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RBSE 12th Maths Model Paper Set 5 with Answers in English

March 31, 2022 by Prasanna Leave a Comment

Students must start practicing the questions from RBSE 12th Maths Model Papers Set 5 with Answers in English Medium provided here.

RBSE Class 12 Maths Board Model Paper Set 5 with Answers in English

Time : 2 Hours 45 Min.
Maximum Marks : 80

General Instructions to the Examinees:

  • Candidate must write first his/her Roll. No. on the question paper compulsorily.
  • All the questions are compulsory.
  • Write the answer to each question in the given answer book only.
  • For questions having more than one part the answers to those parts are to be written together in continuity.
  • Write down the serial number of the question before attempting it.

Section – A

Question 1.
Multiple Choice Questions
(i) Consider a binary operation * on N defined as a * b = a3 + b3. Choose the correct Answer. (1)
(a) * is both associative and commutative.
(b) * is commutative but not associative.
(c) * is associative but not commutative.
(d) * is neither commutative nor associative.
Answer:
(b) * is commutative but not associative.

RBSE 12th Maths Model Paper Set 5 with Answers in English

(ii) The value of sin-1\(\left(\cos \frac{3 \pi}{5}\right)\) is:
(a) \(\frac{\pi}{10}\)
(b) \(\frac{3 \pi}{5}\)
(c) \(\frac{- \pi}{5}\)
(d) \(\frac{- 3 \pi}{5}\)
Answer:
(c) \(\frac{- \pi}{5}\)

(iii) If A = [2, – 3, 4], B = \(\left[\begin{array}{l}
3 \\
2 \\
2
\end{array}\right]\), X = [1, 2, 3] and Y = \(\left[\begin{array}{l}
2 \\
3 \\
4
\end{array}\right]\), then AB + XY equals:
(a) [28]
(b) [24]
(c) 28
(d) 24
Answer:
(a) [28]

(iv) If Aij is the cofactor of the element aij of the determinant \(\left|\begin{array}{ccc}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\), then the value of a32,A32 is
(a) 5
(b) 100
(c) – 110
(d) 110
Answer:
(d) 110

(v) If x = a cos θ and y = o sin θ, then \(\) is:
(a) cot θ
(b) tan θ
(c) cosec2 θ
(d) – cot θ
Answer:
(d) – cot θ

RBSE 12th Maths Model Paper Set 5 with Answers in English

(vi) \(\int \frac{d x}{\sin ^{2} x \cos ^{2} x}\) equals:
(a) tan x + cot x + C
(b) tan x – cot x + C
(c) tan x cot x + C
(d) tan x – cot 2x + C
Answer:
(b) tan x – cot x + C

(vii) A homogeneous differential equation of the form \(\) = h\(\) can be solved by making the substitution:
(a) y = vx
(b) v – yx
(c) x = vy
(d) x = v
Answer:
(c) x = vy

(viii) The position vector of a point, which divides the join of points with position vectors \(\vec{a}-2 \vec{b}\) and \(2 \vec{a}+\vec{b}\) externally in the ratio 2 : 1 is.
(a) \(3 \vec{a}-4 \vec{b}\)
(b) \(3 \vec{a}+4 \vec{b}\)
(c) \(4 \vec{a}+3 \vec{b}\)
(d) \(4 \vec{a}-3 \vec{b}\)
Answer:
(b) \(3 \vec{a}+4 \vec{b}\)

(ix) From a pack of 52 cards two cards are drawn randomly one by one without replacement. The probability that both of them are red is: (1)
(a) \(\frac{1}{2}\)
(b) \(\frac{25}{51}\)
(c) \(\frac{25}{102}\)
(d) \(\frac{2}{51}\)
Answer:
(c) \(\frac{25}{102}\)

RBSE 12th Maths Model Paper Set 5 with Answers in English

(x) If A = \(\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\), then for what value of α, A is an identity matrix?
(a) 0
(b) \(\frac{\pi}{2}\)
(c) π
(d) \(\frac{\pi}{4}\)
Answer:
(a) 0

(xi) The value of the constant k, so that the given function is continuous at the indicated point, is:
RBSE 12th Maths Model Paper Set 5 with Answers in English 1
(a) 5
(b) 0
(c) 10
(d) 25
Answer:
(c) 10

(xii) The projection of the vector î + 3 ĵ + 7 k̂ on the vector 7 î – ĵ + 8k̂ is :
(a) \(\frac{60}{\sqrt{114}}\)
(b) \(\frac{30}{\sqrt{115}}\)
(c) \(\frac{8}{\sqrt{111}}\)
(d) \(\frac{7}{\sqrt{114}}\)
Answer:
(a) \(\frac{60}{\sqrt{114}}\)

Question 2.
Fill in the blanks :
(i) If A = {1, 2, 3}, then number of equivalence relations containing (1, 2) is ____________ . (1)
Answer:
2

(ii) The principal value of cosec-1 (- √2) is ____________ . (1)
Answer:
\(\left(-\frac{\pi}{4}\right)\)

(iii) If A = \(\left[\begin{array}{ll}
6 & 9 \\
2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
2 & 6 & 0 \\
7 & 9 & 8
\end{array}\right]\), then AB = ____________ . (1)
Answer:
\(\left[\begin{array}{ccc}
75 & 117 & 72 \\
25 & 39 & 24
\end{array}\right]\)

(iv) The value of \(\frac{d}{d x}\) (sin x2) is ____________ . (1)
Answer:
2x cosx2

(v) The value of
RBSE 12th Maths Model Paper Set 5 with Answers in English 2
is ____________ . (1)
Answer:
2

(vi) If \(\vec{a}\) = 2î + ĵ + 3k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂, then |\(\vec{a}\) . \(\vec{b}\)| is ____________ . (1)
Answer:
√507

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 3.
Very Short Answer Type Questions
(i) Show that the function f: R → R given by f(x) = x3 is injective. (1)
Answer:
Here f : R → R
f(x) = x3
then x1, x2 ∈ R (domain)
f(x1) = f(x2)
⇒ x13 = x23
⇒ x1 = x2
Thus, f is injective.

(ii) Write the following function in the simplest form :
tan-1 \(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\), \(\frac{-\pi}{4}\) < x < \(\frac{3 \pi}{4}\)
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 3

(iii) If A = \(\left[\begin{array}{c}
-2 \\
4 \\
5
\end{array}\right]\), B = [1 3 – 6], verify that (AB)’ = B A’.
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 4

RBSE 12th Maths Model Paper Set 5 with Answers in English

(iv) Find the value of x if \(\left|\begin{array}{cc}
x+1 & x-1 \\
x-3 & x+2
\end{array}\right|\) = \(\left|\begin{array}{cc}
4 & -1 \\
1 & 3
\end{array}\right|\)
Answer:
Given
\(\left|\begin{array}{ll}
x+1 & x-1 \\
x-3 & x+2
\end{array}\right|=\left|\begin{array}{cc}
4 & -1 \\
1 & 3
\end{array}\right|\)
On expanding, we get
⇒ (x + 1) (x + 2) – (x – 1) (x – 3) = 4 × 3 – (- 1) × 1
⇒ (x2 + x + 2x + 2) – (x2 – x – 3x + 3) = 12 + 1
⇒ (x2 + 3x + 2) – (x2 – 4x + 3) = 13
⇒ x2 + 3x + 2 – x2 + 4x – 3 = 13
⇒ 7x – 1 = 13
⇒ 7x – 1 + 13
⇒ 7x = 14
Thus, x = 2

(v) Find the derivative of the function sin-1(e-x) with repec1 to x. (1)
Answer:
Let y = sin-1(e-x)
Differentiating both sides w.r.t.’x’,
we get
RBSE 12th Maths Model Paper Set 5 with Answers in English 5

(vi) Integrate the function ∫\(\frac{x^{3}}{\sqrt{1-x^{8}}}\) dx (1)
Answer:
Let I = ∫\(\frac{x^{3}}{\sqrt{1-x^{8}}}\) dx
Putting x4 = t
⇒ 4x3 dx = dt
or x3 dx = \(\frac{1}{4}\) dt
∴ I = \(\frac{1}{4} \int \frac{d t}{\sqrt{1-t^{2}}}\) = \(\frac{1}{4}\) sin-1t + C
= \(\frac{1}{4}\) sin-1 (x4) + C

RBSE 12th Maths Model Paper Set 5 with Answers in English

(vii) Show that the function y = ax + 2a2 is a solution of the differential equation
2\(\left(\frac{d y}{d x}\right)^{2}\) + x\(\left(\frac{d y}{d x}\right)\) – y = 0
Answer:
The given function is
y = ax + 2a2 ……. (1)
Differentiating equation (1) w.r.t. ‘x’, we get
\(\frac{d y}{d x}\) = a
Substituting the value of ‘a’ in equation, we get
RBSE 12th Maths Model Paper Set 5 with Answers in English 6

(viii) Show that: \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 7

(ix) A couple has two children. Find the probability that both children are males, if it is known that at least one of the children is male. (1)
Answer:
If a couple has two children then the sample space is
S = { (b, b),(b, g),(g, b),(g, g)}
Let E and F respectively denote the events that both children are males and at least one of the children is a male.
∴ E ∩ F = {(b, b)} ⇒ P(E ∩ F) = \(\frac{1}{4}\)
P (E) = \(\frac{1}{4}\); P (F) = \(\frac{3}{4}\)
⇒ P(E/F) = \(\frac{P(E \cap F)}{P(F)}\) = \(\frac{\frac{4}{3}}{4}\) = \(\frac{1}{3}\)

(x) Show that point A(a, b + c), B(b, c + a) and C(c, a + b) are collinear. (1)
Answer:
Three joints will be collinear if area of ∆ABC is zero.
Area of triangle
RBSE 12th Maths Model Paper Set 5 with Answers in English 8
(Since C1 and C3 are identical)
Thus, given points are collinear.
Hence Proved

RBSE 12th Maths Model Paper Set 5 with Answers in English

(xi) Verify that the given function is a solution of the corresponding differential
equation x + y = tan-1 y : y2y’ + y2 + 1 = 0 (1)
Answer:
The given function is:
x + y = tan-1y …….. (1)
Differentiating equation (1) w.r.t.
‘x’, we get
1 + y’ = \(\frac{1}{1+y^{2}}\) . y’
⇒ (1 + y2) + (1 + y2)y’ = y’
⇒ 1 + y2 + y’ + y2y’ = y’
⇒ y2y’ + y2 + 1 = 0
verified

(xii) Find the angle between \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\), where \(\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}\) and \(\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}\).
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 9

Section – B
Short Answer Type Questions

Question 4.
Show the function f: R → R defined by f(x) = \(\frac{x}{x^{2}+1}\), as injection, surjection or bijection. (2)
Answer:
(i) Injection test : Let x1, x2 ∈ R
f(x1) = f(x2) = \(\frac{x_{1}}{x_{1}^{2}+1}=\frac{x_{2}}{x_{2}^{2}+1}\)
⇒ x1(x22 + 1) = x2(x12 + 1)
⇒ x1 x22 + x1 = x2 x12 + x2
⇒ x2 x2 (x2 – x1) = (x2 – x1)
⇒ (x2 – x1) (x1x2 – 1) = 0
⇒ x2 = x1 or x1x2 = 1
⇒ x1 = x2 or x1 = \(\frac{1}{x_{2}}\)
Here, f is not an injection.

(ii) Surjection test: Let y ∈ R (co-domain) be any arbitrary element.
Consider, y = f(x)
∴ y = \(\frac{x}{x^{2}+1}\) ⇒ x2y + y = x
⇒ x2y – x + y = 0
⇒ x = \(\frac{1 \pm \sqrt{1-4 y^{2}}\frac{-1}{2}}{2 y}\), which does not exist for 1 – y2 < 0, i.e., for y > \(\frac{1}{2}\) and y < \(\).
In particular for y = 1 ∈ R
(co-domain), there does not exist any x ∈ R (domain) such that f(x) = y.
∴ f is not surjection.
Thus, f is neither injection nor surjection.

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 5.
Express the matrix \(\left[\begin{array}{ccc}
3 & -2 & -4 \\
3 & -2 & -5 \\
-1 & 1 & 2
\end{array}\right]\) as the sum of a symmetric and skew-symmetric matrix and verify your result.
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 10
RBSE 12th Maths Model Paper Set 5 with Answers in English 11
So, P is a symmetric matrix and Q is a skew symmetric matrix.
Now, P + Q = \(\) (A + A’) + \(\frac{1}{2}\) (A – A’)
RBSE 12th Maths Model Paper Set 5 with Answers in English 12
Thus, matrix A is expressed as the sum of symmetric matrix and skew-symmetric matrix.

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 6.
The sum of three numbers is 6. If we multiply the third number by 2 and add the first number to the result, we get 7. By adding second and third numbers to three times the first number, we get 12. Using matrices find the number.
Answer:
Let the three numbers be x, y and z respectively. Then,
x + y + z = 6
x + 2z = 7
3x + y + z =12
We obtain the following system of simultaneous linear equations:
x + y + z = 6
x + 0y + 2z = 7
3x + y + z = 12
The above system of equation can be written in matrix form as:
RBSE 12th Maths Model Paper Set 5 with Answers in English 13
= 1 (0 – 2) – (1 – 6) + 1 (1 – 0)
= – 2 + 5 + 1 = 4 ≠ 0
So, the above system of equations has a unique solutions given by X = A-1B
Cofactors of | A | are :
a11 = -2, a12 = 5, a13 = 1, a21 = 0, a22 = – 2, a23 = 2, a31 = 2, a32 = – 1 and a33 = – 1
Matrix formed by the cofactors of |A| is:
RBSE 12th Maths Model Paper Set 5 with Answers in English 14
x = 3, y = 1, z = 2
Hence, the number are 3, 1 and 2 respectively.

Question 7.
Find the relationship between a and b so that the function f defined by
RBSE 12th Maths Model Paper Set 5 with Answers in English 15
is continuous at x = 3. (2)
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 16
∴ 3a + 1 = 3b + 3
⇒ 3a = 3b + 3 – 1
⇒ 3a = 3b + 2
⇒ a = b + \(\frac{2}{3}\)
If b = k an arbitrary real number, then a = k + \(\frac{2}{3}\)
Thus, a = k + \(\frac{2}{3}\) and b = k

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 8.
Evaluate ∫\(\frac{2 x^{2}+1}{x^{2}\left(x^{2}+4\right)}\) dx
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 17
⇒ 2t + 1 = A(t + 4) + Bt
⇒ 2t + i = (A + B)t + 4A
On comparing the coefficients of t and constant terms, we get
A + B = 2
and 4A = 1
On solving these two equations, we get
A = \(\frac{1}{4}\) and B = \(\frac{7}{4}\)
RBSE 12th Maths Model Paper Set 5 with Answers in English 18

Question 9.
Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces. (2)
Answer:
The number of aces is a random variable.
Let it be denoted by X. Clearly, X can take the values 0,1, or 2.
Now, since the draws are done with replacement, therefore, the two draws form independent experiments.
Therefore, P(X = 0) = P(non-ace and non-ace)
= P(non-ace) × P(non-ace)
= \(\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}\)
P(X = 1) = P(ace and non-ace or non-ace ana ace)
= P(ace and non-ace) + P(non-ace and ace)
= P(ace). P(non-ace) + P(non-ace).P(ace)
= \(\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52}=\frac{24}{169}\)
and P(X = 2) = P(ace and ace)
\(\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}\)
Thus, the required probability distribition is
RBSE 12th Maths Model Paper Set 5 with Answers in English 19

Question 10.
Find x, if [x – 5 – 1] \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]\) = 0 (2)
Answer:
We have
RBSE 12th Maths Model Paper Set 5 with Answers in English 20
[(x – 2) × x – 10 × 4 + (2x – 8) × 1] = [0]
[x2 – 2x – 40 + 2x – 8] = [0]
⇒ x2 – 48 = 0
⇒ x2 = 16 × 3
⇒ x = ±4√3

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 11.
If y = sin-1\(\left\{x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^{2}}\right\}\) and 0 < x < 1, then find \(\frac{d y}{d x}\). (2)
Answer:
Given
RBSE 12th Maths Model Paper Set 5 with Answers in English 21

