CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Maths Set 4 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.
What is the principal value of \(\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\)? [1]
(a) [-π, π]
(b) \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
(c) (-∞, ∞)
(d) \(\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]\)
Solution:
(b) \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
Explanation:

Question 2.
Let A and B be the events associated with the sample space s, then the value of P(\(\frac{A}{B}\)) is: [1]
(a) P(\(\frac{A}{B}\)) = 1
(b) P(\(\frac{A}{B}\)) = P(A)
(c) 0 ≤ P(\(\frac{A}{B}\))
(d) 0 ≤ P(\(\frac{A}{B}\)) ≤ 1
Solution:
(d) 0 ≤ P(\(\frac{A}{B}\)) ≤ 1

Question 3.
For any two vectors \(\vec{a}\) and \(\vec{b}\) if \(\vec{a} \perp \vec{b}\) then the value of \(\vec{a} \cdot \vec{b}\) is: [1]
(a) 0
(b) 1
(c) 2
(d) -1
Solution:
(a) 0
Explanation:
\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)
= \(|\vec{a}||\vec{b}| \cos 90^{\circ}\)
= 0

Question 4.
If \((\hat{i}+3 \hat{j}+8 \hat{k}) \times(3 \hat{i}-\lambda \hat{j}+\mu \hat{k})=0\), then λ and μ are respectively: [1]
(a) 27, -9
(b) 9, 9
(c) -9, 18
(d) -1, 1
Solution:
(a) 27, -9
Explanation:
Given, \((\hat{i}+3 \hat{j}+9 \hat{k}) \times(3 \hat{i}-\lambda \hat{j}+\mu \hat{k})=0\)
∴ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 3 & 9 \\
3 & -\lambda & \mu
\end{array}\right|=\overrightarrow{0}\)
\(\hat{i}(3 \mu+9 \lambda)-\hat{j}(\mu-27)+\hat{k}(-\lambda-9)=0 \hat{i}+0 \hat{j}+0 \hat{k}\)
On comparing the coefficients of \(\hat{i}, \hat{j}\) and \(\hat{k}\) we get,
3µ + 9λ = 0, -µ + 27 = 0 and -λ – 9 = 0
µ = 27 and -λ = 9
or µ = 27 and λ = -9

Question 5.
The interval in which y = x²e^-x is increasing w.r.t x is: [1]
(a) x ∈ (0, 2)
(b) x ∈ [1, 0]
(c) x ∈ (∞, 0]
(d) x ∈ [0, ∞)
Solution:
(a) x ∈ (0, 2)
Explanation:

Question 6.
If y = \(\sqrt{a^2-x^2}\), then y\(\frac{d y}{d x}\) is: [1]
(a) 0
(b) x
(c) -x
(d) 1
Solution:
(c) -x
Explanation:

Question 7.
The value of \(\int_1^2 \frac{d x}{x \sqrt{x^2-1}}\) is: [1]
(a) \(\frac{\pi}{3}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{6}\)
Solution:
(a) \(\frac{\pi}{3}\)
Explanation:

Question 8.
In an LPP, the objective function is always: [1]
(a) linear
(b) quadratic
(c) cubic
(d) biquadratic
Solution:
(a) linear

Question 9.
What is the degree of the differential equation: [1]
\(5 x\left(\frac{d y}{d x}\right)^2-\frac{d^2 y}{d x^2}-6 y=\log x\)
(a) 1
(b) 2
(c) 3
(d) not defined
Solution:
(a) 1
Explanation:
Here, the highest order derivative is \(\frac{d^2 y}{d x^2}\), whose degree is 1.

Question 10.
The area bounded by the shaded region as shown in the figure below is: [1]

(a) \(\frac{3}{2}\) sq. units
(b) \(\frac{9}{4}\) sq. units
(c) 4 sq. units
(d) \(\frac{5}{2}\) sq. units
Solution:
(b) \(\frac{9}{4}\) sq. units
Explanation:

Question 11.
The value of \(\int_2^3 \frac{\log x^2}{x} d x\) is: [1]
(a) log 6 log(\(\frac{3}{2}\))
(b) log(\(\frac{3}{2}\))
(c) 2 log 3
(d) \(\frac{1}{3}\) log 6
Solution:
(a) log 6 log(\(\frac{3}{2}\))
Explanation:

Question 12.
The area bounded by the curve y = f(x), the y-axis, y = c and y = d is: [1]
(a) \(\int_c^d f(x) d x\)
(b) \(\int_a^b f(x) d x\)
(c) \(\int_c^d y \cdot d x\)
(d) \(\int_a^b y \cdot d x\)
Solution:
(a) \(\int_c^d f(x) d x\)
Explanation:
Area = \(\int_c^d f(x) d x\)

Question 13.
If \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\), then the value of x is: [1]
(a) 6
(b) 3
(c) 7
(d) 1
Solution:
(a) 6
Explanation:
As given \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\)
⇒ (4 – 2)² = (3x – 2) – (x + 6)
⇒ 4 = 3x – 2 – x – 6
⇒ 2x = 12
⇒ x = 6

Question 14.
A is a skew-symmetric matrix and a matrix B such that B’AB is defined, then B’AB is a: [1]
(a) symmetric matrix
(b) skew-symmetric matrix
(c) Diagonal matrix
(d) upper triangular symmetric
Solution:
(b) skew-symmetric matrix
Explanation:
A is a skew-symmetric matrix
⇒ A’ = -A
consider (B’AB)’ = (AB)'(B’)’ = B’A'(B’)’
= B’A’B
= B'(-A)B
= -B’AB
As (B’AB) = -B’AB
Hence, B’AB is a skew-symmetric matrix.

Question 15.
If A and B are square matrices of order 3 such that |A| = 1 and |B| = 3, then the value of |3AB| is: [1]
(a) 3
(b) 9
(c) 27
(d) 81
Solution:
(d) 81
Explanation:
As AB is of order 3 and
|3AB| = 3³|AB|
= 27|A||B|
= 27 × 1 × 3
= 81

Question 16.
For the matrices A and B, if multiplication is defined and AB = A and BA = B, then B² is: [1]
(a) A
(b) B
(c) I
(d) AB
Solution:
(b) B
Explanation:
Since, BA = B
⇒ (BA)B = BB
⇒ B(AB) = B²
⇒ BA = B² [∵ AB = A]
⇒ B = B² [∵ BA = B]

Question 17.
For what value of k ∈ N, is \(\left|\begin{array}{ll}
k & 3 \\
4 & k
\end{array}\right|=\left|\begin{array}{cc}
4 & -3 \\
0 & 1
\end{array}\right|\). [1]
(a) 4
(b) 1
(c) 3
(d) 0
Solution:
(a) 4
Explanation:
Given, \(\left|\begin{array}{cc}
k & 3 \\
4 & k
\end{array}\right|=\left|\begin{array}{cc}
4 & -3 \\
0 & 1
\end{array}\right|\)
⇒ k² – 12 = 4 – 0
⇒ k² = 16
⇒ k = ±4
⇒ k = 4 ∈ N

Question 18.
If R is an equivalence relation defined in set A = {1, 2, 3 ….. 10} as R = {(a, b): |a – b| is a multiple of 3}. The equivalence class of {1} is: [1]
(a) {1}
(b) {1, 2}
(c) {1, 4, 10}
(d) {1, 4, 7, 10}
Solution:
(d) {1, 4, 7, 10}
Explanation:
For equivalence class {1}
(a, 1) ∈ R for a ∈ A
⇒ |a – 1| is a multiple of 3
⇒ a – 1 = 3λ
⇒ a = 3λ + 1
a = 1, 4, 7, 10
∴ {1} = {1, 4, 7, 10}

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): A function f: N → N be defined by:
\(f(n)= \begin{cases}\frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even for all } n \in N \text { one-one }\end{cases}\)
Reason (R): A function f: A → B is said to be injective if f(a) = f(b) ⇒ a = b [1]
Solution:
(d) A is false but R is true
Explanation:
For n = 1, f(1) = \(\frac{1+1}{2}\) = 1
For n = 2, f(2) = \(\frac{2}{2}\) = 1
⇒ f(1) = f(2) = 1
⇒ f(x) is not one-one
∴ Assertion is false

But the definition of injectivity is true so, the reason is true.

Question 20.
Assertion (A): \(\frac{d}{d x}\left(x^{x^x}\right)=x^{x^x} \cdot x(1+2 \log x)\)
Reason (R): \(\left(x^x\right)^x=x^{x^2}=e^{x^2}=e^{x^2} \log x \) [1]
Solution:
(d) A is false but R is true
Explanation:

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

Question 21.
Find the value of tan^-1(1) + tan^-1(-√3) [2]
Solution:

Question 22.
Find the area bounded by the curve y = cos x, x ∈ [0, π]
OR
Find \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left|\frac{2-\sin x}{2+\sin x}\right| d x\) [2]
Solution:
Shaded area = \(\int_0^\pi|\cos x| d x\)

Question 23.
Two vectors \(\hat{j}+\hat{k}\) and \(3 \hat{i}-\hat{j}+4 \hat{k}\), represent two side vectors \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) respectively of ΔABC. What is the length of the median through A?
OR
Write the Cartesian equation of the following line given in vector form \(\vec{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})\) [2]
Solution:

OR
The point through which the line passes is (2, 1, -4) and dr’s is 1, -1, -1.
∴ Cartesian equation of line is \(\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z+4}{-1}\) or \(\frac{x-2}{1}=\frac{1-y}{1}=\frac{z+4}{-1}\)

Question 24.
What is the particular solution of the differential equation xy dy = (y + 5) dx, given that y(5) = 0 [2]
Solution:
Given xy dy = (y + 5) dx
⇒ \(\frac{y d y}{y+5}=\frac{d x}{x}\)
On integrating both sides
\(\int\left(\frac{y+5-5}{y+5}\right) d y=\int \frac{d x}{x}\)
⇒ \(\int\left\{1-\frac{5}{y+5}\right\} d y=\int \frac{d x}{x}\)
⇒ y – 5 log(y + 5) = log x + C ……(i)
when x = 5, y = 0
0 – 5log 5 = log 5 + C
⇒ C = -6 log 5
Put it in (i), and we get
y – 5 log(y + 5) = log x – 6 log 5 is the required solution.

Question 25.
If \(\vec{a} \cdot \vec{a}=0\) and \(\vec{a} \cdot \vec{b}=0\), then what could be concluded about the vector \(\vec{b}\)? [2]
Solution:
It is given that \(\vec{a} \cdot \vec{a}=0\) or \(|\vec{a}|^2=0\)
⇒ |a| = 0
⇒ a = 0
i.e., a is a vector because its magnitude is zero
Hence, \(\vec{a} \cdot \vec{b}=0\), whatever b may (∵ a = 0)
This means that nothing can be concluded about \(\vec{b}\), it can be any vector

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

Question 26.
Let f: W → W be defined as f(n) = n – 1, if w is odd and f(n) = n + 1, if n is even. Show that f is one-one and onto. [3]
Solution:
f: W → W
f(n) = n – 1, if n is odd
= n + 1, if n is even
when n is odd
f(n₁) = f(n₂)
⇒ n₁ – 1 = n₂ – 1
⇒ n₁ = n₂
So, f(n) is one-one when n is even
f(n1) = f(n₂)
n₁ + 1 = n₂ + 1
n₁ = n₂
So, f(n) is one-one when n is odd
f(n) = n – 1
y = n – 1
n = y + 1
Put n in f(n)
f(n) = y + 1 – 1 = y
when n is even
f(n) = n + 1
y = n + 1
n = y – 1
Put n in f(n)
f(n) = y – 1 + 1 = y
So, f(n) is onto
So, the function f(n) is bijective i.e. one-one and onto

Question 27.
If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t) show that \(\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}=\frac{b}{a}\)
OR
If y = \(\frac{\log x}{x}\) show that \(\frac{d^2 y}{d x^2}=\frac{2 \log x-3}{x^3}\) [3]
Solution:
Given, x = a sin2t (1 + cos2t)
y = b cos2t(1 – cos2t)
\(\frac{d x}{d t}\) = a[cos2t × 2(1 + cos2t) + sin2t(-sin2t) × 2]
= 2a[cos2t (1 + cos2t) – sin 2t]
= 2a[cos2t + cos²2t – sin²2t)
= 2a(cos2t + cos4t)
\(\frac{d y}{d t}\) = b[-sin2t × 2(1 – cos2t) + cos2t(2 × sin2t)
= 2b[-sin2t + 2sin2t cos2t)
= 2b(sin4t – sin2t)

Question 28.
Using integration, find the area of a ΔPQR, the coordinates of vertices being P(1, 6), Q(2, 8), and R(3, 4). [3]
Solution:
On plotting the points P(1, 6), Q(2, 8), and R(3, 4), we notice, we have to find the shaded area.

∴ ar(PQR) = ar(∠PQM) + ar(MQRN) – ar(∠PRN) ……..(i)
For ar(∠PQM): Curve PQ, x-axis between x = 1 and x = 2
Equation of PQ:
y – 6 = \(\frac{8-6}{2-1}\) (x – 1)
⇒ y – 6 = 2(x – 1)
⇒ y = 2x + 4
∴ ar(∠PQM) = \(\int_1^2(2 x+4) d x\)
= \(\left[x^2+4 x\right]_1^2\)
= (4 + 8) – (1 + 4)
= 7 ……(ii)
For ar(MQRN): Curve QR, x-axis between x = 2 and x = 3
Equation of QR:
y – 8 = \(\frac{4-8}{3-2}\) (x – 2)
⇒ y – 8 = -4(x – 2)
⇒ y = -4x + 16
∴ ar(MQRN) = \(\int_2^3(-4 x+16) d x\)
= \(\left[-2 x^2+16 x\right]_2^3\)
= (-18 + 48) – (-8 + 32)
= 30 – 24
= 6 …….(iii)
For ar(LPRN): curve PR: x-axis between x = 1 and x = 3
Equation of PR:
y – 6 = \(\frac{4-6}{3-1}\) (x – 1)
⇒ y – 6 = -(x – 1)
⇒ y = -x + 7
∴ ar(∠PRN) = \(\int_1^3(-x+7) d x\)
= \(\left[\frac{-x^2}{2}+7 x\right]_1^3\)
= \(\left(\frac{-9}{2}+21\right)-\left(\frac{-1}{2}+7\right)\)
= \(\frac{33}{2}-\frac{13}{2}\)
= 10 Sq. units …..(iv)
Substituting from (ii), (iii), (iv) in (i), we get
ar(ΔPQR) = 7 + 6 – 10 = 3 Sq. units

Question 29.
Show the differential equation (1 + e^2x) dy + (1 + y²) e^x dx = 0, given that y = 1 when x = 0
OR
Show that the differential equation 2y e^x/y dx + (y – 2x e^x/y) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1. [3]
Solution:

Question 30.
The random variable X can take only the values 0, 1, 2, 3. Given that:
P(X = 0) = P(X = 1) = P and P(X = 2) = P(X = 3) such that \(\Sigma P_i x_i^2=2 \Sigma P_i x_i\), find the value of P.
OR
A problem in Mathematics is given to three students whose chances of solving it are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) is the probability that the problem is solved.
Solution:
Given, X = 0, 1, 2, 3 and P(X = 0) = P(X = 1) = 1, P(X = 2) = P(X = 3)
Such that \(\sum \mathrm{P}_i x_i^2=2 \sum \mathrm{P}_i x_i\)
Now, ΣP_i = 1
⇒ P₀ + P₁ + P₂ + P₃ = 1
⇒ P + P + x + x = 1
⇒ 2P + 2x = 1
⇒ 2x = 1 – 2P
⇒ x = \(\frac{1-2 P}{2}\)
The probability distribution of X is given by

OR
Let A, B, and C be the three students and P(A), P(B), P(C) be the probabilities of solving a problem respectively.
P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\), P(C) = \(\frac{1}{4}\)
P[problem will be solved at least by 1] = 1 – P(\(\overline{\mathrm{A}}\)) P(\(\overline{\mathrm{B}}\)) P(\(\overline{\mathrm{C}}\))
= 1 – [1 – P(A)] [1 – P(B)] [1 – P(C)]
= 1 – \(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\)
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)

Question 31.
Find the points on the line \(\frac{x+2}{3}=\frac{y+1}{2}\) = \(\frac{z-3}{2}\) at a distance of 5 units from the point P(1, 3, 3) [3]
Solution:
Given, the equation of a line is:
\(\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}\) = λ (say)
x = 3λ – 2, y = 2λ – 1, z = 2λ + 3
So, we have a point on the line is:
Q(3λ – 2, 2λ – 1, 2λ + 3) …….(i)
Now, given that distance between two points P(1, 3, 3) and Q(3λ – 2, 2λ – 1, 2λ + 3) is 5 units
i.e. PQ = 5
⇒ \(\sqrt{\left[(3 \lambda-2-1)^2+(2 \lambda-1-3)^2+(2 \lambda+3-3)^2\right]}=5\)
On Squaring both sides, we get
(3λ – 3)² + (2λ – 4)² + (2λ)² = 25
⇒ 9λ² + 9 – 18λ + 4λ² + 16 – 16λ + 4λ² = 25
⇒ 17λ² – 34λ = 0
⇒ 17λ (λ – 2) = 0
Either 17λ = 0 or λ – 2 = 0
∴ λ = 0 or 2
On putting λ = 0 and λ = 2 in equation (i),
we get the required point as (-2, -1, 3) or (4, 3, 7)

Section – D (20 Marks)
This section comprises long answer type questions (LA) of 5 marks each

Question 32.
If \(\vec{a}=3 \hat{i}-\hat{j}\) and \(\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}\), then express \(\vec{b}\) in the form \(\vec{b}=\overrightarrow{b_1}+\overrightarrow{b_2}\), where \(\overrightarrow{b_1} \| \vec{a}\) and \(\overrightarrow{b_2} \perp \vec{a}\). [5]
Solution:

Question 33.
Given A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\) verify that BA = 6I, how can we use the result to find the values of x, y, z from given equations x – y = 3, 2x + 3y + 4z = 17, y + 2z = 17
OR
If A = \(\left[\begin{array}{rrr}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\) find A-1. How we can use A-1 to find x, y, z for the following system of equations:
2x – 3y + 5z = 16, 3x + 2y – 4z = -4, x + y – 2z = -3 [5]
Solution:

Question 34.
Solve the following differential equation:
\(\left[y-x \cos \left(\frac{y}{x}\right)\right] d y\) \(+\left[y \cos \left(\frac{y}{x}\right)-2 x \sin \left(\frac{y}{x}\right)\right] d x=0\)
OR
Sketch the graph of y = |x + 3| and evaluate the area under the curve y = |x + 3| above x-axis and between x = -6 to x = 0. [5]
Solution:


OR
First we sketch the graph of y = |x + 3|
∴ y = |x + 3| =\(\left\{\begin{aligned}
x+3, & \text { if } x+3 \geq 0 \\
-(x+3), & \text { if } x+3<0
\end{aligned}\right.\)
⇒ y = |x + 3| = \(\left\{\begin{array}{r}
x+3, \text { if } x \geq-3 \\
-x-3 \text {, if } x<-3
\end{array}\right.\)
So, we have y = x + 3 for x ≥ -3 and y = -x – 3 for x < -3.
A sketch of y = |x + 3| is shown below:

Here y = x + 3 is the straight line that cuts the X and Y-axis at (-3, 0) and (0, 3) respectively.
Thus, y = x + 3 for x ≥ -3 represents the part of Line which lies on the left side of x = -3 clearly,
required area = Area of region ABPA + Area of region PCOP

Question 35.
Find the area of the greatest rectangle that can be inscribed in an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) [5]
Solution:
Let ABCD be a rectangle having area A inscribed in an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) …..(i)
Let, the coordinates of A be (α, β)

Then, the coordinate of B = (α, -β)
C = (-α, -β)
D = (-α, β)
∴ Area of rectangle = L × B
= 2α × 2β
= 4αβ
⇒ A = \(4 \alpha \sqrt{b^2\left(1-\frac{\alpha^2}{a^2}\right)}\) [∵ \(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}=1\)]

Section – E (12 Marks)

This section comprises 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.
A shopkeeper sells three types of flower seeds A1, A2, and A3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60%, and 35%.

Based on the above information, answer the following questions:
(A) What is the probability that a randomly choosen seed to germinate? [1]
(B) What is the probability that a seed of type A3 does not germinate? [1]
(C) What is the probability that the seed is of type A2 given that a randomly choosen seed does not germinate?
OR
If a randomly choosen seed germinates, what is the probability it is of type A? [2]
Solution:

Question 37.
Let R be the feasible region of a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (max. or min.), when the variable x and y are subject to constraints described by linear inequalities, this optimal value occurs at the corner point (vertex) of the feasible region.
Based on the above information, answer the following questions:
(A) What is an objective function of LPP? [1]
(B) In solving an LPP “minimize f = 6x + 10y subject to constraints x ≥ 6, y ≥ 2, 2x + y ≥ 10, x ≥ 0, y ≥ 0” which among is redundant constraint? [1]
(C) The feasible region for an LPP is shown in the figure. Let Z = 3x – 4y, be the objective function. Then, at which point minimum of Z occurs?

OR
The feasible region for an LPP is shown shaded in the figure. Let F = 3x – 4y be the objective function. Then, what is the maximum value of F. [2]

Solution:
(A) Objective function is a linear function whose maximum or minimum values is to be found.
(B) When x ≥ 6 and y ≥ 2, then
2x + y ≥ 2 × 6 + 2
i.e., 2x + y ≥ 14
Hence, x ≥ 0, y ≥ 0, and 2x + y ≥ 10 are automatically satisfied by every point of the region.
Hence, answer is 2x + y ≥ 10, x ≥ 0, y ≥ 0.
(C)

Minimum of z = -32 at (0, 8)
The feasible region for an LPP is

Hence, the maximum of F = 0

Question 38.
As we know good planning can save energy, time, and money. A farmer wants to construct a circular well and a square garden in his field. He wants to keep their perimeters 600 m.

Based on the above information, answer the following questions:
(A) If the radius of the circular garden is ‘r’ m and the side of the square garden is ‘x’ m, then what is the sum of their areas? And find the number which exceeds its square by the greatest possible. [2]
(B) At what radius, is the sum of their areas is least? [2]
Solution:
(A) Sum of area = πr² + x²
But 2πr + 4x = 600
x = \(\frac{600-2 \pi r}{4}\)
∴ Sum of areas = \(\pi r^2+\left(\frac{600-2 \pi r}{4}\right)^2\)
Assume y is the difference between x and its square
i.e., y = x – x²
For max. difference between the numbers
\(\frac{d y}{d x}\) = 0 = 1 – 2x
⇒ x = \(\frac{1}{2}\)
(B) Sum of areas,