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 c^{2} + a^{2} – b^{2} = 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) = x^{5} + 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) e^{2}

(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^{-1}x

∴ 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 5x^{2} – 9y^{2} = 0 (1)

Answer:

Given, 5x^{2} – 9y^{2} = 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 a_{11} = -1; ∴ N_{11} = 4 and A_{11} = (-1)^{1+1}(4) = 4

a_{12} = 2; ∴ N_{12} = -3 and A_{12} = (-1)^{1+2}(-3) = 3

a_{21} = -3; ∴ N_{21} = 2 and A_{21} = (-1)^{2+1}(2) = -2

a_{22} = -3; ∴ N_{22} = -1 and A_{22} = (-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

3x^{2} + kxy + 2y^{2} = 0 (2)

Answer:

Given. 2x + y = 0

⇒ 2 x = -y …. (i)

Then, 3x^{2} + by + 2y^{2} = 0

⇒ 3x^{2} + kx (-2x) + 2 (4x^{2}) = 0 [from (1)]

⇒ 3x^{2} – 2k^{2} + 8x^{2} = 0

⇒ 11x^{2} – 2kx^{2} = 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.

∴ x_{1} = 1, y_{1} = 2, z_{1} = 3, a = 3, b = 2, c = 5

Equation of a plane in cartesian form is:

a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 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 = 2x^{3} – x^{2} + 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’

⇒ y^{2} + x^{2} = 2C

⇒ x^{2} + y^{2} = 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) = sec^{2} x

But sec^{2} 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 = -m^{2}

∴ y = mx – m^{2}

∴ \(\frac{d y}{d x}\) = 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 ax^{2} + 2hxy + by^{2} = 0 is equal to the angle between the lines represented by 2x^{2} – 5xy + 3y^{2} = 0, then show that 100(h^{2} – 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 + x^{2})\(\frac{d^2 y}{d x^2}\) + (2x – m)\(\frac{d y}{d x}\) = 0

Answer:

Given

To prove,

(1 + x^{2})\(\frac{d^2 y}{d x^2}\) + (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(x^{2} + y^{2}) (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 – x^{2}

∴ \(\frac{d(\mathrm{~A})}{d x}\) = 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 y^{2} = 4ax and x^{2} = 4ay (4)

Answer:

The equations of the parabolas are:

y^{2} = 4ax …….. (i)

x^{2} = 4ay ………. (ii)

\(\left(\frac{x^2}{4 a}\right)^2\) = 4ax [by (ii)]

⇒ x^{4} = 64a^{3}

⇒ x[x^{3} – (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: