NCERT Solutions for Class 12 Maths Chapter 6 Applications of Derivative: Download Exercise with Solutions PDF

Class 12 Math Chapter 6 Application of Derivatives has significant weight in board examinations, contributing between 8–10 marks. Application of Derivatives also has a significant weight in competitive exams directly 1-2 questions are asked from related concepts in JEE Main, and indirectly since it is useful in calculus, 3D and 2D geometry, and Physics chapters. Students can check the key topics and formulae of Application of derivatives chapters below;

Class 12 Application of Derivatives Key Topic

• Rate of Change of Quantities
• Increasing and Decreasing Functions & Determining intervals where a function is increasing or decreasing using differentiation.
• Finding Tangents and Normals using the first-order derivative concept.
• Applications of Derivatives in Two-dimensional geometry and conic section of classes 11 and 12. 
• Finding Maxima and Minima using first and second-order derivatives

Check out the Chapter 6 Application of Derivatives Dropped Topics – 6.4 Tangents and Normals, 6.5 Approximations, Examples 45, 46, Ques. 1, 4–5 and 20–24 (Miscellaneous Exercise), Page 245 Points 4–10 in the Summary

Application of Derivatives Important Formulae for CBSE and Competitive Exams

• Rate of Change: $\frac{dy}{dx}$, where $y$is the dependent variable and $x$ is the independent variable.
• Slope of Tangent: $m = f'(x)$
  
• Slope of Normal: $m = -\frac{1}{f'(x)}$
  
• Equation of the tangent: $y - y_1 = f'(x_1)(x - x_1)$
  
• Equation of the normal: $y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)$
• Approximation Formula: $f(x + h) \approx f(x) + h \cdot f'(x)$
  
• Condition for Increasing Function: $f'(x) > 0$
  
• Condition for Decreasing Function: $f'(x) < 0$
  
• Local Maxima/Minima (First Derivative Test): Solve $f'(x) = 0$and analyze sign changes of $f'(x)$around the critical points.
• Conditions for Minima and Maxima:
  • • $f'(x) = 0$and $f''(x) > 0$for local minima.
    • $f'(x) = 0$and $f''(x) < 0$for local maxima.
• Second Derivative Test: Use $f''(x)$to confirm the nature of critical points.

Students can find the complete exercise-wise solution of the Class 12 Application of Derivatives NCERT PDF in one place designed by our experts. This Class 12 Application of Derivatives Solution PDF for NCERT Textbook is helpful for all students in various aspects. Both CBSE boards and competitive exams aspirants; JEE Main, CUSAT CAT, BITSAT, etc... Students can access the Application of Derivatives Class 12 solutions pdf download for free below;

Class 12 Math Chapter 6 Application of Derivatives Solution PDF: Free PDF Download

Class 12 Application of Derivatives Exercise 6.1 discusses the fundamental usage of Derivative techniques to solve various kinds of problems. Application of Derivatives Exercise 6.1 focuses on the rate of change of quantities, increasing and decreasing functions, tangents and normals, approximations, and finding maxima and minima. The concepts discussed in Class 12 Math Exercise 6.1 Solutions will help students get a good understanding of calculus and even provide tools to solve problems in Class 12 physics and other fields too. Class 12 Ch 6 Exercise 6.1 Solutions includes 18 Questions (10 Long, 6 Short, 2 MCQs). Students can check the complete solutions for Exercise 6.1 below;

Application of Derivatives Exercise 6.1 Solutions

Q1. Find the rate of change of the area of a circle with respect to its radius r when

$\frac{d}{dr}=\frac{d}{dr}$ (πr²) = 2πr.

A.1. (a) r = 3cm

When r = 3 cm,

$\frac{d}{dr}$ = 2 × π × 3 cm = 6π.

Thus, the area of the circle is changing at the rate of 6π cm $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

(b) r = 4cm

whenr = 4cm,

$\frac{d}{dr}$ = 2 × π × 4cm = 8π.

Thus, the area of the circle is changing at the rate of 8 $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

 

Q2. 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?

A.2. Let x be the length of edge,v be the value and s be the surface area of the cube then,

y = x3.

and S = 6x2, where x is a fxn of time.

Now,  $\frac{dY}{dF}=8c{m}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dt}$ (x3) = 8

$\frac{d{x}^3}{dx}·\frac{dx}{dt}=8$ (by chain rule)

$\Rightarrow$ 3x2  $\frac{dx}{dx}=8.$

$\frac{dx}{dt}=\frac{8}{3x^2.}$

Now,  $\frac{dS}{dt}=\frac{d}{dt}$ (bx2) =  $\frac{d\left(6x^2\right)}{dx}-\frac{dx}{dx}$ = 12x  $\frac{8}{3x^2}=\frac{32}{x}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

When x = 12 cm,

$\frac{dS}{dt}=\frac{32}{12}=\frac{8}{3}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

 

Q3. 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.

A.3. Let ‘r’ cm be the radius of the circle which is afxn of time.

Then,  $\frac{dy}{dt}$ = 8 3cm/s as it is increasing.

Now, the area A of the circle is A = πr2.

So, the rate at which the area of the circle change  $\frac{d}{dt}=\frac{d}{dt}$ πr2.

$=\frac{d}{dr}\pi r^2·\frac{dr}{dt}$

= 2πr 3

= 6πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 10cm,

$\frac{d}{dt}$ = 6.π × 10 = 60π ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

 

Q4. 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?

A.4. Let ‘x’ cm be the length of edge of the cube which is a fxⁿ of time t then,

$\frac{dx}{dt}$ = 3cm/s as it is increasing.

Now, volume v of the cube is v = x³

Ø Rate of change of volume of the cube $\frac{dv}{dt}=\frac{d{q}^3}{dt}.$

$=\frac{d{x}^3}{dx}\frac{dx}{dt}$

= 3x².3

= 9x² ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When x = 10cm.

$\frac{dv}{dt}$ = 9 x (10)²= 900 ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

 

Q5. 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?

A.5. The area A of the circle with radius π is A = πr²

Then, rate of change in area  $\frac{\mathrm{dA}}{dt}=\frac{d\stackrel{\dot{}}{x}{r}^2}{dt}$ =  $\frac{dx{r}^2}{dr}·\frac{dr}{dt}$

= 2πr  ${\frac{\mathrm{dr}\mathrm{<br>}}{dt}}_{.}$

Q The wave moves at a rate 5cm/s we have,

$\frac{dr}{\mathrm{dF}}$ = 5cm/s

So,  $\frac{\mathrm{dA}}{dt}$ =r. 5 = 10πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 8 cm.

$\frac{d}{dt}$ = 10.π.8 cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ = 80 × cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

 

Q6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

A.6. The circumference C of the circle with radius r is C = 2πr.

Then, rate of change in circumference is  $\frac{\mathrm{dC}}{dt}=\frac{d}{dt}2\pi r$ =  $2\pi \frac{dx}{df}.$

Q Radius of circle increases at rate 0.7 cm/s we get,

$\frac{dy}{dt}=0.7$ cm/s

So,  $\frac{\mathrm{dC}}{dt}$ = 2.× 0.7 cm/s = 1.4 × cm/s.

 

Q7. 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 = 8cm and y = 6cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

A.7. Since the length x is decreasing and the widthy is increasing with respect to time we have

$\frac{dx}{dt}$ = 5cm/min and  $\frac{dy}{dt}$ = A cm/min

(a) The perimeter P of a rectangle will be, P = 2(x + y)

Q  Rate of change of perimeter, $\frac{dP}{dt}=\frac{d^2}{dt}(x+y)$

$=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)$

= 2( -5 + 4)

= -2 cm/min

(b) The area A of the rectangle is A = x. y.

Rate of change of area is  $\frac{\mathrm{dA}}{dt}=\frac{d}{dt}(x·y)$

$=x\frac{dy}{dt}+\frac{dx}{dt}·y.$

= 4x - 5y

So,  $\frac{\mathrm{dA}}{dt}$ |x = 8ay = 6cm = 4(8) - 5(6) = 32 - 30 = 2 cm²/min spherical balloon

 

Q8. 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

A.8. Let ‘r’ cm be the radius of volume V. measured Then,

$V\; =\frac{4}{3}\pi r^3$

Now, rate at which balloon is being inflated = 900 ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{dv}{dr}=900$ ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$ = 900 ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dr}\left(\frac{4}{3}\times r^3\right)\frac{dr}{dt}=900$

$\frac{4}{3}$ $\pi$ × 3 × r² $\frac{dr}{dr}$ = 900.

$\Rightarrow \frac{dr}{dt}=\frac{900}{4\pi r^2}.$

When r = 15cm,

${\frac{ds}{dt}|}_{r=15}=\frac{900}{4\pi \times {(15)}^2}$ =  $\frac{900}{900\pi }=\frac{1}{\pi }$ cm/s.

Q Radius of balloon increases by $\frac{1}{\pi }$ per second.

 

Q9. 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

A.9. The volume v of a spherical balloon with radius r is V. $\frac{4}{3}\pi r^3.$

with respect its radius.

Then, the rate of change of volume  $|\frac{\mathrm{dV}}{dr}=\frac{d}{dr}\left(\frac{4}{3}·\pi r^3\right)$

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

$=\frac{4}{3}\pi \times 3\times {\pi }^2$

= 4 $\pi$ r²

Whenx = 10 cm,

$\frac{\mathrm{dV}}{dr}$ = 4 $\pi$10)² = 400 $\pi$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 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

A.10. Since, the bottom of ground is increasing with time t,

$\frac{dx}{dt}$ = 2cm/s

From fig, Δ ABC, by Pythagorastheorem

AB² + BC² = AC²

$\Rightarrow$ x² + y² = 5²

x² + y² = 25 ____ (1)

Differentiating eqⁿ (1) w. r. t. time t we get,

$\frac{d}{dt}\left(x^2+y^2\right)=\frac{d}{dt}(25)$

$\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.$

$\Rightarrow 2x\times 2+2y\frac{2y}{dy}=0$

$\Rightarrow \frac{dy}{dt}=-\frac{2x}{y}$ m/s

When x = 4m, the rate at which its height on the wall decreases is

$\frac{dy}{dt}=-\frac{2\times 4}{3}$  $\{\begin{array}{l}\therefore 4^2+y^2=5^2\\ \Rightarrow y^2=25-16\\ \Rightarrow y\text{\hspace{0.17em}√9}\left(L\text{\hspace{0.17em}engthcan'tbenegetive}\right)\\ \Rightarrow y=3\end{array}$

$\Rightarrow \frac{dy}{dt}=-\frac{8}{3}$ room

 

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

A.11. Given eqⁿ of the curve is 6y = x³ + 2.______ (1)

Wheny coordinate change s 8 times as fast as x-coordinate

$\frac{dy}{dx}$ = 8 _____ (2)

Now, differentiating eqⁿ (1) wrt.x we get,

$\frac{d}{dx}(6y)=\frac{d}{dx}\left(x^3+2\right)$

$\Rightarrow 6\frac{dy}{dx}=3x^2+0.$

6 × 8 = 3x² (using eqⁿ(2))

$\Rightarrow x^2=\frac{48}{3}=16$

$\Rightarrow +\surd x=\pm \surd 16$

x = $\pm$4.

When x = 4, we have, 6y = 4³+ 2 = 64 + 2 + 66

$\Rightarrow y=\frac{66}{6}$ y =11.

And when x = -4, we have, 6y = ( -4)3 + 2 = -64 + 2 = -62

$\Rightarrow y=\frac{62}{6}=\frac{31}{3}$

The tequired point s are (4, 11) and  $\left(-4,\frac{31}{3}\right)$

 

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?

A.12. Let x be the radius of the bubble with volume .V. then,

$\frac{dr}{dt}=\frac{1}{2}$ cm/s

andV =  $\frac{4}{3}\pi n^3$

Rate of change of volume  $\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}{\pi }^3\right)$ =  $\frac{4}{3}\pi \frac{d}{d{t}^4}{r}^3$

$=\frac{4}{3}\pi \times 3r^2\frac{dr}{dt}.$

= 4πr² × $\frac{1}{2}$

= 2πr².

${\frac{dv}{dt}|}_{r=cm}$ = 2x(1)² 2π. ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

 

Q13. A balloon, which always remains spherical, has a variable diameter 3 (2 1) 2 x + . Find the rate of change of its volume with respect to x.

A.13. Given, diameter of the spherical balloon =  $\frac{3}{2}$ (2x + 1)

So, radius of the spherical r =  $\frac{1}{2}\times \frac{3}{2}(2x+1)$

$=\frac{3}{4}(2x+1)$

Then, volume of the spherical V =  $\frac{4}{9}\pi r^3$

$=\frac{4}{3}\pi \times [\frac{3}{4}{\left(\frac{3}{(2x+1)}\right]}^3$

$=\frac{9\pi }{16}{(2x+1)}^3.$

Q Rate of change of volume wrt.tox, $\frac{dV}{dx}=\frac{d}{dx}\left[\frac{\pi }{16}{(2x+1)}^3\right]$

$=\frac{9\pi }{16}\times 3\times {(2x+1)}^2·\frac{d}{dx}(2x+1)$

$=\frac{27\pi }{16}{(2x+1)}^2\times 2=\frac{27\pi }{8}{(2x+1)}^2.$

 

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?

A.14. Let r cm and h cm be the radius and the height of the cone. Then,

h =  $\frac{1}{6}$ r. H = 6h

So, volume, V of the cone =  $\frac{1}{3}$ πr²h

$=\frac{1}{3}\pi {(6h)}^2\times h.$  $\frac{4d}{dt}.$

= 12 × h³

Rate of change of volume of the cone wrt the height is

$\frac{\mathrm{dV}}{dh}=\frac{d}{dh}\left(12\pi h^3\right)$ = 12 × π × 3 × h².

As the sand is pouring from the pipe at rate of 12 ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

we have

$\frac{d}{dt}=12$

$\Rightarrow \frac{dv}{dh}\times \frac{dh}{dt}=12$

$\Rightarrow 36\pi h^2\frac{dh}{dt}=12.$

$\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi h^2}=\frac{1}{3\pi h^2}.$

${\therefore \frac{dh}{dt}|}_{h=4 }=\frac{1}{3\pi \times {(4)}^2}=\frac{1}{48\pi }.$

Hence, the height is increasing at the rate of  $\frac{1}{48x}$ 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.

A.15. Given, c(x) = 0.007 x³- 0.003x² + 15x + 400.

Since the marginal cost is the rate of change of total cost wrt the output we have,

Marginal cost, MC, =  $\frac{\mathrm{dC}}{dx}(x)$

= 0.007 × 3x²- 0.003 × 2x + 15.

When x = 17,

Then, MC = 0.007 × 2. (17)² - 0.003 2(17) + 15.

= 6.069 - 0.102 + 15.

= 20.967

Hence, the required marginal cost = ` 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.

A.16. Given, R(x) = 13x² + 26x + 15.

Marginal revenue is the rate of change of total revenue with respect to the number of units sold Marginal revenue (MR) =  $\frac{dR(x)}{dx}$

$=\frac{d}{d}\left(13x^2+26x+15\right)$

= 13 × 2x + 26

= 26x + 26

When x = 7,

MR = 26 × 7 + 26 = 182 + 26 = 208.

Hence, the required marginal reverse = ` 208.

Choose the correct answer for questions 17 and 18.

 

Q17. 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

A.17. The area A of the isle with radius r is given by with respect to radius r A = πr².

Then, rate of change of area of the circle  $\frac{d·A}{dx}=\frac{d\pi r^2}{dr}$

= 2πr.

When r = 6 cm

$\frac{\mathrm{dA}}{dt}=2\pi \times 6=12\pi .$

Q option (B) is correct.

 

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

(A) 116 (B) 96 (C) 90 (D) 126

A.18. Given, R(x) = 3x² + 36x + 5.

Marginal revenue,  $\frac{d}{dx}R(x)=\frac{d}{dx}\left(3x^2+36x+5\right)$

= 3 × 2x + 36

= 6x + 36

When x = 15.

$\frac{d}{dx}R(x)=6\times 15+36$ = 90 + 36 = 126

Option (D) is correct.

Class 12 Application of Derivatives Exercise 6.2 Solutions will be a comprehensive guide for students covering all concepts, formulae, and more. Class 12 Math NCERT Exercise 6.2 focuses on advanced applications of derivatives, including determining the monotonicity of functions, identifying points of local maxima and minima, and solving related problems. These topics will be very helpful for the students preparing for competitive exams such as JEE, BITSAT, and more. Application of Derivatives Exercise 6.2 Solutions will include explanations of 19 Questions (10 Long, 7 Short, 2 MCQs). Students can check the complete solution of Exercise 6.2 below;

Application of Derivative Exercise 6.2 Solutions

Q1. Show that the function given by f (x) = 3x + 17 is increasing on R.

