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## TN State Board 12th Maths Model Question Paper 5 English Medium

Instructions:

- The question paper comprises of four parts.
- You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
- questions of Part I, II. III and IV are to be attempted separately
- Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
- Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
- Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
- Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours

Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.

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

2 & 0 \\

1 & 5

\end{array}\right]\) and B = \(\left[\begin{array}{ll}

1 & 4 \\

2 & 0

\end{array}\right]\)

then adj (AB)| = ……….

(a) – 40

(b) -80

(c) -60

(b) -20

Answer:

(b) -80

Question 2.

i^{n} + 1^{n+1} + i^{n + 2} + i^{n +3} is ……….

(a) 0

(b) 1

(c)- 1

(d) i

Answer:

(a) 0

Question 3.

If ω = \(\text { cis } \frac{2 \pi}{3}\) , then the number of distinct roots of

\(\left|\begin{array}{ccc}

z+1 & \omega & \omega^{2} \\

\omega & z+\omega^{2} & 1 \\

\omega^{2} & 1 & z+\omega

\end{array}\right|=0\)

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(a) 1

Question 4.

sin^{-1}(cos x) = \(\frac{\pi}{2}-x\) is valid for ……….

(a) -π ≤ x ≤0

(b) 0 ≤ x ≤ π

(c) \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)

(d) \(-\frac{\pi}{4} \leq x \leq \frac{3 \pi}{4}\)

Answer:

(b) 0 ≤ x ≤ π

Question 5.

tan^{-1} x + cot^{-1} = …….

(a) 1

(b) -π

(c) π/2

(d) π

Answer:

(c) π/2

Question 6.

The equation of the normal to the circle x^{2} + y^{2} – 2x – 2y + 1 = 0 which is parallel to the line 2x + 4y = 3 is …………..

(a) x + 2y = 3.

(b) x + 2y + 3 = 3

(c) 2x + 4y + 3 = 0

(d) x – 2y + 3 = 0

Answer:

(a) x + 2y = 3.

Question 7.

The axis of the parabola x^{2} = 20y is ……….

(a) y = 5

(b) x = 5

(c) x = 0

(d) y = 0

Answer:

(c) x = 0

Question 8.

If \(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}\) for non-coplanar vectors \(\vec{a}, \vec{b}, \vec{c}\) then…….

(a) \(\vec{a}\) parallel to \(\vec{b}\)

(b) \(\vec{b}\) parallel to \(\vec{c}\)

(c) \(\vec{c}\) parallel to \(\vec{a}\)

(d) \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

Question 9.

The vector equation of a plane whose distance from the origin is p and perpendicular to a unit vector n̂ is ….

(a) \(\vec{r} \cdot \vec{n}=p\)

(b) \(\begin{aligned}

&-n\\

&r \cdot n=q

\end{aligned}\)

(c) \(\vec{r} \times \vec{n}=p\)

(d) \(\vec{r} \cdot \hat{n}=p\)

Question 10.

The point of inflection of the curve y=(x – 1)^{3} is …………..

(a) (0,0)

(b) (0,1)

(c) (1,0)

(d) (1,1)

Answer:

(c) (1,0)

Question 11.

The asymptote to the curve y^{2}(1 + x) = x^{2} (1 – x) is …………..

(a) x = 1

(b) y = 1

(c) y = -1

(d) x = -1

Answer:

(d) x = -1

Question 12.

The solution of a linear differential equation ax + Px = Q where P and Q are function of y, is ……….

(a) y (I.F) = ∫(I.F) Q dx + c

(b) y (I.F) = ∫(1.F) Q dy + c

(c) y (I.F) = ∫(1.F) Q dy + c .

(d) x (I.F) = ∫(1.F) Q dx + c

Question 13.

A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is ……..

(a) 0.2%

(b) 0.4%

(c) 0.04%

(d) 0.08%

Answer:

(b) 0.4%

Question 14.

