Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 10 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Maths Set 10 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.
The range of the function f(x) = \(\frac{|x-1|}{(x-1)}\) is: [1]
(a) {1, 3}
(b) {1, 2}
(c) {-1, 1}
(d) {-1, 3}
Solution:
(c) {-1, 1}
Explanation:
Question 2.
The principal value of tan-1(-√3) is: [1]
(a) \(\frac{\pi}{3}\)
(b) \(\frac{-\pi}{3}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{-\pi}{6}\)
Solution:
(b) \(\frac{-\pi}{3}\)
Explanation:
tan-1(-√3) = -tan-1(√3) = \(\frac{-\pi}{3}\)
Question 3.
Using principal values, the value of sec-1 √2 + cosec-1 √2 is: [1]
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{4}\)
(c) 0
(d) 1
Solution:
(a) \(\frac{\pi}{2}\)
Explanation:
sec-1 √2 + cosec-1 √2 = \(\frac{\pi}{2}\) [∵ sec-1 x + cosec-1 x = \(\frac{\pi}{2}\)]
Question 4.
The minor of the element of the third row and second column in the following determinant is: [1]
\(\left|\begin{array}{ccc}
3 & -3 & 1 \\
4 & 0 & 4 \\
2 & 5 & -7
\end{array}\right|\)
(a) 8
(b) -8
(c) 16
(d) 4
Solution:
(a) 8
Explanation:
M32 = \(\left|\begin{array}{ll}
3 & 1 \\
4 & 4
\end{array}\right|\)
= 12 – 4
= 8
Question 5.
For A = \(\left[\begin{array}{rr}
1 & 4 \\
-1 & 3 \\
0 & 5
\end{array}\right]\), (AT)T is equal to [1]
(a) AT
(b) (A-1)T
(c) A
(d) \(\frac{1}{A}\)
Solution:
(c) A
Explanation:
Question 6.
The value of \(\left|\begin{array}{ccc}
4^2 & 4^3 & 4^4 \\
4^3 & 4^4 & 4^5 \\
4^4 & 4^5 & 4^6
\end{array}\right|\) is: [1]
(a) 4
(b) 16
(c) 32
(d) 0
Solution:
(d) 0
Explanation:
Question 7.
The maximum value of \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1 & 1 & 1+\cos \theta
\end{array}\right|\) is: [1]
(a) 1
(b) 0
(c) \(\frac{\pi}{2}\)
(d) 2
Solution:
(a) 1
Explanation:
\(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1 & 1 & 1+\cos \theta
\end{array}\right|\)
= 1(1 + sin θ)(1 + cos θ) – 1] – 1[1 + cos θ – 1] + 1[1 – 1 – sin θ)]
= [1 + sin θ + cos θ + sin θ cos θ – 1] – cos θ – sin θ)]
= sin θ + cos θ + sin θ cos θ – cos θ – sin θ
= sin θ cos θ
The maximum value of sin θ is 1 and cos θ is 1.
Hence, maximum value of \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1 & 1 & 1+\cos \theta
\end{array}\right|\) is 1 × 1 = 1.
Question 8.
The value of \(\int_{\pi / 4}^{\pi / 2} \cot x d x\) is: [1]
(a) log√2
(b) log(cosec \(\frac{\pi}{2}\))
(c) 1
(d) 0
Solution:
(a) log√2
Explanation:
Question 9.
The value of \(\int_{-1}^1 e^{|x|} d x\) is: [1]
(a) \(\frac{1}{e}+e-2\)
(b) \(e+\frac{1}{e}\)
(c) \(e-\frac{1}{e}-1\)
(d) \(\frac{e^2-1}{e}+2\)
Solution:
(a) \(\frac{1}{e}+e-2\)
Explanation:
Question 10.
The sum of the order and degree of the differential equation is: [1]
\(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\)
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3
Explanation:
Given, differential equation is \(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\)
The order of the differential equation is 2, the degree is 1.
Sum of the order and degree of the differential equation = 2 + 1 = 3
Question 11.