A.1. We have,.

f (x) = 3x + 17.

So, f (x) = 3 > 0 $\forall x\in R$

∴f (x) is strictly increasing on R.

 

Q2. Show that the function given by f (x) = e2x is increasing on R.

A.2. We have, f (x) = e^2x

So, f (x) =  $\frac{d}{dx}{e}^{2x}$ =  $e^{2x}\frac{d}{dx}2x$ = 2e^2x> 0  $\forall x\in R.$

∴f (x) is strictly increasing on R.

 

Q3. Show that the function given by f (x) = sin x is (a)

increasing in  $\left(0,\frac{\pi }{2}\right)$ (b) decreasing in  $\left(\frac{\pi }{2},\pi \right)$

(c) neither increasing nor decreasing in  $\left(0,\pi \right)$

A.3. We have, f (x) = sin x.

So, f (x) = cosx.

(a) when,x ∈ $\left(0,\frac{\pi }{2}\right)$ i e, x in 1st quadrat.

f(x) = cos.x> 0

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

(b) when,x ∈ $\left(\frac{\pi }{2},\pi \right)$ in IInd quadrat

f(x) = cosx< 0.

∴f(x) is strictly decreasing(π

(c) When ,x ∈(0,π).

f(x) = cosx is increasing in  $\left(0,\frac{\pi }{2}\right)$ and decreasing

in  $\left(\frac{\pi }{2},\pi \right)$ and f  $\left(\frac{\pi }{2}\right)$ = cos  $\frac{\pi }{2}=0.$

∴f (x) is neither increasing not decreasing in (0,π).

 

Q4. Find the intervals in which the function f given by f(x) = 2x² – 3x is

(a) increasing (b) decreasing

A.4. We have, f (x) = 2x² 3x

So, f(x) =  $\frac{d}{dx}\left(2x^2-3x\right)=4x-3.$

Atf (x) = 0.

$\Rightarrow$ 4x - 3 = 0

i e,  $x=\frac{3}{4}$ divides the real line into two

disjoint interval  $\left(-\infty ,\frac{3}{4}\right)\left(\frac{3}{4},\infty \right)$

(a) Now,

f(x) = 4x - 3 > 0 $\forall x\in \left(\frac{3}{4},\infty \right)$

So, f (x) is strictly increasing in  $\left(\frac{3}{4},\infty \right)$

(b) Now, f (x) = 4x - 3 < 0 $\forall x\left(-\infty ,\frac{3}{4}\right)$

So, f (x) is strictly decreasing in  $\left(-\infty ,\frac{3}{4}\right)$

 

Q5. Find the intervals in which the function f given by f(x) = 2x³ – 3x² – 36x + 7 is

(a) increasing (b) decreasing

A.5. We have, f (x) = 2x³- 3x²- 36x + 7.

So, f (x) =  $\frac{d}{dx}\left(2x^3-3x^2-36x+7\right)=6x^2-6x-36.$

= 6 (x²-x- 6).

= 6 (x²- 3x + 2x- 6)

= 6 [x (x- 3) + 2 (x- 3)]

= 6 (x- 3) (x + 2).

At, f (x) = 0

$\Rightarrow$ 6 (x- 3) (x + 2) = 0.

So, when x- 3 = 0 or x + 2 = 0.

$\Rightarrow$ x = 3 or x = -2.

Hence we an divide the real line into three disjoint internal

I  $(-\infty ,-2)\; II(-2,3)andIII(3,\infty )$

At x ∈ $(-\infty ,-2),$

f(x) = ( + ve) ( -ve) ( -ve) = ( + ve) > 0.

So, f (x) is strictly increasing in  $(-\infty ,-2)$

At, x∈( -2,3),

f(x) = ( + ve) ( + ve) ( -ve) = ( -ve) < 0.

So, f (x) is strictly decreasing in ( -2,3).

At, x ∈ $(3,\infty )$

f (x) = ( + ve) ( + ve) ( + ve) = ( + )ve> 0.

So, f (x) is strictly increasing in  $(3,\infty )$

∴ (a) f (x) is strictly increasing in  $(-\infty ,-2)and(3,-\infty )$

(b) f (x) is strictly decreasing in ( -2,3)

Q.6. Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x² + 2x – 5 (b) 10 – 6x – 2x² (c) –2x³ – 9x² – 12x + 1 (d) 6 – 9x – x²

(e) (x + 1)³ (x – 3)³ 

A.6. (a) f (x) = x² + 2x - 5.

f(x) = 2x + 2 = 2 (x + 1).

At, f(x) = 0

2 (x + 1) = 0

$\Rightarrow$ x = -1.

At, x  $(-\infty ,-1),$

f(x) = (- ) ve< 0.

So, f (x)is strictly decreasing or  $(-\infty ,-1).$

At x ∈ $(-1,\infty )$

f(x) = ( + ve) > 1

f(x) is strictly increasing on  $(-1,\infty ).$

(b) f(x) = 10 - 6x- 2x²

So, f(x) = - 6 - 4x = - 2 (3 + 2x).

Atf(x) = 0

$\Rightarrow$ 2 (3 + 2x) = 0.

$\Rightarrow$ x =  $\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

At x  $(-\infty ,\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.),$

∴f(x) is strictly increasing on  $\left(-\infty ,\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

At x  $\left(\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\infty \right)$

f(x) = ( -ve) ( + ve) = ( - ) ve< 0.

∴f (x) is strictly decreasing on  $\left(\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\infty \right)$

(c) f (x) = 2x³- 9x²- 12x.

So, f (x) =- 6x²- 18x- 12 = - 6 (x² + 3x + 2).

= -6 [x² + x + 2x + 2]

= -6 [x (x + 1) + 2 (x + 1)]

= -6 (x + 1) (x + 2)

At, f (x) = 0.

$\Rightarrow$ 6 (x + 1) (x + 2) = 0

$\Rightarrow$ x = -1 and x = -2.

At, x  $(-\infty ,-2)$ , f(x) = ( -ve) ( -ve) ( -ve) = ( -ve) < 0

So, f(x) is strictly decreasing.

At, x( -2, -1), f(x) = ( -ve) ( -ve) ( -ve) = ( -ve) < 0.

So, f(x) is strictly decreasing

(d)f(x) = 6 - 9x-x²

So, f(x) = - 9 - 2x

At, f(x) = 0

$\Rightarrow$ 9 - 2x = 0

$\Rightarrow$ x = $\raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

At, x ∈ $\left(-\infty ,\raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right),$

f(x) > 0

So, f(x) is strictly increasing.

At x ∈ $\left(\raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\infty \right).$

f(x) < 0.

So, f(x) is strictly decreasing.

(e) f(x) = (x + 1)³ (x- 3)³.

So, f(x) = (x + 1)³ $\frac{d}{dx}{(x-3)}^3+{(x-3)}^3\frac{d}{dx}{(x+1)}^3$

= (x + 1)³. 3 (x- 3)² + (x- 3)³. 3 (x + 1)²

= 3 (x + 1)² (x- 3)² (x + 1 + x- 3).

= 3 (x + 1)² (x- 3)² (2x- 2).

= 6 (x + 1)² (x- 3)² (x- 1).

For strictly increasing,

f(x) > 0.

As 6 > 0,(x + 1)²> 0 and (x- 3)²> 0. We need

(x 1) > 0.

$\Rightarrow$ x> 1.

∴f (x) is strictly increasing on  $(1,\infty ).$

And strictly decreasing on  $(-\infty ,1).$

Q7. Show that 2 log(1 ) 2 x y x x = +− + , x > – 1, is an increasing function of x throughout its domain.

A.7. We have, y = log  $(1+x)-\frac{2x}{2+x},x>-1$

Differentiating the above wrt.x.we get,

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

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{(2+x)\cdot 2-2x}{{(2+x)}^2}.$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{4+2x-2x}{{(2+x)}^2}=\frac{1}{1+x}-\frac{4}{{(2+x)}^2}.$

$=\frac{{(2+x)}^2-4(1+x)}{(1+x){(2+x)}^2}$

$=\frac{4+x^2+4x-4-4x}{(1+x){(2+x)}^2}$

$\frac{dy}{dx}=\frac{x^2}{(1+x){(2+x)}^2}.$

The given domain of the given function isx> -1.

