JEE Exam > JEE Notes > Mathematics (Maths) Class 12 > Sample Solution Paper 5 - Math, Class 12
Sample Solution Paper 5 - Math, Class 12 | Mathematics (Maths) Class 12 - JEE PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
CBSE XII | Mathematics
Sample Paper 5 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5
SECTION – A
1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??
For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
2. We have,
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
Page 2
CBSE XII | Mathematics
Sample Paper 5 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5
SECTION – A
1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??
For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
2. We have,
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
CBSE XII | Mathematics
Sample Paper 5 Solution
OR
? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is
x 2 y 1 z 3
3 5 6
? ? ?
??
Page 3
CBSE XII | Mathematics
Sample Paper 5 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5
SECTION – A
1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??
For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
2. We have,
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
CBSE XII | Mathematics
Sample Paper 5 Solution
OR
? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is
x 2 y 1 z 3
3 5 6
? ? ?
??
CBSE XII | Mathematics
Sample Paper 5 Solution
Section-B
5. We shall prove by principle of mathematical induction
Here, let
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??
? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?
Assuming result to be true for n = k i.e. P(k) to be true
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??
We have to prove P (k +1) is true,
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??
? P (k + 1) is true.
Thus by principle of mathematical induction
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
for all nN ?
Page 4
CBSE XII | Mathematics
Sample Paper 5 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5
SECTION – A
1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??
For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
2. We have,
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
CBSE XII | Mathematics
Sample Paper 5 Solution
OR
? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is
x 2 y 1 z 3
3 5 6
? ? ?
??
CBSE XII | Mathematics
Sample Paper 5 Solution
Section-B
5. We shall prove by principle of mathematical induction
Here, let
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??
? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?
Assuming result to be true for n = k i.e. P(k) to be true
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??
We have to prove P (k +1) is true,
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??
? P (k + 1) is true.
Thus by principle of mathematical induction
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
for all nN ?
CBSE XII | Mathematics
Sample Paper 5 Solution
OR
? Since A is skew symmetric matrix.
??
? ? ?
? ? ?
? ? ?
? ? ? ? ?
??
??
T
T
T3
T
T
Therefore,A A
AA
A ( 1) A
AA
A A ......(Since A A )
2 A 0
A0
6. Let P(x, y) be any point on the given curve x
2
+ y
2
– 2x – 3 = 0.
Tangent to the curve at the point (x, y) is given by
dy
dx
.
Differentiating the equation of the curve w .r. t. x we get
dy
2x 2y 2 0
dx
dy 2 2x 1 x
dx 2y y
? ? ?
??
??
Let P(x1, y1) be the point on the given curve at which the tangents are parallel to the x
axis
? ?
dy
dx
x ,y
11
?
?
?
?
= 0
?
1x
1
y
1
?
= 0
? 1 – x1 = 0
? x1 = 1
To get the value of y1 just substitute x1 = 1 in the equation x
2
+ y
2
– 2x – 3 = 0, we get
? ?
2
2
1 y 2 1 3 0
1
2
y 4 0
1
2
y4
1
y2
1
? ? ? ? ?
? ? ?
??
? ? ?
So the points on the given curve at which the tangents are parallel to the x-axis are
(1, 2) and (1, -2).
Page 5
CBSE XII | Mathematics
Sample Paper 5 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5
SECTION – A
1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??
For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
2. We have,
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
CBSE XII | Mathematics
Sample Paper 5 Solution
OR
? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is
x 2 y 1 z 3
3 5 6
? ? ?
??
CBSE XII | Mathematics
Sample Paper 5 Solution
Section-B
5. We shall prove by principle of mathematical induction
Here, let
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??
? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?
Assuming result to be true for n = k i.e. P(k) to be true
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??
We have to prove P (k +1) is true,
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??
? P (k + 1) is true.
Thus by principle of mathematical induction
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
for all nN ?
CBSE XII | Mathematics
Sample Paper 5 Solution
OR
? Since A is skew symmetric matrix.
??
? ? ?
? ? ?
? ? ?
? ? ? ? ?
??
??
T
T
T3
T
T
Therefore,A A
AA
A ( 1) A
AA
A A ......(Since A A )
2 A 0
A0
6. Let P(x, y) be any point on the given curve x
2
+ y
2
– 2x – 3 = 0.
Tangent to the curve at the point (x, y) is given by
dy
dx
.
Differentiating the equation of the curve w .r. t. x we get
dy
2x 2y 2 0
dx
dy 2 2x 1 x
dx 2y y
? ? ?
??
??
Let P(x1, y1) be the point on the given curve at which the tangents are parallel to the x
axis
? ?
dy
dx
x ,y
11
?
?
?
?
= 0
?
1x
1
y
1
?
= 0
? 1 – x1 = 0
? x1 = 1
To get the value of y1 just substitute x1 = 1 in the equation x
2
+ y
2
– 2x – 3 = 0, we get
? ?
2
2
1 y 2 1 3 0
1
2
y 4 0
1
2
y4
1
y2
1
? ? ? ? ?
? ? ?
??
? ? ?
So the points on the given curve at which the tangents are parallel to the x-axis are
(1, 2) and (1, -2).
CBSE XII | Mathematics
Sample Paper 5 Solution
7. Here,
33
x asin t , y bcos t ??
Differentiating (1) wrt t
dx
2
3asin t cost
dt
?? and
dy
2
3bcos t sint
dt
? ? ?
dy
2
dy 3bcos t sint b
dt
cot t
2
dx
dx a
3asin t cost
dt
??
? ? ? ?
?
? Slope of the tangent at t
2
?
?
dy b
cot 0
dx a 2
2
? ?
? ? ?
?
?
?
Hence, equation of tangent is given by
3
y bcos 0 or y 0
2
?
? ? ?
8. The volume of a sphere(V) with radius (r) is given by,
4
3
Vr
3
??
?Rate of change of volume (V) w.r.t. (t) is given by,
?
??
??
??
??
??
?
? ? ?
??
??
?
?
dV dV dr
.
dt dr dt
d 4 dr
3
r.
dr 3 dt
dr
2
4 r .
dt
dV
3
It is given that 900 cm / s
dt
dr 900 225
22
dt
4 r r
Therefore, when radius = 15 cm
dr 225 1
2
dt
(15)
Hence,the rate at which the radius of the balloon
increases whe
?
1
n the radius is 15 cm is cm / s.
Read More
|
204 videos|290 docs|139 tests
|
FAQs on Sample Solution Paper 5 - Math, Class 12 - Mathematics (Maths) Class 12 - JEE
| 1. What are the different types of integrals in calculus? |  |
Ans. In calculus, there are mainly two types of integrals: indefinite integrals and definite integrals. Indefinite integrals give the antiderivative of a function, while definite integrals calculate the area under a curve between two specific points.
| 2. How do you find the derivative of a function? | |
Ans. To find the derivative of a function, you can use differentiation rules such as the power rule, product rule, quotient rule, and chain rule. These rules allow you to find the rate of change of a function at a given point.
| 3. What is the relationship between integrals and derivatives? | |
Ans. The relationship between integrals and derivatives is given by the Fundamental Theorem of Calculus. It states that if a function f(x) is continuous on an interval [a, b], and F(x) is its antiderivative, then the definite integral of f(x) from a to b is equal to F(b) - F(a). In other words, the integral of a function gives you its antiderivative.
| 4. How do you solve a limit in calculus? | |
Ans. To solve a limit in calculus, you can use various techniques such as direct substitution, factoring, rationalization, L'Hôpital's rule, and special limit theorems. These methods help determine the value that a function approaches as it gets arbitrarily close to a certain point.
| 5. What is the concept of continuity in calculus? | |
Ans. Continuity in calculus refers to the smoothness and unbroken nature of a function. A function is continuous at a point if it is defined at that point, and the limit of the function as x approaches that point exists and is equal to the value of the function at that point. Continuity is an important concept in calculus as it allows for the application of many fundamental theorems and techniques.
About this Document
|
4.90/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Sample Solution Paper 5 - Math, Class 12 for JEE 2024 is part of Mathematics (Maths) Class 12 preparation.
The notes and questions for Sample Solution Paper 5 - Math, Class 12 have been prepared according to the JEE exam syllabus. Information about Sample Solution Paper 5 - Math, Class 12 covers topics
like and Sample Solution Paper 5 - Math, Class 12 Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Sample Solution Paper 5 - Math, Class 12.
Introduction of Sample Solution Paper 5 - Math, Class 12 in English is available as part of our Mathematics (Maths) Class 12
for JEE & Sample Solution Paper 5 - Math, Class 12 in Hindi for Mathematics (Maths) Class 12 course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: Sample Solution Paper 5 - Math, Class 12 | Mathematics (Maths) Class 12 - JEE
Description
Full syllabus notes, lecture & questions for Sample Solution Paper 5 - Math, Class 12 | Mathematics (Maths) Class 12 - JEE - JEE | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 12 | Best notes, free PDF download
Information about Sample Solution Paper 5 - Math, Class 12
In this doc you can find the meaning of Sample Solution Paper 5 - Math, Class 12 defined & explained in the simplest way possible. Besides explaining types of
Sample Solution Paper 5 - Math, Class 12 theory, EduRev gives you an ample number of questions to practice Sample Solution Paper 5 - Math, Class 12 tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Sample Solution Paper 5 - Math
,Viva Questions
,shortcuts and tricks
,Class 12 | Mathematics (Maths) Class 12 - JEE
,mock tests for examination
,ppt
,MCQs
,past year papers
,Summary
,video lectures
,Previous Year Questions with Solutions
,Free
,Class 12 | Mathematics (Maths) Class 12 - JEE
,study material
,Objective type Questions
,Class 12 | Mathematics (Maths) Class 12 - JEE
,practice quizzes
,Important questions
,Sample Solution Paper 5 - Math
,Sample Paper
,Exam
,Sample Solution Paper 5 - Math
,Extra Questions
,Semester Notes
;
Additional Information about Sample Solution Paper 5 - Math, Class 12 for JEE Preparation
Sample Solution Paper 5 - Math, Class 12 Free PDF Download
The Sample Solution Paper 5 - Math, Class 12 is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Sample Solution Paper 5 - Math, Class 12 now and kickstart your journey towards success in the JEE exam.
Importance of Sample Solution Paper 5 - Math, Class 12
The importance of Sample Solution Paper 5 - Math, Class 12 cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Sample Solution Paper 5 - Math, Class 12 Notes
Sample Solution Paper 5 - Math, Class 12 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Sample Solution Paper 5 - Math, Class 12.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Sample Solution Paper 5 - Math, Class 12 Notes on EduRev are your ultimate resource for success.
Sample Solution Paper 5 - Math, Class 12 JEE Questions
The "Sample Solution Paper 5 - Math, Class 12 JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Sample Solution Paper 5 - Math, Class 12 on the App
Students of JEE can study Sample Solution Paper 5 - Math, Class 12 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Sample Solution Paper 5 - Math, Class 12,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Sample Solution Paper 5 - Math, Class 12 is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup