Maharashtra Board Class 12 Maths Sample Paper Set 2 with Solutions

Maharashtra State Board Class 12th Maths Sample Paper Set 2 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Maths Model Paper Set 2 with Solutions

Section A

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

(i) If p ∧ q is F. p → q is F then the truth values of p and q are ………………
(a) T.T
(b) T.F
(c) ET
(d) EF
Answer:
(b) T.F

(ii) The principal solutions of equation cot θ = √3 are:
(a) \(\frac{\pi}{6} \cdot \frac{7 \pi}{6}\)
(b) \(\frac{\pi}{6}, \frac{5 \pi}{6}\)
(c) \(\frac{\pi}{6}, \frac{8 \pi}{6}\)
(d) \(\frac{7 \pi}{6}, \frac{\pi}{3}\)
Answer:
(a) \(\frac{\pi}{6} \cdot \frac{7 \pi}{6}\)

(iii) If acute angle between tines ax² + 2hxy + by² = 0 is, π/4, then 4h² =
(a) a² + 4ab + b²
(b) a² + 6ab + b²
(c) (a + 2b)(a + 3b)
(d) (a – 2b)(2a + b)
Answer:
(b) a² + 6ab + b²

(iv) The direction ‘ratios of the line which is perpendicular to the two lines \(\frac{x-7}{2}\) = \(\frac{y+17}{-3}\) = \(\frac{z-6}{1}\) and \(\frac{x+5}{1}\) = \(\frac{y+3}{2}\) = \(\frac{z-4}{-2}\) are:
(a) 4,5, 7
(b) 4, – 5, 7
(c) 4, – 5, – 7
(d) – 4, 5,8
Answer:
(a) 4,5, 7

v) If LPP has two optimal solutions, then it has infinite number of optimal solutions

(vi) Let f(x) and g(x) be differentiable for 0 ≤ x ≤ 1 such that f(0) = 0, g(0) = 0. f(1) = 6. Let there exists a real number c in (0,1) such that f'(c) = 2g'(c). Then the value of g(1) must be:
(a) 1
(b) 3
(c) 2.5
(d) -1
Answer:
(b) 3

(vii) \(\int_0^{\frac{\pi}{2}} \sin ^6 x \cos ^2 x . d x\) =
(a) \(\frac{7 \pi}{256}\)
(b) \(\frac{3 \pi}{256}\)
(c) \(\frac{5 \pi}{256}\)
(d) \(\frac{-5 \pi}{256}\)
Answer:
(c) \(\frac{5 \pi}{256}\)

(viii) The area bounded by the curve y = x³, the X-axis and the lines x = – 2 and x = 1 is:
(a) -9 sq. units
(b) \(\frac{-15}{4}\) sq. units
(c) \(\frac{15}{4}\) sq.units
(d) \(\frac{17}{4}\)sq.units
Answer:
(c) \(\frac{15}{4}\) sq.units

Question 2.
Answer the following questions:

(i) Check whether the following matrix is invertible or not : \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, A^-1 exists.

(ii) State whether the equation 2 sin θ = 3 has a solution or not?
Solution:
2 sin θ = 3
∴ sin θ = \(\frac{3}{2}\)
This is not possible because – 1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iii) Elvaluate: \(\int_0^{\frac{\pi}{4}} \cot ^2 x \cdot d x\)
Solution:
\(\int_0^{\frac{\pi}{4}} \cot ^2 x \cdot d x\)

(iv) Determine the order and degree of the foll.owirig differential. equation:
\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
Solution:
The given Differential equation is
\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
The D.E. has highest order derivative — with power 2.
∴ \(\left(\frac{d y}{d x}\right)^2\) = 2 sin x + 3
The D.E has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ The given D.E. is of order 1 and degree 2.

Section B
Attempt any EIGHT of the following questions:

Question 3.
Examine whether the following statement pattern is a tautology or o contradiction or a .contingency.
(p V q) → (q ∧ p).
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → (q V p) is a tautology.

Question 4.
Transform \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\) into an upper trianguLar matrix by suitable row transformations.
Solution:
Given matrix is :
\(\left[\begin{array}{ccc}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\)

Question 5.
Find k, if the sum of the slopes of the lines represented by x² + kxy – 3y² = 0 is twice their product.
Answer:
Comparing the equation x² + kxy- 3y2 = 0 with ax² + 2hxy – by² = 0, we get a = 1.2h = k. b = – 3.
Let m₁ and m₂ be the slopes of the lines represented by x² + kxy – 3y2 = 0
∴ m₁ + m₂ = \(\frac{-2h}{b}\) = \(– \frac{k}{-3}\) = \(\frac{k}{3}\)
and m₁ m₂ = \(\frac{a}{b}\) = \(\frac{1}{-3}\) = \(– \frac{1}{3}\)
Now, m₁ + m₂ = 2(m₁ m₂)
∴ \(\frac{k}{3}\) = 2(\(\frac{1}{3}\))
∴ k = -2

Question 6.
Find the separate equation of the tine represented bg the following equation: 3x² – 2√3xy – 3y² = 0.
Answer:
3x² – 2√3xy – 3y² = 0
∴ 3x² – 3√3xy + √3xy – 3y² = 0
∴ 3x(x – √3y) + √3y(x – √3y) = 0
∴ (x – √3y)(3x +√3y) = 0
The separate equations of the lines are
x – √3y = 0 and 3x + √3y = 0

Question 7.
If \(\overrightarrow{A B}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) and initial. Point A (1, 5, 0). Find the terminal. point B.
Answer:
Let \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) be the position vectors of A and B.
Given :A(1,5.0)
a = \(\hat{i}+5 \hat{j}\)
Now, \(\overrightarrow{A B}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)
\(\overline{\mathrm{b}}\) – \(\overline{\mathrm{a}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)
\(\overline{\mathrm{b}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) + \(\overline{\mathrm{a}}\)
= \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) + \(\hat{i}+5 \hat{j}\)
Hence, the terminal points is B(3,1,7).

Question 8.
Find the vector equation of line passing through the point having position vector \(5 \hat{i}+4 \hat{j}+3 \hat{k}\) and having direction ratios -3, 4, 2.
Answer:
Let A be the point whose position vector is a = \(5 \hat{i}+4 \hat{j}+3 \hat{k}\)
Let \(\overline{\mathrm{b}}\) be the vector parallel to the line having direction ratios = – 3,4,2
Then, b = \(-3 \hat{i}+4 \hat{j}+2 \hat{k}\)
The vector equation of the line passing through A(\(\overline{\mathrm{a}}\)) and parallel to \(\overline{\mathrm{b}}\) is \(\overline{\mathrm{r}}\) = \(\overline{\mathrm{a}}\) + λ\(\overline{\mathrm{b}}\) , where λ is a scalar.
∴ The required vector equation of the line is \(\overline{\mathrm{r}}\) = \(5 \hat{i}+4 \hat{j}+3 \hat{k}\) + λ(\(-3 \hat{i}+4 \hat{j}+2 \hat{k}\))

Question 9.
Find the Cartesian equations of the Line passing through A(2,2,1) and B(1, 3,0).
Answer:
The cartesian equations of the line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are:
\(\frac{x-x_1}{x_2-x_1}\) = \(\frac{y-y_1}{y_2-y_1}\) = \(\frac{z-z_1}{z_2-z_1}\)
Here, (x₁, y₁, z₁) = (2, 2,1) and (x₂, y₂, z₂) = (1,3,0)
The required cartesian equation is
\(\frac{x-2}{1-2}\) = \(\frac{y-2}{3-2}\) = \(\frac{z-1}{0-1}\)
i.e., \(\frac{x-2}{-1}\) = \(\frac{y-2}{1}\) = \(\frac{z-1}{-1}\)

Question 10.
Solve graphically: x ≥ 0 and y ≥ 0.
Answer:
Consider the lines whose equations are x = 0, y = 0.
These represents the equations of Y-axis and X-axis respectively, which divide the plane into four parts.
Since x ≥ 0, y ≥ 0, the solution set is in the first quadrant which is shaded in the graph.

Question 11.
Differentiate the following w.r.t x. : cos (x² + a²)
Answer:
Let y = cos (x² + a²)
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [cos (x² + a²)]
= -sin(x² + a²).(2x + 0)
= -2x sin(x²+a²).

Question 12.
Find the area of the ellipse \(\frac{x^2}{25}\) + \(\frac{y^2}{16}\) = 1 using integration.
Answer:
By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO. Clearly for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
Img 5
In the first quadrant y > 0
∴ y = \(\frac{4}{5}\) \(\sqrt{25-x^2}\)
∴ Area of the ellipse = 4 (Area of the region OABO)
Img 6

Question 13.
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
Tablee 1
Answer:
P.m.f. of random variable should satisfy the following conditions:
(a) 0 < p, < 1
(b) Σp_i= 1
Tablee 2
P(X = 3) = – 0.1, i.e., p_i < 0 which does not satisfy 0 ≤ p_i ≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

Question 14.
In a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Answer:
This is a problem of Bernoulli trials. As for each question, there are 2 outcomes. Let X represent the number of correct answers by guessing in the set of 5 multiple- choice questions.
Probability of getting a correct answer is, p = \(\frac{1}{3}\)
q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{1}{2}\)
Clearly, X has a binomial distribution with n = 5
and p = \(\frac{1}{3}\)
The p.m.f of X is given by
P(X = x) = ⁿC_x p^xq^n-x, x = 0,1,2,4,5
i.e., p(x) = ⁿC_x (\(\frac{1}{3}\))^x (\(\frac{2}{3}\))^5-x x = 0,1,2,4,5
P(four or more correct answers) = P[X ≥ 4] = P(4) + P(5)
Img 7
Hence,’the probability of getting four or more correct answer is \(\frac{11}{243}\)

Section C
Attempt any EIGHT of the followings questions:

Question 15.
Find the matrix of the co-factors for the following matrix.
\(\left[\begin{array}{ccc}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ccc}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Here,
Img 8

Question 16.
Find the general solution of the following equation:
cos θ + sin θ = 1.
Answer:
cos θ + sin θ = 1
∴ sin θ = 1 – cos θ
Img 9
The general solution of sin θ = 0 is θ = nπ, n ∈ Z and tan θ = tan α is 0 = nπ + α, n ∈ Z
The required general solution is
\(\frac{θ}{2}\) = nπ, n ∈ Z or \(\frac{θ}{2}\) = nπ + \(\frac{π}{2}\), n ∈ Z
i.e., θ = 2nπ, n ∈ Z or θ = 2njt + \(\frac{\pi}{2}\), n ∈ Z

Question 17.
Show that the following points are collinear: P = (4,5,2), Q = (3,2,4), R = (5,8,0).
Answer:
Let a, b, c be position vectors of the points.
P = (4,5,2), Q = (3,2,4), R = (5,8,0) respectively.
Then \(\overline{\mathrm{a}}\) = \(4 \hat{i}+5 \hat{j}+2 \hat{k}\), \(\bar{b}\) = \(3 \hat{i}+2 \hat{j}+4 \hat{k}\), \(\bar{c}=5 \hat{i}+8 \hat{j}+0 \hat{k}\)
AB = \(\bar{b}\) – \(\bar{a}\)
= \(3 \hat{i}+2 \hat{j}+4 \hat{k}\) – \(4 \hat{i}+5 \hat{j}+2 \hat{k}\)
= \(– \hat{i}-3 \hat{j}+2 \hat{k}\)
= -(\(\hat{i}+3 \hat{j}-2 \hat{k}\)) …(1)
and BC = \(\bar{c}\) – \(\bar{b}\)
= \(5 \hat{i}+8 \hat{j}+0 \hat{k}\) – \(3 \hat{i}+2 \hat{j}+4 \hat{k}\)
= \(2 \hat{i}+6 \hat{j}-4 \hat{k}\)
= 2(\(\hat{i}+3 \hat{j}-2 \hat{k}\))
= 2.AB ………….[By (1)]
∴ BC is a non-zero scalar multiple of AB
∴ They are parallel to each other.
But they have point B in common.
∴ BC and AB are collinear vectors.
Hence, the points A, B and C are collinear.

Question 18.
Find the position vector of point R which divides the line joining the points P and Q whose position vectors are \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) in the ratio 3:2 is internally.
Answer:
It is given that the points P and Q have position vectors
\(\bar{p}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\bar{q}\) =-\(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) respectively.
If R(\(\bar{r}\)) divides the line segment PQ internally in the ratio 3:2, by section formula for internal division.
\(\bar{r}\) = \(\frac{3 q+2 p}{3+2}\)
Img 10
coordinates of R = \(\left(-\frac{11}{5}, \frac{4}{5},-\frac{9}{5}\right)\)
Hence, the position vector of R is \(\frac{1}{5}\)(\(-11 \hat{i}+4 \hat{j}-9 \hat{k}\)) and the coordinates of R are \(\left(-\frac{11}{5}, \frac{4}{5},-\frac{9}{5}\right)\)

Question 19.
Differentiate the following w.r.t. x : \(x^{\tan ^{-1} x}\)
Answer:
Let y = \(x^{\tan ^{-1} x}\)
Then log y = log(\(x^{\tan ^{-1} x}\)) = (tan^-1 x) (log x)
Differentiating both sides w.r.t. x, we get
Img 11

Question 20.
Find the equations of tangents and normals to the following curves at the indicated points on them : x³ + y³ – 9xy = 0 at (2,4)
Solution:
x³ + y³ – 9xy = 0
Differentiating both sides w.r.t. x, we get
Img 12
= Slope of the tangent at (2,4)
∴ The equation of the tangent at (2,4) is
y – 4 = \(\frac{4}{5}\)(x – 2)
∴ 5y – 20 = 4x – 8
4x – 5y +12 = 0
The slope of normal at (2,4) = \(\frac{4}{5}\)
= \(-\frac{5}{4}\)
∴ The equation of the normal at (2,4) is
y – 4 = \(-\frac{5}{4}\)(x-2)
∴ 4y – 16 = -5x + 10
∴ 5x + 4y – 26 = 0
Hence, the equation of tangent and normal are 4x – 5y + 12 = 0 and 5x – 4y – 26 = 0 respectively.

Question 21.
Evaluate the following integrals: \(\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x\)
Answer:
Img 13

Question 22.
Integrate the following function w.r.t. :
Answer:
Img 14
= log |cosec t – cot t| + c
= log |cosec (x + log x) – cot (x + log x)| + c.

Question 23.
Evaluate the following integral as limit of a sum:
Answer:
Let f[x) = 3x² – 1, for 0 ≤ x ≤ 2.
Divide the closed interval [0,2] into n subintervals each of length h at the points.
0,0 + h, 0+ 2h,…..,0 + rh,……..,0+nh = 2
i.e. 0, h, 2h, …,rh,…….,nh = 2
∴ h = \(\frac{2}{n}\) and as n → oo, h → 0
Here, a = 0
∴ f(a + rh) = f(0 + rh) = f(rh) = 3(rh)² – 1 = 3r²h² – 1
Img 15
= 4(1+0)(2+0) – 2
= 8 – 2 = 6

Question 24.
Solve the following differential equation:
\(x \sin \left(\frac{y}{x}\right) d y\) = \(\)
Answer:
Img 16
Intergrating both sides, we get
\(\int \sin v d v\) = \(-\int \frac{1}{x} d x\)
∴ -cos v = -log x – c
cos(\(\frac{y}{x}\)) = log x + c
This general solution.

Question 25.
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than 4 appears on at least one die.
Answer:
When a die is tossed twice, the sample space S has 6×6 = 36 sample points.
∴ n(S) = 36
Trial will be a success if the number on at least one die is 5 or 6.
Let X denote the number of dice on which 5 or 6 appears. Then X can take values 0,1,2,
When X = 0 i.e., 5 or 6 do not appear on any of the dice, then
X = {(1,1), (1, 2), (1, 3), (1, 4), (2,1), (2, 2), (2, 3). (2, 4), (3,1), (3, 2), (3, 3), (3,4). (4,1), (4, 2), (4. 3), (4,4)}.
∴ n(X) = 16.
∴ P(X = 0) = \(\frac{n(X)}{n(S)}\) = \(\frac{16}{36}\) = \(\frac{4}{9}\)
When X = 1, i.e., 5 or 6 appear on exactly one of the dice, then
X = {(1, 5), (1, 6), (2. 5), (2, 6), (3, 5). (3, 6), (4, 5), (4, 6), (5,1), (5, 2), (5, 3), (5,4), (6,1). (6.2). (6.3), (6,4)}
∴ n(X) =16
∴ P(X = 1) = \(\frac{n(X)}{n(S)}\) = \(\frac{16}{36}\) = \(\frac{4}{9}\)
When X = 2, i.e., 5 or 6 appear on both of the dice, then X = {(5,5), (5, 6), (6, 5), (6, 6)}
∴ n(X) = 4.
∴ P(X = 1) = \(\frac{n(X)}{n(S)}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

Question 26.
In a large school, 80% of pupil like mathematics. A visitor. to the school asks each of 4 pupils, chosen at random, whether they like mathematics.
Find the probability that the visitor obtains answer yes from at least 2 pupils:
(a) when the number of pupils questioned remains at 4.
(b) when the number of pupils questioned is increased to 8.
Answer:
Let X = number of pupils like Mathematics
p = probability that pupils like Mathematics
∴ p = 80% = \(\frac{80}{100}\) = \(\frac{4}{5}\)
and q = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Given, n = 4
∴ X – B(4, \(\frac{4}{5}\))
The p.m.f of X is given by
P(X = x) = ⁿC_x p^xq^n-x
i.e., p(x) = ⁴C_x (\(\frac{4}{5}\))^x (\(\frac{1}{5}\))^4-x

Section D
Attempt any FIVE of the following questions:

Question 27.
Using the truth table prove the following logical equivalence. – (p v q)v (~ p ∧ q) = ~ p
Solution:
Img 17
The entries in columns 3 and 7 are identical.
∴ ~ (\((p \vee q)\)) v (- p ∧ q) = ~ p

Question 28.
In ∆ABC prove that (b + c – a) tan A/2 = (c + a – b) tan B/2 = (a + b – c) tan C/2.
Solution:
(b + c – a) tan A/2
Img 18

Question 29.
Express \(-\hat{i}-3 \hat{j}+4 \hat{k}\) as the linear combination of the vectors \(2 \hat{i}+\hat{j}-4 \hat{k}\), \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(3 \hat{i}+\hat{j}-2 \hat{k}\).
Solution:
Let \(\bar{a}\) = \(2 \hat{i}+\hat{j}-4 \hat{k}\)
\(\bar{b}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\)
\(\bar{c}\) = \(3 \hat{i}+\hat{j}-2 \hat{k}\)
\(\bar{p}\) = \(-\hat{i}-3 \hat{j}+4 \hat{k}\)
Suppose \(\bar{p}\) = x\(\bar{a}\) + y\(\bar{b}\) + z\(\bar{c}\)
Then,
\(-\hat{i}-3 \hat{j}+4 \hat{k}\) = x(\(2 \hat{i}+\hat{j}-4 \hat{k}\)) + y(\(2 \hat{i}-\hat{j}+3 \hat{k}\)) + z(\(3 \hat{i}+\hat{j}-2 \hat{k}\))
\(-\hat{i}-3 \hat{j}+4 \hat{k}\) = \((2 x+2 y+3 z) \hat{i}\) + \((x – y + z) \hat{j}\) + \((-4 x+3 y-2 z) \hat{k}\)
By equality of vectors
2 x+2 y+3 z = -1
x – y + z = -3
-4 x+3 y-2 z = 4
We have to solve these equations by using Cramer’s Rule.
Img 19
Img 20
\(\bar{p}\) = 2\(\bar{a}\) + 2\(\bar{b}\) – 3\(\bar{c}\)

Question 30.
Find the values of λ so that the lines \(\frac{1-x}{3}\) = \(\frac{7 y-14}{\lambda}\) = \(\frac{z-3}{2}\) and \(\frac{7-7 x}{3 \lambda}\) = \(\frac{y-5}{1}\) = \(\frac{6-z}{5}\) are the right angles.
Solution:
The equations of the given lines are
\(\frac{1-x}{3}\) = \(\frac{7 y-14}{\lambda}\) = \(\frac{z-3}{2}\)
and \(\frac{7-7 x}{3 \lambda}\) = \(\frac{y-5}{1}\) = \(\frac{6-z}{5}\)
Equation (1) can be written as:
\(\frac{-(x-1)}{3}\) = \(\frac{7(y-2)}{2 \lambda}\) = \(\frac{z-3}{2}\)
i.e., \(\frac{x-1}{-3}=\frac{y-2}{\frac{2 \lambda}{7}}=\frac{z-3}{2}\)
The direction ratios of this line are
a₁ = -3, b₁ = \(\frac{2 \lambda}{7}\), c₁ = 2
Equation (2) can be written as:
\(\frac{-7(x-1)}{3 \lambda}=\frac{y-5}{1}=\frac{-(z-6)}{5}\)
The direction ratios of this line are
a₂ = \(\frac{-3 \lambda}{7}\) , b₂ = 1, c₂ = – 5
since the lines (1) and (2) are at right angles,
a₁a₂ + b₁b₂ + c₁c₂ = 0
∴ (-3) (\(\frac{-3 \lambda}{7}\)) + (\(\frac{2 \lambda}{7}\))(1) + 2(-5) = 0
∴ (\(\frac{9 \lambda}{7}\)) + (\(\frac{2 \lambda}{7}\)) – 10 = 0
∴ \(\frac{11 \lambda}{7}\) = 10
∴ λ = \(\frac{70}{11}\).

Maharashtra Board Class 12 Maths Previous Year Question Papers