For any value of n∈z, \(\int_{0}^{\pi} e^{\cos ^{2} x} \cos ^{3}[(2 n+1) x] d x \) is……….

(a) π/2

(b) π

(c) 0

(d) 2

Answer:

(c) 0

Question 15.

If n is odd then \(\int_{0}^{\pi / 2} \sin ^{n} x d x\) is …. ….

Answer:

(d) \(\frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdot \frac{n-5}{n-4} \dots \frac{2}{3} \cdot 1\)

Question 16.

The solution of \(\frac{d y}{d x}=2^{y-x}\)is ……….

(a) 2^{x} + 2^{y} = c

(b) 2^{x} – 2^{y} = c

(c) \(\frac{1}{2^{x}}-\frac{1}{2^{y}}=c\)

(d) x + y = c

Answer:

(c) \(\frac{1}{2^{x}}-\frac{1}{2^{y}}=c\)

Question 17.

If p and q are the order and degree of the differential equation \(y \frac{d y}{d x}+x^{3}\left(\frac{d^{2} y}{d x^{2}}\right)+x y=\cos x\) when…..

(a) p < q

(b) p = q

(c) p > q

(d) p exists and q does not exist

Answer:

(c) p > q

Question 18.

A pair of dice numbered 1, 2, 3, 4, 5, 6 of a six-sided die and 1, 2, 3, 4 of a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. Then the number of elements in the inverse image of 7 is ……….

(a) 1

(6) 2

(c) 3

(d) 4

Answer:

(d) 4

Question 19.

If in 6 trials, X is a binomial variate which follows the relation 9P(X = 4) = P(X = 2),

then the probability of success is …….

(a) 0.125 .

(b) 0.25

(c) 0.375

(d) 0.75

Answer:

(b) 0.25

Question 20.

Which one of the following statements has truth value F ?

(a) Chennai is in India or √2 is an integer

(b) Chennai is in India or √2 is an irrational number

(c) Chennai is in China or √2 is an integer

(d) Chennai is in China or √2 is an irrational number

Answer:

(c) Chennai is in China or √2 is an integer

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 x 2 = 14)

Question 21.

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

8 & -4 \\

-5 & 3

\end{array}\right]\), verify that A(adj A) = (adj A)A = |A|I_{2},

Question 22.

Find the value of \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)

Answer:

We know that the 9th roots of unity are 1, 6, 62, …, 08 Sum of the roots:

1 + ω + ω^{2} + … + ω^{8} = 0

⇒ ω + ω^{2} + … + ω^{8} = -1

Question 23.

Solve the equation 😡^{4} – 14x^{2} + 45 = 0.

Answer:

The given equation is x^{4} – 14x^{2} + 45 = 0

Let x^{2} = y

y^{2} – 14y + 45 = 0 ⇒ (y – 9)(y – 5) = 0

y = 9,5

∴ x^{2} = 9, x^{2}= 5

x = ±3, x = ±√5

∴ The roots are ±3, ±√5

Question 24.

Find the principal value of sin \(\sin ^{-1}\left(-\frac{1}{2}\right)\) (in radians and degrees).

Answer:

Question 25.

Show that the points (2,3,4),(-1,4,5) and (8,1,2) are collinear.

Answer:

Given points are (2, 3, 4), (-1,4,5) and (8,1,2). Equation of the line joining of the first and

second point is

\(\frac{x-2}{-3}=\frac{y-3}{1}=\frac{z-4}{1}=m(\text { say })\)

(-3m + 2, m + 3, m + 4)

On putting m = -2, we get the third point is (8, 1, 2)

∴ Given points are collinear.

Question 26.

Evaluate: \(\int_{0}^{2 \pi} \frac{\cos x}{\sqrt{4+3 \sin x}} d x\)

Answer:

I = \(\int_{0}^{2 \pi} \frac{\cos x}{\sqrt{4+3 \sin x}} d x\)

Put 4 + 3 sin x = t, so that 3 cos x dx = dt = cosx dx = \(\frac{1}{3} dt\) When x = 0, t = 4 + 3(0) = 4 and when x = 2π, t = 4+ 3(0) = 4

Question 27.

Find the order and degre of the diferential equation center equation \(\frac{d^{2} y}{d x^{2}}-y+\left(\frac{d y}{d x}+\frac{d^{3} y}{d x^{3}}\right)^{\frac{3}{2}}=0\)

Answer:

Question 28.

Suppose the amount of milk sold daily at a milk booth is distributed with a minimum

of 200 litres and a maximum of 600 litres with probability density function

\(f(x)=\left\{\begin{array}{lr}

k & 200 \leq x \leq 600 \\

0 & \text { otherwise }

\end{array}\right.\) Find the probability that daily sales will fall between 300 otherwise litres and 500 litres?

Answer:

The probability that daily sales will fall between 300 litres and 500 litres is

Question 29.

Let A = \(\left(\begin{array}{llll}

1 & 0 & 1 & 0 \\

0 & 1 & 0 & 1 \\

1 & 0 & 0 & 1

\end{array}\right)\) ,B = \(\left(\begin{array}{llll}

0 & 1 & 0 & 1 \\

1 & 0 & 1 & 0 \\

1 & 0 & 0 & 1

\end{array}\right)\) , C = \(\left(\begin{array}{llll}

1 & 1 & 0 & 1 \\

0 & 1 & 1 & 0 \\

1 & 1 & 1 & 1

\end{array}\right)\) be any three boolean matrices of the same type. Find A V B

Answer:

Question 30.

Find the general equation of a circle with centre (-3,- 4) and radius 3 units.

Answer:

Equation of the circle in standard form is (x – h)^{2} + (y – k)^{2} = p^{2}

(x-(-3))^{2} + (y-(-4))^{2} = 3^{2} .

(x + 3)^{2} + (y + 4)^{2} = 3^{2}

x^{2} + y^{2} + 6x + 8y + 16 = 0.

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 x 3 = 21]

Question 31.

If A = \(-\left[\begin{array}{lll}

8 & 1 & 4 \\

4 & 4 & 7 \\

1 & 8 & 4

\end{array}\right]\) prove that A^{-1} = A^{T}.

Answer:

Question 32.

If z_{1} = 2 + 5i, z_{2 }= -3 – 4i, and z_{3} = 1 + i, find the additive and multiplicative inverse of z_{1}, z_{2}, and z_{3}.

Answer:

Question 33.

Solve the equation 3x^{3} – 26x^{2} + 52x – 24 = 0 if its roots form a geometric progression

Answer:

The given equation is 3x^{3} – 26x^{2} + 52x – 24 = 0

Given that the root are GP

∴ The roots are \(\frac{\alpha}{r}\) , α, αr

\(\frac{\alpha}{r}\) x α x αr = (-8)

α^{3} = 8

α = 2

Question 34.

Find the value of \(\cot \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}\right)\)

Answer:

Question 35.

If the normal at the point“ t_{1}‘ on the parabola y^{2} = 4ax meets the parabola again at the point ‘t_{2}’, then prove that \(t_{2}=-\left(t_{1}+\frac{2}{t_{1}}\right)\)

Answer:

Equation of normal to y^{2} = 4ax at ‘t’ is y + xt = 2at + at^{3}.

So equation of normal at ‘t_{1},’ is y + xt_{1}, = 2at_{1} + at_{1}^{3} ……….(1)

The normal meets the parabola y^{2} = 4ax at “t_{2}‘ (ie.,) at (at_{2}^{2}, 2at_{2})

Question 36.

Find the coordinates of the foot of the perpendicular drawn from the point (-1, 2, 3) to the straight line \(\vec{r}=(\hat{i}-4 \hat{j}+3 \hat{k})+t(2 \hat{i}+3 \hat{j}+\hat{k})\). Also, find the shortest distance from the given point to the straight line.

Answer:

Comparing the given equation \(\vec{r}=(\hat{i}-4 \hat{j}+3 \hat{k})+t(2 \hat{i}+3 \hat{j}+\hat{k})\) with \(\vec{r}=\vec{a}+t \vec{b}\) we get \(\vec{a}=\hat{i}-4 \hat{j}+3 \hat{k} \hat{k}\) and \(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\). We denote the given point (-1,2,3) by D, and the point (1,-4,3) on the straight line by A . If F is the foot of the perpendicular from D to the straight line, then F is of the form (2t+1, 3t – 4, t + 3) and \(\overrightarrow{\mathrm{DF}}=\overrightarrow{\mathrm{OF}}-\overrightarrow{\mathrm{OD}}=(2 t+2) \hat{i}+(3 t-6) \hat{j}+t \hat{k}\) .

Since \(\vec{b}\) is perpendicular to \(\overrightarrow{\mathrm{DF}}\), we have

Line \(\vec{b} \cdot \overrightarrow{\mathrm{DF}}\) = 0 = 2(2t + 2) + 3(3-6) +1(t) = 0 = t=1

Therefore, the coordinate of F is (3,-1, 4).

Now, the perpendicular distance from the given point to the given line is

\(\mathrm{DF}=|\overrightarrow{\mathrm{DF}}|=\sqrt{4^{2}+(-3)^{2}+1^{2}}=\sqrt{26}\) units.

Question 37.

Find the asymptotes of the curve f(x) = \(\frac{x^{2}+6 x-4}{3 x-6}\)

Answer:

Since the numerator is of highest degree than the denominator. We have a slant

asymptote to find it we have to divide the numerator by the denominator.

So the equation of asymptote is \(y=\frac{x}{3}+\frac{8}{3}\) and 3x – 6 = 0

x = 2

Question 38.

Evaluate: \(\int_{0}^{1} x(1-x)^{n} d x\)

Answer:

Question 39.

For the distribution function given by \(\mathbf{F}(x)=\left\{\begin{array}{ll}

0 & x<0 \\ x^{2} & 0 \leq x \leq 1 \\ 1 & x>1

\end{array}\right.\), find the density function. Also evaluate

(i) P(0.5<x<0.75), (ii) P(X ≤ 50.5) (iii) P(x > 0.75)

Answer:

Question 40.

Let w (x, y) = \(x y+\frac{e^{y}}{y^{2}+1}\) for all (x, y) ∈ R^{2}. Calculate \(\frac{\partial^{2} w}{\partial y \partial x}\) and \(\frac{\partial^{2} w}{\partial x \partial y}\)

Answer:

Part – IV

IV. Answer all the questions. [7 x 5 = 35]

Question 41.

(a) An amount of ₹ 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is ₹ 5,000. The income from the third bond is ₹ 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)

Answer:

Let the amount invested in 6% bond be ₹ x

and the amount invested in 8% bond be ₹y

and the amount invested in 10% bond be ₹z

Now x + y + z = 65000

(i.e) 6x + 8y + 10z = 500000

(÷by 2) 3x + 4y + 5z = 250000

Also given that

\frac{10}{100} z-\frac{8}{100} y = 800 (i.e) -8y + 10z = 80000

(÷ by 2) -4y + 5z = 40000

(i.e,) x + y + z = 65000…..(1)

y + 2z = 55000…..(2)

13z = 260000 …..(3)

Form (3) z = \(\frac{260000}{13}\)

Substituting z = ₹ 20000 in (2) we get

y + 40000 = 55000 = y ⇒ 55000 – 40000 = ₹ 15000

Substituting z = ₹ 20000, y = ₹ 15000 in (1) we get

x + 15000 + 20000 = 65000

x = 65000 – 35000 = ₹ 30000

So the amount invested in

6% bond x = ₹ 30000

8% bond y = ₹ 15000

and 10% bond z = ₹ 20000

[OR]

(b) Find the equation of the curve whose slope is \(\frac{y-1}{x^{2}+x}\) and which passes through the point (1, 0).

Answer:

Question 42.

(a) If arg (z – 1) = \(\frac{\pi}{6}\) and arg (z + 1) = 2\(\frac{\pi}{3}\) , then prove that |z| = 1.

Answer:

Given arg (z-1) = \(\frac{\pi}{6}\) = 30° and arg (z + 1) = \(\frac{2 \pi}{3}\) = 120°

So, arg (z + 1) – arg (z – 1) = 120° – 30° = 90°

ie, ang \(\frac{z+1}{z-1}=90^{\circ}=\frac{\pi}{2}\)

\(\operatorname{Re} \frac{z+1}{z-1}=0\) [∴arg = \(\tan \theta=\frac{I P}{R P}\)

Therefore x= -1 and x=1 are vertical asymptotes.

The rough sketch of the curve is shown on the right side.

Question 43.

(a) Solve the equation x^{3} – 9x^{2} + 14x + 24 = 0 if it is given that two of its roots are in the ratio 3:2.

Answer:

The given equation is x^{3} – 9x^{2} +14x + 24 = 0…

Since, the two roots are in the ratio 3:2.

The roots are α , 3λ, 2λ

α + 3λ + 2λ = -b = 9

= a + 5λ = 9 (a) ….. (1)

α (3λ) (2λ) = -24

6λ^{2}a = -24 => λ^{2}α = -4 …….(2)

α = 9 – 5λ

λ^{2} (9 – 52) = -4

9λ^{2} – 5λ^{3} + 4 = 0

5λ^{3} – 9λ^{2} – 4 = 0

(λ – 2) (5λ^{2} + λ + 2) = 0

λ = 2,5λ^{2} + λ+ 2 = 0 has only Imaginary roots ∆ < 0

∴ when λ = 2, α = 9-5 (2) = 9 – 10 = -1

∴ The roots are α, 3λ, 2λ i.e., -1, 6, 4

[OR]

(b) Show that the lines \(\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\) and \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\)

are coplanar. Also -3 .. 1 5 -1 2 find the equation of the plane containing these two lines.

Answer:

From the lines we have,

(x_{1}, y_{1}, z_{1}) = (-3, 1, 5); (b_{1}, b_{2}, b_{3}) = (-3, 1, 5)

(x_{2}, y_{2}, z_{2}) = (-1, 2, 5); (d_{1}, d_{2}, d_{3}) = (-1, 2,5)

Conditions for coplanar

\(\left|\begin{array}{ccc}

2 & 1 & 0 \\

-3 & 1 & 5 \\

-1 & 2 & 5

\end{array}\right|\) = 2(5 – 10) – 1(-15 + 5) + ((-6 + 1) = -10 + 10 = 0

Given two lines are coplanar

(x + 3)(5 – 10) – (y – 1)[-15 + 5) + (2 – 5)(-6 + 1] = 0

5(x + 3) + 10(y – 1) – 5(z – 5) = 0

(÷ by -5) ⇒ (x + 3) – 2 (y – 1) + (z – 5) = 0

x + 3 – 2y + 2 + 2 – 5 = 0 .

or x – 2y + z = 0

(x + 1)(5 – 10) – (y – 2)(-15 + 5) + (z – 5)(-6 + 1) = 0

-5(x + 1) + 10(y – 2) – 5(2 – 5) = 0

(x + 1) – 2(y – 2) + (2 – 5) = 0

x + 1 – 2y + 4 + 2 – 5 = 0

x – 2y + z = 0

Question 44.

(a) Find the area of the region bounded by the parabola y^{2} = x and the line y = x -2.

Answer:

To find the points of intersection solve the two equations y^{2} = x and y=x-2

(b) Solve \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x}\)

Answer:

Question 45.

(a) Solve \(\cos \left(\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)\right)=\sin \left\{\cot ^{-1}\left(\frac{3}{4}\right)\right\}\)

Answer:

(b) Find the mean and variance of a random variable X, whose probability density x for x 20 function is \(f(x)=\left\{\begin{array}{cc}

\lambda e^{-\lambda x} & \text { for } x \geq 0 \\

0 & \text { otherwise }

\end{array}\right.\)

Answer:

Observe that the given distribution is continuous

Question 46.

(a) Show that the equation of the normal to the curve x= a cos^{3}θ, y = a sin^{3} θ at ‘θ’ is

x cos θ – y sin θ= a cos 2 θ.

Answer:

x = a cos^{3}θ

\(\frac{d x}{d \theta}\) = a[ 3cos^{2}θ](-sin θ) = -3acos^{2}θ sin θ

y= a sin^{3}θ

Slope of the normal = \(\frac{-1}{m}=\frac{\cos \theta}{\sin \theta}\)Point = (x1, yy) = (a cos^{3} θ , a sin^{3} θ ) sin θ

So, equation of the normal is

Y-a sin^{3} θ = \(\frac{\cos \theta}{\sin \theta}\)(x- a cos^{3} θ)

i.e. y sin θ – a sin^{4} θ = x cos θ – a cos^{4} θ

x cos θ – sin θ = a cos^{4} θ – a sin^{4} θ

i.e x cos θ – y sin θ = a [cos^{2}θ+ sin^{2}θ] [cos^{2}θ – sin^{2}θ]

= a [cos 2 θ]

So, the equation of the normal is x cos θ – y sin θ = a cos 2θ

[OR]

(b) Verify (i) closure property (ii) commutative property (iii) associative property (iv) existence of identity and (v) existence of inverse for the operation +_{5} on Z_{5} using table corresponding to addition modulo 5.

Answer:

It is known that Z_{5} = {[O], [1], [2], [3], [4]} . The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0],[1],[2],[3],[4]}.

(i) Since each box in the table is filled by exactly one element of Z_{5}, the output a +_{5}b is unique

and hence +_{5} is a binary operation.

(ii) The entries are symmetrically placed with respect to the main diagonal. So +_{5} has commutative

property.

(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual.

For instance, (2 +_{5}3) +_{5} 4 = 0 +_{5} 4 = 4 (mod 5)

and 2 +_{5} (3 +_{5} 4) = 2 +_{5} 2 = 4 (mod 5). Hence (2 +_{5}3) +_{5} 4 = 2 +_{5} (3 +_{5}4).

Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +_{5} is associative.

(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.

(v) The existence of inverse is guaranteed provided the identity O exists in each row and each column. From Table-A, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z_{5} , say 2 is outlined below.

First find the position of the identity element 0 in the III row headed by 2. Move horizonta along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2 +_{5} 3 = 0 (mod 5). In this way, the inverse of each and every element of Z_{5} can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2 , and that of 4 is 1.

Question 47.

(a) Identify the type of conic and find centre, foci, vertices, and directrices the following:

18x^{2} + 12y^{2} – 144x + 48y + 120 = 0

Answer:

18x^{2} + 12y^{2} – 144x + 48y + 120 = 0

(18x^{2} – 144x) + (12y^{2} + 48y) = -120

18(x^{2} – 8x) + 12 ( 12 + 4y) = -120

18(x^{2} – 8x + 16 – 16) + 12 (12+ 4y + 4 – 4) = -120

18(x – 4)^{2} – 288 + 12(y + 2)^{2} – 48 = -120

18(x – 4)^{2} + 12(y + 2)^{2} = -120 + 288 + 48 = 216

[OR]

(b) Prove that g(x, y) = x log \(\left(\frac{y}{x}\right)\) is homogeneous; what is the degree? Verify Euler’s Theorem for g.

Answer:

g (x, y) = x log(y/x)

8 (tx, ty) = (tx) log \(\left(\frac{t y}{t x}\right)\)

= txlog \(\left(\frac{y}{x}\right)\)

g(tx, ty) = t g(x, y)

∴ ‘g’is a homogeneous function of degree 1. By Euler’s theorem, we have

∴ Euler’s theorem verified.