$\Rightarrow$ (x + 1) > 0.

Also,  $x^2\ge 0$

(2 + x)²> 0.

Hence,  $\frac{dy}{dx}=\frac{(+ve)}{(+v)(+ve)}=(+ve)>0.$

∴ y is an increasing function of x throughout its domain.

Q8. Find the values of x for which y = [x(x – 2)]2 is an increasing function.

A.8. We have, y = [x (x- 2)]².

Differentiating the above w rt. x we get,

$\frac{dy}{dx}=\frac{d}{dx}[x]{(x-2)}^2$

$=2[x(x-2)]\frac{d}{dx}[x(x-2)]$

$=2[x(x-2)]\left[x\frac{d}{dx}(x-2)+(x-2)\frac{dx}{dx}\right]$

= 2 [x (x- 2)] (x + x- 2)

= 2x (x - 2) (2x - 2)

$\frac{dy}{dx}$ = 4x (x - 2) (x - 1).

Now,  $\frac{dy}{dx}=0$

$\Rightarrow$ 4x (x - 2) (x - 1) = 0.

i e, x = 0, x = 2, x = 1 divides the real line into

four disjoint interval.  $(-\infty ,0],$ [0,1],[1,2] and  $\left[2,\infty \right].$

when x  $\left[-\infty ,0\right].$

$\frac{dy}{dx}=(-ve)(-ve)(-ve)=$  $(-ve)\le 0,0x=0$

∴f (x) is decreasing in  $\left[-\infty ,0\right].$

When  $x\in [0,1]$

$\frac{dy}{dx}=(+ve)(-ve)(-ve)$  $=(+ve)\ge 0,x=0x=1$

∴f (x) is increasing in [0,1].

When x  $\in [1,2]$

$\frac{dy}{dx}=(+ve)(-ve)(+ve)=(-ve)\le 0,$  $\mathrm{0for}x=1+x=2$

∴f (x) is decreasing.

When  $x\in \left(2,\infty \right).$

$\frac{dy}{dx}=(+ve)(+ve)(+ve)=(+ve)\ge \mathrm{0,}0,forx=2$

∴f (x) is in creasing

Hence, f (x) is increasing for x  $\in [0,1)x\in [2,\infty )$

Q9. Prove that  $y=\frac{4sin\theta }{\left(2+cos\theta \right)}-\theta$ is an increasing function of θ  $\left(0,\frac{\pi }{2}\right)$ .

A.9. We have, y =  $\frac{4sin\theta }{(2+cos\theta )}-\theta$

Differentiating w rt.Ø we get,

$\frac{dy}{d\theta }=\frac{d}{d\theta }\left[\frac{4sin\theta }{2+cos\theta }-\theta \right]$

$=\frac{(2+cor\theta )\frac{d}{d\theta }(4sin\theta )-4sin\theta \frac{d}{d\theta }(2+cor\theta )}{{(2+cos\theta )}^2}-\frac{d\theta }{d\theta }.$

$=\frac{(2+cos\theta )(4cos\theta )-4sin\theta (-sin\theta )}{{(2+cos\theta )}^2}-1$

$=\frac{8cos\theta +4co{s}^2\theta +4si{n}^2\theta \cdot }{{(2+cos\theta )}^2.}-1$

$=\frac{8cos\theta +4\left(co{s}^2\theta +si{n}^2\theta \right)-{(2+cos\theta )}^2}{{(2+cos\theta )}^2}$

$=\frac{8cor\theta +4-\left[4+co{s}^2\theta +4cos\theta \right]}{{(2+cor\theta )}^2}.$

$=\frac{8cos\theta +4-4-co{s}^2\theta -4cos\theta }{{(2+cos\theta )}^2}$

$=\frac{4cos\theta -co{s}^2\theta }{{(2+cos\theta )}^2}=\frac{cos\theta (4-cos\theta )}{{(2+cos\theta )}^2}$

When  $\theta \in \left[0,\frac{\pi }{2}\right]$ we know that,  $0\le cos\theta \le 1.$

So,  $4-cos\theta >0$

And also, (2 + cos $\theta$)²> 0.

$\therefore \frac{dy}{d\theta }\ge 0\forall \theta \in \left[0,\frac{\pi }{2}\right]$

Hence, y is an increasing fxⁿ of $\theta$ in $\left[0,\frac{\pi }{2}\right]$

Q10. Prove that the logarithmic function is increasing on (0, ∞).

A.10. We have, f (x) = log x.

So, f (x) =  $\frac{d}{dx}(logx)=\frac{1}{x}.$

i e, f (x)  $=\frac{1}{x}$ > 0. When  $x\ne 0x\in (0,\infty )$

Hence, the logarithmic fx is strictly increasing on $(0,\infty ).$

Q11. Prove that the function f given by f(x) = x² – x + 1 is neither strictly increasing nor decreasing on (– 1, 1).

A.11. We have, f(x) = x²-x + 1.

So, f(x) =  $\frac{d}{dx}\left(x^2-x+1\right)=2x-1$

Atf(x) = 0.

$\Rightarrow$ 2x - 1 = 0

I e, x =  $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ divides the real line into two disjoint interval the interval ( -1, 1) into

Two disjoint interval

$\left(-1,\frac{1}{2}\right)\left(\frac{1}{2},1\right)$

And f(x) is strictly increasing in  $\left(\frac{1}{2},1\right)$ and strictly decreasing in  $\left(-1,\frac{1}{2}\right)$

Hence, f(x) is with a increasing or decreasing on ( -1,1).

 

Q12. Which of the following functions are decreasing on  $\left(0,\frac{\pi }{2}\right)$ ?

(A) cos x (B) cos 2x (C) cos 3x (D) tan x

A.12. (A) We have,

f(x) = cosx

So, f(x) = -sin x

When x  $\in \left[0,\frac{\pi }{2}\right)$ we know that sin x> 0.

$\Rightarrow$ -sinx< 0. f(x) < 0  $\forall x\in \left(0,\frac{\pi }{2}\right)$

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

(B) We have, f(x) = cos 2x

So, f(x) = -2sin 2x

When  $x\in \left(0,\frac{\pi }{2}\right)$ we know that sin x> 0.

i e, 0  $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.<br>$=> 0<2x <π < x<>

So, sin 2x> 0 (sinØ is ( +ve) in 1st and 2ndquadrant).

$\Rightarrow$ -2sin 2x< 0.

$\Rightarrow$ f (x) < 0.

∴f (x) is strictly decreasing on  $\left(0,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right).$

(c) We have, f(x) = cos 3x

$\Rightarrow$ f(x) = -3 sin 3x.

As  $0<x<\frac{\pi }{2}$

$\to 0<3x<\frac{3\pi }{2}$

We can divide the interval into two

Case I At, 0 < 3x

sin 3x> 0.

$\Rightarrow$ -3 sin 3x< 0  $\{\begin{array}{l}\because 0<Bx<\pi \hfill \end{array}$

$\begin{array}{l}\to 0<x<\frac{\pi }{3}.\hfill \end{array}$

$\Rightarrow$ f(x) < 0.

∴f(x) is strictly decreasing on  $\left(0\text{'}\frac{\pi }{3}\right)$

case II. At  $\pi <3x<\frac{3\pi }{2}weget(ii{i}^{rd}quadrout).$

sin 3x< 0.

$\Rightarrow$ -3 sin 3x> 0.

$\Rightarrow$ f(x) > 0.  $\left\{\because \pi <3x<\frac{3\pi }{2}\Rightarrow \frac{\pi }{3}<x<\frac{\pi }{2}\right\}$

∴f(x) is strictly increasing on  $\left(\frac{\pi }{3},\frac{\pi }{2}\right)$

Hence, f(x) is with a increasing or decreasing on  $\left(0,\frac{\pi }{2}\right).$

(d)we have, f(x) = tan x.

So, f(x) = sec²x.

$\Rightarrow$ f(x) = sec²x> 0  $\forall x\in \left(0,\frac{\pi }{2}\right).$

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

Hence, the fxⁿcosx and cos 2x are strictly decreasing on $\left(0,\frac{\pi }{2}\right).$

 

Q13. On which of the following intervals is the function f given by f(x) = x¹⁰⁰ + sin x –1 decreasing ?

(A) (0,1) (B) ,  $\left(\frac{\pi }{2},\pi \right)$ (C)  $\left(0,\frac{\pi }{2}\right)$ (D) None of these

A.13. We have, f (x) = x 100 + sin x - 1.

So, f (x) = 100x⁹⁹ + cosx.

(A) When x (0,1). We get.

x>0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100 x⁹⁹> 0.

Now, 0 radian = 0 degree

and 1 radian =  $\frac{180^{°}\times 1}{\pi }=\frac{180^{°}}{\left(\raisebox{1ex}{$22$}\!\left/ \!\raisebox{-1ex}{$7$}\right.\right)}=57^{°}.$

So, cosx> 0 for x∈ (0,1) radian = (0,57)

∴f (x) > 0 for x∈ (0,1).

(B) When x ∈ $\left(\frac{\pi }{2},\pi \right)$ we get,

So, x> 1

$\Rightarrow$ x⁹⁹> 1

$\Rightarrow$ 100x⁹⁹> 100.  $\left\{\begin{array}{cc}\hfill \because \left(\frac{\pi }{2},\pi \right)=(1.57,3.14)\hfill & \hfill which is great than 1\hfill \end{array}\right\}$

And cosx is negative between -1 and 0.

So, f (x) = 100x⁹⁹ + cosx> 100 - 1 = 99 > 0.

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

(c) When x ∈ $\left(0,\frac{\pi }{2}\right),$ we get,

$\Rightarrow$ x> 0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100x⁹⁹> 0

andcosx> 0. (firstquadrant).

I e, f(x) > 0.

∴f(x) is strictly increasing on  $\left(0,\frac{\pi }{2}\right)$

Hence, option (D) is correct.

Q14. For what values of a the function f given by f(x) = x² + ax + 1 is increasing on [1, 2]?

A.14. We have, f(x) = x² + x + 1

So, f(x) = 2x + a

If f(x) is strictly increasing onx  $(1,2),f^{\prime }(x)>0.$

So, 1

$\to 2\times 1<2\times x<2\times 2$

$\Rightarrow 2<2x<4$

$\Rightarrow 2+a<2x+a<4+a.$

$\Rightarrow 2+a<f^{\prime }(x)<4+a.$

The minimum value of f (x) is 2 + a and that of men value is 4 + a.

∴ 2 +a> 0and 4 + a> 0

$\Rightarrow$ a> -2 and a> -4.

$\Rightarrow$ a> -2.

∴The best value of a is -2.

 

Q15. Let I be any interval disjoint from [–1, 1]. Prove that the function f given by

$f\left(x\right)=x+\frac{1}{x}$ is increasing on I.

A.15. We have,f(x) =  $\stackrel{\dot{}}{x}+\frac{1}{x}$

So, f(x) =  $1-\frac{1}{x^2}=\frac{x^2-1}{x^2}=\frac{(x-1)(x+1)}{x^2}$

So, for every x∈ I, where I is disjoint from [-1,1] $$

f(x) =  $\frac{(+ve)(+ve)}{(+ve)}=(+ve)>0whenx>1.$

and f(x) =  $\frac{(-ve)(-ve)}{(+ve)}$ = ( + ve) > 0 when x< -1.

∴f(x) is strictly increasing on I .

 

Q16. Prove that the function f given by f(x) = log sin x is increasing on  $\left(0,\frac{\pi }{2}\right)$ and decreasing on  $\left(\frac{\pi }{2},\pi \right)$ .

A.16. We have, f(x) = log sin x

So, f(x) =  $\frac{1}{sinx}\frac{dsinx}{dx}=\frac{cosx}{sinx}=cotx$

When  $x\in \left(0,\frac{\pi }{2}\right)$

f(x) = cot x> 0 (Ist quadrat )

So, f(x) is strictly increasing on  $\left(0,\frac{\pi }{2}\right)$

When x ∈ $\left(\frac{\pi }{2},\pi \right)$

f(x) = cot x< 0. (IIndquadrant ).

So, f (x) is strictly decreasing on  $\left(\frac{\pi }{2},\pi \right)$

 

Q17. Prove that the function f given by f (x) = log |cos x| is decreasing on  $\left(0,\frac{\pi }{2}\right)$ and increasing on  $\left(3\frac{\pi }{2},2\pi \right)$

A.17. We have, f (x) = log  $\left|cosx\right|.$

f(x) =  $\frac{1}{cosx}\frac{d}{dx}x=-\frac{sinx}{cosx}=-tanx$

whenx  $\left(0,\frac{\pi }{2}\right),$ we get.

tanx> 0 (Ist quadrant).

$\Rightarrow$ tanx< 0

$\Rightarrow$ f(x) < 0.

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

When x ∈ $\left(\frac{3\pi }{2},2\pi \right)$ we get,

tanx|< |0 (ivth quadrant).

$\Rightarrow$ -tanx|>| 0

$\Rightarrow$ f(x) > 0

∴f(x) is increasing on  $\left(\frac{3\pi }{2},2\pi \right).$

 

Q18. Prove that the function given by f (x) = x³ – 3x² + 3x – 100 is increasing in R.

A.18. We have, f(x) = x³- 3x² 3x- 100.

So, f(x) = 3x²- 6x + 3 = 3 (x²- 2x + 1) = 3 (x- 1)²

For  $x\in R.$

(x- 10)² $\ge 0$ , 0 for x = 1.

$\Rightarrow$ 3 (x- 1)² $\ge 0$

$\Rightarrow$ f(x)  $\ge 0$

∴f(x) is increasing on  $ℝ$

 

Q19. The interval in which y = x² e^–x is increasing is (A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)

A.19. We have, f(x) = x² e^–x

So, f(x) =  $x^2\frac{d}{dx}{e}^{-x}+e^{-x}\frac{d{x}^2}{dx}$

$=x^2{e}^{-x}\frac{d}{dx}(-x)+e^{-x}2\cdot x\cancel{\frac{dx}{dx}}$

= -x² e^-x + e^-x 2x.

= x e^-x( x + 2).

$=\frac{x(2-x)}{e^{ex}}$

If f (x) = 0.

$\Rightarrow \frac{x(2-x)}{e^x}=0.$

$\Rightarrow$ x = 0, x = 2.

Hence, we get there disjoint interval

$[-\infty ,0),(0,2);(2,\infty )$

When, x  $(-\infty ,0),$ we have, f (x) = ( -ve) ( + ve) = ( -ve) < 0.

So, f is strictly decreasing.

When x ∈(0,2), f (x) = ( + ve) ( + ve) = ( + ve) > 0.

So, f is strictly increasing.

And when x ∈ $(2,\infty ),$ f (x) = ( +ve) ( -ve) = ( -ve) < 0.

So, f is strictly decreasing.

Hence, option (D) is correct.

 

Class 12 Application of Derivatives Exercise 6.3 focuses on several important concepts such as finding tangents and normals, their equations, and the geometric interpretation of slopes at specific points on a curve using the application of derivatives. Application of Derivatives Exercise 6.3 Solutions will have the explanation of 27 Questions including 14 Long, 11 Short, and 2 MCQs. Students can check the complete Class 12 Math Exercise 6.3 NCERT solutions below;

Application of Derivatives Exercise 6.3 Solution

Q1. Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.

A.1. The given eqⁿ of the curve is $y=3x^4-4x.$

Slope of the tangent at x = 4 is given by

${{\frac{dy}{dx}]}_{x=4}=12x^3-4]}_{x=4}=12{(4)}^3-4=12\times 64-4$

$=768-4$

= 764

 

Q2. Find the slope of the tangent to the curve  $y=\frac{x-1}{x-2},x\ne 2,at,x=10$

A.2. The given eqⁿ of the curve is $y=\frac{x-1}{x-2}$

Slope of tangent at x = 10 is given by,

${\frac{dy}{dx}|}_{x=10}={\frac{(x-2)\frac{d}{dx}(x-1)-(x-1)\frac{d}{dx}(x-2)}{{(x-2)}^2}|}_{x=10}$

$={\frac{(x-2)-(x-1)}{{(x-2)}^2}|}_{x=10}={\frac{x-2-x+1}{{(x-2)}^2}|}_{x=10}$

$={\frac{-1}{{(x-2)}^2}|}_{x=10}$

$=\frac{-1}{{(10-2)}^2}=\frac{-1}{8^2}=-\frac{1}{64}$

 

Q3. Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x- coordinate is 2.

A.3. Slope of tangent to the given curve  $y=x^3-x+1$ is

$\frac{dy}{dx}=3x^2-1.$

So,  ${\frac{dy}{dx}|}_{x=2}=3{(2)}^2-1=12-1=11.$

 

Q4. Find the slope of the tangent to the curve y = x³ –3x + 2 at the point whose x-coordinate is 3.

A.4. Slope of tangent to the given curve $y=x^3-3x+2$ is  $\frac{dy}{dx}=3x^2-3.$

so,  ${\frac{dy}{dx}|}_{x=3}=3{(3)}^2-3=27-3=24.$

 

Q5. Find the slope of the normal to the curve  $x=aco{s}^3\theta ,y=asi{n}^3\theta ,at,\theta =\frac{\pi }{4}$

A.5. The Equation of the given curve are

$x=aco{s}^3\theta y=asi{n}^3\theta$

So,  $\frac{dx}{d\theta }=3aco{s}^2\theta (-sin\theta )=-3aco{s}^2\theta sin\theta .$

and  $\frac{dy}{d\theta }=3asi{n}^2\theta cos\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{3asi{n}^2\theta cos\theta }{-3aco{s}^2\theta sin\theta }=-tan\theta$

So,  ${\frac{dy}{dx}|}_{x=\pi /4}=-tan\frac{\pi }{4}=-1$ which is the slope of the tanget to the curve.

Now, required slope of normal to the curve = $\frac{-1}{<br>\mathrm{Slopeoftangent}}$  $\frac{<br>}{}=\frac{-1}{-1}=1$

 

Q6. Find the slope of the normal to the curve  $x=1-sin\theta ,y=bco{s}^2\theta ,at,\theta =\frac{\pi }{2}$

A.6. The given eqⁿ of the curves are

$x=1-asin\theta y=bco{s}^2\theta$

so,  $\frac{dx}{d\theta }=-acos\theta \frac{dy}{d\theta }=-2bcos\theta sin\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{-2bcos\theta sin\theta }{acos\theta }=\frac{2b}{a}sin\theta$

Slope of tangent to curve at  $\theta =\frac{\pi }{2}$ is  ${\frac{dy}{dx}|}_{\theta =\frac{\pi }{2}}$

$=\frac{2b}{a}sin\frac{\pi }{2}$

$=\frac{2b}{a}$

Hence, slope of normal to curve  $=\frac{-1}{2b/a}=-\frac{a}{2b}$

 

Q7. Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.

A.7. The given eqⁿ of the curve is $y=x^3-3x^2-9x+7.$

slope of tangent to the given curve,  $\frac{dy}{dx}=3x^2-6x-9$

when the tangent is parallel to x-axis  $\frac{dy}{dx}=0$

$\Rightarrow 3x^2-6x-9=0$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow x^2+x-3x-3=0$

$\to x(x+1)-3(x+1)=0$

$\to (x+1)(x-3)=0$

$$ x = 3 or x = -1

When x = 3,  $y=3^3-3{(3)}^2-9(3)+7=27-27-27+7=-20$

And when x = -1 $y={(-1)}^3-3{(-1)}^2-9(-1)+7=-1-3+9+7=12$

Hence, the required points are  $(3,-20)(-1,12)$

 

Q8. 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)

A.8. Let the point joining the chord be  $(2,0)(4,4)$

Then slope of the chord  $=\frac{4-0}{4-2}\left\{\because Slope=\frac{y_2-y_1}{x_2-x_1}\right\}$

$=\frac{4}{2}$

= 2

The given eqⁿ of the curve $y={(x-2)}^2$ 

slope of the tangent to the curve  $\frac{dy}{dx}=2(x-2).$

Given that, the tangent is parallel to the chord PQ.

slope of tangent = slope of PQ.

$\Rightarrow 2(x-2)=2.$

$\Rightarrow x=1+2$

$\Rightarrow x=3.$

and  $y={(3-2)}^2=1^2=1.$

The required point on curve is  $(3,1)$

 

Q9. Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.

A.9. The given eqⁿ of the curve is $y=x^3-11x+5$

slope of tangent to the curve  $\frac{dy}{dx}=3x^2-11$

Then eqⁿ of tangent is $y=x-11$  $\Rightarrow x-y-11=0$ which gives us slope  $=\frac{-1}{-1}=1$

So,  $3x^2-11=1$

$\Rightarrow 3x^2=1+11=12$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

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

And when x = 2,  $y={(-2)}^3-11(-2)+5=-8+22+5=19.$

The point  $(2,-9)$ when put into  $y=x-11.$ we get

$-9=2-11$

$\Rightarrow -9=-9$ which is true.

and the point  $\left(-2,19\right)$ when put into  $y=x-11$ gives,

$19=-2-11$

$\Rightarrow 19=-13$ which is not true.

Hence, the required point is  $(2,-9)$ .

 

Q10. Find the equation of all lines having slope – 1 that are tangents to the curve  $y=\frac{1}{x-1},x\ne 1$

A10. The given eqⁿ of curve is $y=\frac{1}{x-1}$

Slope of tangent to the given curve is  $\frac{dy}{dx}=\frac{-1}{{(x-1)}^2}$

Given that, slope of tangent = 1.

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

$\Rightarrow {(x-1)}^2=1.$

$\Rightarrow x-1=\pm 1$

$\to x=1\pm 1.$

ie, X=1+1 $$ or  $x=1-1$

$\Rightarrow x=2$ or  $x=0$

When  $x=2,y=\frac{1}{2-1}=1$

and when  $x=0,y=\frac{1}{0-1}=-1.$

Hence, the point of contact of the tangents are  $(2,1)and(0,-1)$

The reqd. eqⁿ of line are $y-1=(-1)(x-2)\{\begin{array}{cc}\hfill \because y-y_0=m\left(x-x_0\right)\hfill & \hfill eqn{}^{} of line \hfill \end{array}$

and  $y-(-1)=(-1)(x-0).$

$\Rightarrow y-1=-x+2$ and  $y+1=-x$

$\to x+y-3=0$ and  $x+y+1=0.$

 

Q11. Find the equation of all lines having slope 2 which are tangents to the curve

$y=\frac{1}{x-3},x\ne 3$

A.11. The given eqⁿ of curve is $y=\frac{1}{x-3}$

Slope of tangent to the curve is  $\frac{dy}{dx}=\frac{-1}{{(x-3)}^2}.$

Given,  $\frac{dy}{dx}=2$

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

$\Rightarrow {(x-3)}^2=-\frac{1}{2}$ which is not possible

we conclude that there is no possible tangent to the given curve with slope = 2.

 

Q12. Find the equations of all lines having slope 0 which are tangent to the curve

$y=\frac{1}{x^2-2x+3}$

A.12. The given eqⁿ of the curve is $y=\frac{1}{x^2-2x+3}$

Slope of tangent to the curve is  $\frac{dy}{dx}=\frac{-1}{\left(x^2-2x+3\right)}=\frac{d}{dx}\left(x^2-2x+3\right)$

$=\frac{-(2x_1-2)}{{\left(x^2-2x+3\right)}^2}$

Given,  $\frac{dy}{dx}=0$

$\to \frac{-(2x-2)}{{\left(x^2-2x+3\right)}^2}=0$

$\Rightarrow -2(x-1)=0$

$\Rightarrow x=1$

When  $x=1,y=\frac{1}{1^2-2\times 1+3}=\frac{1}{1-2+3}=\frac{1}{2}$

The point of contact of the tangent to the curve is  $\left(1,\frac{1}{2}\right)$

The eqⁿ of the line is $y-\frac{1}{2}=0(x-1)$

$\to y=\frac{1}{2}$

 

Q13. Find points on the curve  $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are

(i) parallel to x-axis (ii) parallel to y-axis.

A.13. Diffrentiating  $\frac{x^2}{9}+\frac{y^2}{16}=1.$ wrt. X we get,

$\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$

$\Rightarrow \frac{2ydy}{16dx}=-\frac{2x}{9}$

$\Rightarrow \frac{dy}{dx}=\frac{-16x}{9y}$

(i) When the tangent is || to x-axis, the slope of tangent is 0

ie,  $\frac{dy}{dx}=0$

$\Rightarrow \frac{-16x}{9y}=0$

$\Rightarrow x=0$ putting this in the eqⁿ of curve. We get,

$\to \frac{0^2}{9}+\frac{y^2}{16}=1$

$\Rightarrow y^2=16$

$\Rightarrow y=\pm 4.$

The point at which the tangents are parallel to x-axis are  $(0,4)and(0,-4)$

(ii) when the tangent is parallel to y-axis, the slope of the normal is 0.

ie,  $\frac{-1}{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}=0$

$\Rightarrow -\frac{dx}{dy}=0$

$\Rightarrow \frac{9y}{16x}=0$

$\to y=0$ , putting this in the eqⁿ of curve we get,

$\frac{x^2}{9}+\frac{y^2}{16}=1.$

$\to x^2=9$

$\to x=\pm 3$

The point at which the tangents are parallel to y-axis are  $(3,0)and(-3,0)$

 

Q14. Find the equations of the tangent and normal to the given curves at the indicated points:

$\begin{array}{l}(i)y=x^4-6x^3+13x^2-10x+5,at,(0,5)\\ (ii)y=x^4-6x^3+13x^2-10x+5,at,(1,3)\\ (iii)y=x^3,at,(1,1)\\ (iv)y=x^2,at,(0,0)\\ (v)x=cosy,y=sint,at,t=\frac{\pi }{4}\end{array}$

A.14. (i) we have,  $y=x^4-6x^3+13x^2-10x+5$

slope of tangent,  $\frac{dy}{dx}=4x^3-18x^2+26x-10$

${\frac{dy}{dx}|}_{(x,y)=(0,5)}=-10.$

slope of normal  $=\frac{-1}{-10}=\frac{1}{10}$

Hence eqⁿ of tangent at (0, 5) is

$y-5=-10(x-0)\to 10x+y-5=0$

And eqⁿ of normal at (0, 5) is

$y-5=\frac{1}{10}(x-0)$

$\Rightarrow 10y-50=x$

$\Rightarrow x-10y+50=0$

(ii) We have,  $y=x^4-6x^3+13x^2-10x+5$

Slope of tangent,  $\frac{dy}{dx}=4x^3-18x^2+26x-10.$

${\frac{dy}{dx}|}_{(x,y)=(1,3)}=4{(1)}^3-18{(1)}^2+26(1)-10$

$=4-18+26-10$

= 30 28

= 2

Slope of normal  $=-\frac{1}{2}$

Hence eqⁿ of tangent at (1, 3) is

$y-3=2(x-1)$

$\Rightarrow y-3=2x-2$

$\to 2x-y+1=0$

And eqⁿ of normal at (1,3) is

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

$\Rightarrow 2y-6=-x+1$

$\to x+2y-7=0$

(iii) we have,  $y=x^3$

Slope of tangent,  $\frac{dy}{dx}=3x^2$

${\frac{dy}{dx}|}_{(1,1)}=3{(1)}^2=3.$

And slope of normal  $=-\frac{1}{3}$

Hence, eqn of tangent at (1, 1) is

$y-1=3(x-1)$

$\Rightarrow y-1=3x-3$

$\to 3x-y-2=0$

And eqⁿ of normal at (1,1) is

$y-1=-\frac{1}{3}(x-1)$

$\to 3y-3=-x+1$

$\Rightarrow x+3y-4=0.$

(iv) We have,  $y=x^2$

Slope of tangent  $\frac{dy}{dx}=2x$

${\frac{dy}{dx}|}_{(0,0)}=0.$

So, eqⁿ of the tangent at (0,0) is

$(y-0)=0(x-0)$

$\Rightarrow y=0.$

ie, x- axis

Hence, the eqⁿ of normal is x = 0 ie, y-axis

(v) We have,  $x=costy=sint$

$\frac{dx}{dt}=-sint·\frac{dy}{dt}=cost$

So, slope of tangent  $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{cost}{-sint}=-cott$

${\frac{dy}{dx}|}_{t=\frac{\pi }{4}}=-cot\frac{\pi }{4}=-1.$

And slope of normal  $=\frac{-1}{-1}=1$

 

Q15. 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

A.15. The eqⁿof the given curve is $y=x^2-2x+7$

Slope of tangent,  $\frac{dy}{dx}=2x-2$

(a) The line $2x-y+9=0\Rightarrow y=2x+9$ compared to  $y=mx+c$ gives,

Slope of line = 2.

If the tangent of the curve is parallel to the line

$\frac{dy}{dx}=slope\; of\; line$

$\to 2x-2=2$

$\to x=\frac{4}{2}\Rightarrow x=2$

When  $x=2,y={(2)}^2-2(2)+7=7$

Hence, the point of contact of the tangent is (2, 7)

The eqⁿ of tangent is $y-7=2(x-2)$

$\Rightarrow y-7=2x-4$

$\to 2x-y+3=0$

(b) The line $5y-15x=13\to y=\frac{15}{5}x+\frac{13}{5}\Rightarrow y=3x+\frac{13}{5}$

compared to  $y=mx+c$ gives

slope of line = 3

As the tangent to the curve is ⊥ to the line.

$\frac{dy}{dx}=\; -1/slope\; opf\; line$

$\Rightarrow 2x-2=-\frac{1}{3}$

$\Rightarrow 6x-6=-1$

$\to 6x=5$

$\to x=\frac{5}{6}$

When  $x=\frac{5}{6}$ we get  $y={\left(\frac{5}{6}\right)}^2-2\times \frac{5}{6}+7$

$=\frac{25}{36}-\frac{5}{3}+7$

$=\frac{25-60+252}{36}=\frac{217}{36}$

Hence, the point of contact of the tangent is  $\left(\frac{5}{6},\frac{217}{36}\right)$

And eqⁿ of the tangent is

$y-\frac{217}{36}=\frac{-1}{3}\left(x-\frac{5}{6}\right)$

$\Rightarrow 3y-\frac{217}{12}=-x+\frac{5}{6}$

$\Rightarrow x+3y-\frac{217}{12}-\frac{5}{6}=0$

$\Rightarrow x+3y-\frac{217-10}{12}=0$

$\Rightarrow x+3y-\frac{227}{12}=0$

$\Rightarrow 12x+36y-227=0.$

 

Q16. Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = – 2 are parallel.

A.16. The given eqⁿ of the curve is $y=7x^3+11$ .

Slope of tangent  $\frac{dy}{dx}=21x^2$

${\frac{dy}{dx}|}_{x=2}=21{(2)}^2=21\times 4=84.$

and  ${\frac{dy}{dx}|}_{x=-2}=21{(-2)}^2=21\times 4=84$

The tangent to the given curve at x = 2 and x = -2 are parallel.

 

Q17. Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point.

A.17. The given eqⁿ of the curve is $y=x^3$ .

Slope of tangent,  $\frac{dy}{dx}=3x^2$

As, slope of tangent = y – coordinate of the point.

$\frac{dy}{dx}=y$

$\Rightarrow 3x^2=x^3$

$\to 3x^2-x^3=0$

$\Rightarrow x^2(3-x)=0$

$\Rightarrow x=0x=3$

When  $x=0,y=0$

and when  $x=3·,y=3^3=27.$

The required points are  $(0,0)and(3,27)$

 

Q18. For the curve y = 4x³ – 2x⁵ , find all the points at which the tangent

passes through the origin.

A.18. The given eqⁿ of the curve is $y=4x^3-2x^5$

Slope of tangent,  $\frac{dy}{dx}=12x^2-10x^4.$ ________(1)

Let P(x, y) be the required point at the tangent passing through the origin (0,0)

Then,  $\frac{dy}{dx}=\frac{y-0}{x-0}=\frac{y}{x}$ _________(2)

So, from (1) and (2) we get,

$\frac{y}{x}=12x^2-10x^4.$

$\Rightarrow y=12x^3-10x^5.$

Putting this value of y in the eqⁿ of curve we get,

$12x^3-10x^5=4x^3-2x^5.$

$\to 8x^3-8x^5=0$

$\Rightarrow 8x^3\left(1-x^2\right)=0$

$\Rightarrow x^3=0$ or  $1-x^2=0$

$\Rightarrow x=0$ or  $x=\pm 1.$

When,  $x=0,y=4{(0)}^3-2{(0)}^5=0$

$x=1,y=4{(1)}^3-2{(1)}^5=4-2=2$

$x=-1,y=4{(-1)}^3-2{(-1)}^5=-4+2=-2$

The required points are (0,0), (1,2) and (1,2)

 

Q19. Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangents are parallel to the x-axis.

A.19. The given eqⁿof the curve is $x^2+y^2-2x-3=0$ ________(1)

Differentiating the given curve wrt.x we get,

$2x+2y\frac{dy}{dx}-2=0.$

$\Rightarrow 2y\frac{dy}{dx}=2-2x$

$\to \frac{dy}{dx}=\frac{1-x}{y}$ slope of tangent

Given, tangent is || to x-axis

ie,  $\frac{dy}{dx}=0$

$\Rightarrow \frac{1-x}{y}=0$

$\Rightarrow x=1$

Putting x = 1 in eqⁿ (1) we get,

$1^2+y^2-2\left(1_i\right)-3=0$

$\Rightarrow y^2-4=0$

$\Rightarrow y^2=4$

$\Rightarrow y=\pm 2$

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

 

Q20. Find the equation of the normal at the point (am² ,am³ ) for the curve ay² = x³

A.20. The given eqⁿ of curve is $a{y}^2=x^3$ _____(1)

Differentiating eqⁿ (1) wrt.x we get,

$2ay\frac{dy}{dx}=3x^2.$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2ay}$ , slope of tangent

${\frac{dy}{dx}|}_{(x,y)=\left(a{m}^2,a{m}^3\right)}=\frac{3{\left[a{m}^2\right]}^2}{2a\left[a{m}^3\right]}=\frac{3a^2{m}^4}{2a^2{m}^3}=\frac{3m}{2}$

corresponding slope of normal  $=\frac{-1}{(3m/2)}=-\frac{2}{3m}$

Hence, eqⁿ of normal at $\left(a{m}^2,a{m}^3\right)$ is

$y-a{m}^3=-\frac{2}{3m}\left(x-a{m}^2\right).$

$\Rightarrow 3my-3a{m}^4=-2x+2a{m}^2$

$\Rightarrow 2x+3my-3a{m}^4-2a{m}^2=0$

$\Rightarrow 2x+3my-a{m}^2\left(2+3m^2\right)=0$

 

Q21. Find the equation of the normals to the curve y = x³ + 2x + 6 which are

parallel to the line x + 14y + 4 = 0.

A.21. The given eqⁿ of the curve is $y=x^3+2x+6$ _____(1)

slope of tangent to the curve,  $\frac{dy}{dx}=3x^2+2$

so, slope of normal to the curve  $=\frac{-1}{3x^2+2}$

Now, the line  $x+14y+4=0\to y=-\frac{1}{14}x-\frac{4}{14}$ compared to  $y=mx+c$ gives

slope of line =  $-\frac{1}{14}$

As the normal is parallel to the line

$\Rightarrow \frac{-1}{3x^2+2}=-\frac{1}{14}$

$\to 3x^2+2=14$

$\Rightarrow 3x^2=12$

$\to x^2=4$

$\Rightarrow x=\pm 2$

When  $x=2,y={(2)}^3+2(2)+6=8+4+6=18$

and when  $x=-2,y={(-2)}^3+2(-2)+6=-8-4+6=-6$

The point of contact of the normal are (2, 18) and (-2, -6)

Hence the eqⁿ of normal are

$y-18=-\frac{1}{14}(x-2)$ and  $y-(-6)=\frac{-1}{14}[x-(-2)]$

$\Rightarrow 14y-252=-x+2$  $\Rightarrow y+6=\frac{-1}{14}(x+2)$

$\Rightarrow x+14y-254=0$  $\Rightarrow 14y+84=-x-2$

$\Rightarrow x+14y+86=0.$

 

Q22. Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at² , 2at).

A.22. The given of the parabola is  $y^2=4ax$

slope of tangent is given by

$\frac{d}{dx}{y}^2=\frac{d}{dx}4ax$

$\to 2y\frac{dy}{dx}=4a$

$\Rightarrow \frac{dy}{dx}=\frac{2a}{y}.$

${\frac{dy}{dx}|}_{\left(a{t}^2,2at\right)}=\frac{2a}{2at}=\frac{1}{t}$

so, slope of normal

so, slope of normal  $=\frac{-1}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$t$}\right.\right)}=-t$

Hence eqⁿ of tangent at point $\left(a{t}^2,2at\right)$ is

$y-2at=\frac{1}{t}\left(x-a{t}^2\right)$

$\to ty-2a{t}^2=x-a{t}^2$

$\to x-ty-a{t}^2+2a{t}^2=0$

$x-ty+a{t}^2=0$

And eqⁿ of normal at point $\left(a{t}^2,2at\right)$ is

$\left(y-2at\right)=\left(-t\right)\left(x-a{t}^2\right)$

$\to y-2at=-tx+a{t}^3$

$\Rightarrow tx+y-2at+a{t}^3=0$

$\Rightarrow tx+y-at\left(2+t^2\right)=0$

 

Q23. Prove that the curves x = y² and xy = k cut at right angles* if 8k² = 1.

A.23. The given eqⁿ of the curves are

$x=y^2$ ________(1)

and  $xy=x$ ___________(2)

Differentiating eqn (1) and (2) wrt ‘x’ we get,

$$  $\frac{d{y}^2}{dx}=\frac{dx}{dx}$

$\to 2y\frac{dy}{dx}=1$

$\to \frac{dy}{dx}=\frac{1}{2y}$ _________(3)

and  $\frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(x\right)$

$\to x\frac{dy}{dx}+y=0$

$\to \frac{dy}{dx}=-\frac{y}{x}$ _________(4)

Since the two curves cut each other at right angles we get,

$\left(\frac{1}{2y}\right)\times \left(\frac{-y}{x}\right)=-1.$

$\to 2x=1$ ________(5)

The point of intersection can be solve from eqn (1) and (2), 

$y^2.y=k$

$\to y^3=k$

$\to y=k^{1/3}$

$\therefore x=y^2=r^{2/3}.$

Hence, using eqⁿ (5) we get,

$2{\left(r\right)}^{2/3}=1.$

$\Rightarrow {\left[2r^{2/3}\right]}^3=1^3$

$\Rightarrow 8r^2=1$

Hence proved

 

Q24. 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  $(x_0,y_0)$

A.24. The given eqⁿ of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ______(1)

Differentiating eqn (1) wrt ‘x’ we get,

$\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0$

$\Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{2x}{a^2}$

$\Rightarrow \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$

${\frac{dy}{dx}|}_{\left(x,y\right)}=\left(x_0,y_0\right)\frac{b^2{x}_0}{a^2{y}_0}$ is the reqd slope of tangent to the curve

So, eqⁿ of tangent at point $\left(x_0,y_0\right)$ is

$y-y_0=\frac{b^2{x}_0}{a^2{y}_0}\left(x-x_0\right).$

$\Rightarrow \frac{y{y}_0}{b^2}-\frac{{y_0}^2}{b^2}=\frac{x{x}_0}{a^2}-\frac{{x_0}^2}{a^2}$

$\Rightarrow \frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=\frac{{x_0}^2}{a^2}-\frac{{y_0}^2}{b^2}$

As  $\left(x_0,y_0\right)$ lies on the parabola given by eqn (1) we write,

$\frac{{x_0}^2}{a^2}-\frac{{y_0}^2}{b^2}=1$

Hence,  $\frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1$

 

Choose the correct answer in Exercises 26 and 27.

 

Q26. The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is

(A) 3 (B)  $\frac{1}{3}$ (C) -3 (D) -  $\frac{1}{3}$

A.26. Given,  $y=2x^2+3sinx$

Slope of tangent,  $\frac{dy}{dx}=4x+3cosx$

${\frac{dy}{dx}|}_{x=0}=4\times 0+3\times cos0=3$

So, slope of normal  $=\frac{-1}{3}$

Option (D) is correct.

 

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)

A.27. The given eqn of curve is  $y^2=4x$ .

Then, differentiating wrt x we get,

$2y\frac{dy}{dx}=4$

$\Rightarrow \frac{dy}{dx}=\frac{2}{y}$ which is the slope of the tangent to the curve.

The line  $y=x+1$ compared to  $y=mx+c$ gives slope of line = 1.

Since, tangent is the line we have,

$\frac{2}{y}=1.$

$\Rightarrow y=2$

Putting y = 2 in  $y^2=4x$ we get,  $4x=2^2\Rightarrow x=1.$

Hence, the required point is  $\left(1,2\right)$

Option (A) is correct.