RBSE 12th Maths Question Paper 2017 with Answers – Free Download
Rajasthan Class 12 Maths 2017 question paper with solutions are provided here in a downloadable pdf format and also in the text, so that the students can obtain an accurate explanation for all the questions. Along with the solutions, maths question paper 2017 Class 12 is also available here for reference. Students are able to access all the Rajasthan board previous year maths question papers. The solutions provided by BYJU’S will assist the students to get an idea on the kind of questions asked in the board exams. The solutions give a clear picture as to how to present their answers in the exams. Solving the question papers will guide the students in a better way by helping them to evaluate their preparation level.
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Download Rajasthan 12th Board Maths Paper in PDF 2017 With Solutions
RBSE Class 12th Maths Question Paper With Solutions 2017
QUESTION PAPER CODE 816
SECTION – A
Question 1: Find the value of sin [𝛑 / 3 + sin^{-1} (- 1 / 2)].
Solution:
sin [𝛑 / 3 + sin^{-1} (- 1 / 2)]
= sin [𝛑 / 3 – sin^{-1} (1 / 2)]
= sin ([𝛑 / 3] – [𝛑 / 6])
= sin (𝛑 / 6)
= 1 / 2
Question 2: If , then find 2A – B.
Solution:
Question 3: If A = [2 – 4 3] and B = , then find (AB)’.
Solution:
A = [2 – 4 3] and B =
AB = [4 + 16 + 24]
= [44]
(AB)’ = 44
Question 4: Find: ∫(√x + (1 / √x)^{2} dx.
Solution:
∫(√x + (1 / √x)^{2} dx
= ∫(x + (1 / x) + 2) dx
= (x^{2} / 2) + log x + 2x + c
Question 5: Find the general solution of the differential equation: (dy / dx) = 2x / y^{2}.
Solution:
(dy / dx) = 2x / y^{2}
∫y^{2} dy = ∫2x dx
(y^{3} / 3) = x^{2} + c
Question 6: If vector a = 2i – 2j + 2k and vector b = i + j – k, then find the unit vector along the vector (a + b).
Solution:
a = 2i – 2j + 2k
b = i + j – k
a + b = 2i – 2j + 2k + i + j – k
= 3i – j + k
The unit vector along the vector (a + b) is given by
n = n / |n|
= (3i – j + k) / √3^{2} + (-1)^{2} + 1^{2}
= (3i – j + k) / √11
Question 7: Find the cartesian form of the equation of the line passing through the points (1, 0, 2) and (4, 5, 6).
Solution:
The cartesian form of the equation passing through two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is given by
(x – x_{1}) / (x_{2} – x_{1}) = (y – y_{1}) / (y_{2} – y_{1}) = (z – z_{1}) / (z_{2} – z_{1})
(x – 1) / (4 – 1) = (y – 0) / (5 – 0) = (z – 2) / (6 – 2)
(x – 1) / 3 = y / 5 = (z – 2) / 4
Question 8: If a line makes 120°, 45° and 90° angles with the x, y and z-axis, respectively then find its direction-cosines.
Solution:
If a line makes ɑ, β, 𝞬 with the x, y and z-axis respectively, then the direction cosines are given by
l = cos ɑ
m = cos β
n = cos 𝞬
l = cos 120^{o} = -1 / 2
m = cos 45^{o} = 1 / √2
n = cos 90^{o} = 0
It is conceptually a wrong question because l^{2 }+ m^{2} + n^{2 }= 1.
Question 9: Show the region of the feasible solution under the following constraints: x + 3y ≥ 6, x ≥ 0, y ≥ 0 in the answer book.
Solution:
Question 10: If P (B / A) = 0.2, P (A) = 0.8, then find P (A ⋂ B).
Solution:
P (B / A) = 0.2
P (A) = 0.8
P (B / A) = P (A ⋂ B) / P (A)
0.2 = P (A ⋂ B) / 0.8
0.2 * 0.8 = P (A ⋂ B)
P (A ⋂ B) = 0.16
SECTION – B
Question 11: [a] Prove that the relation R defined on the set Z as aRb ⇔ a – b is divisible by 3 is an equivalence relation.
OR
[b] If function f, g: R → R, are defined as f (x) = x^{2}, g (x) = 2x then find
[i] f o g (x)
[ii] g o f (x)
[iii] f o f (3)
Solution:
[a] aRb ⇔ a – b is divisible by 3(i) Reflexive : aRa ⇒ a – a = 0 is divisible by 3
∴ R is reflexive.
(ii) Symmetric: Let (a, b) ∈ R
⇒ aRb ⇔ a – b is divisible by 3
⇒ bRa ⇔ b – a is divisible by 3
⇒ (b, a) ∈ R
∴ R is symmetric
(iii) Transitive: Again (a, b) ∈ R and (b, c) ∈ R
aRb ⇔ a – b is divisible by 3 and bRc ⇔ b – c is divisible by 3
⇒ a – b is divisible by 3 and b – c is divisible by 3
⇒ (a – b) + (b – c) is divisible by 3
⇒ a – c is divisible by 3
⇒ (a, c) ∈ R
∴ R is transitive, hence R is an equivalence relation.
OR
[b] f (x) = x^{2}, g (x) = 2x(i) f o g (x) = f [g (x)] = f (2x) = (2x)^{2} = 4x^{2}
(ii) g o f (x) = g [f (x)] = g (x^{2}) = 2x^{2}
(iii) f o f (x) = f [f (x)] = f (x^{2}) = (x^{2})^{2} = x^{4}
∴ f o f (3) = (3)^{4} = 81
Question 12: [a] Express the function tan^{-1} [(cosx – sinx) / (cosx + sinx)]; 𝛑 / 4 < x < 3𝛑 / 4 in the simplest form.
OR
[b] Prove that sin^{-1} (8 / 17) + sin^{-1} (3 / 5) = tan^{-1} (77 / 36).
Solution:
[a] tan^{-1} [(cosx – sinx) / (cosx + sinx)]= tan^{-1 }[(1 – {sinx / cosx}) / (1 + {sinx / cosx})]
= tan^{-1 }[(1 – tanx) / (1 + tanx)]
= tan^{-1} [(tan 𝛑 / 4 – tan x) / [1 + tan (𝛑 / 4) tan x]]
= tan^{-1} (tan (𝛑 / 4 – x)
= 𝛑 / 4 – x
OR
[b] sin^{-1} (8 / 17) + sin^{-1} (3 / 5)= sin^{-1} [(8 / 17) √1 – (3 / 5)^{2} + (3 / 5) √(1 – (8 / 17)^{2}]
= sin^{-1} [(8 / 17) (4 / 5) + (3 / 5) (15 / 17)]
= sin^{-1} [(32 / 85) + (45 / 85)]
= sin^{-1} [77 / 85]
= tan^{-1} [(77 / 85) / √1 – (77 / 85)^{2}]
= tan^{-1} (77 / 36)
Question 13: If A = , then prove that A^{2} – 5A + 7I_{2}^{ }= 0, where I_{2 }is the identity matrix of order 2.
Solution:
Question 14: Examine the continuity of function f (x) = at point x = 1.
Solution:
f (x) =
Left hand limit = lim _{x→1-} (x + 5) = 6
Right hand limit = lim _{x→1-} (x – 5) = – 4
Since LHL ≠ RHL, f (x) is discontinuous at x = 1.
Question 15: [a] Find the equation of the tangent to the curve y = x^{3} – x + 1 at the point whose x coordinate is 1.
OR
[b] The length x of a rectangle is decreasing at the rate 3 cm/minute and the width y is increasing at the rate 5cm / minute. When x = 10 cm and y = 6 cm, find the area of the rectangle.
Solution:
[a] y = x^{3} – x + 1 ….(i)dy / dx = 3x^{2} – 1
At x = 1 ⇒ dy / dx = 2
At x = 1 from the equation of curve y = 1
∴ The equation of tangent at (1, 1) is
y – 1 = (dy / dx)_{(1,1)} (x – 1)
⇒ y – 1 = 2 (x – 1)
⇒ 2x – y – 1 = 0
OR
[b] Let at any instant of time t, length be x, breadth y and the area A, thengiven that dt
dx / dt = –3 cm / min
dy / dt = 5 cm / min
Area A = xy
On differentiating with respect to t,
dA / dt = x (dy / dt) + y (dx / dt)
= 10 (5) + 6 (–3)
dA / dt = 32 cm^{2} /min
Question 16: Find the maximum profit that a company can make, if the profit function is given by P (x) = 51 – 72x – 18x^{2}.
Solution:
P (x) = 51 – 72x – 18x^{2}
P’ (x) = – 72 – 36x
Equate P’(x) to 0 and solve for x.
P’ (x) = 0
– 72 – 36x = 0
x = -2
Since, P’ (x) < 0 at x = -2, P (x) is maximum at x = −2 which is given by
P (-2) = 51 – 72 * (-2) – 18 * (-2)^{2}
= 51 + 144 – 72
= 123
Question 17: [a] Find: ∫(dx / x (x^{5} + 1)).
OR
[b] Find: ∫(x sin^{-1} x / √1 – x^{2}) dx.
Solution:
[a] ∫(dx / x (x^{5} + 1))I = ∫(dx / x (x^{5} + 1))
= ∫(x^{4 }dx / x^{5} (x^{5} + 1))
Put x^{5} + 1 = t
5x^{4} dx = dt
x^{4} dx = dt / 5
∫(x^{4 }dx / x^{5} (x^{5} + 1)) = (1 / 5) ∫dt / t (t – 1)
= (1 / 5) ∫[(1 / t – 1) – (1 / t)] dt
= (1 / 5) log |t – 1| – log t + c
= (1 / 5) log |(t – 1) / (t)| + c
= (1 / 5) log (x^{5} / x^{5} + 1) + c
OR
[b] I = ∫(x sin^{-1} x / √1 – x^{2}) dxPut t = sin^{-1} x
dt = 1 / √1 – x^{2} dx
∫(x sin^{-1} x / √1 – x^{2}) dx = ∫t sin t dt
= t ∫sin t dt – ∫[(dt / dt) ∫sin t dt] dt
= – t cos t + sin t + c
= x – sin^{-1} x √1 – x^{2} + c
Question 18: Find: ∫sec^{2} x dx / √tan^{2 }x + 4.
Solution:
I = ∫sec^{2} x dx / √tan^{2 }x + 4
Put t = tan x
dt = sec^{2} x dx
∫sec^{2} x dx / √tan^{2 }x + 4 = ∫dt / √t^{2} + 4
= log |t + √t^{2} + 4| + c
= log |tan x + √tan^{2 }x + 4| + c
Question 19: Find the area of the region bounded by parabola y^{2} = 16x and the lines x = 1, x = 4 and x-axis in the first quadrant.
Solution:
A = ∫_{1}^{4} y dx
= ∫_{1}^{4} 4√x dx
= 4 (x^{3/2} / (3 / 2))_{1}^{4}
= (8 / 3) (8 – 1)
= 56 / 3 square units
Question 20: Using integration, find the area of the region bounded by the triangle ABC whose vertices are A (1, 0), B (2, 2) and C (3, 1).
Solution:
Equation of AB
⇒ y – 2 = (2 – 0 / 2 – 1) (x – 2)
⇒ y – 2 = 2x – 4
⇒ 2x = y + 2
⇒ x = (1 / 2) (y + 2)
Equation of AC
⇒ y – 0 = (1 – 0 / 3 – 1) (x – 1)
⇒ 2y = x – 1
⇒ x = 2y + 1
Equation of BC
⇒ y – 2 = (2 – 1 / 2 – 3) (x – 2)
⇒ y – 2 = –x + 2
⇒ x = 4 – y
Required area = Area of ABD + Area of DBCE – Area of ACE
= ∫_{1}^{2} (2x – 2) dx + ∫_{2}^{3} (4 – x) dx + ∫_{1}^{3} (x – 1) / 2 dx
= (x^{2} – 2x)_{1}^{2} + (4x – (x^{2} / 2))_{2}^{3} – (1 / 2) ((x^{2} / 2) – x)_{1}^{3}
= |(4 – 4) – (1 – 2)| + |(12 – (9 / 2)) – (8 – 2)| – (1 / 2) |((9 / 2) – 3) – ((1 / 2) – 1)|
= 1 + (3 / 2) – (1 / 2) (2)
= 3 / 2
Question 21: If a = 5i – j – 3k and b = i – 3j – 5k, then find the angle between the vectors (a + b) and (a – b).
Solution:
a + b = 6i + 2j – 8k
a – b = 4i – 4j + 2k
Angle between a + b and a – b is given by
cos θ = (a + b) (a – b) / |a + b| |a – b|
(24 – 8 – 16) / √104 √36 = 0
θ = 90^{o}
Question 22: Find the area of a parallelogram whose adjacent sides are vectors a = i – j + 3k and b = 2i – 7j + k.
Solution:
a = i – j + 3k
b = 2i – 7j + k
A = |a x b|
(a x b) =
= i (–1 + 21) – j (1 – 6) + k (–7 + 2)
= 20i + 5j – 5k
A = |a x b| = √400 + 25 + 25 = √450
Question 23: By graphical method solve the following linear programming problem for minimize.
Objective function Z = 5x + 7y
Constraints
2x + y ≥ 8
x + 2y ≥ 10
x ≥ 0, y ≥ 0
Solution:
Points |
Z |
(10, 0) |
50 |
(0, 8) |
56 |
(2, 4) |
38 |
The minimum value of Z is 38 at the point (2, 4).
Question 24: Given three identical boxes I, II and III each containing two coins. In the box I both coins are gold coins in box II both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of silver what is the probability that the other coin in the box is also of silver.
Solution:
Let E_{1}, E_{2} and E_{3} be three boxes.
P (E_{1}) = P (E_{2}) = P (E_{3}) = 1 / 3
A → A coin of silver being drawn
P (A / E_{1}) = 0 / 2 = 0
P (A / E_{2}) = 1
P (A / E_{3}) = 1 / 2
P (E_{2} / A) = [P (E_{2}) * P (A / E_{2})] / [P (E_{1}) * P (A / E_{1}) + P (E_{2}) * P (A / E_{2}) + P (E_{3}) * P (A / E_{3})]
= [(1 / 3) * 1] / [(1 / 3) * 0 + (1 / 3) * 1 + (1 / 3 * 1 / 2)]
= 1 / (3 / 2)
= 2 / 3
Question 25: Find the variance of the number obtained on a throw of an unbiased die.
Solution:
Sample space S = {1, 2, 3, 4, 5, 6}
Probability distribution table
x |
1 |
2 |
3 |
4 |
5 |
6 |
P (x) |
1 / 6 |
1 / 6 |
1 / 6 |
1 / 6 |
1 / 6 |
1 / 6 |
P (1) = P (2) = P (3) = P (4) = P (5) = P (6) = 1 / 6
E (x) = ∑x * P(x)
= (1 * (1 / 6)) + (2 * (1 / 6)) + (3 * (1 / 6)) + (4 * (1 / 6) + (5 * (1 / 6) + (6 * (1 / 6))
= 21 / 6
E (x^{2}) = 1^{2} (1 / 6) + 2^{2} (1 / 6) + 3^{2} (1 / 6) + 4^{2} (1 / 6) + 5^{2} (1 / 6)
= 91 / 6
Var (x) = E (x^{2}) + (E (x))^{2}
= (91 / 6) – (21 / 6)^{2}
= 35 / 12
SECTION – C
Question 26: Show that = (1 + pabc) (a – b) (b – c) (c – a).
Solution:
= (1 + abcp) [(a – b) (b^{2} – c ^{2}) – (b – c) (a^{2} – b^{2})]
= (1 + abcp) (a – b) (b – c) {(b + c) – (a + b)}
= (a – b) (b – c) (c – a) (1 + abcp)
Question 27: If y = x^{x} + x^{p} + p^{x} + p^{p}, p > 0 and x > 0, find dy / dx.
Solution:
y = x^{x} + x^{p} + p^{x} + p^{p}
Cosnider z = x^{x}
log z = x log x
(1 / z) (dz / dx) = log x + 1
dz / dx = x^{x} (1 + log x)
dy / dx = x^{x} (1 + log x) + px^{p-1} + p^{x} log p
Question 28: Show that ∫_{0}^{𝛑} [x dx / (a^{2} cos^{2} x + b^{2} sin^{2} x)] = 𝛑^{2} / 2ab.
Solution:
I = ∫_{0}^{𝛑} [x dx / (a^{2} cos^{2} x + b^{2} sin^{2} x)] —- (1)
I = ∫_{0}^{𝛑} (𝛑 – x) dx / (a^{2} cos^{2} x + b^{2} sin^{2} x) —- (2)
Adding (1) and (2)
2I = ∫_{0}^{𝛑} (𝛑) dx / (a^{2} cos^{2} x + b^{2} sin^{2} x)
2I = 2𝛑 ∫_{0}^{𝛑/2} dx / (a^{2} cos^{2} x + b^{2} sin^{2} x)
I = 𝛑 ∫_{0}^{𝛑/2} sec^{2} x dx / a^{2} + b^{2} tan^{2} x
Let tan x = t
sec^{2} x dx = dt
I = 𝛑 ∫_{0}^{∞} dt / a^{2} + b^{2}t^{2}
I = 𝛑 / b^{2} ∫_{0}^{∞} dt / [(a^{2} / b^{2}) + t^{2}]
I = (𝛑 / b^{2}) * (1 / (a / b)) [tan^{-1} t / (a / b)]_{0}^{∞}
I = (𝛑 / ab) * [(𝛑 / 2) – 0]
I = 𝛑^{2} / 2ab
Question 29: [a] Find the solution of the differential equation (x – y) dy – (x + y) dx = 0.
OR
[b] Find the solution of the differential equation cos^{2} x * (dy / dx) + y = tan x [0 ≤ x ≤ 𝛑 / 2]
Solution:
[a] (x – y) dy – (x + y) dx = 0dy / dx = (x + y) / (x – y)
Let y = vx
dy / dx = v + x (dv / dx)
v + x (dv / dx) = (x + vx) / (x – vx)
x (dv / dx) = (1 + v) / (1 – v) – v
x (dv / dx) = (1 + v – v + v^{2}) / (1 – v)
∫(1 – v) / (1 + v^{2}) dv = ∫dx / x
∫1 / (1 + v^{2}) dv – ∫v / (1 + v^{2}) dv = ∫dx / x
tan^{-1} v – (1 / 2) log (1 + v^{2}) = log x + c
tan^{-1} (y / x) – (1 / 2) log (1 + [y^{2} / x^{2}]) = log x + c
tan^{-1} (y / x) – (1 / 2) log [y^{2} + x^{2}] = c
OR
[b] cos^{2} x * (dy / dx) + y = tan xdy / dx + y / cos^{2} x = tan x / cos^{2} x which is linear equation in ‘y’
Here, P = 1 / cos^{2} x = sec^{2} x; Q = sec^{2} x . tan x
IF = e^{∫P dx} = e^{tanx}
ye^{tanx} = ∫e^{tanx} tanx sec^{2} x dx + c
Put tanx = z in R.H.S.
sec^{2} x dx = dz
∫e^{tanx} tanx sec^{2} x dx + c
= ∫e^{z} z dz
= ze^{z} – ∫1 e^{z} dz
= ze^{z} – e^{z}
= (z – 1)e^{z}
= (tan x – 1) e^{tanx}
ye^{tanx} = (tan x – 1) e^{tanx} + c
y = (tan x – 1) + ce^{-tanx}
Question 30: [a] Find the shortest distance between the lines r = (i – 2j + 3k) + ƛ (- i + j – 2k) and r = (i – j – k) + 𝛍 (i + 2j – 2k).
OR
[b] Find the equation of the plane that contains the point (2, -1, 3) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8.
Solution:
[a] a_{1} = i – 2j + 3ka_{2} = i – j – k
b_{1} = (- i + j – 2k)
b_{2} = (i + 2j – 2k)
Shortest distance = |(a_{2} – a_{1}) (b_{1} * b_{2}) / |(b_{1} * b_{2})|
(b_{1} * b_{2}) = i (–2 + 4) – j (2 + 2) + k (–2 –1)
= 2i – 4j – 3k
(a_{2} – a_{1}) = j – 4k
SD = |(- 4 + 12) / √4 + 16 + 9| = 8 / √29
OR
[b] Let the equation of planea (x – 2) + b (y + 1) + c (z – 3) = 0 ….(1)
The plane (1) is perpendicular to the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8
∴ 2a + 3b – 2c = 0 … (2)
a + 2b – 3c = 0 … (3)
Solving (2) & (3)
From (1)
– 5 (x – 2) + 4 (y + 1) + (z – 3) = 0
–5x + 4y + z + 11 = 0
5x – 4y – z – 11 = 0