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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Continuity and Differentiability chapter are the following:

• Continuity and Differentiability
• Introduction
• Algebra of continuous functions
• Differentiability
• Derivatives of composite functions
• Derivatives of implicit functions
• Derivatives of inverse trigonometric functions
• Exponential and Logarithmic Functions
• Logarithmic Differentiation
• Derivatives of Functions in Parametric Forms
• Second Order Derivative
• Mean Value Theorem
• Summary

There are total eight exercises and one misc exercise(144 Questions fully solved) in the class 12th maths chapter 5 Continuity and Differentiability.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Find \(\\ \frac { dy }{ dx } \) in the following

Ex 5.3 Class 12 Maths Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,
\(2+3\frac { dy }{ dx } =cosx \)
=>\(\frac { dy }{ dx } =\frac { 1 }{ 3 } (cosx-2)\)

Ex 5.3 Class 12 Maths Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
\(2+3.\frac { dy }{ dx } =cosy\frac { dy }{ dx } \)
=>\(\frac { dy }{ dx } =\frac { 2 }{ cosy-3 } \)

Ex 5.3 Class 12 Maths Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,
\(a+2\quad by\quad \frac { dy }{ dx } =-siny\frac { dy }{ dx } \)
=>\(or\quad (2b+siny)\frac { dy }{ dx } =-a=>\frac { dy }{ dx } =-\frac { a }{ 2b+siny } \)

Ex 5.3 Class 12 Maths Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tanx + y
Differentiating w.r.t. x,

Ex 5.3 Class 12 Maths Question 5.
x² + xy + y² = 100
Solution:
x² + xy + xy = 100
Differentiating w.r.t. x,

Ex 5.3 Class 12 Maths Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
Given that
x³ + x²y + xy² + y³ = 81
Differentiating both sides we get

Ex 5.3 Class 12 Maths Question 7.
sin² y + cos xy = π
Solution:
Given that
sin² y + cos xy = π
Differentiating both sides we get
\(2\quad sin\quad y\frac { d\quad siny }{ dx } +(-sinxy)\frac { d(xy) }{ dx } =0\)

Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get

Ex 5.3 Class 12 Maths Question 9.
\(y={ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right) \)
Solution:
\(y={ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right) \)
put x = tanθ
\(y={ sin }^{ -1 }\left( \frac { 2tan\theta }{ { 1+tan }^{ 2 }\theta } \right) ={ sin }^{ -1 }(sin2\theta )=2\theta \)
\(y={ 2sin }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \)

Ex 5.3 Class 12 Maths Question 10.
\(y={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 3 } } <x<\frac { 1 }{ \sqrt { 3 } } \)
Solution:
\(y={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) \)
put x = tanθ

Ex 5.3 Class 12 Maths Question 11.
\(y={ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1 \)
Solution:
\(y={ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1 \)
put x = tanθ
\(y={ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta \)
\(y={ 2tan }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \)

Ex 5.3 Class 12 Maths Question 12.
\(y={ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1\)
Solution:
\(y={ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1\)
put x = tanθ
we get

Ex 5.3 Class 12 Maths Question 13.
\(y={ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ,-1<x<1 \)
Solution:
\(y={ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ,-1<x<1 \)
put x = tanθ
we get

Ex 5.3 Class 12 Maths Question 14.
\(y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } } \)
Solution:
\(y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } } \)
put x = tanθ
we get
\(y=sin^{ -1 }\left( 2sin\quad \theta \sqrt { 1-{ x }^{ 2 } } \right) \)
\(y=sin^{ -1 }\left( 2sin\theta \quad cos\theta \right) \quad ={ sin }^{ -1 }(sin2\theta )\quad =2\theta \)
\(y=2sin^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ \sqrt { { 1-x }^{ 2 } } } \)

Ex 5.3 Class 12 Maths Question 15.
\(y=sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } } \)
Solution:
\(y=sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } } \)
put x = tanθ
we get
\(y=sec^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) ={ sec }^{ -1 }\left( \frac { 1 }{ cos2\theta } \right) \)
\(y=sec^{ -1 }(sec2\theta )=2\theta ,\quad y=2{ cos }^{ -1 }x \)
\(\therefore \frac { dy }{ dx } =\frac { -2 }{ \sqrt { { 1-x }^{ 2 } } } \)

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Hindi Medium Ex 5.3

NCERT Class 12 Maths Solutions

• Chapter 1 Relations and Functions
• Chapter 2 Inverse Trigonometric Functions
• Chapter 3 Matrices
• Chapter 4 Determinants
• Chapter 5 Continuity and Differentiability
• Chapter 6 Application of Derivatives
• Chapter 7 Integrals Ex 7.1
• Chapter 8 Application of Integrals
• Chapter 9 Differential Equations
• Chapter 10 Vector Algebra
• Chapter 11 Three Dimensional Geometry
• Chapter 12 Linear Programming
• Chapter 13 Probability Ex 13.1

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Continuity and Differentiability chapter are the following:

• Continuity and Differentiability
• Introduction
• Algebra of continuous functions
• Differentiability
• Derivatives of composite functions
• Derivatives of implicit functions
• Derivatives of inverse trigonometric functions
• Exponential and Logarithmic Functions
• Logarithmic Differentiation
• Derivatives of Functions in Parametric Forms
• Second Order Derivative
• Mean Value Theorem
• Summary

There are total eight exercises and one misc exercise(144 Questions fully solved) in the class 12th maths chapter 5 Continuity and Differentiability.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

Differentiate the functions with respect to x in Questions 1 to 8.

Ex 5.2 Class 12 Maths Question 1.
sin(x² + 5)
Solution:
Let y = sin(x2 + 5),
put x² + 5 = t
y = sint
t = x²+5
\(\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(\frac { dy }{ dx } =cost.\frac { dt }{ dx } =cos({ x }^{ 2 }+5)\frac { d }{ dx } ({ x }^{ 2 }+5)\)
= cos (x² + 5) × 2x
= 2x cos (x² + 5)

Ex 5.2 Class 12 Maths Question 2.
cos (sin x)
Solution:
let y = cos (sin x)
put sinx = t
∴ y = cost,
t = sinx
∴\(\frac { dy }{ dx } =-sin\quad t,\frac { dt }{ dx } =cos\quad x \)
\(\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =(-sint)\times cosx\)
Putting the value of t, \(\frac { dy }{ dx } =-sin(sinx)\times cosx\)
\(\frac { dy }{ dx } =-[sin(sinx)]cosx\)

Ex 5.2 Class 12 Maths Question 3.
sin(ax+b)
Solution:
let = sin(ax+b)
put ax+bx = t
∴ y = sint
t = ax+b
\(\frac { dy }{ dt } =cost,\frac { dt }{ dx } =\frac { d }{ dx } (ax+b)=a\)
\(Now\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =cost\times a=acos\quad t\)
\(\frac { dy }{ dx } =acos(ax+b)\)

Ex 5.2 Class 12 Maths Question 4.
sec(tan(√x))
Solution:
let y = sec(tan(√x))
by chain rule
\(\frac { dy }{ dx } =sec(tan\sqrt { x } )tan(tan\sqrt { x } )\frac { d }{ dx } (tan\sqrt { x } )\)
\(\frac { dy }{ dx } =sec(tan\sqrt { x } ).tan(tan\sqrt { x } ){ sec }^{ 2 }\sqrt { x } .\frac { 1 }{ 2\sqrt { x } } \)

Ex 5.2 Class 12 Maths Question 5.
\(\\ \frac { sin(ax+b) }{ cos(cx+d) } \)
Solution:
y = \(\\ \frac { sin(ax+b) }{ cos(cx+d) } \) = \(\\ \frac { v }{ u } \)
u = sin(ax+b)

Ex 5.2 Class 12 Maths Question 6.
cos x³ . sin²(x⁵) = y(say)
Solution:
Let u = cos x³ and v = sin² x⁵,
put x³ = t

Ex 5.2 Class 12 Maths Question 7.
\(2\sqrt { cot({ x }^{ 2 }) } =y(say)\)
Solution:
\(2\sqrt { cot({ x }^{ 2 }) } =y(say)\)

Ex 5.2 Class 12 Maths Question 8.
cos(√x) = y(say)
Solution:
cos(√x) = y(say)
\(\frac { dy }{ dx } =\frac { d }{ dx } cos\left( \sqrt { x } \right) =-sin\sqrt { x } .\frac { d\sqrt { x } }{ dx } \)
\(=-sin\sqrt { x } .\frac { 1 }{ 2 } { (x) }^{ -\frac { 1 }{ 2 } }=\frac { -sin\sqrt { x } }{ 2\sqrt { x } } \)