The general solution of the differential equation \(\frac{d y}{d x}=2^{-y}\) is: [1]
(a) 2y = x log 2 + C log 2
(b) 2y = x log 3 – C log 3
(c) y = x log 2 – C log 2
(d) None of these
Solution:
(a) 2y = x log 2 + C log 2
Explanation:
Question 12.
If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), then vectors is \(\vec{a}\) and \(\vec{b}\) are: [1]
(a) parallel
(b) perpendicular
(c) coplanar
(d) collinear
Solution:
(b) perpendicular
Explanation:
⇒ θ = \(\frac{\pi}{2}\)
∴ \(\vec{a} \& \vec{b}\) are perpendicular
Question 13.
If \(\vec{a}\) is any non-zero vector, then the value of \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}\) is: [1]
(a) 1
(b) 0
(c) \(\vec{a}\)
(d) |\(\vec{a}\)|
Solution:
(c) \(\vec{a}\)
Explanation:
Question 14.
For the curve √x + √y = 1, the value of \(\frac{d y}{d x}\) at \(\left(\frac{1}{4}, \frac{1}{4}\right)\) is: [1]
(a) 1
(b) -1
(c) \(\frac{1}{4}\)
(d) \(\frac{-1}{4}\)
Solution:
(b) -1
Explanation:
Question 15.
The direction ratios of the line 6x – 2 = 3y + 1 = 2z – 2 are: [1]
(a) 6, 3, 2
(b) 1, 1, 2
(c) 1, 2, 3
(d) 1, 3, 2
Solution:
(c) 1, 2, 3
Explanation:
Given the equation of a line is
6x – 2 = 3y + 1 = 2z – 2
⇒ \(6\left(x-\frac{1}{3}\right)=6\left(y+\frac{1}{3}\right)=2(z-1)\)
⇒ \(\frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-1}{3}\)
This shows that the given line passes through (1/3, -1/3, 1), and has direction ratios 1, 2, and 3.
Question 16.
If P(A) = \(\frac{3}{10}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = \(\frac{3}{5}\), then P(A/B) + P(B/A) is: [1]
(a) \(\frac{7}{12}\)
(b) \(\frac{6}{17}\)
(c) \(\frac{6}{25}\)
(d) \(\frac{3}{10}\)
Solution:
(a) \(\frac{7}{12}\)
Explanation:
Question 17.
The value of λ such that the vector \(\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\) are orthogonal is: [1]
(a) \(\frac{3}{2}\)
(b) \(\frac{-5}{2}\)
(c) \(\frac{-1}{2}\)
(d) \(\frac{1}{2}\)
Solution:
(b) \(\frac{-5}{2}\)
Explanation:
Since, two non-zero vectors \(\vec{a}\) and \(\vec{b}\) are orthogonal
Question 18.
The vectors having initial and terminal points as (2, 5, 0) and (-3, 7, 4) respectively: [1]
(a) \(-\hat{i}+12 \hat{j}+4 \hat{k}\)
(b) \(5 \hat{i}+2 \hat{j}-4 \hat{k}\)
(c) \(-5 \hat{i}+2 \hat{j}+4 \hat{k}\)
(d) \(\hat{i}+\hat{j}+\hat{k}\)
Solution:
(c) \(-5 \hat{i}+2 \hat{j}+4 \hat{k}\)
Explanation:
Required vector = \((-3-2) \hat{i}+(7-5) \hat{j}+(4-0) \hat{k}\)
= \(-5 \hat{i}+2 \hat{i}+4 \hat{k}\)
Assertion-Reason Based Questions
In the following questions, a statement of Assertion (A) is followed by a statement of 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.
Consider the function
f(x) = \(\left\{\begin{array}{cl}
\frac{x^2+3 x-10}{x-2}, & \text { if } x \neq 0 \\
k, & \text { if } x=2
\end{array}\right.\)
Which is continuous at x = 2
Assertion (A): The value of k is 0.
Reason (R): f(x) is continuous at x = a, if \(\lim _{x \rightarrow a} f(x)=f(a)\). [1]
Solution:
(d) A is false but R is true.
Explanation:
Question 20.
Assertion (A): If y = \(\sin ^{-1}\left(6 x \sqrt{1-9 x^2}\right)\), then \(\frac{d y}{d x}=\frac{6}{\sqrt{1-9 x^2}}\)
Reason (R): \(\sin ^{-1}\left(6 x \sqrt{1-9 x^2}\right)=3 \sin ^{-1} 2 x\). [1]
Solution:
(c) A is true but R is false.
Explanation:
Section – B (10 Marks)
This section comprises of very short answer type-questions (VSA) of 2 marks each.
Question 21.
If \(\cot ^{-1}\left(-\frac{1}{5}\right)\) = x, then find the value of sin x.
OR
Find the value of {tan-1(1) – cot-1(-1)}. [2]
Solution:
Question 22.
Evaluate the determinant ∆ = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right|\) by expanding it along (A) first row (B) first column.
OR
Evaluate ∆ = \(\left|\begin{array}{ccc}
1 & \sin x & 1 \\
-\sin x & 1 & \sin x \\
-1 & -\sin x & 1
\end{array}\right|\) and show that 2 ≤ ∆ ≤ 4. [2]
Solution:
∆ = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right|\)
(A) Expanding along first row \(\left|\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right|\)
= 1(45 – 48) – 2(36 – 42) + 3(32 – 35)
= -3 + 12 – 9
= 0
(B) Expanding along first column \(\left|\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right|\)
= 1(45 – 48) – 4(18 – 24) + 7(12 – 15)
= -3 + 24 – 21
= 0
OR
∆ = \(\left|\begin{array}{ccc}
1 & \sin x & 1 \\
-\sin x & 1 & \sin x \\
-1 & -\sin x & 1
\end{array}\right|\)
= 1(1 + sin2x) – sin x (-sin x + sin x) + 1(sin2x + 1)
= 2(1 + sin2x)
The minimum value of sin2x is 0 and the maximum value is 1.
0 ≤ sin2x ≤ 1
⇒ 0 + 1 ≤ 1 + sin2x ≤ 1 + 1
⇒ 1 ≤ 1 + sin2x ≤ 2
⇒ 2 ≤ 2(1 + sin2x) ≤ 4
⇒ 2 ≤ ∆ ≤ 4
Question 23.
Find the values of x for which y = [x(x – 2)]2 is an increasing function. [2]
Solution:
We have, y = [x(x – 2)]2
Differentiate both sides w.r.t. x
\(\frac{d y}{d x}\) = 2[x(x – 2)][x + x – 2]
⇒ \(\frac{d y}{d x}\) = 2[x(x – 2)][2x – 2]
⇒ \(\frac{d y}{d x}\) = 4x(x – 2)(x – 1)
Putting \(\frac{d y}{d x}\) = 0
⇒ x = 0, 1, 2
The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals (-∞, 0), (0, 1), (1, 2), and (2, ∞).
For x ∈ (-∞, 0), \(\frac{d y}{d x}\) < 0
Therefore, y is strictly decreasing in the interval (-∞, 0)
For x ∈ (0, 1), \(\frac{d y}{d x}\) > 0
Therefore, y is strictly increasing in the interval (0, 1)
For x ∈ (1, 2), \(\frac{d y}{d x}\) < 0
Therefore, y is strictly decreasing in the interval (1, 2)
For x ∈ (2, ∞), \(\frac{d y}{d x}\) > 0
Therefore, y is strictly increasing in the interval (2, ∞)
Hence, y is an increasing function for 0 < x < 1 and 2 < x ∞.
Question 24.
Find \(\int \frac{\sin ^{-1} x}{x^2} d x\). [2]
Solution:
Question 25.
Find the value of λ for which the four points A, B, C, and D with position vectors \(4 \hat{i}+5 \hat{j}+\lambda\) \(\hat{k},-\hat{j}-\hat{k}, 3 \hat{i}+9 \hat{j}+4 \hat{k}\) and \(-4 i+4 j+4\)\(\hat{k}\) are coplanar. [2]
Solution:
Section – C (18 Marks)
This section comprises of short answer type questions (SA) of 3 marks each.
Question 26.
Show that the relation R defined by (a, b) R (c, d) ⇔ a + d = b + c on A × A, where A = {1, 2, 3, 4, ……. 10} is an equivalence relation. Hence write the equivalence class (3, 4); a, b, c, d ∈ A.
OR
Show that the relation R on N × N, defined by (a, b) R (c, d) a + d = b + c. Is an equivalent relation. [3]
Solution:
R is an equivalence relation if R is reflexive, symmetric, and transitive.
For reflexive:
Given R in A × A and (a, b) R (c, d) such that a + d = b + c
For reflexive, consider (a, b) R (a, b), (a, b) ∈ A and apply the given condition
⇒ a + b = b + a; which is true for all A.
Therefore, R is reflexive.
For symmetric:
Given (a, b) R (c, d) such that a + d = b + c
Consider (c, d) R (a, b) on A × A applying the given condition
⇒ c + b = d + a which satisfies given condition
Hence, R is symmetric.
For transitive:
Let (a, b) R (c, d) and (c, d) R (e, f) and (a, b), (c, d), (e, f) ∈ A × A applying given condition:
⇒ a + d = b + c …..(i)
and c + f = d + e ……….(ii)
equation (i) ⇒ a – c = b – d
Now add equations (i) and (ii);
⇒ a – c + c + f = b – d + d + e
⇒ a + f = b + e
∴ (a, b) R (e, f ) also satisfies the condition
Hence R is transitive.
Therefore, by the above inspection, we can say that R is an equivalence relation.
Finding equivalence class [3, 4]:
Let (3, 4) R (a, b) on A × A where A = {1, 2, 3, ………., 10}
⇒ 3 + b = 4 + a
Let a = 1
⇒ b = 2
Therefore, one pair (a, b) = (1, 2)
Similarly, we can find the other pairs (a, b)
Therefore, the equivalence class of [3, 4] = {(1, 2),(2, 3),(3, 4),(4, 5),(5, 6), (6, 7), (7, 8), (8, 9), (9, 10)}
OR
Reflexivity
Consider (a, b) as an arbitrary element of N × N
Here, (a, b) ∈ N × N where (a, b) ∈ N
It can be written as
a + b = b + a
We get (a, b) R (a, b) for all (a, b) ∈ N × N
Hence, R is reflexive on N × N
Symmetry
Consider (a, b), (c, d) e N × N such that (a, b) R (c, d)
It can be written as
a + d = b + c and c + b = d + a
We get (c, d) R (a, b)
(a, b) R (c, d)
⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
Hence, R is symmetric on N × N
Transitive
Consider (a, b), (c, d), (r, f) ∈ N × N such that (a, b) R (c, d) and (c, d) R (e, f)
It can be written as
a + d = b + c and c + f = d + e
By adding both
(a + d) + (c + f) = (b + c) + (d + e)
On further calculation
a + f = b + e where (a, b) R (e, f)
So (a, b) R (c, d) and (c, d) R (e, f) we get
(a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N
Hence, R is transitive on N × N
Therefore, T is reflexive, symmetric, and transitive is an equivalence relation on N × N
Question 27.
If y = xx, prove that \(\frac{d^2 y}{d x^2}-\frac{1}{y} \frac{d y}{d x}-\frac{y}{x}=0\). [3]
Solution:
Question 28.
Find \(\int \frac{3 x+1}{(x-1)^2(x+3)} d x\)
OR
Find \(\int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) d x\). [3]
Solution:
Question 29.
The area between x = y2 and x = 4 is divided into two equal parts by the line x = a. Find the value of ‘a’. [3]
Solution:
The equation of the parabola is x = y2 and lines are given by x = 4, x = a
It is given that
Area of shaded region 2(ABCFA) = Area of shaded region 2(AFOA)
Question 30.
Find the direction cosines of the sides of a triangle whose vertices are (3, 5, -4), (-1, 1, 2), and (-5, -5, -2). [3]
Solution:
The vertices of ∆ABC are A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2).
The direction ratios of the sides AB are (-1 – 3, 1 – 5, 2 + 4) i.e. (-4, -4, 6).
So direction cosines of line AB are
Question 31.
Coloured balls are distributed in three bags as shown below in the table:
A bag is selected at random and then two balls are randomly drawn from the selected bag. They happen to be black and red. What is the probability that they came from bag 2?
OR
In a class, 5% of boys and 10% of the girls have an I.Q. of more than 150. In the class, 60% of the students are boys and are rest are girls. If a student is selected at random and found to have an I.Q. of more than 150, then find the probability that the student is a boy. [3]
Solution:
Given, those bags are selected at random,
P(bag 1) = P(bag 2) = P(bag 3) = \(\frac{1}{3}\)
Let A be the event that the two balls are 1 black and 1 red.
OR
Consider the following events.
E1: Selected Student is a boy.
E2: Selected Student is a girl.
A: The Student has an IQ of hence than 150.
Section – D (20 Marks)
This section comprises of tong answer type questions (LA) of 5 marks each
Question 32.
Find the absolute maximum value and absolute minimum value of the function f(x) = 3 + |x + 1| in [-2, 3].
OR
Find the intervals in which the function f(x) = (x + 1)3 (x – 3)3 is strictly increasing or decreasing. [5]
Solution:
f(x) = 3 + |x + 1|
⇒ f(x) = \(\left\{\begin{array}{l}
3+(x+1) \text { if } x \geq-1 \\
3-(x+1) \text { if } x<-1
\end{array}\right.\)
⇒ f(x) = \(\begin{cases}4+x & \text { if } x \geq-1 \\ 2-x & \text { if } x<-1\end{cases}\)
Therefore, f(x) = 1 or -1
Now, we evaluate the value of f at the endpoints of the interval [-2, 3].
f(-2) = 2 – (-2)
= 2 + 2
= 4
f(3) = 4 + 3 = 7
Hence, the absolute maximum value of f on [-2, 3] is 7 occurring at x = 3
and the absolute minimum value of f on [-2, 3] is 4 occurring at x = -2.
OR
Given function f(x) = (x + 1)3 (x – 3)3
Differentiating both sides w.r.t. x, we get
f'(x) = 3(x + 1)2 (x – 3)3 + 3(x + 1)3 (x – 3)2 [using product rule]
⇒ f'(x) = 3(x + 1)2 (x – 3)2 [x – 3 + x + 1]
⇒ f'(x) = 6(x + 1)2 (x – 3)2 (x – 1)
For maximum and minimum values f'(x) = 0
Putting f'(x) = 0, we get
⇒ x = -1, 3 and 1
So, the points x = -1, x = 1 and x = 3 divides the real line into four disjoint intervals
so possible intervals are: (-∞, -1), (-1, 1), (1, 3), (3, ∞)
Case I: f'(x) < 0 in (-∞, -1), (-1, 1)
So, the function is strictly decreasing in (-∞, -1) ∪ (-1, 1)
Case II: f'(x) > 0 in (1, 3) and (3, ∞)
So, the function is strictly increasing in (1, 3) ∪ (3, ∞)
Question 33.
Find \(\frac{d y}{d x}\), if y = \(x^{(x \cos x)}+\frac{x^2+1}{x^2-1}\). [5]
Solution:
Question 34.
