CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

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CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions :

  1. This question paper contains five sections – A. B. C. D and E. Each section is compulsory. However, there are internal choices in some questions.
  2. Section – A has 18 MCQ’s and 02 Assertion-Reason based questions of 1 mark each.
  3. Section – B has 5 Very Short Answer (VSA)-type questions of 2 marks each.
  4. Section – C has 6 Short Answer (SA)-type questions of 3 marks each.
  5. Section – D has 4 Long Answer (LA)-type questions of 5 marks each.
  6. Section – E has 3 source based/case based/passage based/integrated units of assessment (4 marks each) with sub parts.

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

Question 1.
The value of Cos-1 (cos\(\frac { 3π }{ 2 }\)) is equal to:
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { 3π }{ 2 }\)
(c) \(\frac { 5π }{ 2 }\)
(d) \(\frac { 7π }{ 2 }\)
Solution:
(a) \(\frac { π }{ 2 }\)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-1

Question 2.
Find the vector components of a vector having the initial point (6, 4) and the terminal point (- 2, 7):
(a) 27\(\hat{i}\),\(\hat{j}\)
(b) \(\hat{i}\), -7\(\hat{j}\)
(c) -8\(\hat{i}\), 3\(\hat{j}\)
(d) -3\(\hat{i}\), 8\(\hat{j}\)
Sol.
(c) -8\(\hat{i}\), 3\(\hat{j}\)
Explanation: Let, the initial point of the vector is A (6,4) and the terminal point of the vector is B( – 2, 7).
∴ The required vector is,
\(\vec{AB}\) = ( – 2 – 6)\(\hat{i}\) + (7 – 4) \(\hat{j}\)
= – 82\(\hat{i}\) + 37 \(\hat{j}\)
Hence, the required vector components of the vector are: – 8\(\hat{i}\) and 3 \(\hat{j}\).

Question 3.
Find the magnitude of the vector 3\(\hat{i}\)+ 2\(\hat{j}\) + 12\(\hat{k}\):
(a) \(\sqrt{157}\)
(b) 4\(\sqrt{11}\)
(c) \(\sqrt{213}\)
(d) 9\(\sqrt{3}\)
Solution:
(a) \(\sqrt{157}\)

Explanation:
Magnitude of 3\(\hat{i}\)+ 2\(\hat{j}\) + 12\(\hat{k}\) = \(\sqrt{3^2+2^2+12^2}\) = \(\sqrt{157}\)

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 4.
Let f(x) = |sinx|.Then
(a) f is everywhere differentiable
(b) f is everywhere continuous but not differentiable at x = nπ, n ∈ Z
(c) f is everywhere continuous but not differentiable at x = (2n + 1)\(\frac { π }{ 2 }\),n ∈ Z.
(d) none of these
Solution:
(b) f is everywhere continuous but not differentiable at x = nπ, n ∈ Z

Explanation: We have,f(x) = |sin x |
We know that |x| and sin x are continuous for all real x.
So, |sin x| is also continuous for all real x.
|x| is non-differentiable at x = 0
So, |sin x| is non-differentiable when sin x = O
or x = nπ, n ∈ Z
Hence f(x) is continuous everywhere but not differentiable at x – nit, n∈Z.

Question 5.
By the inspection method, evaluate: ∫(au2 + bu + c)du
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-2
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-3
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-4

Question 6.
The solution of the differential equation \(\frac { dy }{ dx }\) = ex + cos x + x + tan x is:
(a) y = ex + sinx + \(\frac{x^2}{2}\) + log cos x + c
(b) y = ex + sinx + \(\frac{x^2}{2}\) + log sec x + c
(c) y = ex – sinx + \(\frac{x^2}{2}\) + log cos x + c
(d) y = ex – sinx + \(\frac{x^2}{2}\) + log sec x + c
Solution:
(b) y = ex + sinx + \(\frac{x^2}{2}\) + log sec x + c

Explanation:
\(\frac { dy }{ dx }\) = ex + cos x + x + tan x
On integrating both sides, we get
y = ex + sinx + \(\frac{x^2}{2}\) + log sec x + c

Question 7.
The graph of the inequality 2x + 3y > 6 is:
(a) half plane that contains the origin
(b) half plane that neither contains the origin nor the points on the line 2x + 3y =6
(c) whole XOY – plane excluding the points on the line 2x + 3y = 6
(d) entire XOY plane.
Solution:
(b) half plane that neither contains the origin nor the points on the line 2x + 3y = 6

Explanation: Since, origin (0, 0) does not satisfy the given inequality (i.e., 0 + 0 > 6 is not true). Also, 2x + 3y > 6 (i.e., the points on the line 2x + 3y = 6 are not included).
∴The graph is half plane that neither contains the origin nor the points on the line 2x + 3y = 6.

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 8.
∫x2 ex3 dx equals:
(a) \(\frac { 1 }{ 3 }\) ex3 + C
(b) \(\frac { 1 }{ 3 }\) ex4 + C
(c) \(\frac { 1 }{ 2 }\) ex3 + C
(d) \(\frac { 1 }{ 2 }\) ex2 + C
Solution:
(a) \(\frac { 1 }{ 3 }\) ex3 + C

Explanation:
∫x2 ex3
Put x3 = t
3x2dx = dt
x2dx = \(\frac { 1 }{ 3 }\)dt
I = \(\frac { 1 }{ 3 }\)∫etdt
I = \(\frac { 1 }{ 3 }\)et + C
I = \(\frac { 1 }{ 3 }\)ex3 + C

Question 9.
The orthogonal projection of \(\vec{a}\) on \(\vec{b}\) is:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-5
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-6
Explanation: Projection of \(\vec{a}\) on \(\vec{b}\)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-7

Question 10.
The optimal value of the oblective function is attained at the points:
(a) Comer points of the feasible region
(b) Ahy point of the feasible region
(c) On x-axis
(d) On y-axis
Solution:
(a) Comer points of the feasible region

Explanation: By putting the comer points in objective function, we get the optimal value.

Question 11.
The area of triangle with vertices (x1, y1), (x2, Y2) and (x3, y3) is:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-8
Solution: (a)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-9
Explanation: Area of a triangle with vertices (x1, y1), (x2, Y2), (x3, y3) is =
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-9

Question 12.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-10
and Aij is cofactor of aij, then value of ∆ is given by:
(a) a11A31 + a12A32 + a13A33
(b) a11A11 + a12A21 + a13A31
(c) a21A11 + a22A12 + a23A13
(d) a11A11 + a21A21 + a31A31
Solution:
(d) a11A11 + a21A21 + a31A31

Explanation: ∆ = Sum of product of elements of any row (or column) with their correspocling cofactors.
∴∆ a11A11 + a21A21 + a31A31

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 13.
The matrix
A = CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-11 is a :
(a) Identity matrix
(b) Scalar matrix
(c) Skew-symmetric matrix
(d) Diagonal matrix
Solution:
(d) Diagonal matrix

Explanation: Matrix with all elements zero except diagonal is called diagonal matrix.

Question 14.
If P(A) = \(\frac { 4 }{ 5 }\) and P(A∩B) = \(\frac { 7 }{ 10 }\), then P(B|A) is equal to:
(a) \(\frac { 1 }{ 10 }\)
(b) \(\frac { 1 }{ 8 }\)
(c) \(\frac { 7 }{ 8 }\)
(d) \(\frac { 17 }{ 20 }\)
Solution:
(c) \(\frac { 7 }{ 8 }\)

Explanation:
P(A) = \(\frac { 4 }{ 5 }\), P(A∩B) = \(\frac { 7 }{ 10 }\)
P(B|A) = \(\frac { P(A∩B) }{ P(A) }\)
= \(\frac { 7/10 }{ 4/5 }\)
= \(\frac { 7 }{ 8 }\)

Question 15.
Number of arbitrary constants in general solution of differential equation of fourth order is:
(a) 0
(b) 2
(c) 4
(d) 3
Solution:
(c) 4

Explanation: Number of arbitrary constants in the general solution of differential equation is equal to the order of differential equation.

Question 16.
If y = sin-1(\(\frac{2 x}{1+x^2}\)), then \(\frac { dy }{ dx }\) is equal to:
(a) \(\frac{1}{1+x^2}\)
(b) \(\frac{2}{1+x^2}\)
(c) \(\frac{2}{1-x^2}\)
(d) \(\frac{-2}{1+x^2}\)
Solution:
(b) \(\frac{2}{1+x^2}\)

Explanation:
put x = tanθ
y = sin-1(\(\frac{2 \tan \theta}{1+\tan ^2 \theta}\))
= sin-1 (sin 2θ)
y = 2θ = 2 tan-1 x
\(\frac { dy }{ dx }\) = \(\frac{2}{1+x^2}\)

Question 17.
If \(\vec{a}\) . \(\vec{b}\) = \(\frac { 1 }{ 2 }\)|\(\vec{a}\)| |\(\vec{b}\)|, then the angle between \(\vec{a}\) and \(\vec{b}\) is:
(a) 00
(b) 300
(c) 600
(d) 900
Solution:
(c) 600

Explanation:
\(\vec{a}\) . \(\vec{b}\) = \(\frac { 1 }{ 2 }\)|\(\vec{a}\)| |\(\vec{b}\)|
⇒|\(\vec{a}\)| |\(\vec{b}\)| cos θ = \(\frac { 1 }{ 2 }\)|\(\vec{a}\)| |\(\vec{b}\)|
Where θ is the required angle
⇒ cos θ = \(\frac { 1 }{ 2 }\)
⇒ θ = 600

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 18.
Equation of x-axis is:
(a) \(\frac { x }{ 1 }\) = \(\frac { y }{ 1 }\) = \(\frac { z }{ 1 }\)
(b) \(\frac { x }{ 0 }\) = \(\frac { y }{ 1 }\) = \(\frac { z }{ 1 }\)
(c) \(\frac { x }{ 1 }\) = \(\frac { y }{ 0 }\) = \(\frac { z }{ 0 }\)
(d) \(\frac { x }{ 0 }\) = \(\frac { y }{ 0 }\) = \(\frac { z }{ 1 }\)
Solution:
(c) \(\frac { x }{ 1 }\) = \(\frac { y }{ 0 }\) = \(\frac { z }{ 0 }\)

Explanation: Since, On x-axis coordinates of y and z are zero.
∴ Equation of x-axis becomes
\(\frac { x-0 }{ 1-0 }\) = \(\frac { y-0 }{ 0-0 }\) = \(\frac { z-0 }{ 0-0 }\)
\(\frac { x }{ 1 }\) = \(\frac { y }{ 0 }\) = \(\frac { z }{ 0 }\)

Assertion-Reason Based Question
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.
Assertion(A): The matrix
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-12 is a diagonal matrix.
Reason (R): A = [aij] is a square matrix such that aij = 0, ∀ i ≠ j, then A is called diagonal matrix.
Solution:
(d) A is false but R is true.

Explanation:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-12 is not a diagonal matrix.
Since, A is not a square matrix.

Question 20.
Assertion (A): If |A| =5, then |A-1| = \(\frac { 1 }{ 5 }\)
Reason : AA-1 = I.
Solution:
(a) Both A and R are true and R is the correct explanation of A.

Explanation: : AA-1 = A-1 .A = I
∴Reason is true.
On applying this property in assertion.
If |A| – 5 then |A-1| = \(\frac { 1 }{ |A| }\) = \(\frac { 1 }{ 5 }\)
Thus, assertion is true and the reason is the correct explanation of the assertion.

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

Question 21.
Solve for x : cos (2sin-1x) =\(\frac { 1 }{ 9 }\), x > 0.
OR
The value of tan1 tan-1(tan \(\frac { 3π }{ 4 }\)) is ………..
Solution:
Given equation is cos(2 sin-1x) = \(\frac { 1 }{ 9 }\), x > 0
We put sin-1x = y
x = sin y
Equation (i) becomes
cos 2y = \(\frac { 1 }{ 9 }\)
⇒ 1 – 2sin2y = \(\frac { 1 }{ 9 }\)
2sin2y = 1 – \(\frac { 1 }{ 9 }\)
= \(\frac { 8 }{ 9 }\)
sin2 y = \(\frac { 4 }{ 9 }\) (∵x = siny)
x2 = \(\frac { 4 }{ 9 }\)
∴x = ±\(\frac { 2 }{ 3 }\)
But given that x > 0
∴ x = \(\frac { 2 }{ 3 }\)
OR
Principal value of tan-1 θ is (-\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
∴ The principal value of tan-1 (tan \(\frac { 3π }{ 4 }\))
= tan-1 [tan(π – \(\frac { π }{ 4 }\))]
= tan-1 (-tan\(\frac { π }{ 4 }\)) [∵ tan (π – θ) = – tan θ]
= tan-1 tan(-\(\frac { π }{ 4 }\)) [∵ – tan θ = tan (-θ)]
= – \(\frac { π }{ 4 }\) ∈ (-\(\frac { π }{ 2 }\),\(\frac { π }{ 2 }\))

Question 22.
If O is the origin. OP = 3 with direction ratios – 1, 2, – 2 then the coordinates of P are ……………

OR

Write the vector equation of the ime given by \(\frac { x-5}{ 3 }\) = \(\frac { y+4 }{ 7 }\) = \(\frac { z-6 }{ 2 }\)
Solution:
Direction ratios of \(\vec{a}\) are -1, 2, -2.
Therefore, direction cosines of \(\vec{OP}\) are
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-13
Therefore, coordinates of P are ( – 1, 2, – 2).