Ex 5.2 Class 12 Maths Question 9.
Prove that the function f given by f (x) = |x – 1|,x ∈ R is not differential at x = 1.
Solution:
The given function may be written as
\(f(x)=\begin{cases} x-1,\quad if\quad x\ge 1 \\ 1-x,\quad if\quad x<1 \end{cases} \)
\(R.H.D\quad at\quad x=1\quad =\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h } \)

Ex 5.2 Class 12 Maths Question 10.
Prove that the greatest integer function defined by f (x)=[x], 0 < x < 3 is not differential at x = 1 and x = 2.
Solution:
(i) At x = 1
\(R.H.D=\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h } \)

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Hindi Medium Ex 5.2

NCERT Class 12 Maths Solutions

• Chapter 1 Relations and Functions
• Chapter 2 Inverse Trigonometric Functions
• Chapter 3 Matrices
• Chapter 4 Determinants
• Chapter 5 Continuity and Differentiability
• Chapter 6 Application of Derivatives
• Chapter 7 Integrals Ex 7.1
• Chapter 8 Application of Integrals
• Chapter 9 Differential Equations
• Chapter 10 Vector Algebra
• Chapter 11 Three Dimensional Geometry
• Chapter 12 Linear Programming
• Chapter 13 Probability Ex 13.1

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Get Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Determinants Exercise 4.6 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 4 Class 12 Determinants Ex 4.6 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Determinants chapter are the following:

Section Name	Topic Name
4	Determinants
4.1	Introduction
4.2	Determinant
4.3	Properties of Determinants
4.4	Area of a Triangle
4.5	Adjoint and Inverse of a Matrix
4.6	Applications of Determinants and Matrices
4.7	Summary

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Examine the consistency of the system of equations in Questions 1 to 6:

Ex 4.6 Class 12 Maths Question 1.
x + 2y = 2
2x + 3y = 3
Solution:
x + 2y = 2,
2x + 3y = 3
=> \(\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] \)
=> AX = B
Now |A| = \(\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}\)
= 3 – 4
= – 1 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 2.
2x – y = 5
x + y = 4
Solution:
2x – y = 5,
x + y = 4
=> \(\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 4 \end{matrix} \right] \)
=> AX = B
Now |A| = \(\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}\)
= 2 + 1
= 3 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 3.
x + 3y = 5,
2x + 6y = 8
Solution:
x + 3y = 5,
2x + 6y = 8
=> \(\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 8 \end{matrix} \right] \)
=> AX = B
Now |A| = \(\begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix}\)
= 6 – 6
= 0.

Hence, equations are consistent with no solution

Ex 4.6 Class 12 Maths Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
x + y + z = 1
2x + 3y + 2z = 2
x + y + z = \(\\ \frac { 4 }{ a } \)

Ex 4.6 Class 12 Maths Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:
\(\left[ \begin{matrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \)
=> AX = B

Ex 4.6 Class 12 Maths Question 6.
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
Solution:
Given
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
\(\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 2 \\ -1 \end{matrix} \right] \)
\(AX=B|A|=\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right] \)
= 5(18 + 10)+1(12 – 25)+4(-4-15)
= 140-13-76
= 51 ≠ 0
Hence equations are consistent with a unique
solution.

Solve system of linear equations using matrix method in Questions 7 to 14:

Ex 4.6 Class 12 Maths Question 7.
5x + 2y = 4
7x + 3y = 5
Solution:
The given system of equations can be written as

Ex 4.6 Class 12 Maths Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written

Ex 4.6 Class 12 Maths Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
The given system of equations can be written as
\(\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 7 \end{matrix} \right] i.e,,AX=B\)
where \(A=\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}\)

Ex 4.6 Class 12 Maths Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
The given system of equations can be written as
\(\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] i.e,,AX=B\)
where \(A=\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}\)

Ex 4.6 Class 12 Maths Question 11.
2x + y + z = 1,
x – 2y – z = 3/2
3y – 5z = 9
Solution:
The given system of equations are
2x + y + z = 1,
x – 2y – z = 3/2,
3y – 5z = 9
We know AX = B => X = A^-1B