Solve the following L.P.P. graphically:
Minimise Z = 16x + 20y
subject to constraints
x + 2y ≥ 10, x + y ≥ 6, 3x + y ≥ 8, x, y ≥ 0
OR
Solve the following L.P.P. graphically:
Minimise Z = 2.5x + 1.5y + 410
subject to constraints
x + y ≤ 100, x + y ≥ 60, x ≤ 60, y ≤ 50, x, y ≥ 0. [5]
Solution:
We have, subject to the constraints
x + 2y ≥ 10, x + y ≥ 6, 3x + y ≥ 8, x, y ≥ 0
Changing inequations to equations
For x + 2y = 10
For x + y = 6
For 3x + y = 8
Plotting the equation on a graph paper, we get the bounded feasible region (Shaded region) with corners A(0, 8), B(1, 5), C(2, 4), D(10, 0)
Z = 16x + 20y
ZA = 16 × 0 + 20 × 8 = 160
ZB = 16 × 1 + 20 × 5 = 116
ZC = 16 × 2 + 20 × 4 = 112
ZD = 16 × 10 + 20 × 0 = 160
Z is minimum at (2, 4) and the minimum value is 112.
OR
We have, subject to the constraints
x + y ≤ 100, x + y > 60, x ≤ 60, y ≤ 50, x, y ≤ 0
Changing inequations to equations
For x + y = 100
For x + y = 60
Plotting the equation on a graph paper, we get the bounded feasible region (Shaded region) with corners A(10, 50), B(50, 50), C(60, 40), D(60, 0)
Z = 2.5x + 1.5y + 410
ZA = 2.5 × 10 + 1.5 × 50 + 410 = 510
ZB = 2.5 × 50 + 1.5 × 50 + 410 = 610
ZC = 2.5 × 60 + 1.5 × 40 + 410 = 620
ZD = 2.5 × 60 + 1.5 × 0 + 410 = 560
Z is minimum at (10, 50) and the minimum value is 510.
Question 35.
(A) Solve the differential equation:
x dx + y dy = (x dy – y dx) given that y = 1, when x = 1.
(B) Evaluate \(\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x\). [5]
Solution:
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 each.
Question 36.
The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others, and some others (say z) for supervising the workers to keep the colony neat and clean.
Below are 3 statements given by the committee.
| (I) The sum of all awardees is 12. |
| (II) Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. |
| (III) The sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others. |
Based on the above information answer the following questions:
(A) Form the group of equations for the statements (I), (II), and (III). [1]
(B) Write the above-obtained group of equations in matrix form PX = Q. [1]
(C) Calculate adj (P):
OR
What is the number of awardees in each category? [2]
Solution:
(A) The equations are respectively:
x + y + z = 12;
2x + 3(y + z) = 33;
x + z = 2y
(B) Equations are
x + y + z = 12;
2x + 3(y + z) = 33;
x + z = 2y
In matrix form, PX = Q
Question 37.
Let d1, d2, d3 be three mutually exclusive diseases.
Let S = {S1, S2, S3, S4, S5, S6} be the set of observable symptoms of these diseases. For example, S1 is the shortness of breath, S2 is the loss of weight, S3 is the fatigue, etc. Suppose a random sample of 10,000 patients contains 3200 patients with disease d1, 3500 patients with disease d2, and 3300 patients with disease d3. Also, 3100 patients with disease d1, 3300 patients with disease d2, and 3000 patients with disease d3 show the symptom S.
Based on the above information answer the following questions:
(A) A person is chosen at random from the sample of 10,000. What is The probability that the person chosen does not suffer from disease d3? [1]
(B) Find the conditional probability that the patient shows the symptom S given that he suffers from disease d1 and also calculate the conditional probability that the patient shows the symptom S given that he suffers from disease d2.
OR
If a person chosen at random shows the symptom S, then what is the probability that he does suffer from disease d1? [2]
(C) Let Di denote the event that the patient has disease di (i = 1, 2, 3) and S be the event that the patient shows the symptom S. Then find the value of \(\sum_1^3 \mathrm{P}\left(\frac{d_i}{s}\right)\). [1]
Solution:
Question 38.
ABCD is a parallelogram whose adjacent sides are represented by the vector \(\vec{a}\) & \(\vec{b}\). Three of its vertices are A(2, 2, -3), B(2, 0, 4) and D (-1, 4, 1)
Based on the above information answer the following questions:
(A) What is the vector \(\vec{a}\)? And What is the area of Parallelogram ABCD? [2]
(B) Find the length of the diagonals \(\vec{c}\) and \(\vec{d}\)? [2]
Solution:
