Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 11 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

  • This Question paper contains five sections A, B, C, D, and E. Each section is compulsory. However, there are internal choices in some questions.
  • Section A has 18 MCQs and 2 Assertion-Reason-based questions of 1 mark each.
  • Section B has 5 Very Short Answer (VSA) type questions of 2 marks each.
  • Section C has 6 Short Answer (SA) type questions of 3 marks each.
  • Section D has 4 Long Answer (LA) type questions of 5 marks each.
  • Section E has 3 sources based/case based/passage based/integrated units of assessment (4 marks each) with subparts.

Section – A (20 Marks)
(Multiple Choice Questions Each question carries 1 mark)

Question 1.
How many relations are possible in set A such that n(A) = 2? [1]
(a) 4
(b) 8
(c) 16
(d) 32
Solution:
(c) 16
Explanation:
As n(A) = 2
⇒ n(A × A) = 4
∴ Possible relations are, 24 = 16

Question 2.
The domain, for which tan-1x > cot-1x holds true, is: [1]
(a) x = 1
(b) x > 1
(c) x < 1
(d) Not defined
Solution:
(b) x > 1
Explanation:
The graphs of tan-1x and cot-1x indicate that tan-1x > cot-1x for x > 1.

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 3.
The value of {tan-1(1) + cos-1(\(-\frac{1}{2}\)) + sin-1(\(-\frac{1}{2}\))} is: [1]
(a) \(\frac{2 \pi}{3}\)
(c) \(\frac{3 \pi}{4}\)
(b) \(\frac{3 \pi}{5}\)
(d) \(\frac{2 \pi}{5}\)
Solution:
(c) \(\frac{3 \pi}{4}\)
Explanation:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q3

Question 4.
The matrix A = \(\left[\begin{array}{ccc}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\) is a symmetric matrix. Then the value of a and b respectively are: [1]
(a) \(\frac{-2}{3}, \frac{3}{2}\)
(b) \(\frac{-1}{2}, \frac{1}{2}\)
(c) -2, 2
(d) \(\frac{3}{2}, \frac{1}{2}\)
Solution:
(a) \(\frac{-2}{3}, \frac{3}{2}\)
Explanation:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q4

Question 5.
If one root of the equation \(\left[\begin{array}{lll}
7 & 6 & x \\
2 & x & 2 \\
x & 3 & 7
\end{array}\right]\) = 7 is x = -9, then the other two roots are: [1]
(a) 6, 3
(b) 6, -3
(c) -2, -7
(d) 2, 6
Solution:
(c) -2, -7
Explanation:
\(\left|\begin{array}{lll}
7 & 6 & x \\
2 & x & 2 \\
x & 3 & 7
\end{array}\right|\) = 7(7x – 6) – 6(14 – 2x) + x(6 – x2)
= -x3 + 67x – 126
= (x + 9)(-x2 + 9x – 14)
= (x + 9)(-x – 2) (x + 7)
Hence the other two roots are -2 and -7.

Question 6.
The cofactor of (-1) in the matrix \(\left[\begin{array}{ccc}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{array}\right]\) is: [1]
(a) 1
(b) 2
(c) -1
(d) 0
Solution:
(c) -1
Explanation:
Cofactor of (-1) = \((-1)^{2+3}\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)
= (-1) × 1
= -1

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 7.
\(\int x^x(1+\log x) d x\) is: [1]
(a) xx + c
(b) \(\frac{1}{x}\) + c
(c) log x + c
(d) x + c
Solution:
(a) xx + c
Explanation:
Let xx = z
Then x log x = log z
(1 + log x) dx = \(\frac{1}{z}\) dz
∴ I = \(\int z \cdot \frac{1}{z} d z\)
= ∫1 dz
= z + c
= xx + c

Question 8.
The value of \(\int_0^a \frac{\sqrt{a}}{\sqrt{x}+\sqrt{a-x}} d x\) is: [1]
(a) \(\frac{a}{2}\)
(b) a
(c) a2
(d) 0
Solution:
(a) \(\frac{a}{2}\)
Explanation:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q8

Question 9.
The area (in sq. m) of the shaded region as shown in the figure is: [1]
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q9
(a) \(\frac{32}{3}\) sq. units
(b) \(\frac{16}{3}\) sq. units
(c) 4 sq. units
(d) 16 sq. units
Solution:
(a) \(\frac{32}{3}\) sq. units
Explanation:
Given curves are x = y2 and x = 4.
So, their points of intersection are (4, 2) and (4, -2).
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q9.1

Question 10.
The order and the degree of the differential equation \(2 x^2 \frac{d^2 y}{d x^2}-3 \frac{d y}{d x}+y=0\) are: [1]
(a) 1, 1
(b) 2, 1
(c) 1, 2
(d) 3, 1
Solution:
(b) 2, 1
Explanation:
The highest order is 2 and the degree of the highest order is 1.
Hence, the order is 2 and the degree is 1.

Question 11.
The integrating factor of the differential equation \(x \frac{d y}{d x}-y=2 x^2\) is: [1]
(a) log x
(b) \(\frac{1}{x^2}\)
(c) e-x
(d) \(\frac{1}{x}\)
Solution:
(d) \(\frac{1}{x}\)
Explanation:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q11

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 12.
For any vector \(\vec{a}\), the value of \(|\vec{a} \cdot \hat{i}|^2+|\vec{a} \cdot \hat{j}|^2+|\vec{a} \cdot \hat{k}|^2\) is: [1]
(a) a
(b) a2
(c) 1
(d) 0
Solution:
(b) a2
Explanation:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q12

Question 13.
The projection of the vector \(\hat{i}-\hat{j}\) on the vector \(\hat{i}+\hat{j}\) is: [1]
(a) 1
(b) -1
(c) 0
(d) 2
Solution:
(c) 0
Explanation:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q13

Question 14.
A point that lies on the line \(\frac{x-1}{-2}=\frac{y+3}{4}\) = \(\frac{1-z}{7}\) is: [1]
(a) (1, -3, 1)
(b) (-2, 4, 7)
(c) (-1, 3, 1)
(d) (2, -4, -7)
Solution:
(a) (1, -3, 1)
Explanation:
The equation of the Line can be written as \(\frac{x-1}{-2}=\frac{y+3}{4}=\frac{z-1}{-7}\)
So, it passes through (1, -3, 1).

Question 15.
The equation that best describes the given graph is: [1]
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q15
(a) x > y
(b) x < y
(c) x ≤ y
(d) x ≥ y
Solution:
(b) x < y
Explanation:
As pot (0, 1) which Lies in this is satisfied by x < y.

Question 16.
A class consists of 80 students, 25 of them are girls and 55 are boys, 10 of them are rich and the remaining are poor, 20 of them are fair complexioned. The probability of selecting a fair-complexioned rich girl is: [1]
(a) \(\frac{8}{653}\)
(b) \(\frac{3}{256}\)
(c) \(\frac{5}{512}\)
(d) \(\frac{1}{8}\)
Solution:
(c) \(\frac{5}{512}\)
Explanation:
P(E) = \(\frac{25}{80}\), for the event E “student, is a girl”
P(F) = \(\frac{10}{80}\), for the event F “student is rich”
P(G) = \(\frac{20}{80}\) for the event G “student is fair complexioned”
∴ Req = P(E) × P(F) × P(G)
= \(\frac{25}{80} \times \frac{10}{80} \times \frac{20}{80}\)
= \(\frac{5}{512}\)
as E, F, G are independent events.

