12th Maths Question Paper 2023 Maharashtra Board Pdf

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 c² + a² – b² = ac, then ∠B = ………….. (2)
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{6}\)
Answer:
\(\frac{\pi}{3}\)
c² + a² – b² = ac
ie., a² + c² – b² = 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) ∫cos³xdx = ____ (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) = ⁴C_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 x² < y² (2)
Answer:
Given:
p → q: if x < y then x² < y²
Inverse: (~ p → ~q)
If x ≥ y then x² < y²
Contrapositive ~q → ~p
If x² ≥ y² 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 ax² + 2hxy + by² = 0 represents a pair of lines and h² = ab = 0 then find the ratio of their slopes. (2)
Answer:
As h² = 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) = x³ + 10x + 7, x ∈ R is strictly increasing. (2)
Answer:
Given, Differentiating w.r. to x,
f(x) = x³ + 10x + 7
f'(x) = 3x² + 10
Hence 3x² > 0, ∀ x ∈ R, 10 > 0
∴ 3x² + 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 y² = 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 = x² 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^-1x = 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^-1x = \(\frac{\pi}{2}\) – θ
∴ θ + cos^-1x = \(\frac{\pi}{2}\)
∴ sin^-1x + cos^-1x = \(\frac{\pi}{2}\)

Question 17.
If θ is the acute angle between the Lines represented by ax² + 2hxy + by² = 0 then prove that tanθ = \(\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|\) (3)
Answer:
Let m₁ and m₂ be the slopes of the lines represented by the equation
ax² + 2 hxy + by² = 0 ……(1)
Then their separate equations are
y = m₁x and y = m₂x
∴ Their combined equation is (m₁x – y) (m₂x – y) = 0
i.e., m₁m₂ x² – (m₁ + m₂) xy + y² = 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^-1x 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^2ycos 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 S₁
q: Switch S₂
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 = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
∴ A₁₁ = (+1)(4) = 4
A₁₂ = (-1)(3) = -3
A₂₁ = (-1)(2) = -2
A₂₂ = (+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 – x²)\(\frac{d^2 y}{d x^2}\) – x\(\frac{d y}{d x}\) + m²y = 0 (4)
Answer:
Let y = cos(m cos^-1 x)
cos^-1 y = m cos^-1x
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. = \(\int_0^a f(x) d x\) + \(\int_0^a f(2 a-x) d x\)
= I₁ + I₂
Consider I₂ = \(\int_0^a f(2 a-x) d x\)
Put 2a – x = t ∴ x = 2a – t
∴ dx = -dt
As x varies from 0 to a.
t Varies from 2a to a

Maharashtra Board Class 12 Maths Previous Year Question Papers