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CBSE Sample Papers for Class 12 Maths Set 4 with Solutions
Time : 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 MCQ’s and 02 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 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) \(\frac { π }{ 2 }\)
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\(\hat{k}\) = \(\sqrt{3^2+2^2+12^2}\) = \(\sqrt{157}\)
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
Solution:
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.
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:
Solution:
Explanation: Projection of \(\vec{a}\) on \(\vec{b}\)
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:
Solution: (a)
Explanation: Area of a triangle with vertices (x1, y1), (x2, Y2), (x3, y3) is =
Question 12.
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
Question 13.
The matrix
A = 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
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
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:
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
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\(\hat{k}\))
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 \(\vec{c}\)
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
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:
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)
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) =
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
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
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.
The values of the objective function at these points are given in the following table:
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
∴ Required area of shaded region
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
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.
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,
= \(\hat{i}\)(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
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
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:
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.
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).
(ii)
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.
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.
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.
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)