OR

Given equation of line in Cartesian form is
\(\frac { x-5 }{ 3 }\) = \(\frac { y+4 }{ 7 }\) = \(\frac { z-6 }{ 2 }\)
The point on line is (5, -4, 6) and direction ratios are (3, 7, 2).
We know that vector equation of a line passing through \(\vec{a}\) and parallel to \(\vec{b}\) is
\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)
Here, \(\vec{a}\) = (5, -4, 6) and \(\vec{b}\) = (3, 7, 2)
Its equation in vector form is
\(\vec{r}\) = (5\(\hat{i}\) – 4\(\hat{j}\) + 6\(\hat{k}\)) + (3\(\hat{i}\) + 7\(\hat{j}\) + 2\(\hat{k}\))

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 23.
A stone is dropped into a quiet lake and wave move in a circle at a speed of 35 cm/sec. At the instant when the radius of the circular wave is 7.5 cm, how fast is the enclosed area increasing?
Solution:
Let r be the radius and A be the area of the circular wave at any time t.
Then, area of circular wave = A = πr2 and \(\frac { dr }{ dt }\) = 3.5 cm/sec
Now,
On differentiating with respect to t, we get
\(\frac { dA }{ dt }\) = π (2r\(\frac { dr }{ dt }\))
\(\frac { dA }{ dt }\) = 2πr\(\frac { dr }{ dt }\)
\(\frac { dA }{ dt }\) = 2πr(3.5) = 7πr (given)
(\(\frac { dA }{ dt }\))r =7.5 = 7π(7.5)
= 52.5π cm2/sec

Question 24.
Find the general solution of the differential equation \(\frac { dy }{ dx }\)+ y = 1, (y ≠ 1).
Solution:
Given differential equation is,
\(\frac { dy }{ dx }\) + y = 1
⇒ \(\frac { dy }{ dx }\) = 1 – y
\(\frac { dy }{ 1-y }\) = dx
Integrating both sides
– log (1 – y) = x + C
⇒ log(1 – y) = – x – C
⇒ 1 – y = e-x-c = e-x . e-C
⇒ 1 – y = Ae-x, where A = e-C
⇒ y = 1 – Ae-x
which is the required general solution.

Question 25.
If \(\vec{a}\) = \(\hat{i}\) + 2\(\hat{j}\) – \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + \(\hat{j}\) – 5\(\hat{k}\) , find a unit vector in the direction of \(\vec{a}\) – \(\vec{b}\).
Solution:
Here,
\(\vec{a}\) – \(\vec{b}\) = (\(\hat{i}\) + 2\(\hat{j}\) – \(\hat{k}\)) – (\(\hat{i}\) + \(\hat{j}\) – 5\(\hat{k}\))
= -2\(\hat{i}\) + \(\hat{j}\) + 4\(\hat{k}\)
Let \(\vec{a}\) – \(\vec{b}\) = \(\vec{c}\)
Now, unit vector in the direction of \(\vec{c}\)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-14

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

Question 26.
Evaluate: ∫\(\frac{2}{(1-x)\left(1+x^2\right)}\)dx
Solution:
Let \(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac { A }{ 1-x }\) + \(\frac{\mathrm{B} x+\mathrm{C}}{1+x^2}\)
2 = A(1 + x2) + (Bx + C) (1 – x)
Put x = 1
2 = 2A
A = 1
Put x = 0
2 = A + C
2 = 1 + C
c = 1
Put x = 2
2 – 5A + (2B + C)(1 – 2)
2 = 5A – 2B – C
2 = 5 – 2B – 1
2 = 4 – 2B
B = \(\frac { 4-2 }{ 2 }\)
B = 1
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-15

Question 27.
Find : ∫\(\frac{(3 \sin \theta-2) \cos \theta}{5-\cos ^2 \theta-4 \sin \theta}\)dθ.
OR
Find ∫\(\frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)^2}\)dx.
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-16
1 = A(t + 2)2 + B(t + 1) (t + 2) + C(t + 1)
Equating coefficients, t, t and constant terms on both sides, we get
A + B = 0
4A + 3B + C = 0 …………(ii)
4A + 2B + C = 1 …………….(iii)
Subtracting equation (iii) from (ii), we get
B = – 1
Substitute in equation (i)
A = 1
Substitute the value of A and B in (ii), we get
4 – 3 + C = 0
c = – 1
From equation (A)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-17

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 28.
A problem in mathematics is given to three students whose chances of solving it correctly are \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\) and \(\frac { 1 }{ 4 }\) respectively. What is the probability that only one of them solves it correctly?

OR

Two students are trying to solve a problem arid the probability of solving the problem independently by the first and second student are \(\frac { 1 }{ 4 }\) and \(\frac { 1 }{ 5 }\) respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved, (ii) exactly one of them solves the problem.
Solution:
Let A, B, C be the given students and let E1, E2, E3 be the events that the problem is solved by A, B, C respectively. Then, \(\overline{\mathrm{E}}_1\), \(\overline{\mathrm{E}}_2\), \(\overline{\mathrm{E}}_3\) are the events that the given problem is not solved by A, B, C respectively.
Then,
P(E1) = \(\frac { 1 }{ 2 }\), P(E2) = \(\frac { 1 }{ 3 }\) and P(E3) = \(\frac { 1 }{ 4 }\)
P(\(\overline{\mathrm{E}}_1\)) = (1 – \(\frac { 1 }{ 2 }\)) = \(\frac { 1 }{ 2 }\) ; P(\(\overline{\mathrm{E}}_2\)) = (1 – \(\frac { 1 }{ 3 }\)) = \(\frac { 2 }{ 3 }\)
P(\(\overline{\mathrm{E}}_3\)) = (1 – \(\frac { 1 }{ 4 }\)) = \(\frac { 3 }{ 4 }\)
P(exactly one of them solves the problem) = P[(E1 ∩ \(\overline{\mathrm{E}}_2\) ∩ \(\overline{\mathrm{E}}_3\)) or (\(\overline{\mathrm{E}}_1\) ∩ E2 ∩ \(\overline{\mathrm{E}}_3\)) or (\(\overline{\mathrm{E}}_1\) ∩ \(\overline{\mathrm{E}}_2\) ∩ E3)] = P(E1 ∩ \(\overline{\mathrm{E}}_2\) ∩ \(\overline{\mathrm{E}}_3\)) + P(\(\overline{\mathrm{E}}_1\) ∩ E2 ∩ \(\overline{\mathrm{E}}_3\)) + P(\(\overline{\mathrm{E}}_1\) ∩ \(\overline{\mathrm{E}}_2\) ∩ E3)
Now since (E1, \(\overline{\mathrm{E}}_2\), \(\overline{\mathrm{E}}_3\)); (\(\overline{\mathrm{E}}_1\) E2, \(\overline{\mathrm{E}}_3\)); (\(\overline{\mathrm{E}}_1\), \(\overline{\mathrm{E}}_2\) ,E3) are independent events respectively.
∴ P(only one of A, B, C solve the problem correctly) =
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-18
Hence, the required probability is
OR
Let A and B be the events that problem is solved by first and second student respectively.
Probability of solving the problem by A
Probability of solving the problem by B
P(B)
Since the problem is being solved independently
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-19

Question 29.
Find: ∫\(\frac{(3 \sin x-2) \cos x}{13-\cos ^2 x-7 \sin x}\) dx.
Solution:
Let,
I = ∫\(\frac{(3 \sin x-2) \cos x}{13-\cos ^2 x-7 \sin x}\) dx
= ∫\(\frac{(3 \sin x-2) \cos x}{\sin ^2 x-7 \sin x+12}\) dx [∵cos2 x = 1 – sin2 x]
Put sin x = t ⇒ cos x dx = dt
I = ∫\(\frac{(3 t-2)}{t^2-7 t+12}\) dt = ∫\(\frac{3 t-2}{(t-3)(t-4)}\) dt
Let, \(\frac{3 t-2}{(t-3)(t-4)}\) = \(\frac { A }{ t-3 }\) + \(\frac { B }{ t-4 }\)
\(\frac{3 t-2}{(t-3)(t-4)}\) = \(\frac{\mathrm{A}(t-4)+\mathrm{B}(t-3)}{(t-3)(t-4)}\)
3t – 2 = A (t – 4) + B (t – 3)
Equating coefficient of t and constant term on both sides, we get
A + B = 3
– 4A – 3B = – 2 …………(ii)
Solving the above two equations, we get
A = – 7, B = 10
From equation A
\(\frac{3 t-2}{(t-3)(t-4)}\) = \(\frac { -7 }{ t-3 }\) + \(\frac { -10 }{ t-4 }\)
∫\(\frac{3 t-2}{(t-3)(t-4)}\) = -7∫\(\frac { 1 }{ t-3 }\) dt + 10∫\(\frac { 1 }{ t-4 }\) dt
I = – 7 log |t – 3| + 10 log |t – 4| + C
Where C is constant of integration.
∴ I = 10 log |sinx – 4| – 7log|sinx – 3| + C.

Question 30.
Find the particular solution of the following differential equation:
xy\(\frac { dy }{ dx }\) = (x + 2)(y + 2) : y – 1 when x = 1

OR

Find the general solution of the differential equation \(\frac { dy }{ dx }\) = \(\frac { 1-cosx }{ 1+cosx }\)
Solution:
Given differential equation is
xy \(\frac { dy }{ dx }\) = (x + 2)(y + 2)
⇒\(\frac { y }{ y+2 }\) dy = \(\frac { (x+2) }{ x }\) dx
On, integrating both sides, we have
∫\(\frac { y }{ y+2 }\) dy = ∫\(\frac { (x+2) }{ x }\) dx …………(i)
Put y + 2 = t ⇒ y = t – 2
Differential w.r.t.y
1 = \(\frac { dt }{ dy }\)
∴ dy = dt
Equation (i) becomes
∫\(\frac { t-2 }{ t }\) dt = ∫1dx + 2∫\(\frac { dx }{ x }\)
∫(1 – \(\frac { 2 }{ t }\) dt = x + 2 log x + C
t – 2 log t = x + 2 log x + C
y + 2 – 2 log (y + 2) = x + 2 log x + C ………………(ii)
When x = 1 and y = -1
– 1 + 2 – 2 log 1 = 1 + 2 log 1 + C
1 = 1 + C
∴C = 0
Equation (ii) becomes:
y + 2 – 2Iog (y + 2) = x + 2 log x
which is the particular solution of the given differential equation.
OR
Given differential equation is
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-20

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 31.
Solve the following LPP graphically:
Maximize and Minimize Z = 3x + 5y
Subject to constraints:
3x – 4y + 12 ≥ 0
2x – y + 2 ≥ 0
2x + 3y – 12 ≥ 0
0 ≤ x ≤ 4
y ≥ 2.
Solution:
The given LPP cart be re-written as:
Maximize or Minimize
Z = 3x + 5y
Subject to constraints:
3x – 4y ≥ -12
2x – y ≥ – 2
2x + 3y ≥ 12
x ≤ 4
x ≥ 0
Converting the inequations into equations, we obtain the following equations
3x – 4y = – 12, 2x – y = – 2, 2x + 3y – 12, x = 4, y = 2 and x = 0.

These lines are drawn on suitable scale. The shaded region P1 P2 P3 P4 P5 represents the feasible region of the given LPP.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-21
The values of the objective function at these points are given in the following table:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-22
Clearly Z assumes its minimum value 19 at x = 3 and y = 2.
and the maximum value of Z is 42 at x = 4 and y = 6.

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

Question 32.
Find the area of the region enclosed by the parabola x2 = y and the line y = x + 2.
Solution:
Wehave,x2 = y and y = x + 2
x2 = x + 2
x2 – x – 2 = 0
x2 – 2x + x – 2 = 0
x (x – 2) + 1(x – 2) = 0
(x + 1)(x – 2) = 0
x = – 1, 2
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-23
∴ Required area of shaded region
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-24

Question 33.
Show that the function f : R → R given by f (x) = 4x3 + 7 for all x → R is bijective.
OR
Show that the function f: W → W defined by
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-25
is bijective function.
Solution:
f : R →R is given by
f(x) = 4x3 + 7
For injedive:
Let f(x) = f(y)
4x3 + 7 = 4y3 + 7
x3 = y3
x3 – y3 = o
(x – y)(x2 + xy + 9) = 0
x – y = 0 (as the other term cannot be zero)
⇒ x = y
Thus, f (x) = f (y)
x = y ∀ x, y ∈ R
For surjective:
Let y ∈ R
Then, f(x) = y
4x3 + 7 = y
x3 = \(\frac { y-7 }{ 4 }\)
Thus, for any y ∈ R, x = [ \(\frac { y-7 }{ 4 }\) ]1/3 is a real number or x ∈ R.
Since range of f (x) and codomain of f (x) are both real numbers, f (x) is an onto function,
Therefore, f is bijective.
OR
One – one: Let
f(x1) = f(x2), ∀ x1, x2 ∈ W
Case I : When x1, x2 both are even, then
Let f(x1) = f (x2), ∀ x1, x2 ∈ W
⇒ x1 + 1 = x2 + 1
⇒ x1 = x2
⇒ f(x1) = f (x2)
⇒ x1 = x2,∀ x1, x2 ∈ W

Case II: When both x1, x2 are odd, then
Let f(x1) = f(x2), ∀ x1, x2 ∈ W
⇒ x1 – 1 = x2 – 1
⇒ x1 = x2
⇒ f(x1) = f (x2)
⇒ x1 = x2,∀ x1, x2 ∈ W
∴f(x) is a one – one function.
onto:
Case I: Let n = 2m
⇒ f(2m) = 2m + 1 = n + 1
Case II: Let n = 2m + 1
f (2m + 1) = (2m + 1) -1 = 2m
∴f (2m + 1) = 2m = n (even terms)
Also, f(1) = 0
∴f: W → W is noto function.
Since, f(x) is both one – one and onto function, so it is a bijective function.

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 34.
Write the vector equations 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

Find the equation of the line of shortest distance between the lines \(\frac { x-1 }{ 2 }\) = \(\frac { y+1 }{ -1 }\) = \(\frac { z-3 }{ 4 }\) and \(\frac { x-1 }{ 3 }\) = \(\frac { y+6 }{ 4 }\) = \(\frac { z+1 }{ 2 }\) Also, find the shortest distance.
Solution:
Given equation of lines are
\(\frac { x-1 }{ 2 }\) = \(\frac { y-2 }{ 3 }\) = \(\frac { z+4 }{ 6 }\)
and \(\frac { x-3 }{ 4 }\) = \(\frac { y-3 }{ 6 }\) = \(\frac { z+5 }{ 12 }\)
Now, the vector equation of lines are
\(\vec{r}\) = (\(\hat{i}\) + 2\(\hat{j}\) – 4\(\hat{k}\)) + λ(2\(\hat{i}\) + 3\(\hat{j}\) + 6\(\hat{k}\)) …………..(i)
[∵ Vector form of equation of line is = \(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)]
and \(\vec{r}\) = (3\(\hat{i}\) + 3\(\hat{j}\) – 5\(\hat{k}\)) + μ(4\(\hat{i}\) + 6\(\hat{j}\) + 12\(\hat{k}\))
Here, \(\overrightarrow{a_1}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 4\(\hat{k}\),
\(\overrightarrow{b_1}\) = 2\(\hat{i}\) + 3\(\hat{j}\) + 6\(\hat{k}\)
\(\overrightarrow{a_2}\) = 3\(\hat{i}\) + 3 \(\hat{j}\) – 5\(\hat{k}\),
\(\overrightarrow{b_2}\) = 4\(\hat{i}\) + 6\(\hat{j}\) + 12\(\hat{k}\)
Now,
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-26
= \(\hat{i}\)(36 – 36) – \(\hat{j}\)(24 – 24) + \(\hat{k}\)(12 – 12)
= o\(\hat{i}\) – o\(\hat{j}\) + o\(\hat{k}\) = 0
∵ \(\overrightarrow{b_1}\) x \(\overrightarrow{b_2}\) = 0
⇒ Vector b1 is parallel to b2 [∵\(\vec{a}\) x \(\vec{b}\) = 0, then \(\vec{a}\) ||\(\vec{b}\)]
Therefore, the two lines are parallel.
\(\vec{b}\) = (2\(\hat{i}\) + 3\(\hat{j}\) + 6\(\hat{k}\)) ……..(iii)
(Since, direction ratios of given lines are proportional]
Since, the two lines are parallel, we use the formula for shortest distance between two parallel lines.
We know that
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-27
x = 2λ + 1, y = – λ – 1, z = 4λ + 3
Therefpre, a general point on the line (i) is
P(2λ + 1, – λ – 1, 4λ + 3)
Let \(\frac { x-1 }{ 3 }\) = \(\frac { y+6 }{ 4 }\) = \(\frac { z+1 }{ 2 }\) = µ
x = 3µ + 1, y = 4µ – 6, z = 2µ – 1
Therefore, a general point on the line (ii) is
Q(3µ + 1, 4µ – 6, 2µ – 1)
Let PQ be the shortest distance between the two given lines. Then, the line of shortest distance passes
through P and Q.
Direction ratios of PQ are
(3µ – 2λ, 4µ + λ – 5, 2µ – 4λ – 4)
Since, PQ is perpendicular to the line (i),
2(3µ – 2λ) – 1(4µ + λ – 5) + 4(2µ – 4λ – 4) = 0
=> 10µ – 21λ – 11 = 0 ……………(iii)
Also, PQ is perpendicular to the line (ii).
3(3µ – 2λ) + 4(4µ + λ – 5) + 2(2µ – 4λ – 4) = 0
29µ – 10λ – 28 = 0 ………… (iv)
Solving equations (iii) and (iv), we get
\(\frac { λ }{ -280 + 319 }\) = \(\frac { µ }{ 110 – 588 }\) = \(\frac { 1 }{ -609 + 100 }\)
\(\frac { λ }{ 39 }\) = \(\frac { µ }{ -478 }\) = \(\frac { 1 }{ -509 }\)
λ = \(\frac { -39 }{ 509 }\)
λ = \(\frac { 478 }{ 509 }\)
So, the required points, where the line of shortest distance meets the given lines are
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-28

