Maharashtra Board Class 12 Maths Sample Paper Set 1 with Solutions

Maharashtra State Board Class 12th Maths Sample Paper Set 1 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Maths Model Paper Set 1 with Solutions

Section A

Question 1.
Select and write the most appropriate answer from the given alternatives for each questions: (16)

(i) p ∧ q is false and p v q is true, then …….. is not true. (2)
(a) p ∨ q
(b) P ↔ q
(c) -p ∨ -q
(d) q ∨ – p
Answer:
(b) p ↔ q

(ii) The principal solutions of equation sin θ = -1/2 are:
(a) \(\frac{5 \pi}{6}, \frac{\pi}{6}\)
(b) \(\frac{7 \pi}{6}, \frac{11 \pi}{\$ 6}\)
(c) \(\frac{\pi}{6}, \frac{7 \pi}{6}\)
(d) \(\frac{7 \pi}{6} \cdot \frac{\pi}{3}\)
Answer:
(b) \(\frac{7 \pi}{6}, \frac{11 \pi}{6}\)

(iii) The joint equation of the lines through the origin and perpendicular to the pair of lines 3x² + 4xy – 5y² = 0 is: (2)
(a) 5x² + 4xy – 3y² = 0
(b) 3x² + 4xy – 5y² = 0
(c) 3x² – 4xy + 5y² = 0
(d) 5x² + 4xy + 3y² = 0
Answer:
(a) 5x² + 4xy – 3y² = 0

(iv) The vector equation of line 2x – 1 = 3y + 2 = z – 2 is: (2)
(a) r = (\(\frac{1}{2} \hat{i}\) + \(\frac{2}{3} \hat{j}\) + 2\(\hat{k}\)) + λ(3\(\hat{i}\) + 2\(\hat{j}\) + 6\(\hat{k}\))
(b) r = \(\hat{i}\) – \(\hat{j}\) + (2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
(c) (\(\frac{1}{2} \hat{i}\) – \(\hat{j}\)) + λ(\(\hat{i}\) – 2\(\hat{j}\) + 6\(\hat{k}\))
(d) r = (\(\hat{i}\) + \(\hat{k}\)) + λ(\(\hat{i}\) – 2\(\hat{j}\) + 6\(\hat{k}\))
Answer:
(a) r = \(\left(\frac{1}{2} \hat{i}-\frac{2}{3} \hat{j}+2 \hat{k}\right)\) + λ(3\(\hat{i}\) + 2\(\hat{j}\) + 6\(\hat{k}\))

(v) The value of objective function is maximum under linear constraints: (2)
(a) at the centre of feasible region
(b) at (0, 0)
(c) at any vertex of feasible region
(d) the vertex which is of maximum distance from (0, 0).
Answer:
(c) at any vertex of feasible region

(vi) If f(x) = \(\frac{x^2-1}{x^2+1}\), for every real x, then the minimum value of f is: (2)
(a) 1
(b) 0
(c) -1
(d) 2
Answer:
(c) -1

(vii) \(\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x \cdot d x}{(1+\cos x)^2}\)
(a) \(\frac{4-\pi}{2}\)
(b) \(\frac{\pi-4}{2}\)
(c) 4 – \(\frac{\pi}{2}\)
(d) \(\frac{4+\pi}{2}\)
Answer:
(a) \(\frac{4-\pi}{2}\)

(viii) The area bounded by the regions 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by: (2)
(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
Answer:
(a) 12 sq units

Question 2.
Answer the following questions: (4)

(i) Apply the given elementary transformation of the following matrix: (1)
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\). R₁ ↔ R₂
Answer:
A = \(\left[\begin{array}{rr}
1 & 0 \\
-1 & 3
\end{array}\right]\)
By R₁ ↔ R₂, we get
A ~ \(\left[\begin{array}{rr}
-1 & 3 \\
1 & 0
\end{array}\right]\)

(ii) State whether the following equation has a solution or not? (1)
cos²θ = – 1
Answer:
cos²θ = -1
This is not possible because cos²θ ≥ 0 for any θ.
∴ cos²θ = -1 does not have any solution.

(iii) Evaluate: \(\int_0^{\frac{\pi}{2}} x \sin x \cdot d x\) (1)
Answer:

(iv) Determine the order and degree of the following differential equation: (1)
\(\frac{d^2 y}{d x^2}\) + x\(\left(\frac{d y}{d x}\right)\) + y = 2 sin x
Answer:
The given D.E. is
\(\frac{d^2 y}{d x^2}\) + x\(\left(\frac{d y}{d x}\right)\) + y = 2 sin x
This differential equation has highest order derivative \(\frac{d^2 y}{d x^2}\) with power 1.
∴ The given differential equation is of order 2 and degree 1.

Section – B

Attempt any EIGHT of the following questions: (16)

Question 3.
Construct the truth table of the following statement pattern (p ∧ ~ q) ↔ (p → q).
Answer:

Question 4.
Convert \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\) into an identity matrix by suitable row transformations.
Answer:

Question 5.
Find the combined equation of the following pair of line passing through (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0.
Answer:
Let L₁ and L₂ be the lines passing through the point (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 respectively.
Slopes of the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 are \(\frac{-3}{2}\) and \(\frac{-1}{-3}\) = \(\frac{1}{3}\) respectively.
∴ Slopes of the lines L₁ and L₂ pass through the point (2, 3), their equations are
y – 3 = \(\frac{2}{3}\) and y – 3 = -3(x – 2)
∴ 3y – 9 = 2x – 4 and y – 3 = -3x + 6
∴ 2x – 3y + 5 = 0 and 3x + y – 9 = 0
Their combined equation is
(2x – 3y + 5) (3x + y – 9) = 0
∴ 6x² + 2xy – 18x – 9xy – 3y² + 27y + 15x + 5y – 45 = 0
∴ 6x² – 7xy – 3y² – 3x + 32y – 45 = 0

Question 6.
Find the separate equation of the line represented by the following equation:
5x² – 9y² = 0
Answer:
5x² – 9y² = 0
\((\sqrt{5} x)^2\) + (3y)² = 0
(\(\sqrt{5}\)x – 3y) + (\(\sqrt{5}\)x + 3y) = 0
The separate equations of the lines are (\(\sqrt{5}\)x + 3y) = 0 and \(\sqrt{5}\)x – 3y

Question 7.
Find the area of the triangle with vertices (1, 1, 0), (1, 0, 1) and (0, 1, 1).
Answer:

Question 8.
Find the vector equation of the line passing through points having position vector 3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\) and 6\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\).
Answer:
The vector equation of the line passing through the A(\(\bar{a}\)) and B(\(\bar{b}\)) is:
r = \(\bar{a}\) + λ(\(\bar{b}\) – \(\bar{a}\)), λ is a scalar.
∴ The vector equation of the line passing through the points having position vector 3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\) and 6\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) is
\(\bar{r}\) = (3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\)) + λ[(6\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) – (3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\))]
i.e. \(\bar{r}\) = (3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\)) + λ(3\(\hat{i}\) – 5\(\hat{j}\) + 8\(\hat{k}\))

Question 9.
Find the Cartesian equations of the line passing through A(-1, 2, 1) and having direction ratios 2, 3, 1.
Answer:
The Cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x-x_1}{a}\) = \(\frac{y-y_1}{b}\) = \(\frac{z-z_1}{c}\)
∴ The Cartesian equation of the line passing through the point (- 1, 2, 1) and having direction ratios 2, 3, 1 are
\(\frac{x-(-1)}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-1}{1}\)
i.e. \(\frac{x+1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-1}{1}\)

Question 10.
Solve graphically : x ≥ 0.
Answer:
Consider the line whose equation is x = 0. This represents the Y-axis.
To find the solution set we have to check any point other than origin.
Let us check the point (1, 1).
When x = 1, x ≥ 0
∴ (1, 1) lies in the required region.
Therefore, the solution set is the Y-axis and the right side of the V-axis which is shaded in the graph.

Question 11.
Differentiate the following w.r.t x : \(\sqrt{x^2+4 x-7}\)
Answer:

Question 12.
Find the area of the circle x² + y² = 9, using integration.
Answer:
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO. Clearly for this region, the limits of integration are 0 and 3.
From the equation of the circle, y² = 9 – x².
In the first quadrant, y > 0
∴ y = \(\sqrt{9-x^2}\)

Question 13.
State if the following is not the probability mass function of a random variable. Give reasons fór your answer:

Answer:
Probability mass function of random variable should satisfy the following conditions:
(a) 0 ≤ p_i ≤ 1
(b) Σp_i = 1

X	0	1	2
P(X)	0.4	0.4	0.2

(a) Here, 0 ≤ p_i ≤ 1
(b) Σp_i = 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as probability mass function of the random variable X.

Question 14.
A bag consists of 10 balls each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag. What is the probability that none is marked with the digit 0?
Answer:
Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.
Since the balls are drawn with replacement, the trials are Bernoulli trials.
X has a binomial distribution with n = 4 and p = \(\frac{1}{10}\)
and q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
The p.m.f of X is given by
P(X = x) = ⁿC_xp^xq^n-x
i.e. p(x) = ⁴C_x\(\left(\frac{1}{10}\right)^x\left(\frac{9}{10}\right)^{4-x}\), x = 0, 1, ……. 4
P(None of the ball marked with digit 0) = P(X = 0)

Hence, the probability that none of the bulb marked with digit 0 is \(\left(\frac{9}{10}\right)^4\).

Section C

Attempt any EIGHT of the followings questions: (24)

Question 15.
Find the co-factors of the elements of the following matrix.
\(\left[\begin{array}{rrr}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Answer:

Question 16.
Find the general solutions of the following equation:
sin θ = tan θ
Answer:
Given: sin θ = tan θ
⇒ sin θ = \(\frac{\sin \theta}{\cos \theta}\)
⇒ sinθcosθ – sinθ = 1
⇒ sinθ(cosθ – 1) = 0
⇒ sin θ = 0 or cosθ – 1 = 0
⇒ sin θ = O or cos θ = 1
⇒ sin θ = 0 or cosθ = cos 0°
The general solution of sin θ = 0 is θ = nπ, where n ∈ Z and cos θ = cos α is θ = 2nπ + α, where n ∈ Z.
∴ The required general solution is given by
θ = nπ, n ∈ Z or θ = 2nπ ± θ, n ∈ Z
∴ θ = nπ, n ∈ z or θ = 2nπ, n ∈ Z

Question 17.
Show that the following points are collinear:
A = (3, 2,- 4), B = (9, 8, -10), C = (-2, -3, 1)
Answer:
Let a, b, c be the position vectors of the points,
where’A = (3, 2, -4), B = (9, 8, -10) and C = (-2, -3, 1) respectively.
Then,

∴ \(\overrightarrow{\mathrm{BC}}\) is a non-zero scalar multiple of \(\overrightarrow{\mathrm{AB}}\)
∴ They are parallel to each other.
But they have point B in common.
∴ \(\overrightarrow{\mathrm{BC}}\) and \(\overrightarrow{\mathrm{AB}}\) are collinear vectors.
Hence, points A, B and C are collinear.

Question 18.
Are the four points A(1 – 1, 1), B(-1, 1, 1), C(1, 1, 1) and D(2, -3, 4) coplanar ? Justify your answer.
Answer:
The position vectors \(\bar{a}, \bar{b}, \bar{c}, \bar{d}\)of the points A, B, C, D are

By equality of vectors,
y = -2 ……. (i)
2x – 2y = 2 …….(ii)
3y = 0 ……. (iii)
From (i), y = -2
From (iii), y = 0
This is not possible.
Hence, the points A, B, C, D are not coplanar.

Question 19.
Differentiate the following w.r.t.
x : \(\frac{(x+1)^2}{(x+2)^3(x+3)^4}\)
Answer:

Question 20.
Find the equations of tangents and normals to the following curves at the indicated points on them:
y = x² + 2e^x + 2 at (0, 4)
Answer:
y = x² + 2e^x + 2
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x² + 2e^x +2)
= 2x + 2e^x + 0
= 2x + 2e^x
∴ \(\left(\frac{d y}{d x}\right)_{0 t(0.4)}\) = 2(0) + 2e⁰ = 2
= Slope of the tangent at (0, 4).
The equation of the tangent at (0, 4) is
∴ y – 4 = 2(x – 0)
∴ y – 4 = 2x
∴ 2x – y + 4 = 0
The slope of the normal at (0, 4)
= \(\frac{-1}{\left(\frac{d y}{d x}\right)_{a t(0.4)}}\) = \(-\frac{1}{2}\)
∴ The equation of the normal at (0, 4) is
y – 4 = \(-\frac{1}{2}\)(x – 0)
∴ 2y – 8 = -x
∴ x + 2y – 8 = 0
Hence, the equations of tangent and normal are 2x – y + 4 = 0 and x + 2y – 8 = 0 respectively.

Question 21.
Evaluate the following integrals: \(\int \frac{5 x+2}{3 x-4} d x\)
Answer:

Question 22.
Integrate the following functions w.r.t x: \(\frac{x^n-1}{\sqrt{1+4 x^n}}\)
Answer:

Question 23.
Evaluate the following integrals as limit of a sum : \(\int_1^3(3 x-4) d x\)
Answer:
Let f(x) = 3x —4, for 1 ≤ x ≤ 3.
Divide the closed interval [1, 3] into n subintervals each of length h at the points
1.1 + h, 1 + 2h, 1 + rh…..1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 1

Question 24.
Reduce the following differential equation to the variable separable form and hence solve: \(\frac{d y}{d x}\) = cos(x + y)
Answer:

Question 25.
Find the probability distribution of number of tails in the simultaneous tosses of three coins.
Answer:
When three coins are tossed simultaneously, the sample space is
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Let X represent the number of taiLs.
It can be seen that X can take the value of 0, 1, 2 or 3
P(X = 0) = P(HHH) = \(\frac{1}{8}\)
P(X = 1) = P(HHT) + P(HTH) + P(THH) = \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) = \(\frac{3}{8}\)
P(X = 2) = P(HTT) + P(THT) + P(TTH) = \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) = \(\frac{3}{8}\)
P(X = 3) P(TTT) = \(\frac{1}{8}\)
Thus, the probability distribution is as follows.

Question 26.
In a large school 80% of the pupil like Mathematics. A visitor to the school asks each of 4 pupils, chosen at random, whether they like Mathematics. Calculate the’ probabilities of obtaining an answer from 0, 1, 2, 3, 4 of the pupils.
Answer:
Let X = number of pupils like Mathematics
P = probability that pupils like Mathematics

Section D

Attempt any FIVE of the following questions: (20)

Question 27.
Using the truth table prove the following logical equivalence, (p ∨ q) → r ≡ (p → r) ∧ (q → r)
Answer:

Question 28.
In ∆ABC, if cot A, cot B, cot C are in A.P., then show that a², b², c² are also in A.P.
Answer:
By the sine rule,
\(\frac{\sin A}{a}\) = \(\frac{\sin B}{b}\) = \(\frac{\sin C}{c}\) = k
∴ sin A = ka, sin B = kb, sin C = kc
Now, cot A, cot B, cot C are in A.P.
∴ cot C – cot B = cot B – cot A
⇒ cot A + cot C = 2 cot B

Question 29.
If \(\bar{a}\) = \(\hat{i}\) -2\(\hat{j}\), \(\bar{b}\) = \(\hat{i}\) + 2\(\hat{j}\), \(\bar{c}\) = 2\(\hat{i}\) + \(\hat{j}\) – 2\(\hat{k}\), then find

(i) \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\))
(ii) (\(\bar{a}\) × (\(\bar{b}\)) × \(\bar{c}\). Are the results same? Justify.
Answer:
(i)

(ii)

Question 30.
Find the coordinates of the foot of the perpendicular drawn from the point 2\(\hat{i}\) – \(\hat{j}\) + 5\(\hat{k}\) to the line r = (11\(\hat{i}\) – 2\(\hat{j}\) – 8\(\hat{k}\)) + λ(1o\(\hat{i}\) – 4\(\hat{j}\) – 11\(\hat{k}\)). Also find the length of the perpendicular.
Answer:

Hence, the coordinates of the foot of perpendicular are
(1, 2, 3) and length of perpendicular = \(\sqrt{14}\) units.

Question 31.
Find the second order derivatives of the following:
e^4x. cos 5x
Answer:

Question 32.
Find the maximum and minimum values of the function
f(x) = cos² x + sin x.
Answer:
f(x) = cos²x + sin x
∴ f'(x) = \(\frac{d}{d x}\)(cos²x + sin x)
= 2cosx.\(\frac{d}{d x}\)(cosx) + cosx
= 2 cosx(-sinx) + cosx
= -sin 2x. + cos x
and f”(x) = \(\frac{d}{d x}\)(-sin2x) + cosx
= -cos2x.\(\frac{d}{d x}\)(2x) – sinx
= -cos 2x × 2 – sin x
= -2 cos 2x – sin x
For extreme values of f(x), f'(x) = 0
⇒ -sin 2x + cos x = 0
⇒ -2 sin x cos x + cos x = 0
⇒ cosx(-2sin x + 1) = 0
⇒ cos x = 0 or -2sinx + 1 = 0
⇒ cos x = cos\(\frac{\pi}{2}\) or sinx = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\)
∴ x = \(\frac{\pi}{2}\) or x = \(\frac{\pi}{6}\)

(i) f”(\(\frac{\pi}{2}\)) = -2cosπ – sin\(\frac{\pi}{2}\)
= -2 (-1) – 1 = 1 > 0
By the second derivative test, f is minimum at x = \(\frac{\pi}{6}\)
and minimum value of f at x = \(\frac{\pi}{6}\) is

Hence, the maximum and minimum values of the function f(x) are \(\frac{5}{4}\) and 1 respectively.

Question 33.
Evaluate the following integral:
\(\int \frac{7 x+3}{\sqrt{3+2 x-x^2}} d x\)
Answer:

Question 34.
Solve the following differential equation:
\(\frac{\cos ^2 y}{x}\)dy + \(\frac{\cos ^2 x}{y}\)dx = 0
Answer:

Integrating both sides, we get
∫xdx + ∫ydy + ∫xcos2x dx + ∫ycos2ydy = c₁ …….(i)
Using integration by parts

Multiplying throughout by 4, this becomes
2x² + 2y² + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c₁
∴ 2(x² + y²) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0
where c = -4c₂
This is the general solution.

Maharashtra Board Class 12 Maths Previous Year Question Papers