CBSE Sample Papers for Class 12 Maths Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Maths Set 8 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.
Given a matrix A = [a_ij] of order 3 × 3 whose elements a_ij = \(\frac{(2 i-j)^2}{i+j}\), then the element a₃₂ of matrix A is: [1]
(a) 12
(b) 18
(c) \(\frac{16}{5}\)
(d) \(\frac{15}{4}\)
Solution:
(c) \(\frac{16}{5}\)
Explanation:

Question 2.
The value of the n, such that the differential equation \(\frac{x^n d y}{d x}\) = y (log y – log x + 1) is homogeneous is: [1]
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Explanation:
When n = 1, then the equation will be homogeneous.

Question 3.
A card is drawn from a well-shuffled pack of 52 cards, if E is the event “the card drawn is a king or a queen and F is the event” the card drawn is an ace or a queen.’ Then, the probability of conditional event E/F is: [1]
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{2}{5}\)
Solution:
(b) \(\frac{1}{2}\)
Explanation:
E: 4 kings + 4 queens
F: 4 aces + 4 queens
E ∩ F = 4 queens

Question 4.
If \(\int \frac{x^2}{1-x^6} d x\) = f(x) + c, then the value of \(\frac{d}{d x} f(x)\) is: [1]
(a) \(\frac{x^2}{1-x^6}\)
(b) \(\frac{x^2}{1+x^6}\)
(c) \(\frac{1+x^6}{x^2}\)
(d) \(\frac{1-x^6}{x^2}\)
Solution:
(a) \(\frac{x^2}{1-x^6}\)
Explanation:

Question 5.
The value of x for which the vectors \(\hat{i}+3 \hat{j}+\hat{k}\) and \(x \hat{i}-\hat{j}-\hat{k}\) are perpendicular to each other is: [1]
(a) -4
(b) 4
(c) 5
(d) 3
Solution:
(b) 4
Explanation:
If given vectors are perpendicular to each other, then
1 . x + 3 . (-1) + 1 . (-1) = 0
⇒ x – 3 – 1 = 0
⇒ x = 4

Question 6.
If E₁ is an equivalence class with respect to relation R defined on set A and i = 1, 2 … 5. If x ∈ E_i and y ∈ E_j, i ≠ j, then can we say that x and y are related to each other with respect to relation R? [1]
(a) Yes
(b) No
(c) Can’t be decided
(d) Depends on the situation
Solution:
(b) No
Explanation:
In equivalence classes, elements of the equivalence class are only related to each other. If elements of E_i are only related to each other.

Question 7.
The value of 2m – n, where m and n represents the degree and order of the differential equation \(\frac{d^2 y}{d x^2}\) + x = 0. [1]
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
(b) 0
Explanation:
Given equation is \(\frac{d^2 y}{d x^2}+x=0\)
order (n) = 2, degree (m) = 1
∴ 2m – n = 2 × 1 – 2 = 0

Question 8.
The principal value branch of sec-1x for which it is bijective over the domain R – (-1, 1) is: [1]
(a) {0, π}
(b) \(\left\{0, \frac{\pi}{2}\right\}\)
(c) {0, π} – {\(\frac{\pi}{2}\)}
(d) \(\left\{0, \frac{\pi}{4}\right\}\)
Solution:
(c) {0, π} – {\(\frac{\pi}{2}\)}
Explanation:
The principal value branch of sec^-1x is {0, π} – {\(\frac{\pi}{2}\)}

Question 9.
If the matrix B = \(\left[\begin{array}{ccc}
2 & a & 5 \\
-1 & 4 & b \\
c & -4 & 9
\end{array}\right]\) is a symmetric matrix, then a + b + c is: [1]
(a) 0
(b) 5
(c) 4
(d) -1
Solution:
(a) 0
Explanation:
Given, B = \(\left[\begin{array}{ccc}
2 & a & 5 \\
-1 & 4 & b \\
c & -4 & 9
\end{array}\right]\) is a symmetric matrix.
∴ B^T = B
⇒ \(\left[\begin{array}{ccc}
2 & -1 & c \\
a & 4 & -4 \\
5 & b & 9
\end{array}\right]=\left[\begin{array}{ccc}
2 & a & 5 \\
-1 & 4 & b \\
c & -4 & 9
\end{array}\right]\)
∴ a = -1, b = -4, c = 5
∴ a + b + c = -1 – 4 + 5 = 0

Question 10.
If y = log_ax then \(\frac{d y}{d x}\) is: [1]
(a) \(\frac{1}{a}\)
(b) \(\frac{1}{x}\)
(c) \(\frac{1}{x \log a}\)
(d) \(\frac{1}{a \log x}\)
Solution:
(c) \(\frac{1}{x \log a}\)
Explanation:
Given, y = log_ax

Question 11.
The area bounded by the curve given below is: [1]

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

Question 12.
If \(\int \frac{x^2}{x^2+1} \cdot d x\) = x + f(x) + c then f(x) is: [1]
(a) tan^-1x
(b) -tan^-1x
(c) sin^-1x
(d) -sin^-1x
Solution:
(b) -tan^-1x
Explanation:

Question 13.
If \(\left|\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right|\) = P, then what is the value of \(\left|\begin{array}{lll}
a & d & g \\
b & e & h \\
c & f & i
\end{array}\right|\), given P = 17? [1]
(a) 17
(b) -17
(c) \(\frac{1}{17}\)
(d) \(\frac{-1}{17}\)
Solution:
(a) 17
Explanation:
Consider ∆ = \(\left|\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right|\) = 17
We know that value of the determinant is the same if we take transpose.
∴ \(\left|\begin{array}{lll}
a & d & g \\
b & e & h \\
c & f & i
\end{array}\right|\) = P = 17

Question 14.
The function fgiven by f(x) = tan^-1(sin x + cos x) is decreasing in the interval: [1]
(a) \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\)
(b) \(\left(\frac{\pi}{3}, \pi\right)\)
(c) \(\left(\frac{2 \pi}{3}, \pi\right)\)
(d) \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
Solution:
(a) \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\)
Explanation:
Consider f(x) = tan^-1(sin x + cos x)
f'(x) = \(\frac{1}{1+(\sin x+\cos x)^2}\) . (cos x – sin x) ……..(i)
sign of f'(x) depends upon (cos x – sin x)
We know for x ∈ \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), cos x < sin x
⇒ cos x – sin x < 0
⇒ f'(x) < 0
∴ f is decreasing for x ∈ \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), [from (i)]

Question 15.
Name the method by which the differential equation: \(\frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}\) can be solved. [1]
(a) It is of the form \(\frac{d y}{d x}\) = P
(b) Limit of a sum
(c) Variable separable method
(d) None of these
Solution:
(c) Variable separable method
Explanation:
As the given differential equation could be written as:
\(\frac{d y}{\sqrt{1-y^2}}=\frac{d x}{\sqrt{1-x^2}}\)

Question 16.
The value of \(\tan ^2\left(\frac{1}{2} \sin ^{-1} \frac{2}{3}\right)\) is: [1]
(a) \(\frac{7-3 \sqrt{5}}{2}\)
(b) \(\frac{3+\sqrt{5}}{3}\)
(c) \(\frac{3 \sqrt{5}}{2}\)
(d) \(\frac{3 \sqrt{3}-2}{2}\)
Solution:
(a) \(\frac{7-3 \sqrt{5}}{2}\)
Explanation:

Question 17.
The general solution of the differential equation \(\log \left(\frac{d y}{d x}\right)\) = 2x + y is: [1]
(a) \(\frac{1}{e^x}=\frac{1}{e^y}+c\)
(b) \(-\frac{1}{e^y}=\frac{1}{2} e^{2 x}+c\)
(c) \(\frac{1}{e^{2 x}}=\frac{1}{e^{x y}}+c\)
(d) \(\frac{e^{2 x}}{2 e x^{e x}}=\frac{-1}{e^y}+c\)
Solution:
(b) \(-\frac{1}{e^y}=\frac{1}{2} e^{2 x}+c\)
Explanation:

Question 18.
If x = t² and y = t³, then the value of \(\frac{d^2 y}{d x^2}\) is: [1]
(a) \(\frac{3}{4 t}\)
(b) \(\frac{3}{2} t\)
(c) \(\frac{1}{2 t}\)
(d) \(\frac{t}{2}\)
Solution:
(a) \(\frac{3}{4 t}\)
Explanation:

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 value of \(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i}+\hat{k})+\) \(\hat{k} \cdot(\hat{i} \times \hat{j})\) is 1.
Reason (R): Since, \(\hat{i} \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=0\). [1]
Solution:
(c) A is true but R is false.
Explanation:
As \(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot \hat{(i \times} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})\) = \(\hat{i} \hat{i}+\hat{j} \cdot(-j)+\hat{k} \cdot \hat{k}\)
= 1 – 1 + 1
= 1

Question 20.
Assertion (A): If \(\int_0^1\left(3 x^2+2 x+k\right) \cdot d x=0\), then the value if k is -1.
Reason (R): \(\int x^n \cdot d x=\frac{x^{n+1}}{n+1}\). [1]
Solution:
(d) A is false but R is true.
Explanation:
We have, \(\int_0^1\left(3 x^2+2 x+k\right) \cdot d x=0\)
⇒ \(\left[x^3+x^2+k x\right]_0^1\) = 0
⇒ [1 + 1 + k] – 0 = 0
⇒ 2 + k = 0
⇒ k = -2

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

Question 21.
In a linear programming problem, objective function, z = x + 2y. The subjective the constraints
x + 2y ≥ 100
2x- y ≤ 0
2x + y ≤ 200
x ≥ 0, y ≥ 0
The graph of the following equations is shown below.

Name the feasible region, and find the corner point at which the objective function is minimum. [2]
Solution:
Here the feasible region is ABCDEA
So, corner points are A(0, 50), B(20, 40), C(50, 100), E(0, 200)

The minimum value of z is 100 at all the points on the line segment joining the points (0, 50) and (20, 40).

Question 22.
What is the equation of the curve passing through t(1, -1) and satisfying the differential equation \(\frac{d y}{d x}=\frac{y}{x}\).
OR
Find the shaded area covered by the curve shown in the figure. [2]

Solution:

Question 23.
The function f defined by f(x) = \(\left\{\begin{array}{l}
k x+1, \text { if } x \leq \pi \\
\cos x, \text { if } x>\pi
\end{array}\right.\) is continuous at x = π. Then, what is the value of k? [2]
Solution:
The given function is:
\(f(x)=\left\{\begin{array}{l}
k x+1, \text { if } x \leq \pi \\
\cos x, \text { if } x>\pi
\end{array}\right.\)
The given function f is continuous at x = 1

Question 24.
Given, a relation R₁, on the set R₁ of real numbers, as R₁ = J(x, y) : x² – 3xy + 2y² = 0, x, y ∈ R}. If the relation R reflexive or symmetric? [2]
Solution:
Reflexive, as relation R₁ = {(x, y) ∈ R × R : x² = 3xy + 2y² = 0}
(x, x) ∈ R
⇒ x² – 3x² + 2x² = 0, always true, for x ∈ R.
Hence, reflexive.
Let (x, y) ∈ R₁
⇒ x² – 3xy + 2y² = 0 ……..(i)
Let x = 2, y = 1
⇒ 4 – 6 + 2 = 0 true, (2, 1) ∈ R₁
Now, (y, x) ∈ R₁
⇒ y² – 3yx + 2x² = 0 …….(ii)
y = 1, x = 2
⇒ 1 – 6 + 8 ≠ 0, (1, 2) ∉ R₁
(i) & (ii) are not equation i.e. (x, y) ∈ R₁
⇒ (y, x) ∉ R₁
Hence, it is not symmetric.

Question 25.
Evaluate: \(\int \frac{3 a x}{b^2+c^2 x^2} d x\).
OR
Evaluate: \(\int \frac{\sqrt{\cos \theta}}{\sin \theta} d \theta\). [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 S in the set R of real numbers defined as S = {(a, b): a, b ∈ R and a ≤ b³} is neither reflexive nor symmetric and nor transitive.
OR
If f: R → R is the function defined by f(x) = 4x³ + 7, then show that f is a bijection. [3]
Solution:
The given relation is:
S = {(a, b) : a, b ∈ R and a ≤ b³}
Reflexive: As \(\frac{1}{2} \leq\left(\frac{1}{2}\right)^3\) where \(\frac{1}{2}\) ∈ R, is not true
∴ \(\left(\frac{1}{2}, \frac{1}{2}\right)\) ∉ S
Thus, S is not reflexive.
Symmetric: As (-2) ≤ 3³, where -2, 3 ∈ R, is true
but 3 ≤ (-2)³ is not true
i.e. (-2, 3) ∈ S but (3, -2) ∉ S.
Thus, S is not symmetric
Transitive: As 3 ≤ \(\left(\frac{3}{2}\right)^3\) and \(\frac{3}{2} \leq\left(\frac{4}{3}\right)^3\), where 3, \(\frac{3}{2}\), \(\frac{4}{3}\) ∈ R are true but 3 ≤ \(\left(\frac{4}{3}\right)^3\) is not true
i.e., (3, \(\frac{3}{2}\)) ∈ S and \(\left(\frac{3}{2}, \frac{4}{3}\right)\) ∈ S but (3, \(\frac{4}{3}\)) ∉ S
Thus, R is not transitive.
Hence, S is neither reflexive nor symmetric nor transitive.
OR
Given, function is f: R → R such that
f(x) = 4x³ + 7
To show: f is bijective, we have to show that f is one-one and onto.
One-one function: Let, x₁, x₂ ∈ R such that
f(x₁) = f(x₂)
⇒ \(4 x_1^3+7=4 x_2^3+7\)
⇒ \(4 x_1^3=4 x_2^3\)
⇒ \(x_1^3-x_2^3=0\)
⇒ \(\left(x_1-x_2\right)\left(x_1^2+x_1 x_2+x_2^2\right)=0\)

But eq. (ii) gives complex roots as x₁, x₂ ∈ R
∴ x₁ – x₂ = 0
⇒ x₁ = x₂
Thus, f(x₁) = f(x₂)
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ R
Therefore, f(x) is one-one function.
Onto function: Let y ∈ R (co-domain) by any arbitrary number.
Then, f(x) = y
⇒ 4x³ + 7 = y
⇒ 4x³ = y – 7
⇒ x = \(\left(\frac{y-7}{4}\right)^{1 / 3}\)
which is a real number. [∵ y ∈ R]
Thus, for every y ∈ R (co-domain) there exist
x = \(\left(\frac{y-7}{4}\right)^{1 / 3}\) ∈ R (domain) such that
f(x) = \(f\left[\left(\frac{y-7}{4}\right)^{1 / 3}\right]=4\left[\left(\frac{y-7}{4}\right)^{1 / 3}\right]^3+7\)
= \(4\left(\frac{y-7}{4}\right)+7\)
= y – 7 + 7
= y
⇒ f(x) is an onto function.
Since f(x) is both one-one and onto so it is bijective.

Question 27.
Find A^-1, if A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and then show that \(\mathrm{A}^{-1}=\frac{\mathrm{A}^2-3I}{2}\).
OR
The sum of the tree number is 6.
If we multiply the third number by 3 and add the second number to it, we get 11. By adding the first and third numbers we get double the second number. Represent it algebraically and find the numbers using the matrix method. [3]
Solution:
We have, A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)
Clearly |A| = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)
= 0 – 1(0 – 1) + 1(1 – 0)
= 2 ≠ 0
∴ A^-1 exists
Now let us evaluate the cofactors of elements of |A|.
Clearly,



OR
Let, the first second and third numbers are x, y and z respectively.
Then according to the given condition, we have
x + y + z = 6, y + 3z = 11 and x + z = 2y or x – 2y + z = 0
This system of equations can be written in matrix form:

Question 28.
Sand is powered from the pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm. [3]
Solution:
Let V be the volume of the cone and ‘h’ be the height and r be the radius of the base of the cone.
Now, \(\frac{d V}{d t}\) = 12 cm³/sec …….(i)
Height of cone = \(\frac{1}{6}\) × radius of cone
h = \(\frac{1}{6}\) × r
r = 6h …….(ii)
Now, volume of cone, V = \(\frac{1}{3} \pi r^2 h\)

Question 29.
Write the vector equation of the following lines and hence find the distance between them
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\)
\(\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}\)
OR
If O is origin OP = 3 with direction ratios proportional to -1, 2, -2 then what are the coordinates of P? [3]
Solution:


OR
Let the coordinates of point P be (x₁, y₁, z₁)
Given, OP = 3
\(\sqrt{\left(x_1-0\right)^2+\left(y_1-0\right)^2+\left(z_1-0\right)^2}=9\)
\(x_1^2+y_1^2+z_1^2=9\) ……(i)
The direction ratio of OP are -1, 2, -2
Therefore,
x₁ – 0 = -1 ⇒ x₁ = -1
y₁ – 0 = 2 ⇒ y₁ = 2
z₁ – 0 = -2 ⇒ z₁ = -2
Putting the values of x₁, y₁, z₁ in (1) we get,
(-1)² + (2)² + (-2)² = 9
9 = 9
x₁, y₁, z₁ satisfies (1)
Hence, the coordinates of point P are (-1, 2, -2)

Question 30.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19, 20 years one student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of random variable X. [3]
Solution:
Here, the total number of students = 15
The ages of students in ascending order are 14, 14, 15, 16, 16, 17, 17, 17, 18, 19, 19, 20, 20, and 21.

Therefore, the p.a. of random variable X is as follows:

Question 31.
The feasible region of a ∠PR is given as follows:

(A) Write the constraints with respect to the above in terms of x and y.
(B) Find the coordinate of B and C and maximize, z = x + y. [3]
Solution:
(A) Equation of line is: \(\frac{x}{25}+\frac{y}{50}\) ≤ 1
⇒ 2x + y ≤ 50
Equation of second line is: \(\frac{x}{40}+\frac{y}{20}\) ≤ 1
⇒ x + y ≤ 40
∴ Constraint are 2x + y ≤ 50
x + 2y ≤ 40
x ≥ 0, y ≥ 0
(B) Coordinates of B are (20, 10) and C(0, 20)
∴ For z = x + y

Hence, z is the maximum at the point (20, 10).

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

Question 32.
Find the equation of a line passing through the point P(2, -1, 3) and perpendicular to the lines:
\(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\) and \(\vec{r}=(2 \hat{i}-\hat{j}-3 \hat{k})+\mu(\hat{i}+2 \hat{j}+2 \hat{k})\)
OR
Find the equation of a line passing through the point (1, 2, -4) and perpendicular to two lines:
\(\vec{r}=(8 \hat{i}-19 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k})\) and \(\vec{r}=(15 \hat{i}+29 \hat{j}+5 \hat{k})+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})\). [5]
Solution:
Let the line Passing through points (2, -1, 3) is
\(\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(a \hat{i}+b \hat{j}+c \hat{k})\) …….(i)
If line (i) is perpendicular to lines \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\) and \(\vec{r}=(2 \hat{i}-\hat{j}-3 \hat{k})+\mu(\hat{i}+2 \hat{j}+2 \hat{k})\)

Question 33.
Using integration, find the area of the region bounded by the triangle whose vertices are (2, 3), (4, 7), and (6, 2). [5]
Solution:

Question 34.
The sum of surface areas of a sphere and a cuboid with sides \(\frac{x}{3}\), x, and 2x is constant. show that the sum of their volumes is minimum if x is equal to three times the radius of the sphere.
OR
An advertisement firm is supplied with decorative wire pieces of 34 m each and is asked to cut the wire into two pieces, from one piece a circular sign board is to be made and from another a square one and the idea is to keep the sum of the areas enclosed by a circle and square to be minimum for writing slogans. It wire is cut at ‘x’ m from one end and made into a circle and r represents the radius of the circle. What is the minimum combined area of the window? [5]
Solution:
Surface area of cuboid = 2(lb + bh + hl)
= \(2\left(2 x^2+\frac{2 x^2}{3}+\frac{x^2}{3}\right)\)
= \(2\left(\frac{9 x^2}{3}\right)\)
= 6x²
Let the radius of the sphere be r.
The surface area of the sphere = 4πr²
∴ 6x² + 4πr² = k (constant) …….(i)
Now, the sum of volumes of cuboid and sphere is:
V = \(\frac{2}{3} x^3+\frac{4}{3} \pi r^3\) ……(ii)
Putting the value of r from (i) into (ii), we get
V = \(\frac{2}{3} x^3+\frac{4}{3} \pi\left(\frac{k-6 x^2}{4 \pi}\right)^{3 / 2}\) ……(iii)
Differentiating (iii) w.r.t x, we get
\(\frac{d V}{d x}=2 x^2+\frac{9}{3} \pi\left(\frac{1}{4 \pi}\right)^{3 / 2} \times \frac{3}{2}\left(k-6 x^2\right)^{1 / 2}(-12 x)\) ……(iv)
For minimum or maximum value, \(\frac{d V}{d x}\) = 0

Question 35.
(A) Find the general solution of the differential equation \(y-x \frac{d y}{d x}=x+y \frac{d y}{d x}\)
(B) Evaluate: \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}\). [5]
Solution:

Section – E (12 Marks)

This section comprises 3 case-study/passage-based questions of 4 marks each with two Sub-part. 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.
An electric circuit includes a device that gives energy to the charged particles constituting the current, such as a battery or a generator; devices that use current, such as lamps, electric motors, or computers; and the connecting wires or transmission lines.

An electric circuit consists of two subsystems say A and B as shown below:

For previous testing procedures, the following probabilities are assumed to be known.
P(A fails) = 0.2, P(B fails alone) = 0.15, P(A and B fail) = 0.15
Based on the above information answer the following questions:
(A) What is the probability that B fails? [1]
(B) What is the probability that A fails alone? [1]
(C) Find the probability that the whole of the electric system fails?
OR
Find the conditional probability that B fails when A has already failed. [2]
Solution:
(A) Consider the following events
E = A fails, F = B fails
Given P(E) = 0.2, P (\(\overrightarrow{\mathrm{E}}\) ∩ F) = 0.15, F(E ∩ F) = 0.15
Since, P(\(\overrightarrow{\mathrm{E}}\) ∩ F) = 0.15
⇒ P(F) – P(E ∩ F) = 0.15
⇒ P(F) = 0.15 + P(E ∩ F)
⇒ P(F) = 0.15 + 0.15
⇒ P(F) = 0.30
(B) P(E ∩ \(\overrightarrow{\mathrm{F}}\)) = P(E) – P(E ∩ F)
= 0.2 – 0.15
= 0.05
(C) If the electric system fails, we mean that A is also failed and B is also failed.
i.e., we have to find P(E ∪ F), where E is an event when A fails and B is an event when B fails.
∴ P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
= 0.2 + 0.3 – 0.15
= 0.5 – 0.15
= 0.35
OR
Let E = A fail, F = B fail
∴ P(E) = 0.2, P(E ∩ F) = 0.15, P(\(\overrightarrow{\mathrm{F}}\) ∩ F) = 0.15

Question 37.
To promote the usage of house toilets in villages, especially for women, are organisations tried to generate awareness among the villagers through (i) house calls (ii) letters, and (iii) announcements.

The cost for each mode per attempt is given below.
(i) ₹ 50 (ii) ₹ 20 (iii) ₹ 40
The number of attempts made in villages X, Y, and Z is given below:

Also, the chance of making toilets corresponding to one attempt of given modes is:
(i) 2% (ii) 4% (iii) 20%
Let A, B, and C be the cost incurred by organisation in three villages respectively.
Based on the above information answer the following questions:
(A) Form a required matrix on the basis of the given information. [1]
(B) From a matrix, related to the number of toilets expected in villagers X, Y, and Z after the promotion campaign. [1]
(C) What is the total amount spent by the organisation in all three villages X, Y, and Z?
OR
What is the total no.of toilets expected after the promotion campaign? [2]
Solution:
(A) Here, ₹A, ₹B, and ₹C are the cost incurred by the organisation for villages X, Y, and Z respectively then, A and B will be given by the following matrix equation

Question 38.
Shreya had to go shopping mall for purchasing household items. So she starts walking from her house to the shopping. Instead of going to the shopping mall, she first goes to her friend’s house than to her son’s school, and then reaches the mall. The location of her house friend’s house, her son’s school, and the mall is shown on the coordinates axes in 3-D.


Based on the above information answer the following questions:
(A) What is the total distance travelled by Shreya starting at her house and ending at her house? [2]
(B) What is the extra distance had she travelled if we go directly from her to the mall instead of going to all the places? [2]
Solution: