Maharashtra Board Class 12 Maths Sample Paper Set 4 with Solutions

Maharashtra State Board Class 12th Maths Sample Paper Set 4 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Maths Model Paper Set 4 with Solutions

Time : 3 Hrs.
Max. Marks : 80

General Instructions:
The question paper is divided into Four sections.

1. Section A: Q. 1 contains Eight multiple choice type of questions, each carrying Two marks.
   Q. 2 contains Four very short answer type questions, each carrying One mark.
2. Section B: Q. 3 to Q. 14 contain Twelve short answer type questions, each carrying Two marks. (Attempt any Eight)
3. Section C: Q. 15 to Q. 26 contain Twelve short answer type questions, each carrying Three marks. (Attempt any Eight)
4. Section D: Q. 27 to Q. 34 contain Eight long answer type questions, each carrying Four marks. (Attempt any Five)
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. Use of graph paper is not necessary. Only rough sketch of graph is expected.
8. For each multiple choice type of question, only the first attempt will be considered for evaluation.
9. Start answer to each section on a new page.

Section – A

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

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)

Explanation:

~[p ∧ (q → r)]
≡ p ∨ (q → r)
≡ p ∨ (q ∧ ~r)
…………[~(p→q) ≡ p ∧~q]

ii. If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) and A(adj A) = k I, then the value of k is
(A) 2
(B) -2
(C) 10
(D) -10
Answer:
(B) -2

Explanation:

A(adj A) = |A|I, where I is a unit matrix.
|A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right|\) = 4 – 6 = -2

iii. If cos α, cos β, cos γ are the direction cosines of a line, then the value of sin² α + sin² β + sin² γ is
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(B) 2

Explanation:

sin²α + sin²β + sin²γ = 1 – cos²α – cos²β + 1 – cos²γ
= 3 – (cos²α + cos²β + cos²γ)
= 3 – 1 = 2

iv. The angle between the planes \(\overline{\mathbf{r}}\)-(\(\hat{\mathrm{i}}\) – 2\(\hat{\mathrm{j}}\) + 3\(\hat{\mathrm{k}}\)) + 4 = 0 and \(\overline{\mathbf{r}}\)⋅2\(\hat{\mathrm{i}}\) + \(\hat{\mathrm{j}}\) – 3\(\hat{\mathrm{k}}\)) + 7 = 0 is
(A) \(\frac{\pi}{2}\)
(B) \(\frac{\pi}{3}\)
(C) cos^-1\(\left(\frac{3}{4}\right)\)
(D) cos^-1 \(\left(\frac{9}{14}\right)\)
Answer:
(D) cos^-1\(\left(\frac{9}{14}\right)\)

Explanation:

v. \(\int \frac{\mathrm{e}^x(x-1)}{x^2}\)dx =
(A) \(\frac{\mathrm{e}^x}{x}\) + c
(B) \(\frac{\mathrm{e}^x}{x^2}\) + c
(C) (x – \(\frac{1}{x}\))e^x
(D) x e^-x + c (2)
Answer:
(A) \(\frac{e^x}{x}\) + c

Explanation:

vi. The area bounded by y = Vx and line x = 2y + 3, X-axis in first quadrant is
(A) 2\(\sqrt{3}\) sq. units
(B) 9 sq. units
(C) \(\frac{34}{3}\) sq. units
(D) 18 sq. units
Answer:
(B) 9 sq. units

Explanation:

vii.
\(\frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}\) = 1 is a solution of _______
(A) \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) + yx + \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\) = 0
(B) xy\(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) + x\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\) – y\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = 0
(C) y\(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) + 2\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\) + y = 0
(D) xy\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) + y\(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) = 0 (2)
Answer:
(B) xy\(\frac{d^2 y}{d x^2}\) + x\(\left(\frac{d y}{d x}\right)^2\) – y\(\frac{d y}{d x}\) = 0

Explanation:

viii. If p.m.f. of a d.r.v. X is P(x) = \(\frac{\mathrm{c}}{x^3}\), for x = 1, 2, 3 and = 0, otherwise (elsewhere) then E (X ) =
(A) \(\frac{343}{297}\)
(B) \(\frac{294}{251}\)
(C) \(\frac{297}{294}\)
(D) \(\frac{294}{297}\) (2)
Answer:
(B) \(\frac{294}{251}\)

p.m.f. or d.r.v. X is given
∴ P(1) + P(2) + P(3) = 1
∴ \(\frac{c}{(1)^3}+\frac{c}{2^3}+\frac{c}{3^3}\) = 1
∴ c = \(\frac{216}{251}\)
∴ E(X) = \(\sum_{x=1}^n x \cdot P(x)\)
= \(\frac{216}{251}\left[\frac{1}{(1)^3}+\frac{2}{(2)^3}+\frac{3}{(3)^3}\right]\)
= \(\frac{294}{251}\)

Question 2.
Answer the following: [4 Marks]

i. If the statements p, q are true statements and r, s are false statements, then determine the truth value of the following: (1)
p ∨ (q ∧ r)
Answer:
p ∨ (q ∧ r) ≡ T ∨(T ∧ F)
≡ T ∨ F
≡ T
Hence, truth value is T.

ii. In ∆ABC, if ∠A = 30°, ∠B = 60° and b = \(\sqrt{3}\), then find a. (1)
Answer:
By sine rule, \(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\)
∴ \(\frac{a}{\sin 30^{\circ}}\) = \(\frac{\sqrt{3}}{\sin 60^{\circ}}\)
∴ \(\frac{a}{\frac{1}{2}}\) = \(\frac{\frac{\sqrt{3}}{\sqrt{3}}}{2}\)
∴ a = 1

iii. If x = t², y = t³, find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\). (1)
Answer:
x = t² and y = t³
∴ \(\frac{d x}{d t}\) = 2t and \(\frac{d y}{d t}\) = 3t²
∴ \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 t^2}{2 t}\)
= \(\frac{3}{2}t\)

iv. If X ~ B (n, p) where n = 3, p = \(\frac{1}{2}\), q = \(\frac{1}{2}\), then find P(X = 0). (1)
Answer:
P(X = x) = ⁿC_xp^xq^n-x
∴ P(X = 0) = ³C₀\(\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^3\) = \(\frac{1}{8}\)

Section – B

Attempt any EIGHT of the following questions: [16 Marks]

Question 3.
Construct the truth table for each of the following statement patterns.
[(p → q) ∧ q] → p (2)
Answer:
[(p → q) ∧ q] → p

Question 4.
If sin(sin^-1\(\frac{1}{5}\) + cos^-1x) = 1, then find the value of x.
Answer:

Question 5.
Find the acute angle between lines represented by 3x² + 2xy – y² = 0. (2)
Answer:
Given equation of the lines is 3x² + 2xy – y² = 0.
Comparing with ax² + 2hxy + by² = 0, we get
a = 3, b = 1 and b = -1
Let θ be the acute angle between the lines.
∴ tan θ = \(\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|\) = \(\left|\frac{2 \sqrt{1^2-(3)(-1)}}{3-1}\right|\) = \(\left|\frac{2 \sqrt{1+3}}{2}\right|\) = 2
∴ θ = tan^-11(2)

Question 6.
If the vectors 3\(\hat{\mathrm{i}}\) + 5\(\hat{\mathrm{k}}\), 4\(\hat{\mathrm{i}}\) + 2\(\hat{\mathrm{j}}\) – 3\(\hat{\mathrm{k}}\) and 3\(\hat{\mathrm{i}}\) + \(\hat{\mathrm{j}}\) + 4\(\hat{\mathrm{k}}\) are coterminous edges of the parallelopiped, then find the volume of the parallelopiped.
Answer:
Let \(\bar{a}\) = 3\(\hat{\mathbf{i}}\) + 5\(\hat{\mathbf{k}}\), \(\bar{b}\) = 4\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – 3\(\hat{\mathbf{k}}\) and \(\bar{c}\) = 3\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{j}}\) + 4\(\hat{\mathbf{k}}\)
∴ Volume of parallelepiped = \(\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\)
= \(\left|\begin{array}{ccc}
3 & 0 & 5 \\
4 & 2 & -3 \\
3 & 1 & 4
\end{array}\right|\)
= 3(8 + 3) – 0(16 + 9) + 5(4 – 6)
= 3(11) – 0 + 5(-2) = 33 – 10 = 23
∴ The volume of the parallelopiped is 23 cubic units.

Question 7.
The position vectors of points A and B are 6\(\overline{\mathrm{a}}\) + 2\(\overline{\mathrm{b}}\) and \(\overline{\mathrm{a}}\) – 3\(\overline{\mathrm{b}}\) respectively. If the point C divides AB in the ratio 3 : 2, then show that the position vector of C is 3\(\overline{\mathrm{a}}\) – \(\overline{\mathrm{b}}\). (2)
Answer:
Let \(\bar{c}\) be the position vector of point C.
The position vector of point A is 6\(\bar{a}\) + 2\(\bar{b}\) and the position vector of point B is \(\bar{a}\) – 3\(\bar{b}\).
Point C divides AB in the ratio 3 : 2.
∴ By using section formula,

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

Question 9.
Differentiate x sin x w. r. t. tan x.
Answer:
Let u = x sinx
Differentiating w. r. t. x, we get
\(\frac{d u}{d x}\) = x\(\frac{d}{d x}\)(sin x) + sin x\(\frac{d}{d x}\)(x) = x cos x + sin x
Let v = tan x
Differentiating w. r. t. x, we get
\(\frac{d v}{d x}\) = \(\frac{d}{d x}\)(tan x) = sec² x
∴ \(\frac{d u}{d v}\) = \(\frac{\left(\frac{d u}{d x}\right)}{\left(\frac{d v}{d x}\right)}\) = \(\frac{x \cos x+\sin x}{\sec ^2 x}\)

Question 10.
Find the equation of tangent to the curve y = 2x³ – x² + 2 at the point (\(\frac{1}{2}\), 2). (2)
Answer:
Equation of the curve is y = 2x³ – x² + 2
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = 6x² – 2x
Slope of the tangent at (\(\frac{1}{2}\), 2) is

Question 11.
Evaluate: \(\int \sqrt{1+\sin 2 x}\) dx
Answer:

Question 12.
Evaluate: \(\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^2 x d x\)
Answer:

Question 13.
Find the area bounded by the curve y = -x², X-axis and lines x = 1 and x = 4.
Answer:

Question 14.
Evaluate: \(\int \frac{10 x^9+10^x \log 10}{10^x+x^{10}}\)
Answer:
Let I = \(\int \frac{10 x^9+10^x \log 10}{10^x+x^{10}}\)dx
Put x¹⁰ + 10^x = t
Differentiating w.r.t. x, we get
(10x⁹ + 10^x log 10)dx = dt
∴ I = \(\int \frac{d t}{t}\) = log |t| + c
∴ I = log |x¹⁰ + 10^x| + c

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
Construct the switching circuit of the following: (p ∧ q) ∨ (~ p) ∨ (p ∧ ~q)
Answer:
Let p : The switch S₁ is closed.
q : The Switch S₂ is closed.
~p : The Switch S₁‘ is closed.
~q : The Switch S₂‘ is closed.
The statement (p ∧ q) represents a circuit in which S₁ and S₂ are connected in series.
The statement (p ∧ ~q) represents a circuit in which Si and S₂ are connected in series.
The given statement [(p ∧ q) ∨ (~ p) ∨ (p ∧ ~q)] represent the circuits corresponding to (p ∧ q), (~p) and (p ∧ ~q) are connected in parallel.
Hence, the circuit diagram is

Question 16.
Find the general solution of tan³ θ = 3tan θ.
Answer:
tan³ θ = 3 tan θ
∴ tan³ θ – 3 tan θ = 0
∴ tan θ(tan² θ – 3) = 0
∴ tan θ = 0 or tan² θ = 3 = \((\sqrt{3})^2\)
∴ tan θ = 0 or tan²θ = \(\left(\tan \frac{\pi}{3}\right)^2\)
∴ tan θ = 0 or tan² θ = \(\left(\tan \frac{\pi}{3}\right)^2\)
Since tan θ = 0 implies θ = nπ and tan² θ = tan² α implies θ = nπ ± α , n ∈ Z.
∴ θ = nπ or θ = nπ + \(\frac{\pi}{3}\)
∴ The required general solution is θ = nπ or nπ ± \(\frac{\pi}{3}\), where n ∈ Z.

Question 17.
Find the values of c so that for all real x the vectors xc \(\hat{\mathbf{i}}\) – 6\(\hat{\mathbf{j}}\) + 3\(\hat{\mathbf{k}}\) and x\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) + 2cx \(\hat{\mathbf{k}}\) make an obtuse angle. (3)
Answer:
Let \(\bar{a}\) = x\(\widehat{\mathrm{i}}\) – 6\(\widehat{\mathrm{j}}\) + 3\(\widehat{\mathrm{k}}\) and \(\bar{b}\) = x\(\widehat{\mathrm{i}}\) + 2\(\widehat{\mathrm{j}}\) + 2cx\(\widehat{\mathrm{k}}\)
Vector \(\bar{a}\) and \(\bar{b}\) make an obtuse angle.
i.e. cos θ < 0
But cos θ = \(\frac{\bar{a} \cdot \bar{b}}{|\bar{a} \cdot| \bar{b} \mid}\)
∴ \(\frac{\bar{a} \cdot \bar{b}}{|\vec{a}| \cdot|\bar{b}|}\) < 0
∴ \(\bar{a} \cdot \bar{b}\) < 0
∴ x²c + 6xc – 12 < 0
∵ It is quadratic equation,
ax² + bx + c < 0 For all x
∵ a < 0 and ∆ < 0 …(i)
∴ a = c, ∆ = b² – 4ac
= (6c)² – 4(c)(-12)
= 36c² + 48c
∴ From (i)
c < 0 and 36c² + 48c < 0
c < 0 and c(3c + 4) < 0
∴ \(\frac{-4}{3}\) < c< 0

Question 18.
Show that vector area of a quadrilateral ABCD is \(\frac{1}{2}(\overline{\mathrm{AC}} \times \overline{\mathrm{BD}})\), where AC and BD are its diagonals.
Answer:

Question 19.
By computing the shortest distance, determine whether following lines intersect each other.
\(\overline{\mathbf{r}}\) = (\(\hat{\mathbf{i}}\) – \(\hat{\mathbf{j}}\)) + λ(2\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{k}}\)) and \(\overline{\mathbf{r}}\) = (2\(\hat{\mathbf{i}}\) – \(\hat{\mathbf{j}}\)) + μ(\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\))
Answer:

Hence, the given lines do not intersect each other.

Question 20.
Find the distance of the point 4\(\hat{\mathbf{i}}\) – 3\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\) from the plane \(\overline{\mathbf{r}}\) ⋅ (2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – 6\(\hat{\mathbf{k}}\)) = 21. (3)
Answer:

Question 21.
If sin^-1\(\left(\frac{x^5-y^5}{x^5+y^5}\right)\) = \(\frac{\pi}{6}\), show that \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = \(\frac{x^4}{3 y^4}\).
Answer:

Question 22.
Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\), if y = tan^-1 (cosec x + cotx).
Answer:

Question 23.
The surface area of a spherical balloon is increasing at the rate of 2 cm²/ sec. At what rate the volume of the balloon is increasing when radius of the balloon is 6 cm?
Answer:
Let r be the radius, s be the surface area and V be the volume of the spherical balloon.
Then, \(\frac{\mathrm{ds}}{\mathrm{~d} t}\) = 2 cm² /sec, r = 6cm …[Given]
s = 4πr²
Differentiating w.r.t. t, we get
\(\frac{d s}{d t}\) = 4π(2r).\(\frac{d r}{d t}\)
∴ 2 = 8πr • \(\frac{d r}{d t}\)
∴ \(\frac{d r}{d t}\) = \(\frac{1}{4 \pi r}\) …… (i)
Now, V = \(\frac{4}{3} \pi r^3\)nr
Differentiating w.r.t. t, we get

Thus, the volume of the spherical balloon is increasing at the rate of 6 cm³/sec.

Question 24.
Reduce each of the following differential to the variable separable form and hence solve.
(x – y)²\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = a². (3)
Answer:
(x – y)²\(\frac{d y}{d x}\) = a² ……. (i)
Put x – y = u …… …(ii)
Differentiating w.r.t. x, we get

Question 25.
Solve the differential equation: \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) + y = e^-x. (3)
Answer:
\(\frac{d y}{d x}\) + y = e^-x
This equation is of the form \(\frac{d y}{d x}\) + Py = Q, where
P = 1 and Q = e^-x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int d x}\) = e^x
∴ Solution of the given equation is

Question 26.
For the following p. d. f. of r. y. X, find :
i. P (X < 1) and
ii. P(|X| < 1)
if f(x) = \(\frac{x^2}{18}\), for -3 < x < 3
= 0, otherwise
Answer:
i.

ii.

Section – D

Question 27.
If A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]\), find AB and (AB)^-1. Verify that (AB)^-1 = B^-1A^-1. (4)
Answer:

Question 28.
In ∆ABC, prove that tan \(\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)\) = \(\left(\frac{b-c}{b+c}\right)\)cot\(\frac{\mathrm{A}}{2}\) (4)
Answer:

Question 29.
Show that a homogeneous equation of degree 2 in x andy i.e., ax² + 2hxy + by² = 0 represents a pair of lines through the origin if h² – ab ≥ 0. (4)
Answer:

Consider the homogeneous equotion of degree two in x and y.
ax² + 2hxy + by² = 0 ……(i)
Consider two cases b O and b t O. These two cases are exhaustive.
Case I:
If b = 0, then equation (i) becomes ax² + 2hxy = 0
∴ x(ax + 2hy) = 0, which is the combined equation of lines x = 0 and ax + 2hy = 0
These lines pass through the origin.
Case II:
If b ≠ 0, then we multiply equation (i) by b.
abx² + 2hbxy + b²y² = 0
b²y² + 2hbxyz = -abx²
To make L.H.S. complete square we add h²x² to both sides.

From the above cases, we conclude that the equation ax² + 2hxy + by² = 0 represents a pair of lines passing through the origin if h² – ab ≥ 0.

Question 30.
Maximize: Z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥0, y > 0. Also find maximum value of Z. (4)
Answer:
To draw the feasible region, construct table as follows:


Shaded portion OABCD is the feasible region, whose vertices are O(0, 0), A(7, 0), B, C and D(0, 6)
B is the point of intersection of the lines 3x + y = 21 and x + y = 9.
Solving the above equations, we get
x = 6, y = 3
∴ B ≡ (6, 3)
C is the point of intersection of the lines x + 4y = 24 and x + y = 9.
Solving the above equations, we get
x = 4 y = 5
∴ C ≡ (4, 5)
Here, the objective function is Z = 3x + 5y
∴ Z at O(0, 0) = 3(0) + 5(0) = 0
Z at A(7, 0) = 3(7) + 5(0) = 21
Z at B(6, 3) = 3(6) + 5(3)
= 18 + 15 = 33
Z at C(4, 5) = 3(4) + 5(5)
= 12 + 25 = 37
Z at D(0, 6) = 3(0) + 5(6) = 30
∴ Z has maximum value 37 at C(4, 5).
∴ Z is maximum, when x = 4, y = 5

Question 31.
Verify Lagrange’s mean value theorem for the following functions.
f(x) = \(\frac{x-1}{x-3}\) on [4, 5] (4)
Answer:
f(x) = \(\frac{x-1}{x-3}\), x ∈ [4, 5]
As (x – 1) and (x – 3) are polynomial functions,
a. f(x) is continuous on [4, 5]
b. f(x) is differentiable on (4, 5)
Thus, all the conditions of LMVT are satisfied.
To verify LMVT, we need to find c ∈ [4, 5]
such that f'(c) = \(\frac{f(5)-f(4)}{5-4}\) ……… (i)

Thus, Lagrange’s mean value theorem is verified.

Question 32.
Prove that: \(\int \sqrt{\mathrm{a}^2-x^2} \mathrm{~d}\) = \(\frac{x}{2} \sqrt{\mathrm{a}^2-x^2}\) + \(\frac{\mathrm{a}^2}{2} \sin ^{-1}\left(\frac{x}{\mathrm{a}}\right)\) (4)
Answer:

Question 33.
Show that: \(\int_{-a}^a f(x) d x\) = \(2 \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x\), if f(x) is an even function
= 0, if f(x) is an odd function (4).
Answer:
L.H.S. becomes

Case I:
If f (x) is on even function, then f (-x) = f(x).
Equation (ii) becomes
\(\int_{-a}^a f(x) d x\) = \(2 \cdot \int_0^a f(x) d x\)
Case II:
If f(x) is on odd function, then f (-x) = -f(x).
Equation (ii) becomes
\(\int_{-a}^a f(x) d x\) = 0
\(\int_{-a}^a f(x) d x\) = 2 ⋅ \(\int_0^a f(x) d x\)
= 0, if f(x) is an odd function

Question 34.
In a large school, 80% of the pupils 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 the 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. (4)
Answer:
Let X denote the number of pupils who like mathematics.
P(pupils like mathematics) = p = \(\frac{80}{100}\) = \(\frac{4}{5}\) …[Given]
q = 1 – p = 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) = \(\left(\frac{4}{5}\right)^x\left(\frac{1}{5}\right)^{4-x}\), x = 0, 1, ,…, 4

a. P(obtaining an answer yes from at least two pupils)
= P(X ≥ 2) = 1 – P(X < 2)
= 1 – [P(X = 0 or X = 1))
= 1 – \(\left[\frac{1}{5^4}+\frac{16}{5^4}\right]\) ……. (i)
= \(\frac{5^4-17}{5^4}\) = \(\frac{608}{5^4}\)

b.
Here, value of n is increased to 8, i.e., n = 8
∴ X ~ B(8, \(\frac{4}{5}\))
The pm.f. of X is given by
P(X = x) = ⁸C_x\(\left(\frac{4}{5}\right)^x\left(\frac{1}{5}\right)^{8-x}\)
P(obtaining an answer yes form at least two pupils)
= P(X ≥ 2) = 1 – P(X < 2)
= 1 – [P(X = 0 or X = 1)]
= 1 – [P(X = 0) + P(X = 1)]

Maharashtra Board Class 12 Maths Previous Year Question Papers