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## TN State Board 12th Maths Model Question Paper 2 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 Pan 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}

7 & 3 \\

4 & 2

\end{array}\right]\) then 9I_{2} – A =

(a) A^{-1}

(b) \(\frac{A^{-1}}{2}\)

(c) 3A^{-1}

(d) 2A^{-1}

Answer:

(d) 2A^{-1}

Question 2.

If (1+ i) (1+ 2i) (1+ 3i)…(1+ ni) = x + iy, then 2-5T0…(1 + n^{2} ) is ..

(a) 1

(b) i

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

(d) 1 + n^{2}

Answer:

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

Question 3.

If p + iq = (2 – 3i) (4 + 2i) then q is

(a) 14

(b) -14

(c) -8

(d) 8

Answer:

(c) -8

Question 4.

A zero of x^{3} + 64 is

(a) 0

(b) 4

(c) 4i

(d) -4

Answer:

(d) -4

Question 5.

sin^{-1}(2COS^{2}A – 1) + cos^{-1}(1 – 2 sin^{2}x) = .

(a) π/2

(b) π /3

(c) π/4

(d) π/6

Answer:

(a) π/2

Question 6.

If cot^{-1} x = \(\frac{2 \pi}{5}\) for some x ∈ R, the value of tan^{-1} x is

(a) \(-\frac{\pi}{10}\)

(b) \(\frac{\pi}{5}\)

(c) \(\frac{\pi}{10}\)

(d) \(-\frac{\pi}{5}\)

Answer:

(c) \(\frac{\pi}{10}\)

Question 7.

An ellipse has OB as semi minor axes, F and F’ its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is

Answer:

(a) \(\frac{1}{\sqrt{2}}\)

Question 8.

The focus of the parabola x^{2} = 20 y is

(a) (0, 0)

(b) (5, 0)

(c) (0, 5)

(d) (-5, 0)

Answer:

(c) (0, 5)

Question 9.

If the planes \(\vec{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3\) and \((4 \hat{i}+\hat{j}-\mu \hat{k})=5\) are parallel, then the value of λ and μ are

Answer:

(c) \(-\frac{1}{2}\) , -2

Question 10.

The unit normal vectors to the plane 2x – y + 2z = 5 are

Question 11.

The slope of the line normal to the curve f(x) – 2cos 4x at x = \(\frac{\pi}{2}\) is ……………….

(a) -4√3

(b) -4

(c) √3/12

(d) 4√3

Answer:

(c) √3/12

Question 12.

The curve y^{2} = x^{2} ( 1 – x) has

(a) only one loop between x = -1 and x = 0

(b) only one loop between x = 0 and x = 1

(c) two loops between x = -1 and x = 1

(d) no loop

Answer:

(b) only one loop between x = 0 and x = 1

Question 13.

If w (x, y) = x^{y}, x > 0, then \(\frac{\partial w}{\partial x}\) is equal to

(a) x^{y }logx

(b) y logx

(c) yx^{y-1}

(d) xlogy

Answer:

(c) yx^{y-1}

Question 14.

If (x, y, z) = xy + yz + zx, then f_{x} – f_{z} is equal to

(a) z – x

(b) y – z

(c) x – z

(d) y – x

Answer:

(a) z – x

Question 15.

If \(\frac{\Gamma(n+2)}{\Gamma(n)}\) = 90 then n is

(a) 10

(b) 5

(c) 8

(d) 9

Answer:

(d) 9

Question 16.

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

Answer:

(b)

Question 17.

The slope at any point of a curve y =/(x) is given by \(\frac{d y}{d x}=3 x^{2}\) and it passes through (-1,1).

Then the equation of the curve is

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

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

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

(d) y = x^{3} + 5

Answer:

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

Question 18.

The solution of the differential equation \(\frac{d y}{d x}+\frac{1}{\sqrt{1-x^{2}}}=0\) is

(a) y + sin^{-1}x = c

(b) x + sin^{-1}y = 0

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

(d) x^{2} + 2sin^{-1}y = 0

Question 19.

Suppose that x takes on one of the values 0, 1, and 2. If for some constant k,

P(X = i) = k P(x = i – 1) for i = 1, 2 P(x = 0) = 1/7. Then the value of k is

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(b) 2

Question 20.

Determine the truth value of each of the following statements:

(a) 4 + 2 = 5 and 6 + 3 = 9

(b) 3 + 2 = 5 and 6 + 1 = 7

(c) 4 + 5 = 9 and 1 + 2 = 4

(d) 3 + 2 = 5 and4 + 7 = 11

Answer:

(1) (a) F

(b) T

(c) F

(d) T

Part – II

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

Question 21.

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

4 & 3 \\

2 & 5

\end{array}\right]\) , find x and y such that A^{2} + xA + yI_{2} = O_{2}. Hence , find A^{-1}

Answer:

So, we get 22 + 4x + y = 0, 31+ 5x + y – 0, 27 + 3x = 0 and 18 + 2x = 0.

Hence x = -9 and y = 14. Then, we get A^{2} – 9A + 14I_{2} = O_{2}

Post-multiplying this equation by A^{-1}, we get A – 9I + 14A^{-1} – I_{2} = O_{2}. Hence, we get

Question 22.

Find the principal argument arg z, when z = \(\frac{-2}{1+i \sqrt{3}}\)

Answer:

This implies that one of the values of arg z is \(\frac{2 \pi}{3}\)

Since \(\frac{2 \pi}{3}\) lies between -π and π , the principal argument arg z is \(\frac{2 \pi}{3}\)

Question 23.

Verify whether the line \(\frac{x-3}{-4}=\frac{y-4}{-7}=\frac{z+3}{12}\) lies in the plane 5x – y + z = 8 .

Answer:

Here, (x_{1}, y_{1}, z_{1}) = (3,4, -3) and direction ratios of the given straight line are (a, b, c) – (-4, -7,12).

Direction ratios of the normal to the given plane are (A, B, C) = (5, -1,1).

We observe that, the given point (x_{1}, y_{1}, z_{1}) = (3,4, -3) satisfies the given plane 5x – y + z = 8

Next, aA+ bB + cC = (-4)(5) + (-7)(-l) + (12)( 1) = -1 ≠ 0. So, the normal to the plane is not perpendicular to the line. Hence, the given line does not lie in the plane.

Question 24.

Evaluate \(\begin{aligned}

&\lim\\

&x \rightarrow 0^{+} \quad x^{2} \log e^{x}

\end{aligned}\)

Answer:

Question 25.

Find a linear approximation for the function at the indicated points.

g(x) = \(\sqrt{x^{2}+9}\), x_{0} = -4

The required linear approximation L(x) = g (x_{0}) + g’ (x_{0}) (x – x_{0})

Question 26.

Evaluate: \(\int_{0}^{2 \pi} x^{2} \sin n x d x\) , Where n is a positive integer.

Answer:

Taking u = x^{2} and v = sin nx, and applying the Bernoulli’s formula, we get

Question 27.

Form the differential equation from the equation y^{2} = 4a (x – a)

Answer:

y^{2}= 4 a(x-a) …(1)

Differentiating, 2yy’ = 4a …(2)

Eliminating a between (1) and (2) we get

(yy ‘)^{2} – 2xyy’ + y^{2} = 0

Question 28.

Compute P(X = k) for the binomial distribution, B(n,p) where n = 6, p = 1/3, k = 3

Answer:

Given n = 6, p = 1/3,k = 3

∴ q = 1 – p = \(1-\frac{1}{3}=\frac{2}{3}\)

P(X = x) = nC_{x}p^{x}q^{n-x}, x = 0, 1,2,…………n

P (X = k) = P (X = 3)

Question 29.

Let A = {a + √5b : a, b∈Z} . Check whether the usual multiplication is a binary operation on A.

Answer:

Let A = a + √5 b and B = c + √5 d, where a, b,c,d∈ R.

Now A * B = (a + √5 b) (c + √5 d)

= ac + √5 ad + √5 bc + √5 b √5 d

= (ac + 5 bd) + 45 (ad + bc) ∈ A .

Where a, b,c,d ∈ Z.

So * is a binary operation.

Question 30.

Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} s \quad \frac{\pi}{9}\right)\)

Answer:

Part – III

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

Question 31.

Solve the following system of linear equations, using matrix inversion method:

5x + 2y – 3, 3x + 2y – 5.

Answer:

The matrix form of the system is AX = B, where A :\(\left[\begin{array}{ll}

5 & 2 \\

3 & 2

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

x \\

y

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

3 \\

5

\end{array}\right]\)

We find |A| = \(\left|\begin{array}{ll}

5 & 2 \\

3 & 2

\end{array}\right|\) = 10 – 6 = 4 ≠ 0

So A^{-1} exists A^{-1} = \(\frac{1}{4}\left[\begin{array}{rr}

2 & -2 \\

-3 & 5

\end{array}\right]\)

Then, applying the formula X = A 1B, we get

So the solution is (x = -1 ,y = 4).

Question 32.

If z_{1} = 2 – i and z_{2} = -4 + 3i, find the inverse of z_{1} z_{2} and \(\frac{z_{1}}{z_{2}}\)

Answer:

z_{1} = 2 – i, z_{2} = -4 + 3i

Question 33.

If k is real, discuss the nature of the roots of the polynomial equation 2x^{2} + kx + k = 0, in terms of k.

Answer:

The given quadratic equation is 2x^{2} + kx + k = 0

a = 2,b = k,c = k

∆ = b^{2} – 4ac = k^{2} – 4(2) k = k^{2} – 8k

(i) If the roots are equal

k^{2} – 8k = 0 ⇒ k(k -8) = 0

k = 0, k = 8

(it) If the roots are real

k^{2} – 8k > 0

k (k – 8) > 0

k ∈ (-∞, 0) ∪ (8, ∞)

(iii) ‘If this roots are imaginary

k^{2} – 8k < 0 ⇒ f ∈ (0, 8)

Question 34.

Find the number of solution of the equation tan^{-1} (x – 1) + tan^{-1} x + tan^{-1} (x + 1) = tan^{-1}(3x).

Answer:

tan^{-1}(x – 1) + tan^{-1}x + tan^{-1}(x + 1) = tan^{-1}(3x)

tan^{-1}(x – 1) + tan^{-1}(x + 1) = tan^{-1} 3x – tan^{-1} x

So, the equation has 2 solutions.

Question 35.

Find the non-parametric form of vector equation, and Cartesian equations of the plane \(\vec{r}=(6 \hat{i}-\hat{j}+\hat{k})+s(-\hat{i}+2 \hat{j}+\hat{k})+t(-5 \hat{i}-4 \hat{j}-5 \hat{k})\)

Answer:

Non-parametric form of vector equation

Question 36.

Find two positive numbers whose sum is 12 and their product is maximum.

Answer:

Let the two numbers be x, 12 – x.

Their product p = x (12 – x) – 12x – x^{2}

To find the maximum product.

p’ (x) = 12 – 2x

p”(x) = -2

p’ (x) = 0 ⇒ 12 – 2x = 0 2x = 12 => x = 6

at x = 6, p”(x) = -2 = -ve

⇒ p is maximum at x = 6

when x = 6, 12-x = 12 – 6 = 6

So the two numbers are 6, 6

Question 37.

Find the differential equation that will represent family of all circles having centres on the x-axis and the radius is unity.

Answer:

Equation of a circle with centre on x-axis and radius 1 unit is

(x-a)^{2} + y^{2} = 1 …(1)

Differentiating with respect to x,

2 (x-a) + 2yy’ = 0

⇒ 2 (x – a) = – 2yy’

(or) x-a = -yy’ …(2)

Substituting (2) in (1), we get,

(-yy’)^{2} + y^{2} = 1

(i.e. y^{2} (y’)^{2} + y^{2} = 1 ⇒ y^{2} [1 + (y’)^{2}] = 1

Question 38.

The probability that a certain kind of component will survive a electrical test is \(\frac{3}{4}\). Find

the probability that exactly 3 of the 5 components tested survive.

Answer:

Given n = 5

Probability that a component survive in a test =p = \(\frac{3}{4}\)

∴ q = 1 – p = \(1-\frac{3}{4}=\frac{1}{4}\)

Let ‘X’ be the random variable denotes the number of components survived in a test.

Probability of ‘x’ successes in V trials is

P (X = x) = nC_{x}p^{x}q^{n-x}, x = 0, 1,2,………n

Probability that exactly 3 components survive

Question 39.

Fill in the following table so that the binary operation * on A = {a,b,c} is commutative.

Answer:

Given that the binary operation * is Commutative.

To find a * b :

a * b = b * a (∵ * is a Commutative)

Here b * a = c. So a * b = c

To find a * c :

a * c = c * a (∵ * is a Commutative)

c * a = a. (Given)

So a * c = a

To find c * b :

c * b = b * c

Here b * c = a. So c * b = a

Question 40.

Find the area of the region enclosed by y^{2} = x and y = x – 2.

Answer:

The points of intersection of the parabola y^{2} = x and the line y = x – 2 are (1, – 1) and (4, 2)

To compute the region [shown in the figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect to y no splitting is necessary.

Part – IV

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

Question 41.

(a) Use product \(\left[\begin{array}{rrr}

1 & -1 & 2 \\

0 & 2 & -3 \\

3 & -2 & 4

\end{array}\right]\left[\begin{array}{rrr}

-2 & 0 & 1 \\

9 & 2 & -3 \\

6 & 1 & -2

\end{array}\right]\)

to solve the system of equations x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2.

Answer:

The system of equations can be written in the form AX = C, where

Thus x = 0, y = 5 and z = 3

(b) Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Figure.

Answer:

Question 42.

(a) Simplify (-√3 + 3i)^{31}

Answer:

Let – √3 + 3i = r (cos θ + z sin θ). Then, we get

(b) Solve the following equation x^{4} – 10x^{3} + 26x^{2} – 10x + 1 = 0.

Answer:

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as

[(y^{2} – 2) – 10y + 26] = 0 ⇒ (y^{2} – 10y + 24) = 0 ⇒ (y – 6) {y – 4) = 0 ⇒ y = 6 (or) 4

Hence the roots are 3 ± 2 √2, 2 ± √3

Question 43.

(a) Solve

Answer:

(b) Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?

Answer:

Let P be the principal

Rate of interest 5 %

Given, when t = 0, P = 10000

⇒ c = 10000

Question 44.

(a) Find the equation of the circle through the points (1, 0), (-1, 0), and (0, 1).

Answer:

Let the required circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……(A)

The circle passes through (1, 0), (-1, 0) and (0, 1)

(1,0) ⇒ 1 +0 + 2g(1) + 2f(0) + c = 0 -(A)

2g + c = -1

(-1, 0) ⇒ 1 + 0 + 2g (-1) + 2/(0) + c = 0 …(1)

-2g + c = -1

(0,1)⇒ 0+ 1 +2g(0) + 2f(l) + c = 0 …(2)

2 + c = -1

Now solving (1), (2) and (3) …(3)

2g + c = -1 …(1)

-2g + c = -1 (2)

(1) + (2) ⇒ 2c = -2 ⇒ c = -1

Substituting c = -1 in (1) we get

2g – 1 = -1 2g = -1 + 1= 0 ⇒ g = 0

Substituting c = -1 in (3) we get

2f – 1 = -1 ⇒ 2f = -1 + 1 = 0 ⇒ f = 0

So we get g = 0,f= 0 and c = -1

So the required circle will be

x^{2} + y^{2} + 2(0)x + 2(0)y – 1 = 0

(i.e) x^{2} + y^{2} – 1 = 0 ⇒ x^{2} + y^{2} = 1

[OR]

(b) Find the area of the loop of the curve 3ay^{2} = x(x – a)^{2}.

Answer:

Put y = 0; we get x = 0, a

It meets the x – axis at x = 0 and x = a

∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.

Question 45.

Show that the lines \(\frac{x-3}{3}=\frac{y-3}{-1}, z-1=0\) and \(\frac{x-6}{2}=\frac{z-1}{3}, y-2=0\) interest. Also find the point of intersection.

Answer:

(x_{1}, y_{1}, z_{1}) = (3, 3,1) and (x_{2}, y_{2}, z_{2}) = (6,2,1)

(b_{1} b_{2}, b_{3}) = (3, -1, 0) and (d_{1} d_{2}, d_{3}) = (2, 0, 3)

Condition for intersection of two lines

Given two lines are intersecting lines.

Any point on the first line

(3λ + 3, -λ + 3,1)

Any point on the Second line

(2µ + 6, 2, 3µ + 1)

∴ 3µ + 1 = 1

3µ = 0

µ = 0

-λ + 3 = 2

-λ = -1

λ = 1

∴ The required point of intersection is (6, 2, 1)

(b) Prove that \(p \rightarrow(\neg q \vee r) \equiv \neg p \vee(\neg q \vee r)\) using truth table.

Answer:

The entries in the column corresponding to \(p \rightarrow(\neg q \vee r) \text { and } \neg p \vee(\neg q \vee r)\) are identical.

Hence they are equivalent.

Question 46.

(a) A water tank has the shape of an inverted circular cone with base radius 2 metres and height 4 metres. If water is being pumped into the tank at a rate of 2m3/min, find the rate at which the water level is rising when the water is 3m deep.

Answer:

We first sketch the cone and label it as in diagram. Let V, r and h be respectively the volume of the water, the radius of the cone and the height at time t, where t is measured in minutes.

We are given that \(\frac{d \mathrm{V}}{d t}\) = 2m^{3}/min and we are asked to find

\(\frac{d h}{d t}\) where h is 3m

The quantities V and h are related by the equation \(\mathrm{V}=\frac{1}{3} \pi r^{2} h\). But it is very useful to express V as function of h alone.

In order to eliminate r we use similar triangles in diagram to write \(\frac{r}{h}=\frac{2}{4} \Rightarrow r=\frac{h}{2}\) and the expression for V becomes \(\mathrm{V}=\frac{1}{3} \pi\left(\frac{h}{2}\right)^{2} h=\frac{\pi}{12} h^{3}\)

Now we can differentiate each side with respect to t and we have

(b) Let U (x, y) = e^{x} sin y, where x = st^{2}, y = s^{2}t s, t e R. Find \(\frac{\partial \boldsymbol{U}}{\partial \boldsymbol{s}}, \frac{\partial \boldsymbol{U}}{\partial t}\) and evaluate them at s = t = 1.

Answer:

Question 47.

(a) The probability density function of X is given by f(x) = \(\begin{array}{c}

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

k e^{-\frac{x}{3}} & \text { for } x>0 \\

0 & \text { for } x \leq 0

\end{array}\right.

\end{array}\)

Find (i) the value of k

(ii) the distribution function

(iii) P(X < 3)

(iv) P(5 ≤ X) (v) P(X ≤ 4) .

Answer:

[OR]

(b) Solve [y(1 – xtanx) + x^{2}cosx]dx – xdy = 0.