Maharashtra Board Class 12 Maths Sample Paper Set 5 with Solutions

Maharashtra State Board Class 12th Maths Sample Paper Set 5 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Maths Model Paper Set 5 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. Inverse of statement pattern (p ∨ q) → (p ∧ q) is _____.
(A) (p ∧ q) → (p ∨ q)
(B) ~(p ∨ q) → (p ∧ q)
(C) (~ p ∧ ~ q) (~ p ∨ ~ q)
(D) (~ p ∨ ~ q) → (~ p ∧ ~ q)
Answer:
(C) (~ p ∧ ~ q) → (~ p ∨ ~ q)

Explanation:

Inverse of p → q is ~ p → ~ q
∴ Inverse of [(p ∨ q) → (p ∧ q)]
≡ ~ (p ∨ q) → ~ (p ∧ q)
≡ (~ p ∧ ~q) → (~p ∨ ~q)

ii. If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), adj A = \(\left[\begin{array}{cc}
4 & a \\
-3 & b
\end{array}\right]\), then the values of a and b are
(A) a = -2, b = 1
(B) a = 2, b = 4
(C) a = 2, b = -1
(D) a = 1, b = -2
Answer:
(A) a = – 2, b = 1

Explanation:

A₁₁ = (-1)¹⁺¹(4) = 4, A₁₂ = (- 1)¹⁺² (3) = – 3,
A₂₁ = (- 1)²⁺¹ (2) = – 2, A₂₂ = (-1)²⁺² (1) = 1
∴ adj A = \(\left[\begin{array}{cc}
4 & -3 \\
-2 & 1
\end{array}\right]^{\top}\) = \(\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]\)

iii. If α, β, γ are direction angles of a line and α = 60°, β = 45°, then γ =
(A) 30° or 60°
(B) 45° or 60°
(C) 90°or 30°
(D) 60°or 120°
Answer:
(D) 60° or 120°

Explanation:

cos² α + cos² β + cos² γ = 1
∴ cos² 60° + cos² 45° + cos² γ = 1
∴ \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2\) + cos² γ = 1
∴ \(\frac{1}{4}\) + \(\frac{1}{2}\) + cos² γ = 1
∴ cos² γ = 1 – \(\frac{1}{4}\)
∴ cos γ = ±\(\frac{1}{2}\)
∴ γ = 60° or 120°

iv. If the line \(\frac{x}{3}\) = \(\frac{y}{4}\) = z is perpendicular to the line \(\frac{x-1}{\mathrm{k}}\) = \(\frac{y+2}{\mathrm{3}}\) = \(\frac{\mathrm{z}-3}{\mathrm{k}-1}\), then the value of k is
(A) \(\frac{11}{4}\)
(B) –\(\frac{11}{4}\)
(C) \(\frac{11}{2}\)
(D) \(\frac{4}{11}\)
Answer:
(B) \(\frac{-11}{4}\)

Explanation:

\(\frac{x}{3}\) = \(\frac{y}{4}\) = \(\frac{z}{1}\) implies a₁ = 3, b₁ = 4, c₁ = 1
Also, \(\frac{x-1}{k}\) = \(\frac{y+2}{3}\) = \(\frac{z-3}{k-1}\) implies a₂ = k, b₂ = 3, c₂ = k – 1
Since lines are perpendicular,
a₁a₂ + b₁b₂ + c₁c₂ = 0
∴ 3k + 12 + k- 1 = 0
∴ 4k = – 11
∴ k = \(\frac{-11}{4}\)

v. ∫[sin (log x) + cos (log x)] dx =
(A) x cos (log x) + c
(B) sin (log x) + c
(C) cos (log x) + c
(D) x sin (log x) + c (2)
Answer:
(D) x sin (log x) + c

Explanation:

Let I = \(\int[\sin (\log x)+\cos (\log x)] d x\)
Put log x = t
∴ x = e<sup.t
∴ dx = e^t dt
I = ∫ (sin t + cos t)e^t dt
= e^t sin t + c ……….. [∵ \(\int e^{\dagger}[f(t)\) + f ‘(t)] dt = e^t f(t) + c]
∴ I = x sin (log x) + c

vi. The area enclosed between the parabola y² = 4x and line y = 2x is (2)
(A) \(\frac{2}{3}\) sq. units
(B) \(\frac{1}{3}\) sq. units
(C) \(\frac{1}{4}\) sq. units
(D) \(\frac{3}{4}\) sq. units (2)
Answer:
(B) \(\frac{1}{3}\)sq. units

Explanation:

Required area = \(\int_0^1 2 \sqrt{x} d x-\int_0^1 2 x d x\)
= \(\frac{4}{3}\) – 1
= \(\frac{1}{3}\) sq. units

vii. The solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = \(\frac{y+\sqrt{x^2-y^2}}{x}\) is _____
(A) sin¹\(\left(\frac{y}{x}\right)\) = 2 log |x| + c
(B) sin^-1\(\left(\frac{y}{x}\right)\) = 2 log |x| + c
(C) sin\(\left(\frac{x}{y}\right)\) = log |x| + c
(D) sin\(\left(\frac{y}{x}\right)\) = log |y| + c
Answer:
(B) sin^-1\(\left(\frac{y}{x}\right)\) = log |x| + c

Explanation:

\(\frac{d y}{d x}\) = \(\frac{y+\sqrt{x^2-y^2}}{x}\) ……. (i)
Put y = vx …… (ii)
∴ \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) ……. (iii)
Substituting (ii) and (iii) in (i), we get

viii. P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere). If P (X < a) = P (X > a), then a =
(A) 1
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{2}\)

Explanation:

Question 2.
Answer the following: [4 Marks]

i. Write the negation of the following.
∀ n ∈ N, n² + n + 2 is divisible by 4. (1)
Answer:
∃ n ∈ N such that n² + n + 2 is not divisible by 4.

ii. Find the principal value of sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\). (1)
Answer:
Let sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) = x
∴ sin^-1x = \(\frac{1}{\sqrt{2}}\) = sin \(\frac{\pi}{4}\)
The principal value branch of sin^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{4}\) ≤ \(\frac{\pi}{2}\).
∴ The required principal value of x is \(\frac{\pi}{4}\).

iii. If y = e^{tan x} find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\). (1)
Answer:
y = e^{tan x}
∴ \(\frac{d y}{d x}\) = e^{tan x} × \(\frac{d}{d x}\)(tan x)
∴ \(\frac{d y}{d x}\) = e^{tan x} ⋅ sec²x

iv. If X ~ B (n, p), where n = 4, p = \(\frac{1}{2}\), q = \(\frac{1}{2}\), then find SD(X). (1)
Answer:
X ~ B(n, p), n = 4, p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\)
Var(X) = npq = 4\(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\)
∴ Var(X) = 1

Section – B

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

Question 3.
If statements p, q are true and r, s are false, determine the truth value of (p ∧ ~ r) ∧ (~q ∨ s). (2)
Answer:
(p ∧ ~ r) ∧ (~q ∨ s) ≡ (T ∧ ~ F) ∧ (~T ∨ F)
≡ (T ∧ T) ∧ (F ∨ F) ≡ T ∧ F ≡ F
Hence, truth value is F.

Question 4.
In ∆ABC, prove that a(b cos C – c cos B) = b² – c². (2)
Answer:
L.H.S. = a(b cos C – c cos B)
= ab cos C – ac cos B
= \(\frac{1}{2}\)(2ab cos C – 2ac cos B)
= \(\frac{1}{2}\)(a² + b² – c²) – (c² + a² – b²)}
= \(\frac{1}{2}\)(a² + b² – c² – c² – a² + b²)
= \(\frac{1}{2}\)(2b² – 2c²) = b² – c² = R.H.S.

Question 5.
Find k if one of the lines given by 6x² + kxy + y² = 0 is 2x + y = 0. (2)
Answer:
Given equation of the lines is 6.x² + kxy + y² = 0.
Its auxiliary equation is m² + km + 6 = 0
Since 2x + y = 0 is one of the lines given by 6x² + kxy + y² = 0.
6x² + kxy + y² = 0
Slope of the line 2x + y = 0 is -2.
∴ -2 is one of the roots of the auxiliary equation m² + km + 6 = 0
∴ (-2)² + k(-2) + 6 = 0
∴ 4 – 2k + 6 = 0
∴ 2k = 10
∴ k = 5

Question 6.
Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are -2, 1, -1 and -3, -4, 1. (2)
Answer:
Let L₁ and L₂ be the two lines with direction ratios -2, 1, -1 and -3, – 4, 1 respectively.
Let the direction ratios of the vector perpendicular to L₁ and L₂ be a, b, c.
∴ -2a + b – c = 0
and -3a – 4b + c = 0

Question 7.
Check whether the vectors 2\(\hat{\mathrm{i}}\) + 2\(\hat{\mathrm{j}}\) + 3\(\hat{\mathrm{k}}\), -3\(\hat{\mathrm{i}}\) + 3\(\hat{\mathrm{j}}\) + 2\(\hat{\mathrm{k}}\) and 3\(\hat{\mathrm{i}}\) + 4\(\hat{\mathrm{k}}\) form a triangle or not. (2)
Answer:

Question 8.
Find the vector equation of a plane which is at 42 unit distance from the origin and which is-normal to the vector 2\(\hat{\mathrm{i}}\) + \(\hat{\mathrm{j}}\) – 2\(\hat{\mathrm{k}}\). (2)
Answer:

Question 9.
Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\), if x = a cot θ, y = b cosec θ. (2)
Answer:

Question 10.
Given an interval [a, b] that satisfies hypothesis of Rolle’s theorem for the function
f(x) = x⁴ + x – 2. It is known that a = – 1. Find the value of b. (2)
Answer:
f(x) = x⁴ + x² – 2, x ∈ [a, b] satisfies the hypothesis of Rolle’s theorem.
∴ f(a) = f (b)
∴ f (-1) = f (b)
∴ (-1)⁴ + (-1)² – 2 = b⁴ + b² – 2
∴ 1 + 1 – 2 = b⁴ + b² – 2
∴ b⁴ + b² – 2 = 0
∴ (b² – 1) (b² + 2) = 0
∴ b² = 1 or b² = -2
But b² cannot be neqative.
∴ b² = 1
∴ b = 1 …[As a = -1, b ≠ -1]

Question 11.
Evaluate: ∫cos² x dx (2)
Answer:

Question 12.
Integrate the following functions w. r. t. x: \(\frac{x^{\mathrm{n}-1}}{\sqrt{1+4 x^n}}\) (2)
Answer:

Question 13.
Evaluate: \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2+\sin x}{2-\sin x}\right) \mathrm{d} x\)
Answer:

Question 14.
Find the area of the region bounded by the curve y² = x, X-axis and the lines x = 0, x = 4.
Answer:
Given equation of the curve is
y² = x
∴ y = ±\(\sqrt{x}\)
∴ y = \(\sqrt{x}\) ……..[∵ In first quadrant, y > 0]
Required area = area of the region OBSAO
= 2(area of the region OSAO)

Section – C

Attempt arty EIGHT of the following questions:

Question 15.
Write converse, inverse and contra positive of the following statements. If x < y then x² < y² (x, y ∈ R)
Answer:
Let p : x< y
q : x² < y²
∴ The given statement is p → q.
Its converse is q → p.
If x² < y² then x < y.
Its inverse is ~p → ~q.
If x ≥ y then x² ≥ y².
Its contrapositive is ~q → ~p.
If x² ≥ y² then x ≥ y.

Question 16.
Evaluate : cosec^-1\((-\sqrt{2})\) + cot^-1\((\sqrt{3})\)
Answer:
Let cosec^-1 \((-\sqrt{2})\) = x
∴ cosec x = \((-\sqrt{2})\) = – cosec \(\frac{\pi}{4}\) = cosec
(-\(\frac{\pi}{4}\))
The principal value branch of cosec^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0} and \(-\frac{\pi}{2}\) ≤ \(-\frac{\pi}{4}\) ≤ \(\frac{\pi}{2}\).
∴ x = \(-\frac{\pi}{4}\)
∴ cosec^-1 \((-\sqrt{2})\) = –\(\frac{\pi}{4}\) ……(i)
Let cot^-1\((\sqrt{3})\) = y
∴ cot y = \(\sqrt{3}\) = cot \(\frac{\pi}{6}\)
The principal value branch of cot^-1 is (0, π) and 0 < \(\frac{\pi}{6}\) < π.
∴ y = \(\frac{\pi}{6}\)
∴ cot^-1\((\sqrt{3})\) = \(\frac{\pi}{6}\) ……..(ii)
∴ cosec^-1 (\(-\sqrt{2}\)) + cot^-1\((\sqrt{3})\) = \(-\frac{\pi}{4}+\frac{\pi}{6}\) = \(\frac{-3 \pi+2 \pi}{12}\) = \(-\frac{\pi}{12}\) …[From (i) and (ii)]

Question 17.
If A (\(\overline{\mathrm{a}}\)) and B (\(\overline{\mathrm{b}}\)) are any two points in the space and R(\(\overline{\mathrm{r}}\)) be dividing it internally in the ratio m : n, then prove that \(\overline{\mathrm{r}}\) = \(\frac{\mathrm{m} \overline{\mathrm{~b}}+\mathrm{n} \overline{\mathrm{a}}}{\mathrm{~m}+\mathrm{n}}\). (3)
Answer:
R is a point on the line segment AB(A-R-B) and \(\overline{A R}\) and \(\overline{R B}\) are in the same direction.
Point R divides AB internally in the ratio m : n.

Question 18.
Prove that:
.
(3)
Answer:

Question 19.
Find the shortest distance between the lines \(\overline{\mathrm{r}}\) = (4\(\hat{\mathrm{i}}\) – \(\hat{\mathrm{j}}\)) + λ(\(\hat{\mathrm{i}}\) + 2\(\hat{\mathrm{j}}\) – 3\(\hat{\mathrm{k}}\)) and \(\overline{\mathrm{r}}\) = |\(\hat{\mathrm{i}}\) – \(\hat{\mathrm{j}}\) + 2\(\hat{\mathrm{k}}\)) + µ(\(\hat{\mathrm{i}}\) + 4\(\hat{\mathrm{j}}\) – 5\(\hat{\mathrm{k}}\)). (3)
Answer:
The shortest distance between the lines \(\overline{\mathrm{r}}\) = \(\overline{a_1}+\lambda \overline{b_1}\) and \(\bar{r}\) = \(\overline{a_2}+\mu \overline{b_2}\) is d = \(\left|\frac{\left(\overline{a_2}-\overline{a_1}\right) \times\left(\overline{b_1} \times \overline{b_2}\right)}{\left|\overline{b_1} \times \overline{b_2}\right|}\right|\)

Question 20.
Find the vector equation of the plane passing through the points A(1, -2, 1), B(2, -1, -3) and ’
C(0, 1, 5). (3)
Answer:

Question 21.
If y = (sin x)^x, then find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\). (3)
Answer:
y = (sin x)^x
Taking log on both sides, we get
log y = x log (sin x)
Differentiating w. r. t. x, we get

Question 22.
If y = f(x) is a differentiable function, then show that \(\frac{\mathrm{d}^2 x}{\mathrm{~d} y^2}\) = \(-\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^{-3} \cdot \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\) (3)
Answer:

Question 23.
Find the approximate value of log₁₀ (1016), given that log₁₀ e = 0.4343.
Answer:
Let f(x) = log₁₀ x
= \(\frac{\log _e x}{\log _e 10}\)
= (log₁₀ e)(log_e x)
= (0.4343) log x
∴ f'(x) = \(\frac{0.4343}{x}\)
x = 1016 = 1000 + 16 = a + h
Here, a = 1000 and h = 16
f (a) = f(1000) = log₁₀(1000) = log₁₀ (10)³
= 3log₁₀10 ……..[∵ log₁₀ mⁿ = n log₁₀ m]
= 3
f ‘(a) = f ‘(1000) = \(\frac{0.4343}{1000}\) = 0.0004343
f (a + h) ≈ f (a) + h f ‘(a)
∴ log₁₀ (1016) ≈ 3 + 16(0.0004343)
≈ 3 + 0.0069488
∴ log₁₀ (1016) ≈ 3.006949

Question 24.
The slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also the curve passes through the point (0, 1). Find the equation of the curve.
Answer:
Let (x, y) be any point on the curve y = f (x).
The slope of the tangent at point (x, y) is \(\frac{d y}{d x}\)
∴ \(\frac{d y}{d x}\) = x + xy
∴ \(\frac{d y}{d x}\) = x(1 + y)
∴ \(\frac{d y}{1+y}\) = x dx
Integrating on both sides, we get
\(\int \frac{d y}{1+y}\) = \(\int x d x\)
∴ log |1 + y| = \(\frac{x^2}{2}\) + c₁
∴ 1 + y = \(e^{\frac{x^2}{2}} e^{c_1}\)
∴ 1 + y = \(e^{\frac{x^2}{2}} c,\) ………(i)
where c = e^c1
But the required curve passes through the point (0, 1).
∴ 1 + 1 = e° c
∴ c = 2
Substituting c = 2 in (i), we get
1 + y = 2\(e^{\frac{x^2}{2}}\), which is the required equation of the curve.

Question 25.
Solve the differential equation :
x\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = xtan\(\left(\frac{y}{x}\right)\) +y dx (3)
Answer:
x\(\frac{d y}{d x}\) = x tan \(\left(\frac{y}{x}\right)\) + y ……. (i)
Put y = vx ……….(ii)
Differentiating w. r. t. x, we get
\(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) …..(iii)
Substituting (ii) and (iii) in (i), we get
x(v + x\(\frac{d v}{d x}\)) = xtan\(\left(\frac{v x}{x}\right)\) + vx
∴ v + x\(\frac{d v}{d x}\) = tan v + v
∴ x\(\frac{d v}{d x}\) = tan v
∴ \(\frac{d v}{\tan v}\) = \(\frac{d x}{x}\)
Integrating on both sides, we get
∫cot v dv = \(\int \frac{d x}{x}\)
∴ log |sin v| = log |x| + log |c|
∴ log |sin v| = log |xc|
∴ Sin v = cx
∴ sin \(\left(\frac{y}{x}\right)\) = cx

Question 26.
Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f. f(x) = k(4 – x²), for -2 ≤ x ≤ 2 and = 0 otherwise.
Compute: P(-1 < X < 1) (3)
Answer:

Section – D

Question 27.
If three numbers are added, their sum is ‘2’. If 2 times the second number is subtracted from the sum of first and third number, we get ‘8’ and if three times the first number is added to the sum of second and third number, we get ‘4’. Find the numbers using matrices. (4)
Answer:
Let the three numbers be x, y and z respectively.
According to the first condition,
x + y + z = 2
According to the second condition,
(x + z) – 2y = 8 i.e., x – 2y + z = 8
According to the third condition,
3x + y + z = 4
Matrix form of the above system of equations is
\(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
3 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
8 \\
4
\end{array}\right]\)
Applying R₂ → R₂ – R₁ and R₃ → R₃ – 3R₁ we get
\(\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & -3 & 0 \\
0 & -2 & -2
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{r}
2 \\
6 \\
-2
\end{array}\right]\)
Applying R₂ → \(\left(-\frac{1}{3}\right) R_2\) and R₃ → \(\left(-\frac{1}{2}\right) R_3\), we get
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]\)
Applying R₃ → R₃ – R₂, we get
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{r}
2 \\
-2 \\
3
\end{array}\right]\)
Hence, the original matrix is reduced to an upper triangular matrix.
∴ \(\left[\begin{array}{l}
x+y+z \\
0+y+0 \\
0+0+z
\end{array}\right]\) = \(\left[\begin{array}{c}
2 \\
-2 \\
3
\end{array}\right]\)
∴ By equality of matrices, we get
x + y + z = 2 …… (i)
y = -2 …….(ii)
z = 3 …….. (iii)
Substituting y = -2 and z = 3 in equation (i), we get
x – 2 + 3 = 2
∴ x = 1
∴ 1, -2 and 3 are the required numbers.

Question 28.
Find the general solution of cos θ + sin θ = 1. (4)
Answer:

Question 29.
If θ is the measure of acute angle between the pair of lines given by ax² + 2hxy + by² = 0, then prove that tan θ = \(\left|\frac{2 \sqrt{\mathrm{~h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|\), a + b ≠ 0.
Find the condition for coincident lines. (4)
Answer:

Question 30.
Solve the following L.P.P. by graphical method:
Minimize: z = 8x + 10y subject to 2x + y ≥ 7, 2x + 3y ≥ 15, y ≥ 2, x ≥ 0, y ≥ 0. (4)
Answer:
To draw the feasible region, construct table as follows:

Shaded portion EABCY is the feasible region, whose vertices are A, B and C(0, 7)
A is the point of intersection of the lines y = 2 and 2x + 3y = 15,
Substituting y = 2 in 2x + 15, we get
2x + 6 = 15
∴ x = \(\frac{9}{2}\)
∴ A ≡ (\(\frac{9}{2}\), 2)

B iS the point of intersection of the lines 2x + y = 7 and 2x + 3y = 15.
Solving the above equations, we get
x = \(\frac{3}{2}\), y = 4
∴ B ≡ (\(\frac{3}{2}\), 4) ≡ (1.5, 4)
Here, the objective function is Z = 8x + 10y
Z at A (\(\frac{9}{2}\), 2) = 8(\(\frac{9}{2}\)) + 10(2) = 36 + 20 = 56
Z at B (\(\frac{3}{2}\), 4) = 8(\(\frac{3}{2}\)) +10(4) = 12 + 40 = 52
Z at C(0, 7) = 8(0) + 10(7) = 70
∴ Z has minimum value 52 at B[\(\frac{3}{2}\), 4) i.e. at (1.5, 4)
∴ Z is minimum, when x = \(\frac{3}{2}\) i.e 1.5 and y = 4.

Question 31.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\). (4)
Answer:
Let x be the height and y be the radius of the inscribed cone.
In the given figure, consider ΔA BC.
By Pythagoras theorem,

Question 32.
Prove that:

(4)
Answer:

Question 33.
Show that:

(4)
Answer:
Consider R.H.S.: \(\int_0^a f(x) d x\) + \(\int_0^a f(2 a-x) d x\)
Let I = \(\int_0^a f(x) d x+\int_0^a f(2 a-x) d x\)
= I₁ + I₂ …… (i)
Consider I₂ = \(\int_0^a f(2 a-x) d x\)
Put 2a – x = t
∴ -dx = dt
∴ dx = -dt
When x = 0, t = 2a – 0 = 2a
and when x = a, t = 2a – a = a

Question 34.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards.
Find the probability that
i. all the five cards are spades.
ii. only 3 cards are spades.
iii. none is a spade. (4)
Answer:
Let X denote the number of spades.
P(getting spade) = p = \(\frac{13}{52}\) = \(\frac{1}{4}\)
∴ q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Given, n = 5
∴ X ~ B(5, \(\frac{1}{4}\))

Maharashtra Board Class 12 Maths Previous Year Question Papers