therefore, when r = 10 cm,

dA/dt = 2

$π$

x 10 x 4 = 80

$π$

cm²/s.

Thus, the enclosed area is increasing at the rate of 80

$π$

cm²/s, when r = 10 cm.

Q.27 Find the rate of change of the area of a circle with respect to its radius r when r = 5cm.

Ans

The area A of a circle with radius r is given by A =

$π r^{2}$

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

dA/dr = 2

$π$

r = 10

$π$

(Since, r = 5 cm given)

Thus, the area of the circle is changing at the rate of 10

$π$

cm²/s.

Q.28 Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 2%.

Ans

$\begin{array}{l} V = x^{3} \\ or dV = (\frac{dV}{dx}) Δx = (3 x^{2}) Δx \\ = (3 x^{2}) (0 .02 x) = 0 .06 x^{3} m^{3} \end{array}$

Q.29 Show that the function given by f(x) = 5x – 2 is strictly increasing on R.

Ans

$\begin{array}{l} Let x_{1} and x_{1} be any two numbers in R . Then \\ x_{1} < x_{2 ‘} \\ 5 x_{1} 5 < x_{2} \\ 5 x_{1} - 2 < 5 x_{2} - 2 \\ f (x_{1}) < f (x_{2}) \\ Thus, f is strictly increasing on R . \end{array}$

Q.30 Find the slope of the tangent to the curve y = x³ – x at x = 2.

Ans

$\begin{array}{l} The slope of the tangent atx = 2 is given by \\ {\frac{dy}{dx}|}_{x -2} = (3 x^{2} - 1) |_{x -2} =11 \end{array}$

Q.31 If x = at, y = at², then give the equation of tangent at t = 1.

Ans

$\begin{array}{l} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ \frac{dy}{dx} = 2, at t = 1 \\ x = a = 1, y = a \\ equation of tangent at t = 1 \\ y - a = 2 (x - a) \\ 2 x - y = a \end{array}$

Q.32 Find the rate of change of the area of a circle with respect to its radius r when r = 1 cm.

Ans

The area of a circle A =

$π$

r².

$\frac{dA}{dr} = 2 πr = 2 π (∵ r = 1 cm given)$

Thus, the area of the circle is changing at the rate of 2

$π$

cm²/s when its radius is 1 cm.

Q.33 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?

Ans

Volume of cube = a³.

$\frac{dV}{dt} = 3 a^{2} \frac{da}{dt} = 900 (∵ \frac{da}{dt} = 2 cm / s given)$

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

Q.34 Find the rate of change of volume of a spherical body with respect to radius, when the radius is 10 cm.

Ans

$\begin{array}{l} As valume of sphere V = \frac{4}{3} {πr}^{3} \\ \Rightarrow \frac{dV}{dr} = 4 {πr}^{2} = 4 π {(10)}^{2} = 400 π \\ Hence, the valume of the spherical \\ body is increasing at the rate of 400 π {cm}^{3} / s \end{array}$

Q.35 Show that the function f given by f(x) = x³ – 3x² + 4x, x ∈ R is strictly increasing on R.

Ans

f‘(x) = 3x² – 6x + 4
= 3(x² – 2x + 1) + 1
= 3(x – 1)² + 1 > 0 in every inetrval of R

Therefore, the function f is strictly increasing on R.

Q.36

$\begin{array}{l} Show that the volume of the greatest cylinder that can be inscribed \\ in a cone of height ‘ h ‘ and sem - vertical angle ‘ α ‘ is \frac{4}{27} {πh}^{3} \tan^{2} α . \end{array}$

Ans

$\begin{array}{l} Let x and y be the radius and height of the cylinder which is \\ inscribed in the cone of give height ‘ h ‘ \\ Let r and α be the radius and semi - vertical angle of the cone . \\ Then from the figure, we have \\ PO = h, OA = r, LM = ON = x, LO = y, PL = PO - LO = h - y . \\ From similar triangles PLM and POA, we have \\ \frac{PL}{PO} = \frac{LM}{OA} \\ \Rightarrow \frac{h - y}{h} = \frac{x}{r} \\ \Rightarrow x = \frac{r) h - y}{h} . \dots\dots. (I) \\ If V denotes the volume of the cylinder, then \\ V = {πx}^{2} y . \dots\dots\dots (II) \\ \Rightarrow V = π [\frac{r^{2} {(h - y)}^{2}}{h^{2}}] y \\ \Rightarrow V = \frac{{πr}^{2}}{h^{2}} [h^{2} + y^{2} - 2 hy] y \\ \Rightarrow V = \frac{{πr}^{2}}{h^{2}} [h^{2} y + y^{3} - 2 {hy}^{2}] \\ Therefore, \\ \frac{dV}{dy} = \frac{{πr}^{2}}{h^{2}} [h^{2} + 3 y^{2} - 2 hy] \\ and \frac{d^{2} V}{{dy}^{2}} = \frac{{πr}^{2}}{h^{2}} [6 y - 4 h] \\ Put \frac{dV}{dy} = 0, we get \\ \frac{{πr}^{2}}{h^{2}} [h^{2} + 3 y^{2} - 4 hy] = 0 \\ \Rightarrow [h^{2} + 3 y^{2} - 4 hy] = 0 \\ \Rightarrow [h^{2} - 3 hy - hy + 3 y^{2}] = 0 \\ \Rightarrow (h - 3 y) (h - y) = 0 \\ Either h -3 y = 0 or h - y = 0 \\ Either h = 3 y or h = y \\ or y = \frac{h}{3} . \dots\dots (III) \\ Now, \frac{d^{2} V}{{dy}^{2}} = \frac{{πr}^{2}}{h^{2}} [6 (\frac{1}{3} h) - 4 h] \\ = \frac{{πr}^{2}}{h^{2}} (- 2 h) = = \frac{- 2 {πr}^{2}}{h} < 0 . \\ Thus, the volume is maximum when y = \frac{1}{3} h . \\ In triangle POA, we have \\ tanα = \frac{r}{h} \\ \Rightarrow r = htanα . \dots\dots (IV) \\ When y = \frac{1}{3} h, then x = \frac{r (h - \frac{1}{3} h)}{h} = \frac{2}{3} r \\ \Rightarrow x = \frac{2}{3} htanα . \dots\dots (V) \\ From (II), (III) and (V), we obtain \\ Volume of the greatest cylinder \\ = π {[\frac{2}{3} htanα]}^{2} [\frac{1}{3} h] \\ = \frac{4}{27} {πh}^{3} \tan^{2} α . \end{array}$

Q.37

$\begin{array}{l} A car start from a point M at time t = 0 second and stops \\ at point N . The distance x, in metres, covered by it, in seconds \\ is given by x = t^{2} (2 - \frac{t}{3}) . Find the time taken by it to reach N \\ and also find distance between M and N . \end{array}$

Ans

$\begin{array}{l} We have, \\ x = t^{2} (2 - \frac{t}{3}) \\ \Rightarrow x = 2 t^{2} - \frac{t^{3}}{3} \\ \Rightarrow \frac{dx}{dt} = 4 t - t^{2} \\ This gives velocity of the car at any time t . \\ Suppose the car stops at N after time t_{1} . \\ ∴ At t = t_{1}, \frac{dx}{dt} = 0 \\ \Rightarrow {(\frac{dx}{dt})}_{t - t_{1}} = 0 \\ \Rightarrow 4 t_{1} - {t_{1}}^{2} = 0 \\ \Rightarrow t_{1} (4 - t_{1}) = 0 \\ \Rightarrow t_{1} = 4 [∵ t_{1} = 0 is for point M] \\ Thus, the car takes 4 seconds to reach at N . \\ The distance between M and N is the value of x at t = t_{1} i . e . at t = 4. \\ ∴ MN = (Value of x at t = 4) \\ = 2 \times 4^{2} - \frac{4^{3}}{3} \\ = 32 - \frac{64}{3} \\ = 32 m . \end{array}$

Q.38

$\begin{array}{l} Find points on the curve \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 at which the tangents \\ are parallel to the \\ (i) x - axis (ii) y - axis \end{array}$

Ans

$\begin{array}{l} Let P (x_{1}, y_{1}) be a point on the curve \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 . Then, \\ \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 \\ The equation of the curve is \\ \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 \\ Differentiating both sides with respect to x, we get \\ \frac{2 x}{9} - \frac{2 y}{16} \frac{dy}{dx} = 0 \\ \Rightarrow \frac{dy}{dx} = \frac{16 x}{9 y} \\ \Rightarrow (\frac{dy}{dx}) (x_{1}, y_{1}) = \frac{16 x_{1}}{9 y_{1}} \\ (i) If the tangent at P (x_{1}, y_{1}) is parallel to x - axis, then \\ (\frac{dy}{dx}) (x_{1}, y_{1}) = 0 \\ \Rightarrow \frac{16 x_{1}}{9 y_{1}} = 0 \Rightarrow 16 x_{1} = 0 \Rightarrow x_{1} = 0 . \\ Putting x_{1} = 0 in (i), we get {y_{1}}^{2} = - 16, Which is impossible as y_{1} is real . \\ Hence, there is no point on the curve where tangent is parallel to x - axis . \\ (ii) If the tangent at P (x_{1}, y_{1}) is parallel to y - axis, then \\ (\frac{dy}{dx}) (x_{1}, y_{1}) = 0 \\ \Rightarrow \frac{9 y_{1}}{16 x_{1}} = 0 \Rightarrow y_{1} = 0 . \\ Putting y_{1} = 0 in (i), we get x_{1} = \pm 3 . \\ Hence, required points are (3, 0) and (- 3, 0) . \end{array}$

Q.39

$\begin{array}{l} Find the local maximum or local minimum, if any, of the function \\ f (x) = \sin^{4} x + \cos^{4} x, 0 < x < \frac{n}{2} . Using the first derivative test. \end{array}$

Ans

$\begin{array}{l} We have , \\ y = f (x) = \sin^{4} + \cos^{4} x \\ \Rightarrow \frac{dy}{dx} = 4 \sin^{3} xcosx - 4 \cos^{3} xsinx \\ \Rightarrow \frac{dy}{dx} = - 4 sinxcosx (\cos^{2} x - \sin^{2} x) \\ \Rightarrow \frac{dy}{dx} = - 2 \sin 2 xcos 2 x = - sin4x \\ For a local maximum or a local minimum, we have \\ \frac{dy}{dx} = 0 \\ \Rightarrow - \sin 4 x = 0 \\ \Rightarrow \sin 4 x = 0 \\ \Rightarrow 4 x = π [\begin{array}{l} ∵ 0 < x < \frac{π}{2} \\ ∴ 0 < 4 x < 2 π \end{array}] \\ \Rightarrow x = \frac{π}{4} \\ In the left n bd of x = \frac{π}{4}, we have \\ x< \frac{π}{4} \Rightarrow 4 x < π \Rightarrow \sin 4 x > 0 \Rightarrow - \sin 4 x < 0 \Rightarrow \frac{dy}{dx} < 0 \\ In the right nbd of x = \frac{π}{4}, we have \\ x > \frac{π}{4} \Rightarrow 4 x > π \Rightarrow \sin 4 x < 0 \Rightarrow - 4 \sin 4 x > 0 \Rightarrow \frac{dy}{dx} > 0 \\ Thus, \frac{dy}{dx} changes sign from negative to positives as x increases through \frac{π}{4} . \\ So, x = \frac{π}{4} is a point of local minimum. \\ The local maximum value of f (x) at x is \\ f (\frac{π}{4}) = {(\sin \frac{π}{4})}^{4} + {(\cos \frac{π}{4})}^{4} = \frac{1}{2} . \end{array}$

Q.40 Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.

Ans

$\begin{array}{l} Let ABC be a triangle inscribed in a given circle with centre O \\ and radius r . The area of the triangle will be maximum if its \\ vertex A opposite to the base BC ia at a maximum distance \\ from the base BC . This is possible only when A lies on the \\ diameter perpendicular to BC . Thus AD ⊥ BC . \\ So, triangle ABC must be an isosceles triangle . \\ Let OD = x . \\ In right triangle ODB, we have \\ {OB}^{2} = {OD}^{2} + {BD}^{2} \\ \Rightarrow r^{2} = x^{2} + {BD}^{2} \\ \Rightarrow BD = \sqrt{r^{2} - x^{2}} \\ \Rightarrow BC = 2 BD = 2 \sqrt{r^{2} - x^{2} .} \\ \Rightarrow AD = AO + OD = r + x . \\ Let A denote the area of ΔABC . Then, \\ A = \frac{1}{2} (BC \times AD) \\ \Rightarrow A = \frac{1}{2} \times 2 \sqrt{r^{2} - x^{2}} \times (r + x) \\ \Rightarrow A = (r + x) \sqrt{r^{2} - x^{2}} \\ \Rightarrow \frac{dA}{dx} = \sqrt{r^{2} - x^{2}} - \frac{x (r + x)}{\sqrt{r^{2} - x^{2}}} \\ \Rightarrow \frac{dA}{dx} = \frac{r^{2} - rx - 2 x^{2}}{\sqrt{r^{2} - x^{2}}} \\ For maximum or minimum values of A, we must have \\ \frac{dA}{dx} = 0 \\ \Rightarrow \frac{r^{2} - rx - 2 x^{2}}{\sqrt{r^{2} - x^{2}}} = 0 \\ \Rightarrow (r - 2 x) (r + x) = 0 \\ \Rightarrow r - 2 x = 0 \\ \Rightarrow x = \frac{r}{2} [∵ r + x \neq 0] \\ Now, \frac{dA}{dx} = \frac{r^{2} - rx - 2 x^{2}}{\sqrt{r^{2} - x^{2}}} \\ \Rightarrow \frac{d^{2} A}{{dx}^{2}} = \frac{(- r - 4 x)}{\sqrt{r^{2} - x^{2}}} + \frac{(r^{2} - rx - 2 x^{2}) x}{{(r^{2} - x^{2})}^{\frac{3}{2}}} \\ \Rightarrow {(\frac{d^{2} A}{{dx}^{2}})}_{x = \frac{r}{2}} \\ Thus, A is maximum when x = \frac{r}{2} . \\ ∴ BD = \sqrt{r^{2} - x^{2}} \Rightarrow BD = \frac{\sqrt{3} r}{2} \\ In Δ ODB, we have \\ tanθ = \frac{BD}{OD} \Rightarrow tanθ = \frac{\frac{\sqrt{3} r}{2}}{\frac{r}{2}} = \sqrt{3} \Rightarrow θ = 60^{ο} \\ ∴ ∠ BAC = θ {= 60}^{ο} \\ But, AB = AC . Therefore, ∠ B = ∠ C \\ Thus, we have ∠ A = ∠ B = ∠ C {= 60}^{ο} \\ Hence, A is maximum when Δ ABC is equilateral . \end{array}$

Q.41

$\begin{array}{l} Water is running into a conical vessel, 15 cm deep and 5 cm in \\ radius, at the rate of 0.1 {cm}^{3} / \sec . When the water is 6 cm deep, \\ find at what rate is \\ (i) the water level rising ? \\ (ii) the water - surface area increasing ? \\ (iii) the wetted surface of the vessel increasing ? \end{array}$

Ans

$\begin{array}{l} Let v be the volume of the water in the cone i . e . the volume of the \\ water - cone VA ‘ B ‘ at any time t . Let VO ‘ = h, O ‘ A ‘ = r and VA ‘ = l . Let α \\ be the semi - vertical angle of the cone, Then, \\ A = {πr}^{2} \\ \Rightarrow A = π \frac{h^{2}}{9} [∵ 3 r = h] \\ \Rightarrow \frac{dA}{dt} = \frac{2 πh}{9} \frac{dh}{dt} \\ When h = 6, \frac{dh}{dt} = \frac{1}{40 n ‘} We have \\ \frac{dA}{dt} = \frac{2 π \times 6}{9} \times \frac{1}{40 π} = \frac{1}{30} {cm}^{2} / \sec . \\ Thus, the water - surface area is increasing at the rate of \frac{1}{30} {cm}^{2} / \sec . \\ (iii) Let S be the wetted surface area of the vessel at any time t . \\ Then S = πrl . \\ From figure, we have \\ l^{2} = VA ‘^{2} = VO ‘^{2} + O ‘ A ‘^{2} \\ \Rightarrow l^{2} = h^{2} + r^{2} \\ \Rightarrow l^{2} = h^{2} + \frac{h^{2}}{9} [∵ 3 r = h] \\ l = \frac{\sqrt{10 h}}{3} \\ ∴ S = πrl \\ \Rightarrow S = π (\frac{h}{3}) (\frac{\sqrt{10 h}}{3}) \\ \Rightarrow S = \frac{π}{9} \sqrt{10} h^{2} \\ \Rightarrow \frac{ds}{dt} = \frac{2 π \sqrt{10} h}{9} \frac{dh}{dt} \\ Since, h = 6, \frac{dh}{dt} = \frac{1}{40 n} . Therefore, \\ \Rightarrow \frac{ds}{dt} = \frac{2 π \sqrt{10} h}{9} \times 6 \times \frac{1}{40 π} \\ = \frac{\sqrt{10} h}{30} {cm}^{2} / \sec . \\ Thus . the wetted surface area of the vessel is increasing \\ at the rate of \frac{\sqrt{10}}{30} {cm}^{2} / \sec . \end{array}$

Q.42

$Find the intervals in which f (x) = {(x -1)}^{3} {(x -2)}^{2} is increasing or decreasing .$

Ans

$\begin{array}{l} \Rightarrow f ‘ (x) = 3 {(x - 1)}^{2} {(x - 2)}^{2} + 2 {(x - 1)}^{3} (x - 2) \\ \Rightarrow f ‘ (x) = {(x - 1)}^{2} (x - 2) (3 x - 6 + 2 x - 2) \\ = {(x - 1)}^{2} (x - 2) (5 x - 8) \\ For f (x) to be increasing, we must have \\ f ‘ (x) > 0 . \\ \Rightarrow {(x - 1)}^{2} (x - 2) (5 x - 8) > 0 \\ \Rightarrow (x - 2) (fx -8) > 0 and x \neq 1 [∵ {(x - 1)}^{2} > 0 for all x \neq 1] \\ \Rightarrow 5 (x - \frac{8}{5}) (x - 2) > 0 and x \neq 1 \\ \Rightarrow (x - \frac{8}{5}) (x - 2) > 0 and x \neq 1 [∵ 5 > 0] \\ \Rightarrow x < \frac{8}{5} or x > 2 and x \neq 1 \\ \Rightarrow x \in (- \infty, 1) \cup (1, \frac{8}{5}) or x \in (2, \infty) \\ So, f (x) is increasing on (- \infty, 1) \cup (1, \frac{8}{5}) \cup (2, \infty) . \end{array}$

$\begin{array}{l} For f (x) to bedecreasing, we must have \\ f (x) < 0 . \\ \Rightarrow {(x - 1)}^{2} (x - 2) (fx -8) < 0 \\ \Rightarrow (x - 2) (5 x - 8) < 0 and x \neq 1 [∵ {(x - 1)}^{2} > 0 for all x \neq 1] \\ \Rightarrow 5 (x - \frac{8}{5}) (x - 2) < 0 and x \neq 1 \\ \Rightarrow (x - \frac{8}{5}) (x - 2) < 0 and x \neq 1 [∵ 5 > 0] \\ \Rightarrow x \in 5 (\frac{8}{5}, 2) and x \neq 1 \\ \Rightarrow x \in 5 (\frac{8}{5}, 2) \\ So, f (x) is decreasing on (\frac{8}{5}, 2) \end{array}$

Q.43

$The function f given by f (x) = x^{3} - {6x}^{2} + 14x, x \in R is________on R .$

Ans

$\begin{array}{l} The given function, \\ f (x) = x^{3} - 6 x^{2} + 14 x \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 3 x^{2} - 12 x + 14 \\ = 3 x^{2} - 12 x + 12 + 2 \\ = 3 (x^{2} - 4 x + 4) + 2 \\ = 3 {(x - 2)}^{2} + 2 > 0 in every interval of R . \\ Therefore, the function f is \underline{strictly increasing} on R . \end{array}$

Q.44 If x = 4at and y = 3at², then the equation of tangent at t = 2 is _______________.

Ans

$\begin{array}{l} Given, x = 4 at and y = 3 {at}^{2} \\ Differentiating w . r . t . x, we get \\ \frac{dx}{dt} = 4 a and \frac{dy}{dt} = 6 at \\ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{6 at}{4 a} \\ = \frac{3 t}{2} \\ {(\frac{dy}{dx})}_{t = 2} = \frac{3 (2)}{2} \\ = 3 \\ At t = 2, \\ x = 4 a (2) and y = 3 a {(2)}^{2} \\ = 8 a = 12 a \\ The equation of tangent is, \\ y - y_{1} = {(\frac{dy}{dx})}_{t = 2} (x - x_{1}) \\ y - 12 a = 3 (x - 8 a) \\ y - 12 a = 3 x - 24 a \\ \Rightarrow 3 x - y = 12 a \\ Therefore, the required equation of tangent is \underline{3 x - y = 12 a} . \end{array}$

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The chapter 6 mathematics class 12 notes elaborate on the following topics. It gives apt information and a detailed explanation of all important concepts of the chapter. It also includes several CBSE extra questions for students to practice.

Introduction
Rate of change in quantities
Increasing and decreasing functions
Tangent and normal
Approximations
Minima and Maxima

The fundamental ideas and applications of derivatives are very well explained in the notes using simple language. It is advised that each student should use the best NCERT books and notes to enhance their preparation. Our notes provide NCERT solutions to all exercise questions and some practice questions from the textbook. The Class 12 Mathematics Chapter 6 notes prepared by experts will help the students to tackle important questions easily and in a concise manner, without making silly mistakes.

CBSE Class 12 Maths Revision Notes Chapter 6

Class 12 Mathematics Chapter 6 Notes

Key Topics Covered In Class 12 Mathematics Chapter 6 Notes

Rate of change of quantity:

Decreasing and increasing functions:

Tangent and Normal to a Curve:

Slope of the curve:

The angle of intersection of two curves:

Maxima and Minima:

Monotonicity:

Properties of Monotonicity:

Critical Points:

Points of inflection:

Class 12 Mathematics Chapter 6 notes Exercises & Answer Solutions.

NCERT Exemplar Class 12 Mathematics

Key Features of Class 12 Mathematics Chapter 6 notes

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CBSE Class 12 Maths Revision Notes Chapter 6

Class 12 Mathematics Chapter 6 Notes

Key Topics Covered In Class 12 Mathematics Chapter 6 Notes

Rate of change of quantity:

Decreasing and increasing functions:

Tangent and Normal to a Curve:

Slope of the curve:

The angle of intersection of two curves:

Maxima and Minima:

Monotonicity:

Properties of Monotonicity:

Critical Points:

Points of inflection:

Class 12 Mathematics Chapter 6 notes Exercises & Answer Solutions.

NCERT Exemplar Class 12 Mathematics

Key Features of Class 12 Mathematics Chapter 6 notes

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