Question 12.
Using properties of determinants, prove that:
\(\left|\begin{array}{ccc}
(b+c)^{2} & a^{2} & b c \\
(c+a)^{2} & b^{2} & c a \\
(a+b)^{2} & c^{2} & a b
\end{array}\right|\) = (a – b) (b – c) (c – a) (a + b + c) (a2 + b2 + c2)
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 22
Expanding along C1we get
= (b – a) (c – a) (a2 + b2 + c2) [- b(b + a) + c(c + a)]
= (b – a) (c – a) (a2 + b2 + c2) [- b2 – ab + c2 + ac]
= (b – a) (c – a) (a2 + b2 + c2) [(c2 – b2) + (ca – ab)]
= (b – a) (c – a) (a2 + b2 + c2) [(c – b) (c + b) + a(c – b)]
= (b – a) (c – a) (a2 + b2 + c2) (c – b) (a + b + c)
= (a – b) (b – c) (c – a) (a + b + c) (a2 + b2 + c2)
= RHS.

Question 13.
If
RBSE 12th Maths Model Paper Set 5 with Answers in English 23
then , find the value of a. (2)
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 24

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 14.
For the following differential equation, find the general solution: (2)
sec2 x tan y dx + sec2y tan x dy = 0
Answer:
Given differential equation is
sec2x tan y dx + sec2y tan x dy = 0
Dividing equation (1) by tan x tany
\(\frac{\sec ^{2} x}{\tan x}\)dx + \(\frac{\sec ^{2} y}{\tan y}\)dy = 0 …. (2)
Integrating equation (2), we get
∫ \(\frac{\sec ^{2} x}{\tan x}\)dx + ∫ \(\frac{\sec ^{2} y}{\tan y}\)dy = 0
log |tan x| + log |tan y| = log C
or log |tan x tan y| = log C
or tan x tan y = C…(3)
(x, y ∈ R and x, y is not the odd multiple of π/2)
Equation (3) is the required solution of given differential equation.

Question 15.
If î + ĵ + k̂, 2î + 5ĵ, 3î + 2ĵ – 3k̂ and î – 6ĵ – k̂ are the position vectors of points A, B, C and D respectively, then find the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\). Deduce that \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) are collinear. (2)
Answer:
Note that if θ is the angle between AB and CD, then θ is also the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\).
Now, \(\overrightarrow{A B}\) = Position vector of B – Position vector of A
RBSE 12th Maths Model Paper Set 5 with Answers in English 25
Since 0 ≤ θ ≤ π, it follows that θ = π. This shows that \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) are collinear.

Question 16.
A die is thrown. If E is the event “the number appearing is a multiple of 3” and F be the event “the number appearing is even”, then find whether E and F are independent? (2)
Answer:
Sample space,
S= (1, 2, 3, 4, 5, 6)
Now E = (3, 6), F = {2,4,6}
and E ∩ F = {6}
then P(E) = \(\frac{2}{6}=\frac{1}{3}\)
P(F) = \(\frac{3}{6}=\frac{1}{2}\)
and P(E ∩ F) = \(\frac{1}{6}\)
Clearly P(E ∩ F) = P(E).P(F)
= \(\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}\)
Thus, events E and F are independent events.

Section – C
Long Answer Type Questions

Question 17.
If (tan-1 x)2 + (cot-1 x)2 = \(\frac{5 \pi^{2}}{8}\), then find x.
Prove that: sin-1: sin-1\(\left(2 x \sqrt{1-x^{2}}\right)\) = 2 cos-1x, \(\frac{1}{\sqrt{2}}\) ≤ x ≤ 1.
Answer:
(tan-1 x)2 + (cot-1 x)2 = \(\frac{5 \pi^{2}}{8}\)
⇒ (tan-1x + cot-1)2 – 2 tan-1x cot-1x = \(\frac{5 \pi^{2}}{8}\)
⇒ 16(tan-1 x)2 – 8π(tan-1 x) – 3π2 = 0
(on solving by tan-1 x + cot-1 x = \(\frac{\pi}{2}\))
⇒ 16(tan-1x – \(\frac{3 \pi}{4}\)) (tan-1x + \(\frac{\pi}{4}\)) = 0
⇒ tan-1x = – \(\frac{\pi}{4}\)
⇒ x = – 1 [∵ – \(\frac{\pi}{2}\) < tan-1 x < \(\frac{\pi}{2}\)]

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 18.
Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each case find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem: f(x) = x2 + x – 1 on [o, 4] (3)
Or
If y = cosec-1x, x > 1, then show that : x (x2 – 1)\(\frac{d^{2} y}{d x^{2}}\) + (2x2 – 1)\(\frac{d y}{d x}\) = 0. (3)
Answer:
Given,
f(x) = x2 + x – 1 on (0, 4) Every polynomial function is continuous everywhere on (- ∞, ∞) and differentiable for all arguments. Here, f (x) is a polynomial function. So, it is continuous in 10,4] and differentiable on (0,4). So, both the necessary conditions of Lagrange’s mean value theorem is satisfied.
Therefore, there exist a point CE (0, 4) such that:
f'(c) = \(\frac{f(4)-f(0)}{4-0}\) = \(\frac{f(4)-f(0)}{4}\)
Now, f(x) = x2 + x – 1
Differentiating both sides w.r.t.’x’,
we get
f(x) = 2c + 1
For f’(4), put the value of x = 4 in f(x):
f'(c) = 2c + 1
For f(4), put the value of x = 4 in f(x):
f(4) = 42 + 4 – 1 = 16 + 4 – 1 = 19
For) (0), put the value of x = 0 in f(x):
f(0) = 02 + 0 – 1 = 0 + 0 – 1 = – 1
Now, f’(c) = \(\frac{f(4)-f(0)}{4}\)
⇒ 2c + 1 = \(\frac{19-(-1)}{4}=\frac{20}{4}\)
⇒ 2c + 1 = 5 ⇒ 2c = 5 – 1 = 4
⇒ c = \(\frac{4}{2}\) = 2 ∈ (0,4)
Hence, Lagrange’s mean value theorem is verified.

Question 19.
Find : ∫\(\frac{1}{1+\cot x}\) dx
Or
Find : ∫\(\frac{x+3}{\sqrt{5-4 x-x^{2}}}\) dx
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 26
RBSE 12th Maths Model Paper Set 5 with Answers in English 27

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 20.
Show that each of the given three vectors is a unit vector:
\(\frac{1}{7}\) (2î + 3ĵ + 6k̂), latex]\frac{1}{7}[/latex](3î – 6ĵ + 2k̂), \(\frac{1}{7}\)(6î + 2ĵ – 3k̂)
Or
Find the area of the triangle with vertices A(1, 1, 2), B (2, 3, 5) and C (1, 5, 5). 3
Answer:
RBSE 12th Maths Model Paper Set 5 with Answers in English 28
Hence, given three vectors is a unit vector.

Section – D
Essay Type Questions

Question 21.
Evaluate:
RBSE 12th Maths Model Paper Set 5 with Answers in English 29
Or
Evaluate:
RBSE 12th Maths Model Paper Set 5 with Answers in English 30
Answer:
Let
RBSE 12th Maths Model Paper Set 5 with Answers in English 31

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 22.
Show that the given differential equation is homogeneous and solve it:
(x – y)dy – (x + y)dx = 0. (4)
Or
Find the particular solution of the differential equation
\(\frac{d y}{d x}\) + y cot x = 2x + x2 cot x (x ≠ 0); given that y = 0 when x = \(\frac{\pi}{2}\) (4)
Answer:
Given differential equation is
(x – y)dy – (x + y)dx = 0
RBSE 12th Maths Model Paper Set 5 with Answers in English 32
Thus, f(x, y) is a homogeneous function of degree zero.
So, the given differential equation is homogeneous.
Now putting y = vx … (2)
RBSE 12th Maths Model Paper Set 5 with Answers in English 33
RBSE 12th Maths Model Paper Set 5 with Answers in English 34
which is the required solution.

RBSE 12th Maths Model Paper Set 5 with Answers in English

Question 23.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade? (4)
Or
Find the probability of throwing at most 2 sixes in 6 throws of a single die. (4)
Answer:
Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.
In a well shufflied deck of 52 cards, there are 13 spade cards.
⇒ p = \(\frac{13}{52}=\frac{1}{4}\)
∴ q = 1 – \(\frac{1}{4}=\frac{3}{4}\)
X has a binomial distribution with n = 5 and p = \(\frac{1}{4}\)
P(X = x) = nCx qn-x px, where x = 0, 1, …….. n
RBSE 12th Maths Model Paper Set 5 with Answers in English 35

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