12th Maths Question Paper 2022 Maharashtra Board Pdf

Maharashtra State Board Class 12th Maths Question Paper 2022 with Solutions Answers Pdf Download.

Class 12 Maths Question Paper 2022 Maharashtra State Board with Solutions

Time: 3 Hrs.
Max. Marks: 80

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions [16]

(i) The negation of p ∧ (q → r) is
(a) ~ p ∧ (~q → ~ r)
(b) p ∨ (~q ∨ r)
(c) ~P ∧ (~q → r)
(d) P → (q ∧ ~r)
Answer:
(d) p → (q∧~r)

(ii) In ∆ABC if c² + a² – b² = ac, then ∠B = ……..
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{6}\)
Answer:
(b) \(\frac{\pi}{3}\)

(iii) Equation of line passing through the points (0, 0, 0) and (2, 1, -3) is ……… (2)
(a) \(\frac{x}{2}\) = \(\frac{y}{1}\) = \(\frac{z}{-3}\)
(b) \(\frac{x}{2}\) = \(\frac{y}{-1}\) = \(\frac{z}{-3}\)
(c) \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{-3}\)
(d) \(\frac{x}{3}\) = \(\frac{y}{1}\) = \(\frac{z}{2}\)
Answer:
(a) \(\frac{x}{2}\) = \(\frac{y}{1}\) = \(\frac{z}{-3}\)

(iv) The value of \(\hat{i}\). (\(\hat{j}\) × \(\hat{k}\)) + \(\hat{j}\) (\(\hat{k}\) × \(\hat{i}\)) + \(\hat{k}\)(\(\hat{i}\) × \(\hat{j}\)) is ______.
(a) 0
(b) -1
(c) 1
(d) 3
Answer:
(d) 3

(v) If f(x) = x⁵ + 2x – 3, then (f^-1)¹ (-3) = ____.
(a) 0
(b) -3
(c) –\(\frac{1}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(d) \(\frac{1}{2}\)

(vi) The maximum value of the function f (x) = \(\frac{\log x}{x}\) is _____ (2)
(a) e
(b) [laex]\frac{1}{e}[/latex]
(c) e²
(d) \(\frac{1}{e^2}\)
Answer:
(b) \(\frac{1}{e}\)

(vii) If \(\int \frac{d x}{4 x^2-1}\) = A log\(\left(\frac{2 x-1}{2 x+1}\right)\) + c, then A = ____.
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(d) \(\frac{1}{4}\)

(viii) If the p.m.f of a r. y. X is (2)
P(x) = \(\frac{c}{x^3}\), for x = 1, 2, 3 = 0. otherwise, then E(x) = ____.
(a) \(\frac{216}{251}\)
(b) \(\frac{294}{251}\)
(c) \(\frac{297}{294}\)
(d) \(\frac{294}{297}\)
Answer:
(b) \(\frac{294}{251}\)

Question 2.
Answer the following questions: [4]

(i) Find the principal value of cot^-1\(\left(\frac{-1}{\sqrt{3}}\right)\) (1)
Answer:
Let y = cot^-1\(\left(\frac{-1}{\sqrt{3}}\right)\)
Since, cot^-1(-x) = π – cot^-1x
∴ y = π – cot^-1\(\left(\frac{1}{\sqrt{3}}\right)\)
⇒ y = π – \(\frac{\pi}{3}\) (∵ cot\(\frac{\pi}{3}\) = \(\frac{1}{\sqrt{3}}\))
⇒ y = \(\frac{2 \pi}{3}\)
Since, range of cot^-1 is (0, π)
Hence, principal value is \(\frac{2 \pi}{3}\).

(ii) Write the separate equations of tines represented by the equation 5x² – 9y² = 0 (1)
Answer:
Given, 5x² – 9y² = 0
⇒ \((\sqrt{5} x)^2\) – (3y)² = 0
⇒ (\(\sqrt{5}\) – 3y)(\(\sqrt{5}\) + 3y) = 0
\(\sqrt{5}\) – 3y = 0

(iii) If f’(x) = x^-1, then find f(x) (1)
Answer:
Given, f'(x) = \(\frac{1}{x}\)
on integrating both sides, we get
f(x) = log x + C

(iv) Write the degree of the differential equation
(y”’)² + 3(y’’) + 3xy’ + 5y = 0 (1)
Answer:
The given differential equation is:
(y”’)² + 3(y”) + 3xy’ + 5y = 0
Here, highest order derivative is third order, which is raised to second-degree.
Hence, degree of this diffrential equation is 2.

Section – B

Attempt any EIGHT of the following questions: [16]

Question 3.
Using truth table verify that: (2)
(p ∧ q) ∨ ~g ≡ pv – g
Answer:

Question 4.
Find the cofactors of the elements of the matrix (2)
\(\left[\begin{array}{ll}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Answer:
Given, matrix is \(\left[\begin{array}{ll}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Here a₁₁ = -1; ∴ N₁₁ = 4 and A₁₁ = (-1)¹⁺¹(4) = 4
a₁₂ = 2; ∴ N₁₂ = -3 and A₁₂ = (-1)¹⁺²(-3) = 3
a₂₁ = -3; ∴ N₂₁ = 2 and A₂₁ = (-1)²⁺¹(2) = -2
a₂₂ = -3; ∴ N₂₂ = -1 and A₂₂ = (-1)²⁺²(-1) = -1
∴ The required cofactors are 4, 3, -2, -1.

Question 5.
Find the principal solutions of cot θ = 0 (2)
Answer:
Given, cotθ = 0
since, θ ∈ (0, 2π)
∴ cotθ = 0 = cot\(\frac{\pi}{2}\) = cot(π + \(\frac{\pi}{2}\)) [∵ cot(π + θ) = cot θ]
∴ cotθ = cot \(\frac{\pi}{2}\) = cot\(\frac{3 \pi}{2}\)
∴ θ = \(\frac{\pi}{2}\) or θ = \(\frac{3 \pi}{2}\)
Hence, the required principal solutions are {\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)}

Question 6.
Find the value of k. if 2x + y = 0 is one of the lines represented by
3x² + kxy + 2y² = 0 (2)
Answer:
Given. 2x + y = 0
⇒ 2 x = -y …. (i)
Then, 3x² + by + 2y² = 0
⇒ 3x² + kx (-2x) + 2 (4x²) = 0 [from (1)]
⇒ 3x² – 2k² + 8x² = 0
⇒ 11x² – 2kx² = 0
⇒ k = \(\frac{11}{2}\)

Question 7.
Find the cartesian equation of the plane passing through A (1, 2, 3) and the direction ratios of whose normal are 3, 2, 5. (2)
Answer:
The plane passes through the point A (1, 2, 3) and the direction ratios of its normal are 3, 2, 5.
∴ x₁ = 1, y₁ = 2, z₁ = 3, a = 3, b = 2, c = 5
Equation of a plane in cartesian form is:
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
∴ 3(x – 1) + 2(y – 2) + 5(z – 3) = 0
∴ 3x + 2y + 5z – 22 = 0

Question 8.
Find the cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{1}{2}, \frac{\pi}{3}\right)\) (2)
Answer:
Here, r = \(\frac{1}{2}\) and θ = \(\frac{\pi}{3}\)
Let, the cartesian coordinates be (x, y).
Then,
x = rcosθ = \(\frac{1}{2}\)cos\(\frac{\pi}{3}\) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)
and y = rsinθ = \(\frac{1}{2}\)sin\(\frac{\pi}{3}\) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)
∴ The cartesian coordinates of the given point are (\(\frac{1}{4}\), \(\frac{\sqrt{3}}{4}\))

Question 9.
Find the equation of tangent to the curve y = 2x³ – x² + 2at(\(\frac{1}{2}\), 2) (2)
Answer:

Question 10.
Evaluate:
\(\int_0^{\frac{\pi}{4}} \sec ^4 \times d x\)
Answer:

Question 11.
Solve the differential equation
y\(\frac{d y}{d x}\) + x = 0
Answer:
Given differential equation is
y\(\frac{d y}{d x}\) + x = 0
⇒ y\(\frac{d y}{d x}\) = -x
⇒ y dy = -x dx
On integrating both sides, we get
∫ydy = ∫-x dx
⇒ \(\frac{y^2}{2}\) = \(\frac{-x^2}{2}\) + C’
⇒ y² + x² = 2C
⇒ x² + y² = C, where C = 2C is the required solution of differential equation.

Question 12.
Show that function f(x) = tan x is increasing in (0, \(\frac{\pi}{2}\)) (2)
Answer:
Given, f(x) = tanx
f’(x) = sec² x
But sec² x > 0, ∀ x ∈ (0, π/2)
Hence, f(x) = tan x is strictly increasing in (0, π/2)

Question 13.
Form the differential equation of all lines which makes intercept 3 on x-axis. (2)
Answer:
Equation of straight line is y = mx + c
At x-axis, y = 0
So, x = \(\frac{-c}{m}\)
Here, slope = m
If slope and x-intercept are equal
\(\frac{-c}{m}\) = m ⇒ c = -m²
∴ y = mx – m²
∴ \(\frac{d y}{d x}\) = m
Since, m = 3
∴ \(\frac{d y}{d x}\) = 3, which is the required equation.

Question 14.
If X ~ B (n, p) and E(X) = 6 and Var (X) = 4.2, then find n and p. (2)
Answer:
Given, X ~ B(n, p) and E(X) = 6 and var (X) = 4.2
Now, \(\frac{E(X)}{u(X)}\) = \(\frac{n p}{n p q}\)
⇒ \(\frac{6}{4.2}\) = \(\frac{1}{q}\) ⇒ q = \(\frac{4.2}{6}\) = 0.7
Since, p + q = 1
⇒ p + 0.7 = 1
⇒ p = 0.3
Now E(X) = np
⇒ 6 = n × 0.3
⇒ n = \(\frac{6}{0.3}\) = \(\frac{60}{3}\) = 20

Section – C

Attempt any EIGHT of the following questions: [24]

Question 15.
If 2 tan^-1 (cos x) = tan^-1 (2 cosec x), then find the value of x. (3)
Answer:

Question 16.
If angle between the lines represented by ax² + 2hxy + by² = 0 is equal to the angle between the lines represented by 2x² – 5xy + 3y² = 0, then show that 100(h² – ab) = (a + b)². (3)
Answer:

Question 17.
Find the distance between the parallel lines
\(\frac{x}{2}\) = \(\frac{y}{-1}\) = \(\frac{z}{2}\) and \(\frac{x-1}{2}\) = \(\frac{y-1}{-1}\) = \(\frac{z-1}{2}\) (3)
Answer:
Line passes through (O, O, O) and has direction ratios 2, -1, 2
∴ Vector equation of the line is:
\(\vec{r}\) = (O\(\hat{i}\) + O\(\hat{j}\) + O\(\hat{k}\)) + λ(2\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\))
i.e., \(\vec{r}\) = λ(2\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\))
Now, line \(\frac{x-1}{2}\) = \(\frac{y-1}{-1}\) = \(\frac{z-1}{2}\)
Passes through (1, 1, 1) and has direction ratios 2, -1, 2
∴ vector equation of the line is:
\(\vec{r}\) = (2\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\))
The distance between parallel. is:

Question 18.
If A (5, 1, p), B (1, q, p) and C (1, -2, 3) are vertices of a triangle and G(r, \(\frac{-4}{3}\), \(\frac{1}{3}\)) is its centroid, then find the values of p, q, r by vector method. (3)
Answer:
Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be the position vectors
\(\vec{a}\) = 5\(\hat{i}\) + \(\hat{j}\) + p\(\hat{k}\)

Question 19.
If A\((\bar{a})\) and B\((\bar{b})\) be any two points in the space and R\((\bar{r})\) be a point on the line segment AB dividing it internally in the ratio m : n then prove that
\(\bar{r}\) = \(\frac{m \bar{b}+n \bar{a}}{m+n}\) (3)
Answer:
As R is a point on the line segment
AB (A – R – B) and \(\overline{A R}\) and \(\overline{R B}\) are in the same direction
Point R divides AB internally in the ratio m : n.

Question 20.
Find the vector equation of the plane passing through the point A (-1, 2, -5) and parallel to the vectors 4\(\hat{i}\) – \(\hat{j}\) + 3\(\hat{k}\) and \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\).
Answer:
The vector equation of the plane passing through the point A (\(\vec{a}\)) and parallel to the vectors \(\vec{b}\) and \(\vec{c}\) is
\(\vec{r}\).(\(\vec{b}\) × \(\vec{c}\)) = \(\vec{a}\).(\(\vec{b}\) × \(\vec{c}\))

Question 21.
If y = \(e^{m \tan ^{-1} x}\), then show that (3)
(1 + x²)\(\frac{d^2 y}{d x^2}\) + (2x – m)\(\frac{d y}{d x}\) = 0
Answer:
Given
To prove,
(1 + x²)\(\frac{d^2 y}{d x^2}\) + (2x – m)\(\frac{d y}{d x}\) = 0
Proof:

Question 22.
Evaluate: (3)
\(\int \frac{d x}{2+\cos x-\sin x}\)
Answer:
Let I = \(\int \frac{d x}{2+\cos x-\sin x}\)
Put tan\(\frac{x}{2}\) = t
⇒ x = 2tan^-1 t

Question 23.
Solve x + y = sec(x² + y²) (3)
Answer:

Question 24.
A wire of length 36 meters is bent to form a rectangle. Find its dimensions if the area of the rectangle is maximum. (3)
Answer:
Let x metres and y metres be the Length and breadth of the rectangle.
∴ Perimeter = 2(x + y) = 36
∴ x + y = 18
y = 18 – x
∴ Area of the rectangle = xy
= x(18 – x)
= 18x – x²
∴ \(\frac{d(\mathrm{~A})}{d x}\) = 18 – 2x
∴ \(\frac{d^2(\mathrm{~A})}{d x^2}\) = -2 < 0
Now, \(\frac{d(\mathrm{~A})}{d x}\) = 0, if 18 – 2x = 0
i.e., if x = 9
and \(\frac{d^2 A}{d x^2}\) < 0
∴ By the second derivative test area has maximum value at x = 9
when x = 9, y = 18 – 9 = 9
∴ x = 9 cm, y = 9 cm
∴ Rectangle is a square of side 9 cm.

Question 25.
Two dice are thrown simultaneously. If X denotes the number of sixes. find the expectation of X. (3)
Answer:
Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1 or 2.
∴ P (X = 0) = P (not getting six on any of the dice)
= \(\frac{5 \times 5}{6 \times 6}\) = \(\frac{25}{36}\)
P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)
= \(2\left(\frac{1}{6} \times \frac{5}{6}\right)\) = \(\frac{10}{36}\)
P(X = 2) = P(six on both the dice) = \(\frac{1}{36}\)
∴ The required probability distribution is as follows.

Question 26.
If a fair coin is tossed 10 times. Find the probability of getting at most six heads. (3)
Answer:

Section – D

Attempt any FIVE of the following questions:

Question 27.
Without using truth table prove that (4)
(p ∧ q) ∨ (- p ∧ q) ∨ (p ∧ – q) ≡ p ∨ q
Answer:

Question 28.
Solve the following system of equations by the method of inversion (4)
x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2
Answer:

Question 29.
Using vectors prove that the altitudes of a triangle are concurrent.
Answer:
Let, the altitudes AD and BE intersect at O
Join CO and produce to meet AB in F
Let \(\overrightarrow{O A}\) = \(\vec{a}\)
\(\overrightarrow{O B}\) = \(\vec{b}\), \(\overrightarrow{O C}\) = \(\vec{c}\)

Hence, \(\overrightarrow{\mathrm{OC}}\) is perpendicular to \(\overrightarrow{\mathrm{AB}}\), i.e., CF is the third altitude of the triangle through C.
Hence, the 3 attitudes are concurrent at O.

Question 30.
Solve the L. P. P. by graphical method.
Minimize z = 8x + 10y
Subject to 2x + y ≥ 7
2x + 3y ≥ 15.
y ≥ 2, x ≥ 0 (4)
Answer:
First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.

The feasible region is EPQBY which is shaded in the graph. The vertices of the feasibte region are P, Q and B.
P is the point of intersection of lines 2x + 3y = 15 and y = 2.
Substituting y = 2 in 2x + 3y = 15, we get
2x + 3(2) = 15
∴ 2x = 15 – 6 = 9
∴ x = 4.5
∴ P = (4.5, 2)
Q is the point of intersection of the lines
2x + 3y = 15 …… (i)
2x + y = 7 …….(ii)
On subtracting equation (ii) from equation (i), we get
2y = 8 ⇒ y = 4
From (ii), 2x + 4 = 7
∴ x = 1.5
∴ Q = (1.5, 4)
The values of the objective function z = 8x + 10y at these vertices are:
z(P) = 8(4.5) + 10(2) = 36 + 20 = 56
z(Q) = 8(1.5) + 10(4) = 12 + 40 = 52
z(B) = 8(0) + 10(7) = 70
∴ z has minimum value 52, when x = 1.5 and y = 4

Question 31.
If x = f(t) and y = g(t) are differentiable functions of t so that y is differentiable fùnction of x and \(\frac{d x}{d t}\) ≠ 0, then prove that: (4)
\(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
Hence find \(\frac{d y}{d x}\) if x = sin t and y = cos t.
Answer:
x and y are differentiable functions of t.
Let, there be a small, increment δt in the value of t correspondingly, there should be small increments δx, δy in the values of x and y respectively.

Here, LH.S. of (i) exist and are finite.
Hence. Limits on LH.S. of (i) also should exist and be finite.

Question 32.
If u and v are differentiable functions of x, then prove
∫uv dx = u∫v – \(\int\left[\frac{d u}{d x} \int v d x\right] d x\)
Hence evaluate ∫log x dx (4)
Answer:

Question 33.
Find the area of region between parabolas y² = 4ax and x² = 4ay (4)
Answer:
The equations of the parabolas are:
y² = 4ax …….. (i)
x² = 4ay ………. (ii)

\(\left(\frac{x^2}{4 a}\right)^2\) = 4ax [by (ii)]
⇒ x⁴ = 64a³
⇒ x[x³ – (4a)³] = 0
or x = 0 and x = 4a
∴ y = 0 and y = 4a
Point of intersection of curves are 0 (0, 0) and P (4a, 4a)

Question 34.
Show that: \(\int_0^{\frac{\pi}{4}}\)log(1 + tan x)dx = \(\frac{\pi}{8}\)log 2 (4)
Answer:

Maharashtra Board Class 12 Maths Previous Year Question Papers