RBSE 12th Maths Model Paper Set 4 with Answers in English

Students must start practicing the questions from RBSE 12th Maths Model Papers Set 4 with Answers in English Medium provided here.

RBSE Class 12 Maths Board Model Paper Set 4 with Answers in English

Time : 2 Hours 45 Min.
Maximum Marks : 80

General Instructions to the Examinees:

• Candidate must write first his/her Roll. No. on the question paper compulsorily.
• All the questions are compulsory.
• Write the answer to each question in the given answer book only.
• For questions having more than one part the answers to those parts are to be written together in continuity.
• Write down the serial number of the question before attempting it.

Section – A

Question 1.
Multiple Choice Questions
(i) Let f: R → R be defined as f(x) = 3x. Choose the correct Answer : (1)
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Answer:
(a) f is one-one onto

(ii) sin\(\left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]\) is equal to: (1)
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{4}\)
(d) 1
Answer:
(d) 1

(iii) If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\), then A + A’ = I, then the value of α is : (1)
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{6}\)
(c) π
(d) \(3\frac{\pi}{6}\)
Answer:
(b) \(\frac{\pi}{6}\)

(iv) If A is a square matrix of order 3 such that | adj A | = 64, then find |A |. (1)
(a) ± 8
(b) ± 4
(c) ± 16
(d) ± 2
Answer:
(a) ± 8

(v) The value of constant ‘k’ so that the function

is continuous at x = 0 is : (1)
(a) 3
(b) -3
(c) 0
(d) 1
Answer:
(b) -3

(vi) ∫\(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx equals : (1)
(a) 10^x – x¹⁰ + C
(b) 10^x + x¹⁰ + C
(c) (10^x – x¹⁰)^-1
(d) log (10^x + x¹⁰) + C
Answer:
(d) log (10^x + x¹⁰) + C

(vii) The number of arbitrary constants in the general solution of a differential equation of fourth order are : (1)
(a) 0
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

(viii) In triangle ABC (given figure), which of the following is not true. (1)

(a) \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\)
(b) \(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=\overrightarrow{0}\)
(c) \(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=\overrightarrow{0}\)
(d) \(\overrightarrow{A B}-\overrightarrow{C B}+\overrightarrow{C A}=\overrightarrow{0}\)
Answer:
(c) \(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=\overrightarrow{0}\)

(ix) A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is : (1)
(a) 5/64
(b) 27/32
(c) 5/32
(d) 1/2
Answer:
(c) 5/32

(x) If A = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]\), then A – A^T is: (1)
(a) \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-1 & 0 \\
-0 & 1
\end{array}\right]\)
Answer:
(a) \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\)

(xi) If y = x²⁰, then \(\frac{d^{2} y}{d x^{2}}\) is: (1)
(a) 20x¹⁹
(b) 380x¹⁹
(c) 380x¹⁸
(d) 830x¹⁶
Answer:
(c) 380x¹⁸

(xii) A vector in the direction of vector a = î – 2ĵ that has magnitude 7 units is: (1)
(a) \(\frac{1}{\sqrt{5}} \hat{i}-\frac{2}{\sqrt{5}} \hat{j}\)
(b) \(\frac{7}{\sqrt{5}} \hat{i}-\frac{14}{\sqrt{5}} \hat{j}\)
(c) \(\frac{1}{\sqrt{5}} \hat{i}+\frac{3}{\sqrt{5}} \hat{j}\)
(d) \(\frac{14}{\sqrt{5}} \hat{i}+\frac{7}{\sqrt{5}} \hat{j}\)
Answer:
(b) \(\frac{7}{\sqrt{5}} \hat{i}-\frac{14}{\sqrt{5}} \hat{j}\)

Question 2.
Fill in the blanks :
(i) If R = {(a, a³) : a is a prime number less than 5} be a relation, then the range of R is __________________ (1)
Answer:
{8, 27}

(ii) The value of tan[\(\frac{1}{2}\)cos^-1\(\left(\frac{\sqrt{5}}{3}\right)\)] is __________________. (1)
Answer:
\(\frac{3-\sqrt{5}}{2}\)

(iii) If a skew-symmetric matrix A = \(\left[\begin{array}{ccc}
0 & a & 1 \\
-1 & b & 1 \\
-1 & c & 0
\end{array}\right]\), then the value of (a + b + c)² is __________________ (1)
Answer:
0

(iv) If x² + xy + y² = 100, then \(\frac{dy}{dx}\) is __________________ (1)
Answer:
\(-\frac{(y+2 x)}{(x+2 y)}\)

(v) ∫x^ze^x3 dx is equals to __________________ (1)
Answer:
\(\frac{1}{3\)e^x3 + C

(vi) The projection of the vector î – ĵ on the vector î + ĵ is __________________ (1)
Answer:
0

Question 3.
Very Short Answer Type Questions
(i) Show that – a is not the inverse of a ∈ N for the addition operation + on N and – is not the inverse of a ∈ N for multiplication operation × on N, for a ≠ 1. (1)
Answer:
Since, – a ∉ N, – a cannot be inverse of a for addition operation on N, although – a satisfies a + (- a) = 0 = (-a) + a.
Similarly, for a ≠ 1 in N, \(\frac{1}{a}\)∉ N, which implies that other than 1 no element of N has inverse for multiplication operation on N.

Hence Proved

(ii) Find the value of sin^-1[sin \(\left(-\frac{17 \pi}{8}\right)\)]. (1)
Answer:
We have,

(iii) If 2\(\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
7 & 0 \\
10 & 5
\end{array}\right]\), then find (x – y). (1)
Answer:
Given,

On equating the corresponding elements, we get
8 + y = O and 2x + 1 = 5
⇒ y = -8 and x = \(\frac{5-1}{2}\) = 2
Thus, x – y = 2 – (-8) = 10

(iv) Write the value of \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\) (1)
Answer:
Let ∆ = \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
= 2(8 × 86 – 9 × 75) – 7 (3 × 86 – 5 × 75) + 65 (3 × 9 – 5 × 8)
= 2(688 – 675) – 7 (258 – 375) + 65 (27 – 40)
= 2 × 13 – 7 (-117) + 65 (-13)
= 26 + 819 – 845
= 845 – 845 = 0

(v) If f(x) = (x + 7) and g(x) = (x – 7), x ∈ R, then find the value of \(\frac{d}{dx}\)(fog)x. (1)
Answer:
Given/(x) = x + 7, g(x) = x-7, x ∈ R
Now, (fog){x) = f[g(x)] = f(x – 7)
= (x – 7) + 7
(fog)(x) = x
Differentiating both sides w.r.t. x, we get
\(\frac{d}{dx}\)(fog)x = \(\frac{d}{dx}\)(x) = 1

(vi) Integrate the function ∫x sec² x dx. (1)
Answer:

1= x∫sec²xdx – ∫{\(\frac{d}{dx}\)x∫sec² x dx}dx
= xtanx – ∫1.tan x dx
= x tan x – ∫ tan x dx
= x tanx – (- log | cosx |) + C
= x tan x + log | cosx | +C

(vii) Find the general solution of the differential equation \(\frac{d y}{d x}=\frac{x+1}{2-y}\), (y ≠ 2). (1)
Answer:
The given differential equation is:
\(\frac{d y}{d x}=\frac{x+1}{2-y}\) ……….(1)

Separating the variables in equation (1), we get
(2 – y)dy = (x + 1)dx …..(2)

Integrating both sides of equation (2), we get
∫(2 – y)dy = ∫(x + 1)dx
⇒ 2y – \(\frac{y^{2}}{2}\) = \(\frac{x^{2}}{2}\) + x + C₁
⇒ x² + y² + 2x – 4y + 2C₁ = 0
⇒ x² + y² + 2x – 4y + C = 0,
where C = 2C₁
which is the general solution of equation (1).

(viii) If \(\vec{a}\) = 4î – ĵ + k̂ and \(\vec{b}\) = 2î – 2ĵ + k̂, then find a unit vector parallel to the vector \(\vec{a}+\vec{b}\). (1)
Answer:
Given, vectors are
\(\vec{a}\) = 4î – ĵ + k̂, \(\vec{b}\) = 2î – 2ĵ + k̂
\(\vec{a}+\vec{b}\) = (4î – ĵ + k̂) + (2î – 2ĵ + k̂)
= 6î – 3ĵ + 2k̂

(ix) Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E/F) and P(F/E). (1)
Answer:
It is given that,
P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2

(x) Find values of k if area of triangle is 4 sq. units, and vertices are. (k, 0), (4, 0) and (0,2). (1)
Answer:
Area of triangle
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\)

According to question
= \(\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\) = ±4

Expanding along C₂
⇒ \(\frac{1}{2}\){-2(k – 4)} = ±4

Taking positive sign
-(k – 4) = 4 ⇒ k = 0

Taking negative sign
-(k – 4) = -4 ⇒ k = 8

Thus, k = 0 or k =8

(xi) Form the differential equation of the family of curves xy = Ae^x + Be^-x + x², where A and B are arbitrary constants. (1)
Answer:
The given equation is :
xy = Ae^x + Be^-x + x² …(1)

Differentiating both sides of equation (1) w.r.t. ‘x’, we get
dv –
y + x\(\frac{dy}{dx}\) = Ae^x – Be^-x + 2x …………..(2)

Again, differentiating equation (2) w.r.t. ‘X’, we get
\(\frac{d y}{d x}+\frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}\) = Ae^x + Be^-x + 2 ………….(3)

From equations (1) and (3), we get
x\(\frac{d^{2} y}{d x^{2}}\) + 2\(\frac{d y}{d x}\) = (xy – x²) + 2
Which is the required differential equation.

(xii) Find the area of the parallelogram whose adjacent sides are determined by the vectors \(\vec{a}\) = î – ĵ + 3k̂ and \(\vec{b}\) = 2î – 7ĵ + k̂. (1)
Answer:
Given, adjacent sides of the parallelogram are:
\(\vec{a}\) = î – ĵ + 3k̂
and \(\vec{b}\) = 2î – 7ĵ + k̂

= [-1 × 1 – (-7) × 3]i – [1 × 1 – 2 × 3]j + [1 × (-7) – 2 × (-1)]k̂
= (-1 + 21)î – (1 – 6)ĵ + (-7 + 2)k̂
= 20î + 5ĵ – 5k̂

Area of parallelogram = \(|\vec{a} \times \vec{b}|\)

Section – B
Short Answer Type Questions

Question 4.
Show that the function f: N → N defined by f(x) = x² + x + 1 is one-one but not onto. (2)
Answer:
We have,f(x) = x² + x + 1
Now, f(x₁) = x² + x1 + 1
And f(x₂) = x² + x1 +1
Now, f(x₁) f(x₂)
⇒ x²₁ + x₁ + 1 = x²₂ + x₂ + 1
⇒ x²₁ – x²₂ + x₁ – x₂
(x₁ – x₂) (x₁ + x₂) + (x₁ – x₂) = 0,
[Since, a² – b² = (a + b) (a – b)]
(x₁ – x₂) (x₁ + x₂ + 1) = 0

Since, x₁ + x₂ + 1 ≠ 0 for any x ∈ N
So,f is one-one function.
Clearly, f(x) = x² + x + 1 ≥ 3 for all x ∈ N
∴ x₁ – x₂ = 0 ⇒ x₁ = x₂
So, f(x) does not assume values 1 and 2.
∴ f is not an onto function.
Thus,f is one-one but not onto.

Question 5.
If A = \(\left[\begin{array}{rr}
-3 & 2 \\
1 & -1
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find sclar k so that A² + I = kA (2)
Answer:

On equating the corresponding elements, we get
– 3k = 12 ⇒ k = -4

Question 6.
Determine the product of \(\left[\begin{array}{ccc}
-4 & 4 & 4 \\
-7 & 1 & 3 \\
5 & -3 & -1
\end{array}\right]\left[\begin{array}{ccc}
1 & -1 & 1 \\
1 & -2 & -2 \\
2 & 1 & 3
\end{array}\right]\) and then use to solve the system of equations
x – y + z = 4, x – 2y – 2z = 9 and 2x + y + 3z = 1. (2)
Answer:
Let B = \(\left[\begin{array}{ccc}
-4 & 4 & 4 \\
-7 & 1 & 3 \\
5 & -3 & -1
\end{array}\right]\)
and A = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
1 & -2 & -2 \\
2 & 2 & 3
\end{array}\right]\)

On comparing corresponding elements, we get
Thus, x = 3, y = – 2 and z = – 1

Question 7.
Discuss the continuity of the function f(x) at x follows: (2)

Answer:
Here,

∵ LHL ≠ RHL at x = \(\frac{1}{2}\)
Thus, f(x) is discontinuous at x = \(\frac{1}{2}\)

Question 8.
Evaluate : ∫ log (x² + 1 )dx. (2)
Answer:
Let I = ∫ log (x² + 1 )dx

= x log(x² + 1) – 2(x – tan^-1x) + C
= x log(x² + 1) – 2x + 2tan^-1x + C

Question 9.
A die is tossed thrice. Find the probability of getting an odd number at least once. (2)
Answer:
Probability of getting an odd number in a single throw of a die
= \(\frac{3}{6}=\frac{1}{2}\)

Similarly, probability of getting an even number
= \(\frac{3}{6}=\frac{1}{2}\)

Probability of getting an even number three times
= \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}\)

Therefore, probability of getting an odd number at least once
= 1 – Probability of getting an odd number in none of the throws
= 1 – Probability of getting an even number thrice
= 1 – \(\frac{1}{8}=\frac{7}{8}\)

Question 10.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), then show that A³ – 23A – 40I = 0. (2)
Answer:
We have,

Question 11.
if x^myⁿ = (x + y)^m+n, then prove that \(\frac{d y}{d x}=\frac{y}{x}\). (2)
Answer:
Given, x^myⁿ = (x + y)^m+n

Taking log of both sides, we get
log(x^myⁿ) = log(x + y)^m+n
log(x^m) + log(yⁿ)= (m + n) log (x + y)
m log x + n log y= (m + n) log (x + y)

Differentiating both sides w.r.t.’x’, we get

Question 12.
By using properties of determinants, show that : (2)

Answer:
Let

Question 13.
Evaluate : ∫⁴₂\(\frac{x}{x^{2}+1}\)dx. (2)
Answer:
Let I = ∫⁴₂\(\frac{x}{x^{2}+1}\)dx
Put x² + 1 = t ⇒ 2x dx = dt ⇒ x dx = \(\frac{dt}{2}\)

Lower limit
When x = 2, then t = 2² + 1 = 5

Upper limit
When x = 4, then t = 4² + 1 = 17.

Question 14.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis. (2)
Answer:
Equation of parabola having vertex at the origin and axis along positive y-axis is :
x2 = 4ay ………(1)
Differentiating equation (1) w.r.t. ’x’, we get
2x = 4a\(\frac{dy}{dx}\) ⇒ x = 2a\(\frac{dy}{dx}\)

Putting the value of a in equation (1)

Hence, equation (2) is the required differential equation.

Question 15.
If \(\vec{a}\) and \(\vec{b}\) are two unit vectors such that \(\vec{a}\) + \(\vec{b}\) is also a unit vector, then find the angle between \(\vec{a}\) and \(\vec{b}\). (2)
Answer:

Hence, the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{2 \pi}{3}\).

Question 16.
Two candidates A and B had apply for a vacancy. Probability, that A is selected, is \(\frac{1}{3}\) and that B is selected, is \(\frac{1}{5}\). Find the probability that : (2)
(i) any one is selected,
(ii) no one is selected,
Answer:
Given,
P(A) = \(\frac{1}{3}\) and P(B) = \(\frac{1}{5}\)
(i) Required probability that any one either A or B is selected
= P(A.B̄) + P(Ā.B)
= P(A)-P(B̄) + P(Ā).P(B)
= P(A).[1 – P(B)] + [1 – P(A)].P(B)

(ii) Required probability if no one is selected:
= P(Ā.B̄) = P(Ā).P(B̄) .
= [1 – P(A)].[1 – P(B)]
= [1 – \(\frac{1}{3}\)].[1 – \(\frac{1}{5}\)]
= \(\frac{2}{3} \times \frac{4}{5}=\frac{8}{15}\)

Section – C
Long Answer Type Questions

Question 17.
Solve for x, tan^-1(x + 1) + tan^-1 (x – 1) = tan^-1\(\frac{8}{31}\). (3)
Or
Prove that : 2 tan^-1\(\frac{3}{4}\) – tan^-1\(\frac{17}{31}=\frac{\pi}{4}\). (3)
Answer:
Given, tan^-1(x + 1) + tan^-1 (x – 1) = tan^-1\(\frac{8}{31}\)

⇒ 62x = 16 – 8x²
⇒ 8x² + 62x – 16 = 0
⇒ 4x² + 31x – 8 = 0
⇒ 4x² + 32x – x – 8 = 0
⇒ 4x(x + 8) – 1(x + 8) = 0
⇒ (x + 8) (4x – 1) = 0
Either x = -8 or x = 1/4 When, x = -8
LHS = tan^-1(- 7) + tan^-1(- 9)
= – tan^-1(7) – tan^-1(9),
which is negative, while RHS is positive.
So, x = -8 is not possible.
Thus, x = \(\frac{1}{4}\) is the only solution of the given equation.

Question 18.
If y = tan^-1\(\left[\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right]\), x² ≤ 1, then find \(\frac{dy}{dx}\) (3)
Or
Let c be a curve defined parametrically as a; = a cos³θ,y = a sin³ θ, 0 ≤ θ ≤ \(\frac{\pi}{2}\). Determine a point p on c, where the tangent to c is parallel to the Chord joining the points (a, 0) and (0, a). (3)
Answer:
Given, y = tan^-1\(\left[\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right]\)
Putting x² = cos 2θ

On differentiating both sides w.r.t.’x’, we get

Question 19.
Evaluate : ∫\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\) dx. (3)
Or
Evaluate : ∫\(\frac{1}{\sqrt{(x-a)(x-b)}}\) dx. (3)
Answer:
Let
⇒ 3x – 1 = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2)
⇒ 3x – 1 = A(x² – 5x + 6) + B(x² – 4x + 3) + C(x² – 3x + 2)
⇒ 3x – 1 = (A + B + Cx² + (- 5A – 4B – 3C)x + 6A + 3B + 2 C

Comparing the coeffcients of x², x and constant terms in both sides, we have
A + B + C = 0
– 5A – 4B – 3C = 3
and 6A + 3B + 2C = -1

Now, solving these equations, we get
A = 1, B = -5, C = 4

Alternative :
(3x -1) = A(x – 2) (x – 3) + B(x – 1) (x – 3) + C(x – 1)(x – 2)
Putting x = 1
3 × 1 – 1 = A(1 – 2)(1 – 3)
⇒ 2 = A(- 1)(- 2)
⇒ 2 = 2A
⇒ A = 1

Putting x = 2
3 × 2 – 1= B(2 – 1)(2 – 3) = B(- 1)
⇒ 6 – 1 = – B
∴ B = – 5

Putting x = 3
3 × 3 – 1 = C(3 – 1)(3 – 2)
⇒ 9 – 1= C × 2 × 1
⇒ 8 =2C
∴ C = 4
Values of A, B and C also obtain in this method.

Question 20.
Find a unit vector perpendicular to each of the vector \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\), where \(\vec{a}\) = 3î + 2ĵ + 2k̂ and \(\vec{b}\) = î + 2ĵ – 2k̂. (3)
Or
Let the vectors o,b,c be given as a₁î + a₂ĵ + a₃k̂, b₁î + b₂ĵ + b₃k̂, c₁î + c₂ĵ + c₃k̂. Then show that \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\). (3)
Answer:
Given :
\(\vec{a}\) = 3î + 2ĵ + 2k̂ and
\(\vec{b}\) = î + 2ĵ – 2k̂ then \(\vec{a}+\vec{b}\)
= (3î + 2ĵ + 2k) + (î + 2ĵ – 2 k)
= (3 + 1)î + (2 + 2)ĵ + (2 – 2)k̂
⇒ \(\vec{a}+\vec{b}\) = 4î + 4ĵ and \(\vec{a}-\vec{b}\)
= (3î + 2ĵ + 2k̂) – (î + 2ĵ – 2k̂)
= (3 – 1)î+ (2-2)ĵ + (2 + 2)k̂
⇒ \(\vec{a}-\vec{b}\) = 2î + 4k̂

Section – D
Essay Type Questions

Question 21.

Answer:

Put tan x = t ⇒ sec² x dx = dt

Lower limit
When x = 0, then t = tan 0 = 0

Upper limit when x = \(\frac{\pi}{4}\),
then t = tan\(\frac{\pi}{4}\) = 1

Question 22.
Solve the differential equation\(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}\) = 1(x ≠ 0). (4)
Or
In a bank, principal increases continuously at the rate of 5% per year. An amount of ₹ 1000 is deposited with this bank, how much will it worth after 10 years. (e^0.5 = 1.648). (4)
Answer:
Given differential equation is

∴ I.F = e^2√x
On multiplying equation (1) by e^2√x

which is the required solution.

Question 23.
The probability that a bulb produceed by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use. (4)
Or
An experiment succeeds twice as often as it fails. Find the prrobability that in the next six trials, there will be at least 4 successes. (4)
Answer:
Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.
It is given that, p = 0 – 05
∴ q = 1 – p = 1 – 0.05 = 0.95

X has a binomial distribution with n = 5 and p = 0.05
P(X = x) ⁿC_x q^n-xp^x, where x = 1, 2,…. n
= ⁵C_x (0 -95)^5-x. (0 -05)^x

(i) P (none) = P (X = 0)
= ⁵C₀(0.95)⁵. (0 -05)⁰
= 1 x (0 -95)⁵ = (0 -95)⁵

(ii) P (not more than one) = P(X ≤ 1)
= P(X = 0) + P (X = 1)
= ⁵C₀ (0.095)5 X (0 -05)° + ⁵C₁ (0 -95)⁴ x (0 -05)¹
= 1 × (0 -95)⁵ + 5 × (0.95)⁴ × (0 05)
= (0.95)⁵ x (0 -25) (0 -95)⁴
= (0.95)⁴ [0.95 + 0.25]
= (0.95)⁴ × 1.2

(iii) P (more than 1) = P (X > 1)
= 1 – P(X ≤ 1)
= 1 – P (not more than 1)
= 1 – (0.95)⁴ × 1.2

(iv) P (at least one) = P (X ≥ 1)
= 1 – P (X < 1) = 1 – P (X = 0)
= 1 – ⁵C₀ (0.95)⁵ × (0.5)⁰
= 1 – 1 × (0.95)⁵ = 1 – (0 -95)⁵