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CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
Page 2
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
Page 3
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
CBSE XII | Mathematics
Sample Paper 4 Solution
Section-B
5. Let A = IA
1 1 2 3
2 2 1 3 3 1
23
2 3 3 1 0 0
2 2 3 0 1 0 A
3 2 2 0 0 1
R R R R
1 1 4 1 1 1
2 2 3 0 1 0 A
3 2 2 0 0 1
Applying R R 2R ,R R 3R
1 1 4 1 1 1
0 0 5 2 1 2 A
0 5 10 3 3 4
RR
1 1 4
0 5 10
0 0 5
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
???
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ?
?
??
??
? ? ?
??
?? ?
??
1 1 1
3 3 4 A
2 1 2
? ??
??
??
??
?? ??
??
2 2 3 3
1 1 3 2 2 3
11
R R , R R
55
1 1 1
1 1 4
3 3 4
0 1 2 A
5 5 5
0 0 1
2 1 2
5 5 5
R R 4R ,R R 2R ,
3 1 3
5 5 5
1 1 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
??
??
??
?? ?
??
??
?
??
??
?
??
??
??
??
??
?
??
??
? ? ? ?
??
?
??
????
?
????
?
????
????
??
?
??
??
??
Page 4
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
CBSE XII | Mathematics
Sample Paper 4 Solution
Section-B
5. Let A = IA
1 1 2 3
2 2 1 3 3 1
23
2 3 3 1 0 0
2 2 3 0 1 0 A
3 2 2 0 0 1
R R R R
1 1 4 1 1 1
2 2 3 0 1 0 A
3 2 2 0 0 1
Applying R R 2R ,R R 3R
1 1 4 1 1 1
0 0 5 2 1 2 A
0 5 10 3 3 4
RR
1 1 4
0 5 10
0 0 5
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
???
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ?
?
??
??
? ? ?
??
?? ?
??
1 1 1
3 3 4 A
2 1 2
? ??
??
??
??
?? ??
??
2 2 3 3
1 1 3 2 2 3
11
R R , R R
55
1 1 1
1 1 4
3 3 4
0 1 2 A
5 5 5
0 0 1
2 1 2
5 5 5
R R 4R ,R R 2R ,
3 1 3
5 5 5
1 1 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
??
??
??
?? ?
??
??
?
??
??
?
??
??
??
??
??
?
??
??
? ? ? ?
??
?
??
????
?
????
?
????
????
??
?
??
??
??
CBSE XII | Mathematics
Sample Paper 4 Solution
1 1 2
1
R R R
23
0
55
1 0 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
23
0
55
11
Hence A 0
55
2 1 2
5 5 5
?
??
???
??
????
?
????
?
????
????
??
?
??
??
??
???
??
??
?
??
?
??
??
?
??
??
??
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.
? ?
11
x ,y
dy
dx
?
?
?
?
= 0
?
1
1
1x
y
?
= 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
2
1
2
1
1
1 y 2 1 3 0
y 4 0
y4
y2
? ? ? ? ?
? ? ?
??
? ? ?
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 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
CBSE XII | Mathematics
Sample Paper 4 Solution
Section-B
5. Let A = IA
1 1 2 3
2 2 1 3 3 1
23
2 3 3 1 0 0
2 2 3 0 1 0 A
3 2 2 0 0 1
R R R R
1 1 4 1 1 1
2 2 3 0 1 0 A
3 2 2 0 0 1
Applying R R 2R ,R R 3R
1 1 4 1 1 1
0 0 5 2 1 2 A
0 5 10 3 3 4
RR
1 1 4
0 5 10
0 0 5
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
???
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ?
?
??
??
? ? ?
??
?? ?
??
1 1 1
3 3 4 A
2 1 2
? ??
??
??
??
?? ??
??
2 2 3 3
1 1 3 2 2 3
11
R R , R R
55
1 1 1
1 1 4
3 3 4
0 1 2 A
5 5 5
0 0 1
2 1 2
5 5 5
R R 4R ,R R 2R ,
3 1 3
5 5 5
1 1 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
??
??
??
?? ?
??
??
?
??
??
?
??
??
??
??
??
?
??
??
? ? ? ?
??
?
??
????
?
????
?
????
????
??
?
??
??
??
CBSE XII | Mathematics
Sample Paper 4 Solution
1 1 2
1
R R R
23
0
55
1 0 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
23
0
55
11
Hence A 0
55
2 1 2
5 5 5
?
??
???
??
????
?
????
?
????
????
??
?
??
??
??
???
??
??
?
??
?
??
??
?
??
??
??
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.
? ?
11
x ,y
dy
dx
?
?
?
?
= 0
?
1
1
1x
y
?
= 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
2
1
2
1
1
1 y 2 1 3 0
y 4 0
y4
y2
? ? ? ? ?
? ? ?
??
? ? ?
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 4 Solution
OR
Consider the given equation,
3
2
2
1
y x 2x 4
Differentiating the above function with respect to x, we have,
dy
3x 2
dx
m 3x 2
? ? ?
??
? ? ?
2
12
Given that the tangents to the given curve are
perpendicular to the line x+14y+3=0
1
Slope of this line, m
14
Since the given line and the tangents to
the given curve are perpendicular, we have,
m m 1
?
?
? ? ?
? ?
2
2
2
2
3
3
1
3x 2 1
14
3x 2 14
3x 12
x4
x2
If x = 2, y = x 2x 4
y 2 2 2 4
y8
? ??
? ? ? ?
??
??
? ? ?
??
??
? ? ?
??
? ? ? ? ?
??
? ? ? ?
3
3
If x = 2, y = x 2x 4
y 2 2 2 4
y 16
? ? ?
? ? ? ? ? ? ?
? ? ?
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FAQs on Sample Solution Paper 4 - Math, Class 12 - Mathematics (Maths) Class 12 - JEE
| 1. How can I calculate the probability of an event occurring? |  |
Ans. To calculate the probability of an event occurring, you need to divide the number of favorable outcomes by the total number of possible outcomes. For example, if you want to find the probability of rolling a 6 on a fair six-sided die, the favorable outcome is 1 (rolling a 6) and the total number of possible outcomes is 6 (numbers 1 to 6 on the die). Therefore, the probability is 1/6.
| 2. What is the difference between permutations and combinations? | |
Ans. Permutations and combinations are both ways to count the number of possible outcomes, but they are used in different scenarios.
In permutations, the order of the items or events matters. For example, if you want to arrange 3 different books on a shelf, the number of permutations would be 3! (3 factorial), which is equal to 3 x 2 x 1 = 6.
In combinations, the order does not matter. For example, if you want to choose 2 books from a set of 3, the number of combinations would be 3C2 (3 choose 2), which is equal to 3! / (2! x (3-2)!) = 3.
| 3. How can I find the standard deviation of a set of data? | |
Ans. To find the standard deviation of a set of data, you can follow these steps:
1. Calculate the mean (average) of the data set.
2. Subtract the mean from each data point and square the result.
3. Calculate the mean of the squared differences.
4. Take the square root of the mean from step 3.
The result is the standard deviation, which measures the variability or spread of the data set.
| 4. What is the significance of the slope in a linear regression model? | |
Ans. In a linear regression model, the slope represents the rate of change in the dependent variable for every one unit increase in the independent variable. It indicates the direction and steepness of the relationship between the two variables.
A positive slope indicates a positive relationship, meaning that as the independent variable increases, the dependent variable also tends to increase. A negative slope indicates a negative relationship, meaning that as the independent variable increases, the dependent variable tends to decrease.
The magnitude of the slope indicates the strength of the relationship. A larger slope suggests a stronger relationship between the variables, while a smaller slope suggests a weaker relationship.
| 5. How can I calculate the area under a curve using integration? | |
Ans. To calculate the area under a curve using integration, you can follow these steps:
1. Define the function that represents the curve.
2. Determine the limits of integration, which specify the range of x-values over which you want to calculate the area.
3. Integrate the function over the given limits using the definite integral notation.
4. Evaluate the integral to find the numerical value of the area under the curve.
The result is the area under the curve, which represents the accumulated sum of the function's values within the specified range of x-values. Integration allows you to find the exact area, even for curves that are not easily measurable by other means.
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