Ex 4.6 Class 12 Maths Question 12.
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2.
Solution:
The given system of equations can be written
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 2 \end{matrix} \right] i.e,,AX=B\)

Ex 4.6 Class 12 Maths Question 13.
2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
The given system of equations can be written as:
\(\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] i.e,,AX=B \)

Ex 4.6 Class 12 Maths Question 14.
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12.
Solution:
The given system of equations can be written
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] i.e,,AX=B \)

Ex 4.6 Class 12 Maths Question 15.
If A = \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] \) Find A^-1. Using A^-1. Solve the following system of linear equations 2x – 3y + 5z = 11,3x + 2y – 4z = – 5, x + y – 2z = – 3
Solution:
We have AX = B
where \(A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)

Ex 4.6 Class 12 Maths Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method
Solution:
Let cost of 1 kg onion = Rs x
and cost of 1 kg wheat = Rs y
and cost of 1 kg rice = Rs z
4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Hindi Medium Ex 4.6

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 12 Vistas English
• NCERT Solutions Class 12th Flamingo English

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5

Get Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Determinants Exercise 4.5 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 4 Class 12 Determinants Ex 4.5 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Determinants chapter are the following:

Section Name	Topic Name
4	Determinants
4.1	Introduction
4.2	Determinant
4.3	Properties of Determinants
4.4	Area of a Triangle
4.5	Adjoint and Inverse of a Matrix
4.6	Applications of Determinants and Matrices
4.7	Summary

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5

Find the adjoint of each of the matrices in Questions 1 and 2.

Ex 4.5 Class 12 Maths Question 1.
\(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=A(say)\)
Solution:
Let C_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
C₁₁ = (-1)¹⁺¹ (4) = 4; C₁₂ = (-1)¹⁺² (3) = -3
C₂₁ = (-1)²⁺¹ (2)= – 2; C₂₂ = (-1)²⁺² (1) = -1
Adj A = \(\begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\)

Ex 4.5 Class 12 Maths Question 2.
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{matrix} \right] =A(say) \)
Solution:
\({ A }_{ 11 }={ (-1) }^{ 1+1 }M_{ 11 }=\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}=3\)
Similarly,

Verify A (adjA) = (adjA) •A = |A| I in Qs. 3 and 4.
Ex 4.5 Class 12 Maths Question 3.
\(\begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix}=A(say)\)
Solution:
|A| = 24

Ex 4.5 Class 12 Maths Question 4.
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right] =A(say)\)
Solution:
A₁₁ = 0, A₁₂ = – 11, A₁₃ = 0,
A₂₁ = – 3, A₂₂ = 1, A₂₃ = 1, A₃₁ = – 2
A₃₂ = 8, A₃₃ = – 3

Find the inverse of each of the matrices (if it exists) given in Questions 5 to 11:

Ex 4.5 Class 12 Maths Question 5.
\(\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}=A(say)\)
Solution:
\(\left| A \right| =\begin{vmatrix} 2 & -2 \\ 4 & 3 \end{vmatrix}=6+8=14\neq 0\)
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by

Ex 4.5 Class 12 Maths Question 6.
\(\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}=A(say) \)
Solution:
\(\left| A \right| =\begin{vmatrix} -1 & 5 \\ -3 & 2 \end{vmatrix}=-2+15=13\neq 0 \)
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by

Ex 4.5 Class 12 Maths Question 7.
\(\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A\)
Solution:
|A| = 10
\(\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A\)

Ex 4.5 Class 12 Maths Question 8.
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right] =A\)
Solution:
\(\left| A \right| =\left| \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right| =1\left| \begin{matrix} 3 & 0 \\ 2 & -1 \end{matrix} \right| =-3\neq 0\)
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by

Ex 4.5 Class 12 Maths Question 9.
\(\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A\)
Solution:
|A| = \(\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A\)
= 2(-1-0)-1(4-0)+3(8-3)
So, A is non-singular matrix and therefore, it is invertible.

Ex 4.5 Class 12 Maths Question 10.
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A\)
Solution:
|A| = \(\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A\)
= 1(8-6)+1(0+9)+2(0-6)
= 2+9-12
= -1≠0
∴A is invertible and
\({ A }^{ -1 }=\frac { Adj\quad A }{ |A| } \)

Ex 4.5 Class 12 Maths Question 11.
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right] \)
Solution:
A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right] \)
adj A = \(\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -cos\alpha & -sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{matrix} \right] \)
First find |A| = -cos²α-sin²α
=-1≠0

Ex 4.5 Class 12 Maths Question 12.
Let \(A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix},B=\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\), verify that (AB)^-1 = B^-1A^-1
Solution:
Here |A| = \(A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \)
= 15-14
= 1≠0
\(Adj A=\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \)

Ex 4.5 Class 12 Maths Question 13.
If \(A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) show that A² – 5A + 7I = 0,hence find A^-1
Solution:
A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)
A² = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \)

Ex 4.5 Class 12 Maths Question 14.
For the matrix A = \(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \) find file numbers a and b such that A²+aA+bI²=0. Hence, find A^-1.
Solution:
A = \(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \)
A²+aA+bI²=0

Ex 4.5 Class 12 Maths Question 15.
For the matrix \(A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{matrix} \right] \) Show that A³-6A²+5A+11I₃=0.Hence find A^-1
Solution:
A² = \( \left[ \begin{matrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{matrix} \right] \)

Ex 4.5 Class 12 Maths Question 16.
If \(A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \) show that A³-6A²+9A-4I=0 and hence, find A^-1
Solution:
We have
\(A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \)

Ex 4.5 Class 12 Maths Question 17.
Let A be a non-singular square matrix of order 3×3. Then | Adj A | is equal to:
(a) | A |
(b) | A |²
(c) | A |³
(d) 3 | A |
Solution:
Let A = \(\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right] \)


Dividing by | A |, |Adj. A| = | A |²
Hence, Part (b) is the correct answer.

Ex 4.5 Class 12 Maths Question 18.
If A is an invertible matrix of order 2, then det. (A^-1) is equal to:
(a) det. (A)
(b) \(\\ \frac { 1 }{ det.(A) } \)
(c) 1
(d) 0
Solution:
|A|≠0
=> A^-1 exists => AA^-1 = I
|AA^-1| = |I| = I
=> |A||A^-1| = I
\(|{ A }^{ -1 }|=\frac { 1 }{ |A| } \)
Hence option (b) is correct.

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Hindi Medium Ex 4.5

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
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• NCERT Solutions Class 12 Hindi
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• NCERT Solutions Class 12th Flamingo English

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4

Get Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Determinants Exercise 4.4 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 4 Class 12 Determinants Ex 4.4 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Determinants chapter are the following:

Section Name	Topic Name
4	Determinants
4.1	Introduction
4.2	Determinant
4.3	Properties of Determinants
4.4	Area of a Triangle
4.5	Adjoint and Inverse of a Matrix
4.6	Applications of Determinants and Matrices
4.7	Summary

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4

Ex 4.4 Class 12 Maths Question 1.
Write the minors and cofactors of the elements of following determinants:
(i) \(\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\)
(ii) \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\)
Solution:
(i) Let A = \(\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\)
M₁₁ = 3, M₁₂ = 0, M₂₁ = – 4, M₂₂ = 2
For cofactors

Ex 4.4 Class 12 Maths Question 2.
Write Minors and Cofactor of elements of following determinant
(i) \(\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right| \)
(ii) \(\left| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{matrix} \right| \)
Solution:
(i) Minors M₁₁ = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1

Ex 4.4 Class 12 Maths Question 3.
Using cofactors of elements of second row, evaluate
\(\Delta =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right| \)
Solution:
Given
\(\Delta =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right| \)

Ex 4.4 Class 12 Maths Question 4.
Using Cofactors of elements of third column, evaluate
\(\Delta =\left| \begin{matrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{matrix} \right| \)
Solution:
Elements of third column are yz, zx, xy

Ex 4.4 Class 12 Maths Question 5.
If \(\Delta =\left| \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right| \) and Aij is the cofactors of aij? then value of ∆ is given by
(a) a₁₁A₃₁+a₁₂A₃₂+a₁₃A₃₃
(b) a₁₁A₁₁+a₁₂A₂₁+a₁₃A₃₁
(c) a₂₁A₁₁+a₂₂A₁₂+a₂₃A₁₃
(d) a₁₁A₁₁+a₂₁A₂₁+a₃₁A₃₁
Solution:
Option (d) is correct.

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Hindi Medium Ex 4.4

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 12 Vistas English
• NCERT Solutions Class 12th Flamingo English