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## Karnataka 2nd PUC Maths Model Question Paper 4 with Answers

Time: 3.15 Hours

Max Marks: 100

Instructions :

- The question paper has five parts namely A, B, C, D and E. Answer all the parts.
- Use the graph sheet for the question on Linear programming in PART E.

Part -A

I. Answer the following questions. ( 10 × 1 = 10 )

Question 1.

Operation * is defined by a * b = a. Is * is a binary operation on z+?

Answer:

Let a = 2 and b = 1

Then, a * b = 2 * 1 = 2 ∈ z^{+}

⇒ Yes * is a binary operation.

Question 2.

Write the principal value branch of f(x) = sin^{-1}x.

Answer:

\(\)

Question 3.

Define a diagonal matrix.

Answer:

A square matrix [Aij] is said to be a diagonal matrix if the non – diagonal elements are zero.

Ex: A = \(\left[\begin{array}{ll}

{1} & {0} \\

{0} & {2}

\end{array}\right]\)

Question 4.

If A = \(\left[\begin{array}{ll}

{4} & {7} \\

{3} & {5}

\end{array}\right]\) find |3A|

Answer:

3A = \(\left[\begin{array}{ll}

{12} & {21} \\

{9} & {15}

\end{array}\right]\) ⇒ |3 A| = 180 – 189 = -9.

Question 5.

Write the points of discontinuty for the function f [x] = [x], -3 < x < 3

Answer:

Point of discontinuty are x = -2, x = -1, x = 0, x = 1, x = 2.

Question 6.

Evaluate ∫ cos ecx (cos ecx – cot x) dx

Answer:

∫ cosec x (cosec x – cot x) dx

= ∫cosec^{2}x – cosec x cot x )dx

= cosec^{2}x dx – ∫ cosecx cot x dx

=-cotx – (-cosecx)

=-cotx + cosecx

=cosecx – cotx+c

Question 7.

Find the direction ratios of the vector, joining the points P(2, 3, 0) and Q( -1, -2, -3)., direction from P to Q.

Answer:

\(\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}\)

=((-1) î + (-2) ĵ + (-3)k̂) -(2î + 3ĵ + 0k̂)

= (-1 – 2) î+(-2 – 3)ĵ +(-3 – 0)k̂

\(\overrightarrow{\mathrm{PQ}}\) = -3î -5ĵ-3k̂.

Direction Ratios of \(\overrightarrow{\mathrm{PQ}}\) = (-3, -5, -3).

Question 8.

Find the Equation of a plane with the internet 2, 3 and 4 on x_{1} y_{1} and z axes respectively.

Answer:

we have a = 2, b = 3, c = 4.

Therefore \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ⇒ \(\Rightarrow \frac{x}{2}+\frac{y}{3}+\frac{z}{4}\) = 1

Question 9.

Define optional solution in the linear programming problem.

Answer:

Any point in the feasible region that gives the optional value (maximum or minimum ) of the objective function is called an optional solution.

Question 10.

If A and B are independent events with P(A) = 0.3 P(B) = 0.4, find P(A n B). Since A and B are independent,

we have P(A∩B) = P(A). P(B)

= 0.3 × 0.4

P(A∩B) = 0.12

Part – B

Answer any Ten questions: ( 10 x 2 = 20 )

Question 11.

Find the gof and fog iff (x) – 8x^{3} and

Answer:

g (x) = X^{1/3} ,

(gof) (x) = g (f(x) = g ( 8x^{3})

=(8x^{3})^{1/3}=(2^{3}x^{3})^{1/3}

(gof)(x) =2x

(fog)(x) = f(g(x) = f(x^{1/3}) = 8(x^{1/3})^{3}

(fog)(x)= 8x

Question 12.

Write the function \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\) o < x < π in the Simplest form.

Answer:

Question 13.

Prove that 2sin ^{-1}\(\left(\frac{3}{5}\right)\) = tan ^{-1 }\(\left(\frac{25}{7}\right)\)

Answer:

Question 14.

Find the area of a triangle whose vertices are (1, 3), (2, 5) and (7, 5) using determinant.

Answer:

Area = \(\frac{1}{2}\left|\begin{array}{ll}

{x_{1}} & {y_{1}} \\

{x_{2}} & {y_{2}}

\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}

{1} & {3} & {1} \\

{2} & {5} & {1} \\

{7} & {2} & {1}

\end{array}\right|\)

= \(\frac{1}{2}\) {1[5 – 5] – 3[2 – 7] + 1[10 – 35]}

= \(\frac{1}{2}\) {0 – 3(- 5) + 1(-25)

= \(\frac{1}{2}\) {15 – 25}

= \(\frac{1}{2}\) {-10} = – 5

Thus |Area| – 5 sq. units.

Question 15.

Find \(\frac{d y}{d x}\) if 2x+3y=sin y

Answer:

2x + 3y = siny

diff w. r. t ‘x’, we get

2(1) + 3\(\frac{d y}{d x}\)=cosy\(\frac{d y}{d x}\)

⇒ (cosy – 3)\(\frac{d y}{d x}\)=2

\(\frac{d y}{d x}=\frac{2}{\cos y-3}\)

Question 16.

If x = at^{2} and y = 2 at slow that \(\frac{d y}{d x}=\frac{1}{t}\)

Answer:

Question 17.

Find the approximate change in the volume v of a cube of a side x meters caused by increasing by 2%.

Answer:

Given, v = x^{3}

diff sv. r. t .’x’.

sv = 3x^{2}dx

= 3x^{2 }(2)

sv = 6x^{2}

Question 18.

Evaluate ∫sin^{3}x dx

Answer:

Question 19.

Evaluate \(\int_{0}^{\pi / 2} \cos 2 x d x\)

Answer:

Question 20.

Find the orders and degree of different

Equation \(\left(\frac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0\)

Answer:

Order =2 degree = 1

Question 21.

Find a vector in the direction of the vector \(\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}\) that has magnitude 7 units.

Answer:

Question 22.

If \(\overrightarrow{\mathbf{a}}\) =5î – ĵ – 3k̂ and \(\overrightarrow{\mathbf{b}} \) = î + 3ĵ – 5k̂, then show that the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) are perpendicular.

Answer:

Given \(\overrightarrow{\mathbf{a}}\) =5î – ĵ – 3k̂

\(\overrightarrow{\mathbf{b}} \) = î + 3ĵ – 5k̂

\(\vec{a}+\vec{b}\) = 6î + 2ĵ – 8k̂

\(\vec{a}-\vec{b}\) = 4î – 4ĵ + 2k̂

\(\vec{a}+\vec{b}\)] . \(\vec{a}-\vec{b}\) =

=(6)(4) + (2(-4) + (-8)(2)

= 24 – 8 – 16 = 24 – 24=0 .

\(\vec{a}+\vec{b}\) & \(\vec{a}-\vec{b}\) are perpendicular.

Question 23.

Find the Vector equation of the line, passing through the points (-1, 0, 2) and (3,4,6)

Answer:

Let \(\overrightarrow{\mathbf{a}}\) = (-1) î + 0ĵ + (2)k̂

\(\overrightarrow{\mathbf{b}} \) =(3) î + (4)ĵ + (6)k̂

we have \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})\)

But, \(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \mathrm{k}\)

∴ r = [(-1) î+(0) ĵ+(2)k̂] + λ(4 î + 4ĵ +4k̂)

Question 24.

Two coins are topped once, find P (E/ F) where E: no tail appears, F : no head appears.

Answer:

s={HH, HT, TH, TT }

E={HH},F={TT} and E∩F={/}

∴ P(E) = \(\frac{1}{4}\) and P(E∩F)= {0}

Thus, P (E/F) = \(\frac{\mathrm{P}(\mathrm{EnF})}{\mathrm{P}(\mathrm{F})}=\frac{0}{(1 / 4)}=0\)

Part – C

Answer any Ten Questions : ( 10 x 3 = 30 )

Question 25.

Show that the relation R in the set of real number R defined as R = {(a,b):a≤b^{2}} is number reflexive nor symmetric nor transitive

Answer:

It can be observed that \(\left(\frac{1}{2}, \frac{1}{2}\right)=\frac{1}{2} \not<\frac{1}{4}\)

∴ R is not reflexive.

Now (l,2)ERas 1 < 2^{2}

But, 2 is not less than 1^{2}

∴ (2, 1) ∉ R ∴ is not symmetric

Now,(5, 3) (3, 2) ∈ R

( as 5 > 3^{2} = 9 and 3 < (2)^{2} = 4)

But,5 > (2)^{2} = 4

∴ (5,2) ∉ R ,

∴ R is not Transitive

∴ \(\overline{\mathrm{k}}\) is neither reflexive, nor symmetric nor Transitive.

Question 26.

Prove that tan^{-1} x + tan^{-1} \(\left(\frac{2 x}{1-x^{2}}\right)\) = tan ^{-1}\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\) , |x| < \(\frac{1}{3}\).

Answer:

Question 27.

Find the values of x ,y and % in the following matrices

Answer:

\(\left(\begin{array}{ll}

{x+y} & {2} \\

{5+z} & {x y}

\end{array}\right)=\left(\begin{array}{ll}

{6} & {2} \\

{5} & {8}

\end{array}\right)\)

x + y = 6…(L1), 5 + 2 = 5 (L2), xy = 8….(L3)

From (2):

5 + 2=5 ⇒ 2 = 5-5=0

∴ 2 = o

From (1) :y = 6 – x

∴ (3) ⇒ x (6 – x) = 8 ⇒ 6x – x^{2} = 8

⇒ x^{2} – 6x + 8=0

⇒ (x – 4) (x – 2)=0

⇒ x – 4 or x = 2.

If x = 4, y = 6-4 =2

I x = 2, y = 6-2 = 4

Thus , x = 4, y = 2 and z = 0 or x = 2 ,y = 4 and z = 0.

Question 28.

Differentiate \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}}\) with respect to x.

Answer:

Question 29.

If y = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) find \( \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}\)

Answer:

Put x = tanθ .’. θ = tan ^{-1}x

y = \(\sin ^{-1}\left[\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right]\)

y = sin ^{-1}[ sin 2θ] = 2θ

y =2tan ^{-1}x

diff w.r.t. ‘x’

\(\frac{d y}{d x}=\frac{2}{1+x^{2}}\)

Question 30.

Find the intervals in which the function f given by f(x) = 2x^{3} – 3x^{2} – 36x + 7 is (a) Strictly increasing (b)Strictly decreasing.

Answer:

‘Given f (x) = 2x^{3} – 3x^{2} – 36x + 7

f^{1}(X) = 6X^{2} – 6X – 36 = 6(x^{2} – x^{6})

f^{1}(x) = 6(x + 2)(x – 3)

.’.f^{1} (x) = 0 gives x = -2andx = 3.The points x = -2 and x = 3 divides the real line into three disjoint intervals namely,

(-∞ , -2), (-2, 3) and (3, ∞).

Question 31.

Evaluate : \(\int \frac{(1+\log x)^{2}}{x} d x\)

Answer:

Question 32.

Evaluate : \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x d t\)

Answer:

Put sin ^{-1} ∴ x = t x = sin t

Also \(\frac{1}{\sqrt{1-x^{2}}} d x=d t\)

∴ \(\frac{(x \sin -1 x)}{\sqrt{1-x^{2}}} d x=\int(\sin t) d t\)

= ∫ t sin t dt

Apply Integration by parts.

=t(-cost) – ∫(1)( -cost)dt

= t cos t + ∫ cos t dt

= -tcost + sint

= -sin ^{-1}x \(\sqrt{1-x^{2}}\) + sin (sin^{-1})

= -sin ^{-1}x \(\sqrt{1-x^{2}}\) + x + c

= x – sin^{-1}x \(\sqrt{1-x^{2}}\) +c

Question 33.

Find the area between the curves y = x^{2} and y = x.

Answer:

To find P:

y = x^{2} and y = x.

x = x^{2} => x^{2} – x = 0

⇒ x(x – 1)=0 ⇒ x = 0 or x-1

If x = 0,y = 0 ⇒ (0, 0) x = 1,y = 1=>(1, 1)

That points of Intersection are (0,0) and (1, 1).

∴ Area = \(\int[f(x)-g(x)] d x\)

But f(x) = x, Ig (x) = x^{2}

Thus, the required absolute area is = \(\frac{1}{6}\) Sq.units.

Question 34.

Form the differential equation representing the family of curves y = a sin(x+b) where a and b are arbitrary constants.

Answer:

y=asin(x+b)-(1)

diff (1) w . r. t‘x’

\(\frac{d y}{d x}\)= acos(x+b)-(2)

\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = -asm(x + b) -(3)

Eliminating a and b from (1), (2) and (3)

we get \(\frac{d^{2} y}{d x^{2}}+y=0\)

This is the required differential Equation.

Question 35.

Find the area of a triangle having the points A(l,l,l,) and C (2,3,1) and as its vertices using vector method.

Answer:

We have

Question 36.

Prove that \([\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{d}}]=[\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{e}}]+[\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{d}}]\)

Answer:

Question 37.

Find the distance between parallel lines \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}+\mathbf{m}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\) and \(\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+n(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

Answer:

We have \(\vec{a}_{1}=\hat{i}+2 \hat{j}-4 \hat{k}\) , [/latex]\vec{a}_{2}=3 \hat{i}+3 \hat{j}-5 \hat{k}[/latex] and \(\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\)

Question 38.

Let f : N → R be a function defined by f(x) = 4x^{2} + 12x + 5 .Show that f : N → S, where S is the range of function f, is invertible. Find the inverse of f.

Answer:

Let y ∉ s be any arbitrary element

Then, y = 4x^{2}+12x + 5

Part – D

Answer any six questions. ( 6 x 5 = 30 )

Question 39.

Two cards are drawn successively with Replacement from a well shuffled deck of 52 cards. Find the probability distribution of the number of aces.

Answer:

Let x={0,1,2} denote the number of aces. ∴ p(x = 0) = p (non – ace) x p (non – ace)

= \(\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}\)

P (X = 1)=p (ace I non – ace) + p (non – ace) = p(ace).p(non – ace) + p(non – ace).p(ace)

= \(\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52}=\frac{24}{169}\)

p(x=2)= p(ace& ace) = \(\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}\)

∴ Thus, required probability distribution, is

This shows that gof=I_{N} and fog = I_{S}

⇒ is invertible function

and hence, \(f^{-1}(x)=g=\sqrt{\frac{x+4}{4}}-\frac{3}{2}\)

Question 40.

Verify that (A + B) (C) = AC + BC,

If A = \(\left[\begin{array}{ccc}

{0} & {6} & {7} \\

{-6} & {0} & {8} \\

{7} & {-8} & {0}

\end{array}\right]\), B = \(\left[\begin{array}{lll}

{0} & {1} & {1} \\

{1} & {0} & {2} \\

{1} & {2} & {0}

\end{array}\right]\) and C = \(\left[\begin{array}{c}

{2} \\

{-2} \\

{3}

\end{array}\right]\)

Answer:

Question 41.

Solve the following system of equations by matrix method x+y+z=6, x-y-z = -4 and x+ 2y = -1.

Answer:

Consider

AX=B-(1) => X =A^{-1} B – (2)

Here

A = \(\left[\begin{array}{ccc}

{1} & {1} & {1} \\

{1-1} & {-1} \\

{1} & {2} & {-2}

\end{array}\right]\) , X = \(\left[\begin{array}{l}

{\mathrm{X}} \\

{\mathrm{Y}} \\

{\mathrm{Z}}

\end{array}\right]\) , B = \(\left[\begin{array}{r}

{6} \\

{-4} \\

{-1}

\end{array}\right]\)

To find A^{-1}:

= 1 (2 + 2)- 1 (-2 + 1) + 1(2 + 1)

=1(4) – 1(-1) + 1(3)

=4 + 1 + 3

|A| = 8 ≠ 0

Question 42.

If e^{y}(x+1) = 1 Prove that \(\frac{d y}{d x}=-e^{y}\) and hence prove that \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)

Answer:

e^{y}(X + 1) = 1 ⇒ e^{y} = \(\frac{1}{x+1}\) ….. (1)

Differentiating with respect to ‘x’

Question 43.

The length of a rectangle is decreasing at the rate of 3cm / minute and width – y is increasing at the rate of 2 cm / minute, when x= 10 cm and y = 6 cm, find the rate of change of (i) the perimeter and (ii) the area of the rectangle.

Let → length and y → width.

Data: x = 10cm, y = 6cm

\(\frac{d x}{d t} = -3cm/min,\frac{\mathrm{d} y}{\mathrm{dt}}\) + 2cm / min

(i) perimeter, p=2(x + y)

\(\frac{d p}{d t}=2\left[\frac{d x}{d t}\right]\) = 2[-3+2] = 2(-)

\(\frac{d p}{d t}\) = -2 cm/min

(ii) Araea , A = xy

\(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{y} \frac{\mathrm{dx}}{\mathrm{dt}}\)

= (10)(2) + (6)(-3) = 20-18

\(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\) = 2cm^{2}/min

Question 44.

Find the integral of \(\sqrt{x^{2}+a^{2}}\) w . r .t

‘x’ and hence evaluate \(\int \sqrt{x^{2}+4^{x}+6} d x\)

Answer:

Let I = \(\sqrt{x^{2}+a^{2}}\)

= \(\sqrt{x^{2}+a^{2}}(1) dx\)

Apply Integration by parts

Question 45.

Using Integration find the area of triangle bounded by the verticed A (2, 0), B ( 4, 5) and C (6, 3).

Answer:

Let A (2,0), B = (4, 5), C= (6, 3) be the vertical of a triangle ABC.

Area of ∆ ABC = Area of ∆ ABD + Area of trapezium BDEC – Area of ∆ ACE (i) Equation of the side AB,

(ii) Equation of the side BC,

\(\frac{y-3}{x-6}=\frac{5-3}{4-6}=\frac{2}{-2}\) = -1

y – 3 = – (x – 6)

y-3 = -3 – x + 6

y = 3 – x+6

(iii) Equation of side AC,

\(\frac{y-0}{x-2}=\frac{3-0}{6-2}=\frac{3}{4}\)

y = \(\frac{3}{4}(x-2)\)

Hence, Area of ∆ ABC =

.’. Area of ∆ ABC| = 9 s.q. units Thus, the required area will be 9 sq. units.

Question 46.

Solve the differential Equation ydx -(x+2y^{2}) dy

Answer:

y dx – (x + 2y^{2}) dy = 0

( x + 2y^{2})dy = ydx

This is a Linear Differential Equation in ‘x’.with p = \(\frac{1}{y}\) , Q = 2y?

This is the required General Solution of the ordinary Diffrential Equation.

Question 47.

Derive equation perpendicular to a given vector and passing through a given point both in the vector and cartesian form.

Let \(\overrightarrow{\mathrm{r}}\) be the position vector of any point p (x, y, z) in the plane.

Then the point p lies in the plane if

This is the required equation of a plane in vector form

Cartisian form

This is the required equation of a plane in cartesian form.

Question 48.

There are 5 % defective itemas in a large bulk of items, what is the probability that sample of 10 items will include not more than one defective item?

Answer:

Let p= probability of getting defective item = 0.05

q = probability of getting non – defective

item = 0. 95

and n=10

we have, by Binomial distribution.

p(x) = nC_{x} p^{x} q^{n-x}

p(x) = 10C_{x} (0.05)^{x} (0.95)^{10 – x}

To find p =(x ≤ 1): p = p(x = 0) + p( x = 1)

=10 C_{o} (0.05)_{0} (0.95)_{10} +10C_{1} (0.95)_{9} +10C_{1}(o.o5)^{1} (0.95)^{9}

= (0.95)^{9 }[0.95 + 0.5]

p(x ≤ 1) = (0.95)^{9 }[1.95]

Part – E

Answer any one question : ( 1 x 10 = 10 )

Question 49.

(a) Prove that \(\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a}(x) d x\) when f (2a – x) = f(x) I Hence evaluate \(\int_{0}^{\pi}(\cos x) d x\) (6 Marks)

on using p_{2} we have

Let t = 2a-x ∴ dt = -dx

when x = a; t = a

I x = 2a; t = 0

∴ 2^{nd} Integral in RHS becomes

(b) Prove that \(\left|\begin{array}{ccc}

{\mathbf{1}} & {\mathbf{x}} & {\mathbf{x}^{2}} \\

{\mathbf{x}^{2}} & {\mathbf{1}} & {\mathbf{x}} \\

{\mathbf{x}} & {\mathbf{x}^{2}} & {\mathbf{1}}

\end{array}\right|=\left(\mathbf{x}^{3}-\mathbf{1}\right)^{2}\)

Answer:

Expanding along

= (x^{3}-1)^{2} {[l-0]-0[x-0]+0[x^{2}-x^{2}] }

= (X^{3}-1)^{2}(1)

A = (x^{3} -1)^{2}

Question 50.

(a) Solve the following linear programming problem graphically

Maximise , z =3x+2y S.Tx + 2y ≤ 10,

3x + y ≤ 15,x ≥ 0, y ≥ 0. (6 Marks)

Answer:

consider, x + 2y =10 ………. (1)

∴Equation (l) passes through A(0,5) and B (10,0)

3x + y = 15 …………. (2)

∴ Equation (2) passes through c(0,15) and D (5 , 0)

Here ODEA represents feasible region and it is bounded. The feasible solutions exists at all these points. The optiomal solution is obtained as follows.

Thus, (z)_{max} = 18at x = 4 and y = 3

(b) Find the relation between a and b so that the function f defined by f (x) = {ax + 1,if x≤ 3 is continuous bx + 3, if x > 3} at x = 3. (4 Marks)

Answer:

we have , f(x) = 3a + 1

Also, f is continous at x = 3

⇒ 3a +1 = 3b + 3 = 3a +1

consider, 3a +1 = 3b + 3

3a = 3b + 2

a = \(\frac{3 b+2}{3}\)