Maharashtra State Board Class 12th Maths Question Paper 2023 with Solutions Answers Pdf Download.

## Class 12 Maths Question Paper 2023 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) If p ∧ q is F, p → q is F, then the truth values of p and q are _____ respectively. (2)

(a) T, T

(b) T, F

(c) F, T

(d) F, F

Answer:

(b) T, F

(ii) In ∆ABC, if c^{2} + a^{2} – b^{2} = ac, then ∠B = ………….. (2)

(a) \(\frac{\pi}{4}\)

(b) \(\frac{\pi}{2}\)

(c) \(\frac{\pi}{3}\)

(d) \(\frac{\pi}{6}\)

Answer:

\(\frac{\pi}{3}\)

c^{2} + a^{2} – b^{2} = ac

ie., a^{2} + c^{2} – b^{2} = ac ……. (1)

cos B = \(\frac{a^2+c^2-b^2}{2 a c}\) [By cosine rule]

cos B = \(\frac{a c}{2 a c}\) [From (1)]

∴ cos B = \(\frac{1}{2}\)

⇒ B = cos^{-1}\(\frac{1}{2}\)

⇒ B = 60° ⇒ B = \(\frac{\pi}{3}\)

(iii) The area of the triangle with vertices (1, 2, 0) (1, 0, 2) and (0, 3, 1) in sq. units is ……….. (2)

(a) \(\sqrt{5}\)

(b) \(\sqrt{7}\)

(c) \(\sqrt{6}\)

(d) \(\sqrt{3}\)

Answer:

(c) \(\sqrt{6}\)

A(1, 2, 0), B(1, 0, 2), C(0, 3,1)

(iv) If the corner points of the feasible solution are (0, 10), (2, 2) and (4, 0) then the point of minimum z = 3x + 2y is _____ (2)

(a) (2, 2)

(b) (0, 10)

(c) (4, 0)

(d) (3, 4)

Answer:

(a) (2, 2)

z = 3x + 2y

for (0, 10) ⇒ z = 3(0) + 2(10) = 20

for (2, 2) ⇒ z = (3)(2) + 2(2) = 10 MIN

for (4, 0) ⇒ z = (3)(4) + 2(0) = 12

Hence, minimum point of z = 3x + 2y is (2, 2)

(v) If y is a function of x and log (x + y) = 2xy, then the value of y’(0)…….. (2)

(a) 2

(b) 0

(c) -1

(d) 1

Answer:

(d) 1

Log(x + y) = 2xy

∴ log (x + y) – 2xy = 0

at x = 0 we get

log y = 0

∴ y = 1

Now, \(\frac{d}{d x}\)log(x + y) – \(\frac{d}{d x}\)2xy =0

\(\frac{1}{(x+y)}\)(1 + y’) – 2(xy’ + y) = 0

Now, when x = 0, y = 1 (From above)

(1 + y’) – 2(1) = 0

1 + y’ = 2

∴ y’ = 1

∴ at x = 0, y’ = 1

(vi) ∫cos^{3}xdx = ____ (2)

(a) \(\frac{1}{12}\)sin3x + \(\frac{3}{4}\)sinx + c

(b) \(\frac{1}{12}\)sin3x + \(\frac{1}{4}\)sinx + c

(c) \(\frac{1}{12}\)sin3x – \(\frac{3}{4}\)sinx + c

(d) \(\frac{1}{12}\)sin3x – \(\frac{1}{4}\)sinx + c

Answer:

(a) \(\frac{1}{12}\)sin 3x + \(\frac{3}{4}\) sinx + c

(vii) The solution of the differential equation \(\frac{d x}{d t}\) = \(\frac{x \log x}{t}\) is ……….. (2)

(a) x = e^{ct}

(b) x = e^{ct} + t

(c) x + e^{ct} = 0

(d) xe^{ct} = 0

Answer:

(a) x = e^{ct}

\(\frac{d x}{x \log x}\) = \(\frac{1}{t}\)dt

Integrating on both sides

\(\int \frac{1}{x \log x}\)dx = \(\int \frac{1}{t}\)dt …… (1)

put log x = u ⇒ \(\frac{1}{x}\)dx = du

∴ From (1) we get

∫\(\frac{1}{u}\)du = log|u| + c = log(log x) + c …….. (2)

From (1) and (2)

log (log x) = log |t| + c

log (log x) = log ct

∴ x = e^{ct}

(viii) Let the probability mass function (p.m.f) of a random variable X be P(X = x) = ^{4}C_{x} (\(\frac{5}{9}\))^{x} × (\(\frac{4}{9}\))^{4-x}, for x = 0.

1, 2, 3, 4 then E(X) is equal to _______ (2)

(a) \(\frac{20}{9}\)

(b) \(\frac{9}{20}\)

(c) \(\frac{12}{9}\)

(d) \(\frac{9}{25}\)

Answer:

(a) \(\frac{20}{9}\)

Question 2.

Answer the following questions: [4]

(i) Write the joint equation of co-ordinate axes.

Answer:

Equation of x-axis is y = 0

Equation of y-axis is x = O

∴ Joint equation of co-ordinate axes is xy = 0

(ii) Find the values of C which satisfy |c\(\bar{u}\)| = 3 where

\(\bar{u}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\).

Answer:

|c\(\bar{u}\)| = 3

(iii) Write ∫cot x dx

Answer:

I = ∫cot x dx

Let u = sin x

\(\frac{d u}{d x}\) = cos x

du = cos x dx ……. (1)

I = ∫cot x dx

= \(\int \frac{\cos x}{\sin x}\)dx

= \(\int \frac{d u}{u}\) [From (1)]

= log |u| + c

= log |sin x| + c

(iv) Write the degree of the differential equation

\(e^{\frac{d y}{d x}}\) + \(\frac{d y}{d x}\) = x

Answer:

Given differential equaton is,

\(e^{\frac{d y}{d x}}\) + \(\frac{d y}{d x}\) = x

Degree of the given differential equation is not defined

∴ Given differential equation is not in polynomial form x.

**Section – B**

Attempt any EIGHT of the following questions: [16]

Question 3.

Write inverse and contrapositive of the following statement if x < y then x^{2} < y^{2} (2)

Answer:

Given:

p → q: if x < y then x^{2} < y^{2}

Inverse: (~ p → ~q)

If x ≥ y then x^{2} < y^{2}

Contrapositive ~q → ~p

If x^{2} ≥ y^{2} then x ≥ y

Question 4.

If A = \(\left[\begin{array}{lll}

x & 0 & 0 \\

0 & y & 0 \\

0 & 0 & z

\end{array}\right]\) is a non singular matrix, then find A^{-1} by elementary row transformations. (2)

Hence, write the inverse of \(\left[\begin{array}{rrr}

2 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & -1

\end{array}\right]\)

Answer:

Question 5.

Find the cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\sqrt{2}, \frac{\pi}{4}\right)\). (2)

Answer:

Given, r = \(\sqrt{2}\), θ = \(\frac{\pi}{4}\)

Using x = r cos θ, y = r sin θ we get

x = \(\sqrt{2}\)cos θ, y = \(\sqrt{2}\)sin θ we get

x = \(\sqrt{2}\)cos\(\frac{\pi}{4}\), y = \(\sqrt{2}\)sin\(\frac{\pi}{4}\)

∴ x = \(\sqrt{2}\) × \(\frac{1}{\sqrt{2}}\), y = \(\sqrt{2}\) × \(\frac{1}{\sqrt{2}}\)

∴ x = 1, y = 1

∴ The required cartesion co-ordinates are (1, 1)

Question 6.

If ax^{2} + 2hxy + by^{2} = 0 represents a pair of lines and h^{2} = ab = 0 then find the ratio of their slopes. (2)

Answer:

As h^{2} = ab means lines are coincident

∴ their slopes are equal

∴ their ratio is 1 : 1

Question 7.

If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are the position vectors of the points A, B, C respectively and 5\(\bar{a}\) + 3\(\bar{b}\) – 8\(\bar{c}\) = \(\bar{0}\) then find the ratio in which the point C divides the line segment AB. (2)

Answer:

Given, 5\(\bar{a}\) + 3\(\bar{b}\) – 8\(\bar{c}\) = 0

∴ 8\(\bar{c}\) = 5\(\bar{a}\) + 3\(\bar{b}\)

∴ \(\bar{c}\) = \(\frac{3 \bar{b}+5 \bar{a}}{8}\)

∴ \(\bar{c}\) = \(\frac{3 \bar{b}+5 \bar{a}}{3+5}\)

This shows that the point c divides AB internally in the ratio 3: 5.

Question 8.

Solve the fòllowing inequations graphically and write the corner points of the feasible region:

2x + 3y ≤ 6, x + y ≥ 2, x ≥ 0, y ≥ 0

Answer:

Let the equations are

2x + 3y = 6, x + y = 2

Now,

The shaded region is the feasible region ABDA.

∴ The vertices of feasibLe region are A(0, 2), B(3, 0) and D(2,0)

Question 9.

Show that the function f(x) = x^{3} + 10x + 7, x ∈ R is strictly increasing. (2)

Answer:

Given, Differentiating w.r. to x,

f(x) = x^{3} + 10x + 7

f'(x) = 3x^{2} + 10

Hence 3x^{2} > 0, ∀ x ∈ R, 10 > 0

∴ 3x^{2} + 10 > 0

⇒ f'(x) > 0

Thus, f(x) is strictly increasing function.

Question 10.

Evaluate: (2)

\(\int_0^{\frac{\pi}{2}} \sqrt{1-\cos 4 x}\)dx

Answer:

Question 11.

Find the area of the region bounded by the curve y^{2} = 4x, the X-axis and the lines x = 1, x = 4 for y ≥ 0. (2)

Answer:

Let A be the required area

Question 12.

Solve the differential equation

cos x cos y dy – sin x sin y dx = 0

Answer:

Given that

cos x cos y dy – sin x sin y dx = 0

∴ cos x. cos y dy = sin x sin y dx

∴ \(\frac{d y}{d x}\) \(=\frac{\sin x \cdot \sin y}{\cos x \cdot \cos y}\)

∴ \(\frac{d y}{d x}\) = tan x.tan y

∴ \(\frac{d y}{\tan y}\) = tanx.dx

∴ ∫coty dy = ∫tanx dx

∴ log sin y = log sec x + log c

log sin y – log sec x = log c

∴ log\(\frac{\sin y}{\sec x}\) = log c

∴ cosx.sin y = c

Question 13.

Find the mean of number randomly selected from 1 to 15. (2)

Answer:

The sample space of experiment is

S = {1, 2, 3,……..15}

Let x denote the number selected.

Then X is a random variable which can take values 1, 2, 3, …….. 15

Each number selected is equiprobable therefore,

P(1) = P(2) = P(3) = _ = P(15) = \(\frac{1}{15}\)

Question 14.

Find the area of the region bounded by the curve y = x^{2} and line y = 4. (2)

Answer:

**Section – C**

Attempt any EIGHT of the following questions: [24]

Question 15.

Find the general solution of sinθ + sin3θ + sin5θ = 0.

Answer:

We have sin θ + sin 3θ + sin 5θ = 0

∴ (sin θ + sin 5θ) + sin 3θ = 0

2 sin\(\left(\frac{\theta+5 \theta}{2}\right)\)cos\(\left(\frac{\theta-5 \theta}{2}\right)\) + sin 3θ

∴ 2 sin 3θ × cos 2θ + sin 3θ = 0

[∵ sin C + sin D = 2sin\(\left(\frac{C+D}{2}\right)\)cos\(\left(\frac{C-D}{2}\right)\) ]

∴ sin3θ(2 cos 2θ + 1) = 0

∴ sin 3θ = 0 or cos 2θ = –\(\frac{1}{2}\)

⇒ 3θ = nπ

∴ θ = \(\frac{n \pi}{3}\),n

cos 2θ = –\(\frac{1}{2}\)

∴ cos 2θ = cos \(\frac{2 \pi}{3}\)

⇒ 2θ = 2mπ ± \(\frac{2 \pi}{3}\), m ∈ z

∴ θ = mπ ± \(\frac{\pi}{3}\), m ∈ z

∴ θ = \(\frac{n \pi}{3}\) or θ = mx ± \(\frac{\pi}{3}\), m, n ∈ z

is the required general solution.

Question 16.

If – 1 ≤ x ≤ 1, then prove that sin^{-1} x + cos^{-1} x = \(\frac{\pi}{2}\).

Answer:

Let sin^{-1}x = θ, where x ∈ [-1, 1] and θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

∴ \(\frac{\pi}{2}\) – ∈ [0, π] which is the principal domain of the cosine function.

As cos\(\left(\frac{\pi}{2}-\theta\right)\) = sin θ

cos \(\left(\frac{\pi}{2}-\theta\right)\) = x

∴ cos^{-1}x = \(\frac{\pi}{2}\) – θ

∴ θ + cos^{-1}x = \(\frac{\pi}{2}\)

∴ sin^{-1}x + cos^{-1}x = \(\frac{\pi}{2}\)

Question 17.

If θ is the acute angle between the Lines represented by ax^{2} + 2hxy + by^{2} = 0 then prove that tanθ = \(\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|\) (3)

Answer:

Let m_{1} and m_{2} be the slopes of the lines represented by the equation

ax^{2} + 2 hxy + by^{2} = 0 ……(1)

Then their separate equations are

y = m_{1}x and y = m_{2}x

∴ Their combined equation is (m_{1}x – y) (m_{2}x – y) = 0

i.e., m_{1}m_{2} x^{2} – (m_{1} + m_{2}) xy + y^{2} = 0 …….(2)

∴ (1) and (2) represent the same two lines, comparing the coefficients, we get,

Question 18.

Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are -2, 1, – 1 and -3, – 4, 1.

Answer:

The direction ratios of two lines are -2, 1, -1 and -3, -4, 1

Let \(\bar{a}\) = 2\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\), \(\bar{b}\) = 3\(\hat{i}\) – 4\(\hat{j}\) + \(\hat{k}\)

As we know that vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is obtained by \(\bar{a}\) × \(\bar{b}\)

∴ The direction ratios of the required vector are -3, 5, 11.

Question 19.

Find the shortest distance between lines

\(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) and \(\frac{x-2}{3}\) = \(\frac{y-4}{4}\) = \(\frac{z-5}{5}\)

Answer:

The vector equations of given lines are

Question 20.

Lines \(\bar{r}\) = (\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + λ(2\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\)) and \(\bar{r}\) = (4\(\hat{i}\) – 3\(\hat{j}\) + 2\(\hat{k}\)) are coplanar. Find the equation of the plane determined by them. (3)

Answer:

Question 21.

If y = \(\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \ldots+\infty}}}\), then show that \(\frac{d y}{d x}\) = \(\frac{\sec ^2 x}{2 y-1}\) Find \(\frac{d y}{d x}\) at x = 0

Answer:

Question 22.

Find the approximate value of sin (30°30°).

Given that 1° = 0.0175° and cos 30° = 0866. (3)

Answer:

Let f(x) = sin x

Differentiate w.r.t. x

f'(x) = cos x

Here x = 30°, δx = 30

= \(\left(\frac{1}{2}\right)^{\circ}\) = 0.00875 [∵ 1° = 0.0175]

f(x + δx) = f(x) + f'(x)δx

f[30° + \(\frac{1}{2}\)] = sin 30° + cos 30° × \(\frac{1}{2}\) × 0.0175

= \(\frac{1}{2}\) + 0.866 + 0.00875

[∵ cos 30° = 0.866]

= 0.5 + 0.0075775

= 0.5075775

∴ f(30°30′) = 0.5076

∴ sin (30°30′) = 0.5076

Question 23.

Evaluate ∫xtan^{-1}x dx (3)

Answer:

Question 24.

Find the particular solution of the differential equation \(\frac{d y}{d x}\) = e^{2y} cos x, when x = \(\frac{\pi}{6}\), y = 0. (3)

Answer:

Given that \(\frac{d y}{d x}\) = e^{2y}cos x

\(\frac{d y}{d x}\) = cos x dx

On integrating both sides

\(\int e^{-2 y} d y\) = \(\int \cos x d x\)

\(\frac{e^{-2 y}}{-2}\) = sin x + c ……. (1)

When x = \(\frac{\pi}{6}\), y = 0 then equation (1) becomes

\(\frac{e^{\circ}}{-2}\) = sin\(\frac{\pi}{6}\) + c

\(\frac{1}{2}\) = –\(\frac{1}{2}\) + c (∵ c = -1)

put c = -1 in equation (1), we get

\(\frac{e^{-2 y}}{-2}\) = sin x – 1

∴ e^{-2y} = -2 sin x + 2

Question 25.

For the following probabihty density function of a random variable X, find (a) P(X < 1) and (b) P (|X| < 1).

f(x) = \(\frac{x+2}{18}\); for -2 < x < 4

=0. otherwise (3)

Answer:

(a)

(b)

Question 26.

A die is thrown 6 times. If getting an odd number is a success, find the probability of at least 5 successes. (3)

Answer:

A die is thrown 6 times

As success is getting an odd number

∴ Probability of success, ρ = \(\frac{1}{2}\)

∴ q = 1 – p and x = the number of success.

∴ X ~ B(n = 6, p = \(\frac{1}{2}\))

∴ Probability of at (east 5 successes is:

P(x > 5) = P(x = 5) + P(x = 6)

**Section – D**

Attempt any FIVE of the following questions: [20]

Question 27.

Simplify the given circuit by writing its logical expression. Also write your conclusion. (4)

Answer:

Let p: the switch S_{1}

q: Switch S_{2}

The logical expression for the given circuit is

(Conclusion: The tamp wilt nor glow irespective of the status of the switches)

Question 28.

If A = \(\left[\begin{array}{ll}

1 & 2 \\

3 & 4

\end{array}\right]\) verify that

A (adj A) = (adjA)A = |A|I

Answer:

For A = \(\left[\begin{array}{ll}

1 & 2 \\

3 & 4

\end{array}\right]\)

∴ A_{11} = (+1)(4) = 4

A_{12} = (-1)(3) = -3

A_{21} = (-1)(2) = -2

A_{22} = (+1)(1) = 1

∴ Adj A = \(\left[\begin{array}{ll}

\mathrm{A}_{11} & \mathrm{~A}_{21} \\

\mathrm{~A}_{12} & \mathrm{~A}_{22}

\end{array}\right]\) = \(\begin{array}{cc}

{\left[\begin{array}{c}

4 \\

-3

\end{array}\right.} & \left.\begin{array}{c}

-2 \\

1

\end{array}\right]

\end{array}\)

From (1), (2) and (3) we get.

A(adj A) = (adj A) A = |A|

Question 29.

Prove that the volume of a tetrahedron with coterminus edges \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) is \(\frac{1}{6}\)[\(\bar{a}\) \(\bar{b}\) \(\bar{c}\)].

Hence, find the volume of tetrahedron whose coterminus edges are \(\bar{a}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\), \(\bar{b}\) = –\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\) and \(\bar{c}\) = 2\(\hat{i}\) + \(\hat{j}\) + 4\(\hat{k}\) (4)

Answer:

Let \(\overline{O A}\) = \(\bar{a}\), \(\overline{O B}\) = \(\bar{b}\) and \(\overline{O C}\) = \(\bar{c}\) be coterminus edges of Tetrahedron OABCO.

Let AP be the height of Tetrahedron

Volume of Tetrahedron

= \(\frac{1}{3}\)(Area of base ∆OCB) × (Height AP)

But AP = Scatcir projection of \(\bar{a}\) on \(\bar{b}\) × \(\bar{c}\)

Question 30.

Find the length of the perpendicular drawn from the point P(3, 2, 1) to the line:

\(\bar{r}\) = (7\(\hat{i}\) + 7\(\hat{j}\) + 6\(\hat{k}\)) + λ(-2\(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\)) (4)

Answer:

The Length of the perpendicular is same as the distance of P from the given line.

The distance of point P\((\bar{\alpha})\) from the tine \(\bar{\gamma}\) = \(\bar{\alpha}+\lambda \bar{b}\) is

Question 31.

If y = cos (m cos^{-1} x) then show that

(1 – x^{2})\(\frac{d^2 y}{d x^2}\) – x\(\frac{d y}{d x}\) + m^{2}y = 0 (4)

Answer:

Let y = cos(m cos^{-1} x)

cos^{-1} y = m cos^{-1}x

Differentiating both sides w.rt x

Question 32.

Verify Lagrange’s mean value theorem for the function:

f(x) = \(\sqrt{x+4}\) on the interval [0, 5]. (4)

Answer:

Given that

f(x) = \(\sqrt{x+4}\) ….. (1)

(i) f(x) is continuous on [0, 5]

(ii) f(x) is differentiabLe on (0, 5)

So the Lagrange’s mean value theorem is applicable to the function

Diffèrentiate (1) w.r. to x

f’(x) = \(\frac{1}{2 \sqrt{x+4}}\) …… (2)

Let a = 0 b = 5

from (1),

f(a) = f(0) = \(\sqrt{0+4}\) = 2

f(b) = f(5) = \(\sqrt{5+4}\) = 3

Let c ∈ (0, 5) such that

f'(c) = \(\frac{f(b)-f(a)}{b-a}\)

∴ \(\frac{1}{2 \sqrt{c+4}}\) = \(\frac{3-2}{5-0}\) = \(\frac{1}{5}\)

∴ \(\sqrt{c+4}\) = \(\frac{1}{5}\)

∴ c + 4 = \(\frac{25}{4}\)

c = \(\frac{9}{4}\) ∈ (0, 5)

Thus, Lagrange’s mean value theoron is verified

Question 33.

Evaluate \(\int \frac{2 x^2-3}{\left(x^2-5\right)\left(x^2-4\right)} d x\) (4)

Answer:

Question 34.

Prove that:

\(\int_0^{2 a}\)f(x)dx = \(\int_0^a\)f(x)dx + \(\int_0^a\)f(2a – x)dx (4)

Answer:

Consides

RH.S. = \(\int_0^a f(x) d x\) + \(\int_0^a f(2 a-x) d x\)

= I_{1} + I_{2}

Consider I_{2} = \(\int_0^a f(2 a-x) d x\)

Put 2a – x = t ∴ x = 2a – t

∴ dx = -dt

As x varies from 0 to a.

t Varies from 2a to a