Class  XII
Math
TIME: 3 Hrs.
M.M: 80
General Instructions:
1. This question paper contains two parts A and B. Each part is compulsory. Part A carries 24 marks and Part B carries 56 marks.
2. PartA has Objective Type Questions and PartB has Descriptive Type Questions.
3. Both Part A and Part B have choices.
Part – A :
1. It consists of two sections I and II.
2. Section I comprises of 16 very short answer type questions.
3. Section II contains 2 case studies. Each case study comprises of 5 casebased MCQs. An examinee is to attempt any 4 out of 5 MCQs.
Part – B :
1. It consists of three sectionsIII, IV and V.
2. Section III comprises of 10 questions of 2 marks each.
3. Section IV comprises of 7 questions of 3 marks each.
4. Section V comprises of 3 questions of 5 marks each.
5. Internal choice is provided in 3 questions of Section –III, 2 questions of Section IV and 3 questions of SectionV. You have to attempt only one of the alternatives in all such questions.
Question numbers 1 to 16 are very short answer type questions.
Q.1. What is the principal value branch of cos^{–1} x ?
Ans. As we know that the principal value of cos^{–1}x is [0, π].
y = cos^{–1}x
Q.2. If A is a 3 × 3 matrix such that A = 8, then find 3A.
Ans. Here A = 8
Then 3A = 3^{3}A = 27 × 8 = 216
OR
Give an example of a skew symmetric matrix of order 3.
Ans. Skew symmetric matrix : A square matrix A = [a_{ij}] is said be a skew symmetrix matrix if A^{T} = –A i.e., if A = [a_{y}], then if A is skew symmetrix then A^{T} = –[a_{ij}] or A^{T} = –A.
Q.3. If [x 1] = 0, then find the value of x.
Ans.
⇒ [ x  2 0] = [ 0 0]
⇒ x  2 = 0 [ By def. of equality]
⇒ x = 2
OR
If A is a square matrix such that A^{2 }= A, then write the value of 7A – (I + A)^{3}, where I is an identity matrix.
Ans. 7A – (I + A)^{3} = – I
Detailed Solution:
Since, A^{2} = A,
∴ 7A – (I + A)^{3} = 7A – I^{3} – 3A^{2}I – 3AI^{2} – A^{3}
= 7A – I – 3A – 3A – A^{2}A
= 7A – I – 3A – 3A – A·A
= 7A – I – 3A – 3A – A
= 7A – I – 7A = – I
Q.4. If A and B are symmetric matrices of same order, then find AB – BA.
Ans. Given that, A and B are symmetric matrices.
⇒ A = A’ and B = B’,
Now (AB – BA)’ = (AB)’ – (BA)’ (i)
⇒ (AB – BA)’ = B’A’ – A’B’ [By reversal law]
⇒ (AB – BA)’ = BA – AB [From Eq. (i)]
⇒ (AB – BA)’ = –(AB – BA)
⇒ (AB – BA) is a skew matrix.
Q.5. If are unit vectors along three mutually perpendicular directions, then find .
Ans.
= 1 × 1 × 0 = 0
Q.6. On what set the function f(x) = cot x is discontinuous?
Ans. Given that,
It is discontinuous at
sin x = 0
⇒ x = nπ, n ∈ Z
Thus, the given function is discontinuous at {x = nπ : n ∈ Z}.
Q.7. The set of points where the function f given by f(x) = 2x – 1sin x is differentiable.
Ans. Given that,
f (x) = 2x1 sin x
The function sin x is differentiable.
The function 2x – 1 is differentiable, except
2x – 1 = 0
⇒ x = 1/2
Thus, the given function is differentiable
Q.8. Where is the graph of the inequality 2x + 3y > 6 lies?
Ans. Since, the equation 2x + 3y > 6, hence half plane that neither contains the origin nor the points of the line 2x + 3y = 6.
Q.9. The feasible solution for a LPP is shown in given figure. Let Z = 3x – 4y be the objective function. Find the Minimum of Z occurs at what point.
Ans.
He menicneim, um of Z occurs at (0, 8) and its minimum value is (–32).
Q.10. The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y.
Column A  Column B 
Maximum of Z  325 
Compare the quantity in Column A and Column B.
Ans.
Hence, maximum value of Z = 300 < 325
So, the quantity in column B is greater.
Q.11. What is the domain of the function f : R → R defined by f(x) =
Ans.
Here x^{2} – 3x + 2 ≥ 0
⇒ (x – 1)(x – 2) ≥ 0
⇒ x ≤ 1 or x ≥ 2
Hence the domain of f = (– ∞, 1] ∪ [2, ∞)
Q.12. If A = {1, 2, 3} and B = {a, b, c, d, e}, then what is the number of injective functions from A to B ?
Ans. Number of oneone functions = 5P_{3} = 60
Q.13. What is the equation of normal to the curve y = tan x at (0, 0) ?
Ans. y + x = 0
Given that the equation of the curve y = tan x at (0, 0) is x +y= 0 .
∵ y = tan x
∴ Equation of normal to the curve y = tan x at (0, 0) is
y  0 = 1(x  0)
⇒ y+ x = 0
Q.14. What is the integrating factor of x dy/dx  y = x^{4} 3x ?
Ans. Given that,
⇒ dy/dx  y/x = x^{3}  3
Here, p = 1/x, Q = x^{3}  3
∴
OR
What is the general solution of the differential equation log dy/dx = 2x + y ?
Ans.
⇒ e^{2x} + 2_{e}^{–y} + 2C = 0
Q.15. Find the area bounded by the curve y = sin x and the xaxis between x = 0 and x = 2p.
Ans.
= 2 + 2 sq. units
= 4 sq. units
Q.16. Evaluate
Ans. Let
= [–cos π + cos 0] – [–cos 2π + cos π]
= [1 + 1] – [–1 – 1] = 2 + 2 = 4
OR
Find
Ans.
Both the case study based questions are compulsory. Attempt any 4 sub parts from each question 17 and 18. Each question carries 1 mark.
Q.17. Let X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission is x number of colleges. It is given that:
where k is a positive constant.
Then answer the following questions, based on the above information:
(i) What is the sum of all the probabilities of a distribution ?
(a) –1
(b) 0
(c) 1
(d) 2
Ans. c
(ii) What is the value of k ?
(a) 2/9
(b) 3/8
(c) 1/8
(d) 4/9
Ans. c
(iii) Find the probability that you will get admission in exactly one college ?
(a) 2/9
(b) 1/8
(c) 3/8
(d) 4/9
Ans. b
(iv) Find the probability that you will get admission in at most 2 colleges ?
(a) 5/8
(b) 4/9
(c) 17/21
(d) 1/21
Ans. a
(v) Find the probability that you will get admission in at least 2 colleges ?
(a) 2/9
(b) 8/17
(c) 5/9
(d) 7/8
Ans. d
Q.18. Two friends start a game in which both of them have equal chances to throw a pair of dice. Now, they noticed the number of doublets in their game in three throws of a pair of dice. Based on the above information answer the following questions :
(i) The values of p and q are:
(a) 1/2 and 1/2
(b) 1/3 and 2/3
(c) 1/6 and 5/6
(d) 0 and 1
Ans. c
(ii) The value of P(X = 3) is :
(a) 125/216
(b) 75/216
(c) 15/216
(d) 1/216
Ans. d
(iii) The mean of the above information is :
(a) 13
(b) 14
(c) 12
(d) 15
Ans. c
(iv) The variance of the above data is :
(a) 112
(b) 512
(c) 619
(d) 719
Ans. b
(v) The Standard Deviation of the above data is :
(a) √15/6
(b) √15/2
(c) √15/3
(d) √15/4
Ans. a
Q.19. If A and B are two independent events, then prove that the probability of occurrence of atleast one of A and B is given by 1 – P(A').P(B').
Ans. Required Probability = P(A ∪ B)
= P(A) + P(B) – P(A).P(A)
= P(A)[1 – P(B)] + 1 – P(B')
= P(A)P(B') – P(B') + 1
= [1 – P(B')][1 – P(A)]
= 1 – P(A') P(B')
Q.20. Find the area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3.
Ans.
From figure, area of the shaded region,
Q.21. Show that the function f in A = R – {2/3} defined as f(x) = is onto.
Ans.
Let, y ∈ B
∴ y = f(x)
or
or y(6x – 4) = 4x + 3
or 6xy – 4y = 4x+3
or 6xy – 4x = 4y + 3
or x(6y – 4) = 4y + 3
or
or For every value of y except y = {2/3}, there is
a preimage x = = g(y).
or x ∈ A
∴ f is onto.
OR
How many equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) are there in all ? Justify your answer.
Ans. Equivalence relations could be the following:
{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} and
{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}
So, only two equivalence relations.
Q.22. If A = and I = find k so that A^{2} = 5A + kI.
Ans.
Q.23. Differentiate tan^{–1} with respect to x.
Ans.
∴ f'(x) = 1/2
OR
Find dy/dx at x = 1, y = π/4 if sin^{2}y + cos xy = K.
Ans.
From the given equation
Q.24. Evaluate
Ans.
= – 1 + 2 = 1
Detailed Solution:
Let
= – [0 – (–1)] + (2 – 0)
= –1 +2 = 1
Q.25. Find = 12
Ans.
OR
Find the unit vector perpendicular to each of the vectors
Ans.
Unit vector perpendicular to each at the vector
Q.26. Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6, find P(A' ∩ B').