Question 35.
A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2,800 at interest. However, if trust had interchanged money in bonds, they would have got ₹ 100 less as interest. Using matrix method, find the amount invested by the trust.
Solution:
Let the amount invested by the trust in first and second bond be x and y respectively.
Interest from first bond = \(\frac { 10 x x x 1 }{ 100 }\) = \(\frac { 10x }{ 100 }\)
Interest from second bond = \(\frac { 12 x y x 1 }{ 100 }\) = \(\frac { 12y }{ 100 }\)
Interest received by trust = ₹ 2,800
According to the question,
\(\frac { 10x }{ 100 }\) + \(\frac { 12y }{ 100 }\) = 2,800
⇒ 10 x + 12 y = 1,80,000 ……………..(i)
and
\(\frac { 12x }{ 100 }\) + \(\frac { 10y }{ 100 }\) = 2,700
⇒ 12x + 10y = 2,70,000 ………(ii)
This system of equations can be written in matrix form as follows:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-29
x = 10,000 and y = 15,000
A = x + y = 10,000 + 15,000
= ₹ 25,000
Hence, the amount invested by the trust is 25,000

Section – E
This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-parts. First two case study questions have three sub-parts (i), (ii) (iii) of marks 1, 1, 2 respectively. The third case study question has two sub-parts of 2 marks each.)

Question 36.
Raj is playing with a spring by throwing it in the air which is moving along the function:
f (x) = 3x4 – 4x3 – 12x2 + 5.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-30
Find the given case study, give the answer of the following questions:
(i) Find the critical points of x which it touches the x-axis.
(ii) Find the interval in which spring is strictly increasing.
(iii) Find the interval in which spring is strictly decreasing.

OR

(iii) Find the values of x at which spring has local maxima.
Solution:
(i) f (x) = 3x4 – 4x3 – 12x2 + 5
f’(x) = 12x3 – 12x2 – 24x + 10
= 12x(x2 – x – 2)
= 12x(x2 – 2x + x – 2)
= 12x[x(x – 2) + 1(x – 2)]
= 12x(x+1)(x – 2)
For critical points put f’(x) = 0 x = 0, x = – 1, x = 2
∴ Critical points are (0, – 1, + 2).
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-31

(ii)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-32
f(x) is strictly increasing in ( – 1, 0) ∪ (2, ∞).

(iii) Based on the table given in (ii), f(x) is strictly decreasing in the interval ( – ∞, – 1) ∪ (0, 2).

OR

(iii) For local maxima find f’'(x) = 36x2 – 24x – 24
f’'(0) = – 24 < 0 local maxima at x = 0 and Maxima value f(o) = 5 f”(-1) = 36 + 24 – 24 = 36 > 0
Local Minima f’'(2) = 36 x 22 – 24 x 2 – 24
144 – 48 – 24 = 72 > 0 local minima
∴ Local maxima at x =0.

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 37.
Ram and Rahul are playing with a ball by throwing it in air and the one who throws the ball with maximum height wins the game. Ram’s ball height at any time f is given by h = 3 + 14t – 5t2 whereas Rahul’s ball height is given by H = 15T2 – 4T +3.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-33
With the given case study, give the answer of the following questions:
(i) Find the time when Rams ball reaches its maximum height.
(ii) What is the maximum height of Rams ball?
(iii) In how much does Rahul’s ball reach its maximum height?
OR
(iii) How much height is covered by Rahuls ball?
Solution:
(i) Rams height in function h = 3 + 14t – 5t2
Differentiating h, w.r.t. to t,
We get \(\frac { dh }{ dt }\) = 14 – 10 t
Put, \(\frac { dh }{ dt }\) = 0
14 – 10 t = 0
10 t = 14
t = 1.4

(ii) Maximum height covered by Ram’s ball
h = 3 + 14(1.4) – 5(1.4)2
h = 12.8m

(iii) Rahul will take time T, when function H is maximum
H = 15T2 – 4T + 3
\(\frac { dH }{ dT }\) = 30T – 4
Put \(\frac { dH }{ dT }\) = 0
30T —4 = 0
30T = 4
T = \(\frac { 4 }{ 30 }\) = \(\frac { 2 }{ 15 }\)

OR

(iii) Height H = 15 (\(\frac { 2 }{ 15 }\))2 – 4 x \(\frac { 2 }{ 15 }\) + 3 = \(\frac { 41 }{ 15 }\)

Question 38.
A student has 60% chances to take his online classes through Zoom App and 40% chances to attend his classes through Google meet. Further there is an error of network problem of 2%, when he is taking his classes through Zoom App and 1% chances of network problem when he is taking his classes from Google meet.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-34
Give answer of the following questions:
(i) What is the tota’ probability?
(ii) Find the probability of occurring network problem if he is taking his classes from Zoom App.
Solution:
(i) Total probability is P(E1) P(A/E1) + P(E2) P(A/E2)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions img-35