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## Karnataka 2nd PUC Maths Previous Year Question Paper March 2019

Time: 3 Hrs 15 Min

Max. Marks: 100

Part-A

Answer ALL the following questions: (10 × 1 = 10)

Question 1.

Define Binary Operation.

Answer:

A binary operation * on a set ‘A’ is defined as a function.

* : A × A → A

a * b = ab + 1 ∈ Z ∀ a, b, ∈ Z

∴ * is binary operation on Z.

Question 2.

Find the principal value of cos^{-1}\(\left(-\frac{1}{2}\right)\).

Answer:

\(\frac{2 \pi}{3}\)

Question 3.

Define a Scalar Matrix.

Answer:

In a square matrix the principal diagonal elements are same and all the other elements are zeroes.

Question 4.

Find a value of x for which

Answer:

3 – x^{2} = 3 – 8 ⇒ x^{2} = 8 ⇒ x = ±2\(\sqrt{2}\)

Question 5.

If y = sin(x^{2} + 5) then find \(\frac{d y}{d x}\)

Answer:

y = sin(x^{2} + 5)

\(\frac{d y}{d x}\) = cos(x^{2} + 5)(2x)

Question 6.

Find \(\int(1-x) \sqrt{x} d x\)

Answer:

\(\int(1-x) \sqrt{x} d x\)

Question 7.

Find a value of x for which x(î + ĵ + k̂) is a unit vector.

Answer:

x(î + ĵ + k̂) = 1 ⇒ xî+ xĵ+ xk̂ = 1

⇒\(\sqrt{x^{2}+x^{2}+x^{2}}\) = 1 ⇒ \(\sqrt{3 x^{2}}\) = 1 ⇒ x = \(\frac{1}{\sqrt{3}}\)

Question 8.

If a line has direction ratios 2, -1, -2 then determine its direction cosines.

Answer:

\(\left(\frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}\right)\)

Question 9.

Define objective function in Linear Programming Problem.

Answer:

The linear function Z = ax + by, where ‘a’ and ‘b’ are constants which has to be maximised or minimised is called objective function.

Question 10.

If P(E) = 0.6, P(F) = 0.3 and P(E n f) = 0.2 then find P(F/E).

Answer:

\(P(F / E)=\frac{P(F \cap E)}{P(E)}=\frac{0.2}{0.6}=\frac{2}{6}=\frac{1}{3}\)

Part-B

Answer any TEN questions: (10 × 2 = 20)

Question 11.

Show that the function f: N → N given by f(x) = 2x is one-one but not onto.

Answer:

The function/is one-one, for f(X_{1}) = f(x_{2})

⇒ 2x_{1} = 2x_{2} ⇒ x_{1} = x_{2} Further,/is not onto, as for 1 ∈ N, there does not exist any x in N such that f(x) = 2x = 1.

Question 12.

Prove that sin^{-1} x + cos^{-1} x = \(\frac{\pi}{2}\), x ∈ [-1,1].

Answer:

Let sin^{-1} x = θ then.

x = sin θ ⇒ x = cos(\(\frac{\pi}{2}\) – θ)

⇒ Cos^{-1}x = \(\frac{\pi}{2}\) – θ ⇒ θ + cos^{-1} x = \(\frac{\pi}{2}\)

∴ sin^{-1} x + cos^{-1} x = \(\frac{\pi}{2}\)

Question 13.

Write cot^{-1}\(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\), x > 1 simplest form.

Answer:

Let x = sec q ⇒ q = sec^{-1} x.

Question 14.

Find the area of the triangle with vertices (2, 7), (1, 1) and (10, 8) using determinant method.

Answer:

The area of triangle is given by

= \(\frac { 1 }{ 2 }\) |[-2 (1 – 8) -7 (1 – 10) + 1 (8 – 10)]|

= \(\frac { 1 }{ 2 }\) |-14 + 63 – 2| = \(\frac { 47 }{ 2 }\) sq.units

Question 15.

Find \(\frac{d y}{d x}\) if y = (logx)^{cosx}.

Answer:

Let y = (log x)^{cosx}

⇒ log y = cos x log (logx)

⇒ \(\frac{d}{d x}\) log y = \(\frac{d}{d x}\) cos x log(log x)

Question 16.

Find ax + by^{2} = cosy then find \(\frac{dy}{d x}\)

Answer:

ax + by^{2} = cosy

⇒ \(\frac{d}{d x}\)ax + \(\frac{d}{d x}\)by^{2} = \(\frac{d}{d x}\)cosy

⇒ a + 2by \(\frac{dy}{d x}\) = (-sin y)\(\frac{dy}{d x}\) dx dx

(2by + sin y) \(\frac{dy}{d x}\) = -a ⇒ \(\frac{d}{d x}\) = \(-\frac{a}{2 b y+\sin y}\)

Question 17.

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

Answer:

Note that V = x^{3}

or dV = \(\left(\frac{d V}{d x}\right)\) ∆x = (3x^{2}) ∆x

= (3x^{2}) (0.02x) = 0.06x^{3}m^{3}

Thus, the approximate change in volume is 0.06 x^{3}m^{3}

Question 18.

Find \(\int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} d x.\)

Answer:

Put (1 – tan x) = t ⇒ – sec^{2} x dx = dt

Question 19.

Find \(\int \sin 2 x \cos 3 x d x\)

Answer:

W.K.T sin x cosy = \(\frac { 1 }{ 2 }\)[sin(x + y) +sin(x – y)]

Then \(\int \sin 2 x \cos 3 x d x=\frac{1}{2}\left[\int \sin 5 x d x-\int \sin x d x\right]\)

= \(\frac { 1 }{ 2 }\) [\(\frac { 1 }{ 5 }\)cos 5x + cos x] + C = \(\frac { -1 }{ 10 }\) cos5x + \(\frac { 1 }{ 2 }\) cos x + C.

Question 20.

Find the order and degree (if defined) of the differential equation

\(\left(\frac{d^{3} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)\)=0

Answer:

Order = 2, Degree = not defined.

Question 21.

If \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 8 and \(|\vec{a}|\) = 8\((\vec{b})\) then find \(|\vec{b}|\).

Answer:

Question 22.

Find the projection of the vector ā = 2î + 3ĵ + 2k̂ and b̄ = î – ĵ + k̂

Answer:

Projection of vector ā on b̄ is given by

Question 23.

Find the distance of the point (3, -2, 1) from the plane 2x – y + 2Z + 3 = 0.

Answer:

Question 24.

Probability of solving specific problem independently by A and B are \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 3 }\) respectively. If both try to solve the problem independently, find the probability that the problem is solved.

Answer:

Given P(A) = \(\frac { 1 }{ 2 }\) and P(B) = \(\frac { 1 }{ 23}\)

(i) The problem is solved

= 1 – P(not solved) = 1 – P(A’ ∩ B’)

= 1 – P(A’) P(B’) = 1 – [(1 – \(\frac { 1 }{ 2 }\) – (1 – \(\frac { 1 }{ 3 }\)]

= 1 – (\(\frac { 1 }{ 2 }\) × \(\frac { 2 }{ 3 }\)) = 1 – \(\frac { 1 }{ 3 }\) = \(\frac { 2 }{ 3 }\)

(ii) Exactly one of them solves problem

= P(A ∩ B) + P(A ∩ B’)

= P(A’) + P(B) + P(A) P(B’)

\(=\left(1-\frac{1}{2}\right)\left(\frac{1}{3}\right)+\frac{1}{2}\left(1-\frac{1}{3}\right)\)

\(=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+\frac{1}{2}\left(\frac{2}{3}\right)=\frac{1}{2}\)

Part-C

Answer any TEN questions: (10 × 3 = 30)

Question 25.

Check whether the relation R in 1R of real numbers defined by R = {(a, b): a < b^{3}} is reflexive, symmetric or transitive.

Answer:

(i) For (a, a) ∈R, a < a^{3} which is false.

As for a = -2, a^{3} = -8

∴ a \(\nleq\) a^{3} as – 2 \(\nleq\) – 8

∴ R is not reflexive.

(ii) For (a, b), a < b^{3} and (b, a), b < a^{3} which is false

As for a = 1 and b = 2

a < b^{3} as 1 < 8

Now, b \(\nleq\) a^{3} as 2 \(\nleq\) 1

∴ R is not symmetric.

(iii) For (a, b), (b, c) and (a, c)

a < b^{3} and b < c^{3}

Now a = 25, b = 3, c = 2

a < b^{3} and 25 < 27

b < c^{3} and 3 < 8

but a \(\nleq\) c^{3} as 25 \(\nleq\) 8

.’. R is not transitive.

Hence R is neither reflexive, nor symmetric and nor transitive.

Question 26.

Prove that

cos^{-1} (\(\frac { 4 }{ 5 }\)) + cos^{-1} (\(\frac { 12 }{ 13 }\)) = cos^{-1} (\(\frac { 33 }{ 65 }\))

Answer:

Let

cos^{-1} (\(\frac { 4 }{ 5 }\)) ⇒ cos A = \(\frac { 4 }{ 5 }\) ⇒ Sin A = \(\sqrt{1-\frac{16}{25}}=\frac{3}{5}\)

cos^{-1} (\(\frac { 12 }{ 13 }\)) = cos B = \(\frac { 12 }{ 13 }\) ⇒ sin B = \(\sqrt{1-\frac{144}{169}}=\frac{5}{13}\)

cos(A + B) = cosA.cosB – sinA.sinB \(=\frac{4}{5} \cdot \frac{12}{13}-\frac{3}{5} \cdot \frac{5}{13}=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}\)

cos(A + B) = \(\frac { 33 }{ 65 }\) ⇒ A + B = cos^{-1} (\(\frac { 33 }{ 65 }\))

cos^{-1} (\(\frac { 4 }{ 5 }\)) + cos^{-1} (\(\frac { 12 }{ 13 }\)) = cos^{-1} (\(\frac { 33 }{ 65 }\))

Question 27.

By using elementary transformations, find the inverse of the matrix

Answer:

Question 28.

If x = a(θ + sin θ), y = a(1 – cosθ) then show that \(\frac{d y}{d x}=\tan \left(\frac{\theta}{2}\right)\)

Answer:

We have

\(\frac{d x}{d \theta}\) = a(1 + cos θ), \(\frac{d y}{d \theta}\) = a(sin θ)

Question 29.

Verify Rolle’s theorem for the function f(x) = x^{2} + 2x – 8, x ∈ [-4, 2].

Answer:

f(x) is a polynomial in x. Hence it is continuous over [-4, 2] and differentiable over (-4,2).

f(-4) = (-4)^{2} + 2 (-4) – 8 = 16 – 8 – 8 = 0

f(2) = 4 + 4 – 8 = 0

∴ f(-4) = f(2)

∴ All the conditions of the Rolle’s theorem are satisfied.

∴ there exists a c ∈[-4,2] such that

f'(c) = 0 f'(x) = 2x + 2

f(c) = 2c + 2

f(c) = 0

⇒ 2c + 2 = 0

⇒ 2c = -2

⇒ c = -1 ∈[-4, 2]

∴ Rolle’s theorem is verified.

Question 30.

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

Answer:

f(x) = 2x^{3} – 3x^{2} – 36x + 7

f ‘(x) = 6x^{2} – 6x – 36

f ‘(x) = 0 ⇒ 6x^{2} – 6x – 36 = 0

⇒ 6(x^{2} – x – 6) = 0

⇒ 6(x – 3)(x + 2) = 0

6 ≠ 0, x = 3, x = -2

Question 31.

Find \(\int x \log x d x\)

Answer:

Question 32.

Evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)

Answer:

Question 33.

Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3.

Answer:

Given curve is y^{2} = 4x and line is x = 3

y^{2} = 4x

Question 34.

Form the differential equation of the family of curves y = ae^{3x} + be^{-2x} by eliminating arbitrary constants a and b.

Answer:

Given, y = a e^{3x} + b e^{-2x} …. (1)

y’ = (a e^{3x}) 3 – (be^{-2x}) 2

y’ = 3 a e^{3x} – 2 be^{-2x} …. (2)

From (1) and (2)

y = a e^{3x} + b e^{-2x} × 3

y’ = 3a e^{3x} – 2b e^{-2x} × 1

Diff

⇒ 3y’ – y” = 5be^{-2x}

3y’ – y” = -10be^{-2x} ….(4)

From (3) & (4)

3y’ – y” = (-2) (5be^{-2x})

3y’ – y” = -2[3y – y’]

3y’ – y” = -6y + 2y’

y” – y’ – 6y = 0.

Question 35.

Find a unit vector perpendicular to each of the vector \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\), where ā = î + ĵ + k̂, b̄ = î + 2ĵ + 3k̂.

Answer:

Question 36.

Show that the four points with position vectors: 4î + 8ĵ + 12k̂, 2î + 4ĵ + 6k̂, 3î + 5ĵ + 4k̂ and 5î + 8ĵ + 5k̂ are coplanar.

Answer:

Let

= -2(21 + 0) + 4(7 + 8) – 6(0 + 3)

= -42 + 60 – 18 = -60 + 60 = 0

∴ A, B, C, D are coplanar.

Question 37.

Find the vector equation of the plane passing through the points R(2, 5,-3), S(-2, -3, 5) and T(5, 3, -3).

Answer:

Theorem: Derive intercept form of the equation of a plane

Let the equation of a plane be

Ax + By + Cz + 0 = 0 (1)

Let the plane cuts X – axis at p(a, 0, 0), Y-axis at Q(0, b, 0), Z-axis at R(0, 0, c) substituting P(a, 0, 0) in equation (1)

Aa + D = 0 ⇒ Aa = -D ⇒ A = \(\frac{-D}{a}\)

substituting Q(0, b, 0) in equation (1)

Bb + D = 0 ⇒ Bb = D ⇒ B = \(\frac{-D}{b}\)

R(0, 0, c) in equation (1)

Cc + D = 0 ⇒ Cc = -D ⇒ C = \(\frac{-D}{c}\)

substituting A, B, C in (1)

\(\left(\frac{-D}{a}\right) x+\left(\frac{-D}{b}\right) y+\left(\frac{-D}{c}\right) z=-D\)

Dividing through out by -D

\(\Rightarrow \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

is the required equation of the plane in intercept form

Question 38.

An insurance company insured 2000 scooter drivers, 4000 car drivers and, 6000 truck drivers. The probability of an accident is 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Answer:

Let S : be Insured person is scooter driver

C : be Insured person is car driver

T : be Insured person is truck driver

Let ‘A’ be the event “meets with an accident”.

Part-D

Answer any SIX questions: (6 × 5 = 30)

Question 39.

Let f: N → Y be a function defined as

f(x) = 4x + 3, where

y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse of f.

Answer:

f:N → Y

f(x) = 4x + 3

To prove that / is one – one

Let f(x_{1}) = f(x_{2})

4x_{1} + 3 = 4x_{2} + 3

4x_{1} = 4x_{2}

∴ x_{1 }= x_{2}.

∴ f is one – one

To prove that f is onto

Let y ∈ Y there exits x ∈ N such that

f(x) = y

4x + 3 = y

4x = y – 3, x = \(\frac{y-3}{4}\)

∴ f is one-one and onto

⇒ f is bijective. Hence f is invertible

∴ f^{-1}(y) = \(\frac{y-3}{4}\)

Question 40.

If A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{matrix} \right] \) then show that A^{2} – 23A – 40I = 0

Answer:

We have

Question 41.

Solve the following system of linear equations by matrix method:

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

Answer:

The system of equations can be written in the form A X = B where

|A| = 3(2 – 3) + 2(4 + 4) + 3(-6 -4) = -17

Hence, A is non singular matrix and so its inverse exists. Now

A_{11} = -1 A_{12} = -8 A_{13} = -10

A_{21} = -5 A_{22} = -6 A_{23} = +1

A_{31} = -1 A_{32} = +9 A_{33} = +7

Question 42.

If y = (sin^{-1} x). show that

\(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x\left(\frac{d y}{d x}\right)=0\)

Answer:

Given that y = sin^{-1} x, we have

y_{1} = \(\frac{1}{\sqrt{1-x^{2}}}\) i.e., (1 – x^{2})y_{1}^{2} = 1

so (1 – x^{2}). 2y_{1}y_{2} + y_{1}^{2}(0 – 2x) = 0

Hence (1 – x_{2}) y_{2} – xy_{1} = 0

Question 43.

The length x of a rectangle is decreasing at the rate of 3 cm / min and the width y is increasing at the rate of 2 cm / min. When x = 10 cm and y = 6 cm, find the rates of change of

(i) the perimeter and

(ii) the area of the rectangle.

Answer:

Since the length x is decreasing and the width y is increasing with respect to time, we have

\(\frac{d x}{d t}\) = – 3 cm /min and \(\frac{d y}{d t}\) = 2 cm/mm

(a) The perimeter P of a rectangel is given by P = 2 (x + y)

Therefore \(\frac{d P}{d t}\) = \(2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\) = 2 (-3 + 2)

= – 2 cm / min

(b) The area A of the rectangle is given A = x.y

Therefore \(\frac{d A}{d t}\) = \(\frac{d x}{d t}\) . y + x. \(\frac{d y}{d t}\)

= -3(6)+ 10(2) (as x= 10 cm and y = 6 cm)

= 2 cm^{2}/min.