Ans. P(A'∩B') = P(A∪B)' 1
= 1 – P(A∪B)
= 1 – 0.72 = 0.28
Alternate Method:
P(A'∩B') = P(A').P(B')
= [1 – P(A)] [1 – P(B)] 1
= [1 – 0.3] [1 – 0.6]
= (0.7).(0.4)
⇒ P(A'∩B') = 0.28
Q.27. Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive.
Ans.
because
= 1, for each i = 1, 2, 3
l_{i}l_{j} + m_{i}m_{j} + n_{i}n_{j} = 0(i ≠ j) for each i, j = 1, 2, 3
Q.28. If y = (log x)^{x} + x^{log x}, then find dy/dx.
Ans. Let us suppose, y = u + v
where u = [log (x)]^{x}
and v = x^{logx} ...(ii)
First take u = [log(x)]^{x}
Taking log of both the sides
logu = xloglogx
Now differentiating the above w.r. to x
Now take v = x^{logx}
log v = logx.logx
[Taking log on both sides]
= (log x)^{2}
Differentiating w.r.t. x
⇒ dv/dx = 2x^{logx–1}.logx ...(v)
equation (i), (iv) and (v)
OR
If y = log (1 + 2t^{2} + t^{4}), x = tan^{–1}t, find d^{2}y/dx^{2}.
Ans. y = log (1 + 2t^{2} + t4)
y = log (1 + t^{2})^{2}
y = 2log (1 + t^{2})
= 4 × (1 + t^{2}) = 4(1 + t^{2})
Question Numbers 29 to 35 carry 3 marks each.
Q.29. Solve the differential equation x dy/dx + y = x cos x + sin x, given y (π/2) = 1.
Ans.
x dy/dx + y = xcos x + sin x
or dy/dx + 1/x.y = cos x + 1/x sin x
∴ I.F. = = e^{log x} = x
∴ Solution is xy = ∫( xcos x + sin x)dx
or xy = xsin x + C
Solution is y = sin x.
Q.30. If A = Find A^{–1}.
Hence, solve the system of equations :
3x + 3y + 2z = 1
x + 2y = 4
2x – 3y – z = 5
Ans.
A = 3(– 2) – 1(3) + 2(– 4)
= – 6 – 3 – 8 = – 17 ≠ 0
∴ A^{–1} exists.
(A^{t})X = B
⇒ X = (At)–1B
⇒ X = (A^{–1})tB [∵ (A^{t})^{1} = (A^{1})^{t}]
x = 2, y = 1, z = – 4
Q.31. Solve the following L.P.P. graphically:
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0
and x, y ≥ 0
Ans.
Correct graph of 3 lines
Correct Shade of 3 lines
Z = 5x + 10y
Z_{A (60, 0)} = 300
Z_{B (120, 0)} = 600
Z_{C (60, 30}_{)} = 600
Z_{D (40, 20)} = 400
Minimum value of Z = 300 at x = 60, y = 0
Q.32. Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem, independently, then find the probability that (i) the problem is solved ? (ii) exactly one of them solves the problem ?
Ans.
Let E_{1}: Problem solved by A
E_{2}: Problem solved by B
∴ P(E_{1}) = 1/2 and P(E_{2}) = 1/3
or = 2/3
P(E_{1} ∩ E_{2}) = P(E_{1}).P(E_{2}) = 1/2, 1/3 = 1/6
(i) P(Problem is solved)
= 1  1/2 x 2/3 = 2/3
(ii) P(one of them is solved)
= 1/2
OR
Three persons A, B and C apply for a job of Manager in a Private company. Chance of their selection (A, B and C) are in the ratio 1 : 2 : 4. The probability that A, B and C can introduce changes to improve profits of company are 0·8, 0·5 and 0·3 respectively, if the changes does not take place, find the probability that it is due to the appointment of C.
Ans.
Let the events be described as below :
A : No change takes place
E_{1} : Person A gets appointed
E_{2} : Person B gets appointed
E_{3} : Person C gets appointed
The chances of selection of A, B and C are in the ratio 1 : 2 : 4.
Hence,
Probabilities of A, B and C introducing changes to improve profits of company are 0·8, 0·5 and 0·3 respectively.
Hence probability of no changes on appointment of A, B and C are 0·2, 0·5 and 0·7 respectively.
Hence,
Therefore, required probability i.e., (E_{3}/A) is P(E_{3}/A)
∴ If no change takes place, the probability that it is due to appointment of C is 7/10.
Q.33. If y = x^{sinx} + sin(x^{x}), find dy/dx.
Ans. Let u = x^{sinx}
⇒ ln u = sin x ln x
⇒
Since, y = u + v
Q.34. If x = a sin pt, y = b cos pt, then find dy/dx at t = 0.
Ans. Given x = a sin pt
∴ dx/dt = ap cos pt
and y = b cos pt
∴ dy/dt = – bp sin pt
= 0
OR
If y = Pe^{ax} + Qe^{bx}, show that + aby = 0.
Ans. y = Pe^{ax} + Qe^{bx}
= Pe^{ax} {a^{2} – a^{2} – ab + ab} + Qe^{bx}{b^{2} – ab – b^{2} + ab}
= 0 + 0 = 0 = RHS
Q.35. Find:
Ans. Let 2x = t
Question numbers 36 to 38 carry 5 marks each.
Q.36. Find the point on the curve y2 = 4x which is nearest to the point (2, 1).
Ans. Suppose the required point on the curve is K(p, q) and the given point is A(2, 1).
∴ q^{2} = 4p
To find nearest point let us suppose S^{2 }= T
For critical points T' = 0
⇒ q^{3 }= 8
⇒ q = 2
To find the maxima or minima
Therefore, T is least
From (i) q^{2} = 4p
⇒ 2^{2} = 4p
⇒ p = 1
Therefore, the required point is K(1, 2).
OR
If xcos (a + y) = cos y, then prove that dy/dx = Hence, show that
Ans. Given, xcos (a + y) = cos y
On differentiating both sides w.r.t. y, we get
Again, on differentiating both sides of Eq. (i) w.r.t. x, we get
Hence Proved.
Q.37. Find the position vector of foot of perpendicular and the perpendicular distance from the point P with position vector to the plane Also find the image of P in the plane.
Ans. Equation of plane : 2x + y + 3z = 26
Normal to the plane =
∴ < 2, 1, 3 > are direction ratios of the normal to the plane.
Equation of the line through P(2, 3, 4) and perpendicular to the given plane is
∴ Coordinates of point N are
N(2λ + 2, λ + 3, 3λ + 4)
Since N lie s on the plane,
∴ 2(2λ + 2) + (λ + 3) + 3(3λ + 4) = 26
4λ + 4 + λ + 3 + 9λ + 12 = 26
or 14λ = 26 – 19
or 14λ = 7
or λ = 1/2
∴ Coordinates of the foot of the perpendicular i.e., N are
∴ The length of perpendicular from P to given plane is
Now, N is the midpoint of PP’, where P’(α, β, γ) is the image of point P.
or 6 = 2 + α; 7 = 3 + β; 11 = 4 + γ
or α = 4; β = 4; γ = 7
∴ Image of point P is P’(4, 4, 7).
OR
Find the equation of the plane through the line and parallel to the line Hence, find the shortest distance between the lines.
Ans. The two given lines are
Let a, b, c the D.R's of the normal to the plane containing the line (1).
Therefore, equation of plane is
a(x – 1) + b(y – 4) + c(z – 4) = 0 ...(3)
3a + 2b – 2c = 0 ...(4)
(Q Required plane contains line (1))
2a  4b + 1c = 0 ...(5)
(Q line (2) is parallel to the required plane)
Putting, a = 6λ, b = 7λ, c = 16λ in (3), we get
⇒ 6λ(x – 1) + 7λ(y – 4) + 16λ(z – 4) = 0
⇒ 6x + 7y + 16z – 98 = 0, which is the required equation of the plane
Since line (2) is parallel to required plane
∴ SD between two lines = Perpendicular distance of the point (–1, 1 –2) from the plane.
Q. 38. If A = Find A–1. Use it to solve the system of equations
Hence, solve the system of equations :
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
Ans.
A = – 1 ≠ 0 ∴ A^{–1} exists
Cofactors of A are :
A_{11} = 0; A_{12} = 2; A_{13} = 1 1 m for
A_{21} = – 1; A_{22} = – 9; A_{23} = – 5 4 correct
A_{31} = 2; A_{32} = 23; A_{33} = 13 Cofactors
∴ x = 1, y = 2, z = 3
OR
Show that the matrix B^{T} AB is symmetric or skewsymmetric accordingly when A is symmetric or skewsymmetric.
Ans. Case I : Let A be a symmetric matrix. Then A^{T} = A.
Now (B^{T} AB)^{T} = B^{T}A^{T}(B^{T})^{T} [By reversal law]
= B^{T}A^{T}B [∵ (BT)T = B]
or (B^{T}AB)^{T} = B^{T}AB [∵ AT = A]
∴ B^{T} AB is a symmetric matrix.
Case II : Let A be a skewsymmetric matrix.
Then, A^{T} = – A.
Now (B^{T} AB)^{T} = B^{T}A^{T}(B^{T})^{T} [By reversal law]
or (B^{T}AB)^{T} = B^{T}A^{T}B [∵ (B^{T})^{T} = B]
or (B^{T}AB)^{T} = B^{T}( A)B [∵ AT =  A]
or (B^{T}AB)^{T} =  B^{T}AB
∴ B^{T} AB is a skewsymmetric matrix.
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