Question 17.
The angle between the lines x = 1, y = 2 and y = -1, z = 0 is: [1]
(a) 30°
(b) 60°
(c) 90°
(d) 0°
Solution:
(c) 90°
Explanation:
Given lines are \(\frac{x-1}{0}=\frac{y-2}{0}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y-1}{0}=\frac{z}{0}\)
∴ cos θ = 0.1 + 0.0 + 1.0 = 0
⇒ θ = 90°

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 18.
If \(|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=144\) and \(|\vec{a}|=4\), then \(|\vec{b}|\) is equal to: [1]
(a) 16
(b) 8
(c) 3
(d) 12
Solution:
(c) 3
Explanation:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q18

Assertion-Reason Based Questions
In the following questions, a statement of assertion (A) is followed by a statement of the reason (R).
Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 19.
Assertion (A): The area of a parallelogram with diagonals \(\vec{a}\) and \(\vec{b}\) is \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
Reason (R): If \(\vec{a}\) and \(\vec{b}\) represents the adjacent sides of a triangle, then the area of a triangle can be obtained by evaluating \(|\vec{a} \times \vec{b}|\). [1]
Solution:
(c) A is true but R is false.
Explanation:
If \(\vec{a}\) and \(\vec{b}\) represents the adjacent sides of a triangle, then the area of a triangle can be obtained by evaluating \(\frac{1}{2}|\vec{a} \times \vec{b}|\)

Question 20.
Assertion (A): The position vector of a point say P(x, y, z) is \(\overrightarrow{\mathrm{OP}}=\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) and its magnitude is \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)
Reason (R): If \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), then the coefficient of \(\hat{i}, \hat{j}, \hat{k}\) in \(\vec{r}\) i.e. x, y, z are called direction ratios of the vector \(\vec{r}\). [1]
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
If P(x, y, z), then \(\overline{O P}=x \hat{i}+y \hat{j}+z \hat{k}\) and \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)
Then ORs of OP is < x, y, z >.

Section – B (10 Marks)
This section comprises very short answer type-questions (VSA) of 2 marks.

Question 21.
Write the domain and the range of (A) cos-1 (B) cot-1.
OR
Prove that, sec2 (tan-12) + cosec2 (cot-13) = 15. [2]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q21

Question 22.
Using determinants, find the value of ‘a’. So, points (a, 2), (1, 5), and (2, 4) are collinear.
OR
If f(x) = \(\left|\begin{array}{ccc}
0 & x-a & x-b \\
x+a & 0 & x-c \\
x+b & x+c & 0
\end{array}\right|\), then find the value of f(0). [2]
Solution:
Here, ∆ = \(\frac{1}{2}\left|\begin{array}{lll}
a & 2 & 1 \\
1 & 5 & 1 \\
2 & 4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [a(5 – 4) – 2(1 – 2) + 1(4 – 10)]
= \(\frac{1}{2}\) [a + 2 – 6]
= \(\frac{1}{2}\) [a – 4]
Given that the points (a, 2) (1, 5), and (2, 4) are collinear
∆ = 0
⇒ \(\frac{1}{2}\) (a – 4) = 0
⇒ a = 4
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q22
Expanding along R1
f(0) = 0(0 + c2) + a(0 + bc) – b(ac – 0)
= abc – acb
= 0

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 23.
Find the least value of ‘a’ such that the function f defined as, f(x) = x2 + ax + 1 is strictly increasing on (1, 2). [2]
Solution:
Given, f(x) = x2 + ax + 1
⇒ f'(x) = 2x + a
For 1 < x < 2,
2 + a < 2x + a < 4 + a
⇒ 2 + a < f'(x) < 4 + a
Now, f(x) is strictly increasing on (1, 2) only if f'(x) > 0 for 1 < x < 2,
i.e. if 2 + a ≥ 0, i.e. a ≥ -2
Thus, the least value of a is -2.

Question 24.
The scalar product of the vector \(\hat{i}+\hat{j}+\hat{k}\) with a unit vector along with the sum of vectors \(2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\lambda \hat{i}+2 \hat{j}+3 \hat{k}\) is equal to one. Find the value of λ. [2]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q24
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q24.1

Question 25.
Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular. [2]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q25

Section – C (18 Marks)
This section comprises short answer type questions (SA) of 3 marks.

Question 26.
Consider a function f: \(\left[0, \frac{\pi}{2}\right]\) → R given by, f(x) = sin x and g: \(\left[0, \frac{\pi}{2}\right]\) → R given by g(x) = cos x. Show that, both f and g are one-one, but f + g is not one-one. [3]
Solution:
f is one-one
Let x, y be any two elements in \(\left[0, \frac{\pi}{2}\right]\). then
x ≠ y
⇒ sin x ≠ sin y
⇒ f(x) ≠ f(y)
Thus, f is one-one.
g is one-one
x ≠ y
⇒ cos x ≠ cos y
⇒ g(x) ≠ g(y)
Thus, g is one-one.
f + g is not one-one
(f + g)(0) = f(0) + g(0)
= sin 0 + cos 0
= 0 + 1
= 1
\((f+g)\left(\frac{\pi}{2}\right)=f\left(\frac{\pi}{2}\right)+g\left(\frac{\pi}{2}\right)\)
\(\sin \frac{\pi}{2}+\cos \frac{\pi}{2}\)
= 1 + 0
= 1
∴ (f + g)(0) = (f + g) (\(\frac{\pi}{2}\)) even when 0 ≠ 4
So, f + g is not one-one.

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 27.
If \(x \sqrt{1+y}+y \sqrt{1+x}=0\) {-1 ≤ x ≤ 1}, then prove that \(\frac{d y}{d x}=-\frac{1}{(1+x)^2}\). [3]
Solution:
\(x \sqrt{1+y}+y \sqrt{1+x}=0\)
⇒ \(x \sqrt{1+y}=-y \sqrt{1+x}\)
⇒ x2(1 + y) = y2(1 + x)
⇒ x2 + x2y = y2 + xy2
⇒ x2 – y2 + xy(x – y) = 0
⇒ (x – y) (x + y) + xy(x – y) = 0
⇒ (x – y) (x + y + xy) = 0
⇒ x + y + xy = 0 (∵ x ≠ y)
⇒ y = \(-\frac{x}{1+x}\)
⇒ \(\frac{d y}{d x}=-\frac{(1+x) \cdot 1-x(1)}{(1+x)^2}\) = \(-\frac{1}{(1+x)^2}\)

Question 28.
Evaluate: \(\int_0^{\pi / 4} \log (1+\tan x) d x\).
OR
Find \(\int \frac{d x}{\left(x^2+1\right)(x-1)}\). [3]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q28
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q28.1
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q28.2

Question 29.
Using integration, find the area of the region enclosed by y = |x + 3|, x = -6, x = 0, and y = 0.
OR
Solve: 2xy + y2 – 2x2 \(\frac{d y}{d x}\) = 0; y = 2 when x = 1. [3]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q29
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q29.1
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q29.2

Question 30.
Find the shortest distance between the parallel lines whose vector equations are:
\(\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k})\);
\(\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(4 \hat{i}-2 \hat{j}+2 \hat{k})\). [3]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q30
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q30.1

Question 31.
In a hostel, 60% of the students read Hindi newspapers, 40% read English newspapers and 20% read both Hindi and English newspapers. A student is selected at random.
(A) Find the probability that she reads neither Hindi nor English newspapers;
(B) If she reads Hindi newspapers, find the probability that she reads English newspapers also.
OR
A 5 m long ladder is leaned up against a wall. At a rate of 2 cm/s, the ladder’s bottom is drawn away from the wall along the ground. How quickly does the height on the wall drop when the ladder’s foot is 4 cm from the wall?
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q31
OR
Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall.
Then, by Pythagoras’ theorem, we have:
x2 + y2 = 25 [Length of the ladder = 5m]
⇒ y = \(\sqrt{25-x^2}\)
Then, the rate of change of height (y) with respect to time (t) is given by,
\(\frac{d y}{d t}=\frac{-x}{\sqrt{25-x^2}} \cdot \frac{d x}{d t}\)
It is given that \(\frac{d x}{d t}\) = 2 cm/s
\(\frac{d y}{d t}=\frac{-2 x}{\sqrt{25-x^2}}\)
Now, when x = 4m, we have:
\(\frac{d y}{d t}=\frac{-2 \times 4}{\sqrt{25-4^2}}=-\frac{8}{3}\)
Hence, the height of the ladder on the wall is decreasing at the rate of \(\frac{8}{3}\) cm/s.

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Section – D (20 Marks)
This section comprises tong answer-type questions (LA) of 5 marks.

Question 32.
For the matrix A = \(\left[\begin{array}{cc}
2 & 4 \\
1 & -3
\end{array}\right]\), verify that (A-1)’ = (A’)-1
OR
Using the matrix method, solve the following system of equations: [5]
x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q32
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q32.1
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q32.2

Question 33.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \(\frac{8}{27}\) of the volume of the sphere.
OR
Find the intervals in which the function f given by f(x) = x2 – 2 log (x – 2) – 4x – 7 is increasing or decreasing. Given that the domain of the function is [2, ∞). [5]
Solution:
Let R be the radius of the sphere.
Let r and h be the radius of the base and height of the inscribed cone respectively.
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q33
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q33.1
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q33.2
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q33.3

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 34.
Discuss the differentiability of the function f(x) = |x – 1| + |x – 2|. [5]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q34
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q34.1

Question 35.
(A) The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which the area increases when the side is 10 cm.
(B) Show that, the function f(x) = x3 – 3x2 + 6x – 100 is increasing on R. [5]
Solution:
(A) Let ‘a’ be the side of an equilateral triangle
and A be the area of an equilateral triangle.
Then, \(\frac{d a}{d t}\) = 2 cm/sec
We know that area of an equilateral triangle
A = \(\frac{\sqrt{3}}{4} a^2\)
On differentiating both sides w.r.t ‘t’, we get,
\(\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a \times \frac{d a}{d t}\)
\(\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 \times 10 \times 2\) (Given a = 10)
= 10√3 cm2/sec
Hence, the area is increasing at the rate of 20√3 cm2/sec
(B) Given: f(x) = x3 – 3x2 + 6x – 100
On differentiating both sides w.r.t. x, we get
f'(x) = 3x2 – 6x + 6
= 3x2 – 6x + 3 + 3
= 3(x2 – 2x + 1) + 3
= 3(x – 1)2 + 3 > 0
∴ f'(x) > 0
This shows that function f(x) is increasing on R.

Section – E (12 Marks)

This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-part. First, two case-study questions have three sub-parts (A), (B), (C) of marks 1, 1, 2 respectively. The third case-study question has two subparts of 2 marks.

Question 36.
Jyoti CNC is the largest CNC (Computer numerical control) machine manufacturing company in India. Their unit in Bhubaneswar, Odisha has three machine operators A, B, and C. The operators supervise the machines while they execute the tasks and make any necessary adjustments to produce a better result. Their main focus is to minimize defects as it increases the cost of operations. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively.
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q36
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q36.1
Based on the above information, answer the following questions:
(A) What is the probability that the item produced by operator C is defective? [1]
(B) What is the total probability of producing a defective item in the factory? [1]
(C) What is the Conditional probability that the defective item is produced by operator A?
OR
The factory in charge wants to do a quality check. During the inspection, he picks an item from the stockpile at random. If the chosen item is defective, the. What is the probability that it is not produced by operator C? [2]
Solution:
Let A: A is on the job;
B: B is on the job;
C: C is on the job
D: The item is defective
P(A) = 50% = 0.5;
P(B) = 30% = 0.3;
P(C) = 20% = 0.2
\(P\left(\frac{D}{A}\right)\) = 1% = 0.01
\(P\left(\frac{D}{B}\right)\) = 5% = 0.05
\(P\left(\frac{D}{C}\right)\) = 7% = 0.07
(A) (a) \(P\left(\frac{D}{C}\right)=7 \%=\frac{7}{100}\)
(B) P(D) = P(A) \(P\left(\frac{D}{A}\right)\) + P(B) \(P\left(\frac{D}{B}\right)\) + P(C) \(P\left(\frac{D}{C}\right)\)
= 0.5 × 0.01 + 0.3 × 0.05 + 0.2 × 0.07
= 0.005 + 0.015 + 0.014
= 0.034
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q36.2

Question 37.
One very useful application of Linear programming is using a graphical method for solving problems with two variables. Mrs. Meena Wanted to use that concept to help students figure out how the area of the 3D model is composed of 3 straight lines.
In the below diagram (cross-section of the 3D model), O is the origin. The shaded region R is defined by three inequalities. One of the inequality is x + y ≤ 6.
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q37
Based on the above information, answer the following questions:
(A) Given that the point (x, y) is in the region R. What is the maximum value of x + 2y? [1]
(B) Given that the point (x, y) is in the region R, then find the minimum value of 6x – y. [1]
(C) Find the other two inequalities plotted on the graph?
OR
What is the area of the region R? [2]
Solution:
(A) Corner points of R are (4, 2), (2, 1), and (2, 4).
Z = x + 2y
Z(4, 2) = 4 + 2 × 2 = 4 + 4 = 8
Z(2, 1) = 2 + 2 × 1 = 2 + 2 = 4
Z(2, 4) = 2 + 2 × 4 = 2 + 8 = 10
Hence, the maximum value of (x + 2y) is 10.
(B) Corner points of R are (4, 2), (2, 1), and (2, 4).
Z = 6x – y
Z(4, 2) = 6 × 2 – 2 = 2
Z(2, 1) = 6 × 2 – 1 = 11
Z(2, 4) = 6 × 2 – 4 = 8
Hence, the minimum value of (6x – y) is 8.
(C)
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q37.1
Line 1: This Line is parallel to the y-axis and touches the x-axis at point x = 2.
So, the equation of this line is x = 2 or x – 2 = 0.
Since the feasible region R Lies to the right side of x = 1, this inequality is x ≥ 2.
Line 2: For every point tying on line 2, the x-coordinates are double the y-coordinate.
∴ equation of this line is x = 2y or x – 2y = 0
Since, (3, 2) lies in region R, substituting (3, 2) in x – 2y = 0, we get
3 – 2(2) = 3 – 4 = -1 ≤ 0
∴ Inequality of x = 2y is x – 2y ≤ 0 or x ≤ 2y.
Hence, x ≥ 2 and or x ≤ 2y are the other two inequalities.
OR
Corner points of R are (4, 2), (2, 1), and (2, 4)
The area of a triangle with coordinates (4, 2), (2, 1), and (2, 4) is given as,
A = \(\frac{1}{2}\) |4(1 – 4) + 2(4 – 2) + 2(2 – 1)|
= \(\frac{1}{2}\) |-12 + 4 + 2|
= \(\frac{1}{2}\) × 6
= 3 sq. unit

CBSE Sample Papers for Class 12 Maths Set 11 with Solutions

Question 38.
The order of a differential equation is the order of the highest derivative occurring in the differential equation. The degree of a differential equation whose terms are polynomials in the derivatives is defined as the highest power of the highest-order derivative in it.
A differential equation of the form \(\frac{d y}{d x}\) + Py = θ, where P and Q are functions of x, is a first-order linear differential equation. It has various applications, such as in RL circuits.
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q38
Based on the above information, answer the following questions:
(A) Find the solution of x \(\frac{d y}{d x}\) + y = ex. [2]
(B) Write the integrating factor of (1 – x2) \(\frac{d y}{d x}\) – xy = 1. [2]
Solution:
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q38.1
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q38.2
CBSE Sample Papers for Class 12 Maths Set 11 with Solutions Q38.3