Question 44.

Find the integral of \(\frac{1}{x^{2}-a^{2}}\) with respect to x and hence evalauate \(\int \frac{1}{x^{2}-16} d x\)

Answer:

Question 45.

Using the method of integration, find the smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2.

Answer:

Given circle x^{2} + y^{2} = 4 and the line is x + y = 2

x^{2} + (2 – x)^{2} = 4

[∴ y = 2 – x]

2x^{2} – 4x = 0 ⇒ 2x(x – 2) = 0 ⇒ x = 0, 2

Now, x^{2} + y^{2} = 4 and x + y = 2

⇒ y^{2} = 4 – x^{2} y = 2 – x,

⇒ y = \(y=\sqrt{2^{2}-x^{2}}\)

Required Area

Question 46.

Find the general solution of the differential equation \(\frac{d y}{d x}\)+ (secx)y = tanx, ( 0 ≤ x ≤ \(\frac{\pi}{2}\)).

Answer:

Given circle x^{2} + y^{2} = 4 and the line is x + y = 2

x^{2} + (2 – x)^{2} = 4 [∴ y = 2 – x]

2x^{2} – 4x = 0 ⇒ 2x(x – 2) = 0 ⇒ x = 0, 2

Now, x^{2} + y^{2} = 4 and x + y = 2

⇒ y^{2} = 4 – x^{2 } y = 2 – x,

⇒ y = \(\sqrt{2^{2}-x^{2}}\)

Required Area

Question 47.

Derive the equation of a line in space passing through a given point and parallel to a given vector in both vector and Cartesian form.

Answer:

Let \(\overrightarrow{\mathrm{r}}\) be any position vector on the line

‘L’ and \(\overrightarrow{\mathrm{a}}\) given point on L.

clearly, \(\overrightarrow{\mathrm{AP}}\) || \(\overrightarrow{\mathrm{b}}\)

\(\text { ie, } \overrightarrow{\mathrm{AP}}=\lambda \overrightarrow{\mathrm{b}}\)

where λ – some Real number

Question 48.

Five cards are drawn successively with replacement from a we!! shuffled deck of 52 cards. What is the probability that

(i) all the five cards are spades?

(ii) only five three cards are spaces?

(iii) none of spades?

Answer:

Part-E

Answer any ONE question: (1 × 10 = 10)

Question 49.

(a) Prove that \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) and hence evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x\)

Answer:

Proof: a – x = t

– dx = dt

dx = -dt

L.L: when x = 0, t = a

U.L: When x = a, t = 0

(b) Evaluate

Answer:

Taking a, b, c commn from R_{1}, R_{2}, R_{3}

Question 50.

Minimise and Maximise

z = 5x + 10y

Subject to the constraints :

x + 2y ≤ 120

x + y ≥ 60

x – 2y ≥ 0

x ≥ 0 and y ≥ 0.

Answer:

Draw the graph of the line

Put (0, 0) in the inequality x + 2y ≤ 120

0 + 2(0) ≤ 120

0 ≤ 120 (True)

∴ The half plane lies towards the origin

P (0, 0) in the inequality x + y ≥ 0

0 + 0 ≥ 0 (False)

∴ The half plane lies away from the origin

Put (0, 1) in the inequality x – 2y ≥ 0

0 – 2(1) ≥ 0

-2 ≥ 0 (False)

∴ The half plane does not contain (0, 1)

Check the point (1, 0) in the inequality

x – 2y ≥ 0

1 – 2 (0) ≥ 0

⇒ 1 ≥ 0 (True)

∴ The half plane contains (1,0)

So the half plane is towards x – axis

Corner points Max & Min z = 5x+ 10y

A (40,20) z = 400

B (60, 30) z = 600 → Maximum

C (60, 0) z = 300 → Minimum

D (120, 0) z = 600 → Maximum

∴ At C (60, 0) z is minimum

At B (60,30), D (120,0) z is maximum

(b) Find the value of K, if

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

{K x+1,} & {\text { if } x \leq 5} \\

{3 x-5} & {\text { if } x>3}

\end{array}\right.\)

is continuous at x = 5. (4)

Answer:

Since given f(x) is continous at x = 5 => lim f(x) = lim f(x)=/(5)

⇒ K(5) + 1 = 3(5) – 5

⇒ 5K+ 1 = 10 ⇒ 5K = 9

⇒ K = \(\frac { 9 }{ 5 }\)