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## Karnataka 2nd PUC Maths Model Question Paper 2 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.

Let * be the binary operation on N given by a * b = L . C. M. of a and b, find 20 * 16.

Answer:

20 * 16 =L .C .M. of 20 and 16

20 * 16 =8 ∈ N

Question 2.

Write the domain of f (x) = tan – 1x

Answer:

[-1,1] (or) -1≤ x ≤ 1

Question 3.

Consturct a 3×3 matrix A = [aij]

whose elements are given by aij = \(\frac{\mathrm{i}}{\mathrm{j}}\) .

Answer:

Question 4.

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

{x} & {2} \\

{3} & {x}

\end{array}\right|=\left|\begin{array}{cc}

{x} & {2} \\

{-3} & {-x}

\end{array}\right|\) , find the value of x.

Answer:

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

{x} & {2} \\

{3} & {x}

\end{array}\right|=\left|\begin{array}{cc}

{x} & {2} \\

{-3} & {-x}

\end{array}\right|\)

⇒ x^{2} – 6 = -x^{2} + 6

⇒ x^{2} + x^{2} =6 + 6

⇒ 2x^{2} = 12

⇒ x^{2} = 6

x = ± √6

Question 5.

Differentiate log cos e^{x} w.r.t to x

Answer:

Question 6.

Evaluate : ∫ tan^{2}2x d x

Answer:

∫ tan^{2}2x d x = ∫ (sec^{2}2x – 1 )d x

∫ sec^{2}2xdx – ∫1 d x

∫ tan^{2}2x d x = \(\frac{\tan (2 x)}{2}-x+c\)

Question 7.

Find the angle between the two vectors \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) such that \(|\overrightarrow{\mathrm{a}}|\) = 1,\(|\overrightarrow{\mathrm{b}}|\)=1 and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=1\)

Answer:

Question 8.

Find the Equation of the plane having intercept 3 on the y- axis and parallel to zox plane.

Answer:

y = 3

Question 9.

Define Linear objective function in LPP.

Answer:

Linear function z = ax + by, where a, b are constants, which has to be maximised or minimised is called a linear objective function.

Question 10.

A fair die is rolled. Consider the events E{l,3,5}and F{2,3}, find PC(E/F).

Answer:

Let S={1,2,3,4,5,6}

Part – B

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

Question 11.

Show that the relation R in’the set of integers given by R = {(a,b):5divides(a-b)} is Symmetric and transitive.

Answer:

Let (5, 10) ∈ z,as 5 divides 5 – 10=-5 ∈ z

Also, (10, 5) ∈ z, as divides 10 – 5 = 5 ∈ z

(5,15) = 5 divides 5-15 = -10 ∈ z

∴ R is Symmetric

Further, If (5,10) ∈ z and (10, 15) ∈ z,

as 5 divides both (5, 15) &(10, 15)

∴ (5, 15) = 5 divides 5-15 = -10 ∈ z

⇒ (5,15)tz

⇒ R is transitive.

Question 12.

If sin (sin ^{-1} \(\frac{1}{5}\) + cos ^{-1} x ) = 1, then find the value of x.

Answer:

sin ^{-1} \(\frac{1}{5}\) + cos ^{-1} x = sin ^{-1}(1) = \(\frac{\pi}{2}\)

cos ^{-1}x = \(\frac{\pi}{2}\) – sin^{-1} ( \(\frac{1}{5}\) )

cos ^{-1}x = cos^{-1} \(\frac{1}{5}\)

⇒ x = \(\frac{1}{5}\)

Question 13.

Write the function tan^{-1} \(\left(\frac{\sqrt{1+x^{2}}-1}{x}\right) \) ,x ≠ 0 in the simplest form.

Answer:

Question 14.

If the area of the triangle with vertices (-2,-6) and (5,4) and (k,4) is 35 Sq. units. Find the value of k using determinate method

Answer:

Area of ∆ABC = \(\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|\)

⇒ 35 = \(\frac{1}{2}\left|\begin{array}{rrr}

{2} & {-6} & {1} \\

{5} & {4} & {1} \\

{k} & {4} & {1}

\end{array}\right|\)

⇒ ±70 =2[4 – 4] + 6 [ 5 – k] + 1 [20 + 4k]

⇒ 6(5 – k) ± 20 – 4k= ±70

⇒ 30 – 6k ± 20 – 4k =±70

⇒ 50 – 10k = ±70

∴ 50-10k = ±70 (or) 50-10k = -70

⇒10k = 50 – 70

⇒10k = -20

⇒k = -2

50-10k = -70

⇒10k = 50 + 70

⇒10k = 120

⇒k = 12

Thus k = -2 and k = 12

Question 15.

Find the derivative of √x + √y =9 at (4,9).

Answer:

√x + √y = 9

Question 16.

If y=log_{7} (logx), find \(\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} \mathbf{x}}\)

Answer:

Question 17.

Evaluate : \(\int \frac{3 x^{2}}{1+x^{6}} d x\)

Answer:

Let I = \(\int \frac{3 x^{2}}{1+x^{6}} d x\) = \(\int \frac{3 x^{2}}{1+\left(x^{3}\right)^{2}} d x\) …… (1)

Put x^{3} = t, ∴ 3x^{2 }dx = dt

∴ (1) becomes , I = \(\int \frac{d t}{1+t^{2}}=\tan ^{-1} t+c\)

∴ I = tan^{-1}[x^{3}]+ c

Question 18.

Evaluate : \(\int \mathbf{e}^{\mathbf{x}}\left(\frac{\mathbf{x}-\mathbf{1}}{\mathbf{x}^{2}}\right) \mathbf{d} \mathbf{x}\)

Answer:

Question 19.

Find the slope of the targent to the curve y = x^{3} – 3x + 2at the point whose x – coordinate is 3.

Answer:

y =x^{3} -3x + 2

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

when

x = 3, \(\frac{\mathrm{d} y}{\mathrm{d} x}\) = 3 (3)^{2} -3 = 3(9)-3 = 24

∴ slope of the tangent = 24

Question 20.

Find the order and Degree of the Differential Equation.

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

Answer:

order = 2

Degree = 1

Question 21.

Show that the points \(\mathbf{A}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) \(\mathbf{B}(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})\) and \(\mathbf{c}(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})\), are the vertices of a right angled triangle.

Answer:

Question 22.

If \(\overrightarrow{\mathbf{a}}\) is a unit vector and \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 15 then find \(|\vec{x}|\)

Answer:

Question 23.

Find the distance of the point (2, 3, -5) from the plane \(\vec{r} \cdot(\hat{i}+2 y-2 \hat{k})=9\)

Answer:

Question 24.

If the probability distribution of x is

find the value of k

Clearly eact P_{i}0.& k > 0

Now we have, \(\sum_{i=1}^{n} p(x i)=1\)

⇒ p(x_{1}) + p(x_{2}) + p(x_{3}) + p(x_{4}) + p(x_{5}) =1

⇒ p(0) + p(1) + p(2) + p(3) + p(4) = l

⇒ 0.1 + k + 2k + 2k + k =1

⇒ 6k = 1 – 0.1

⇒ 6k= 0.9

k = 0.15

Part – C

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

Question 25.

Verify whether the function f :R-{3} → R-{1}, defined by f(x) = \(\frac{x-2}{x-3}\) is one- one and onto or not. Give reason.

Answer:

we have f(x) = \(\frac{x-2}{x-3}\)

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

⇒ \(\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}\)

⇒ x_{1} + x_{2 }– 3x_{1} – 2x_{2} + 6

⇒ x_{1}x_{2} – 3x_{1} – 2x_{2} = 3x_{1} – 2x_{1}

⇒ x_{2} = x_{1}

∴ f is one-one

Let y ∈B

Now, f(x) = y ⇒ \(\frac{x-2}{x-3}\) = y

⇒ x-2 = y(x – 3)

⇒ x-2 = xy -3y ⇒ 2-3y = x-xy

⇒ 2-3y =x(1 – y)

⇒ x = \(\frac{2-3 y}{1-y}\)∴(y= 1 )∈A

Thus foe every y ∈ B, I x = \(\frac{2-3 y}{1-y}\) ∈A

such that f(x) = y

∴ f is onto

Hence , f is bijective function

Question 26.

Find the value of \(\tan \frac{1}{2}\left[\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-y^{2}}{1+y^{2}}\right)\right]\) |x| < 1, y > 0 and xy < 1

Answer:

Let x = tan θ and y = tan Φ

Then,

Question 27.

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

{1} & {2} \\

{3} & {4}

\end{array}\right]\). find A^{-1} by elementary operations.

Answer:

Question 28.

If y = tan ^{-1} \(\left(\frac{\sin x}{1+\cos x}\right)\) then prove that \(\frac{d y}{d x}=\frac{1}{2}\)

Answer:

Question 29.

Differentiate x^{sin x}+ (sin x)cos ^{x} w.r.t.’x’

Answer:

Let y = x^{sin x}+ (sin x)^{cosx} ….. (1)

Take, u = X^{sin x}

Take log on b . s,

logu=sinx logx

differentiate w. r. t. ‘x’

Take, v =(sin x)^{cosx}

Take log on both sides

logv = cosx log (sinx)

differentiate w. r. t. ‘x’

Question 30.

Find the absolute maximum value and the absolute minimum value of the function f(x) = sinx+cosx, x ∈ [0,π]

Answer:

we have, f (x) = sinx +cosx

f^{1}(x) = cos x – sin x

Now f^{1}(x) = 0 ⇒ cos x = sin x

⇒ cos x = cos \(\left(\frac{\pi}{2}-x\right)\)

⇒ x =\(\left(\frac{\pi}{2}-x\right)\)

⇒ 2x = \(\frac{\pi}{2}\)

⇒ x = \(\frac{\pi}{4}\) ∈ [0,π]

To find f(0), f(π/4) if f(π) :

f(0) = sin(0) + cos(0) = 0+1 = 1

f(π) = sin(π) + cos(π) = 0-1 = -1

Thus, Absolute Maximum value of f on

[0,π] is √2 at x = π/4, and Absolute Minimum value of fon (0,π) is -1 at x = π

Question 31.

Evaluate \(\int \frac{d x}{x\left(x^{n}+1\right)}\)

Answer:

Question 32.

Evaluate ∫e^{x} sin x dx

Answer:

Question 33.

Find the area of the circle x^{2} + y^{2}= 4 bounded by the line x = 0 and x = 2 which is lying in the first quadrant.

Answer:

Question 34.

In a bank, principal “p” increases continuously at the rate of 5 % per year. Find the principal interest of time t.

Answer:

Let p be the principal at any time t. According to the given problem,

Question 35.

Find a unit perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) where \(\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}\) and \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}} 2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\).

Answer:

\(\vec{a}+\vec{b}=4 \hat{i}+4 \hat{j}+0 \hat{k}\)

\(\vec{a}-\vec{b}=2 \hat{i}+0 \hat{j}+4 \hat{k}\)

A vector which is perpendicular to both \(\vec{a}+\vec{b}\) & \(\vec{a}-\vec{b}\) is given by

Question 36.

If \(\overrightarrow{\mathbf{a}}=-4 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-\lambda \hat{\mathbf{k}}\),

and \(\overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)\(\overrightarrow{\mathbf{c}}=-8 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) are coplanar, find λ

Answer:

Given, \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) are coplanar

∴ [ \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\)] = 0 -4 -6 -X => -1 4

⇒ \(\left|\begin{array}{ccc}

{-4} & {-6} & {-\lambda} \\

{-1} & {4} & {3} \\

{-8} & {-1} & {3}

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

⇒ -4 [12 + 3] +6[-3 + 24]-A, [l + 32 ] = 0

⇒ -60 + 126 – 33λ = 0

⇒ 66 – 33 λ = 0

⇒ 33λ = 66 .

λ = 2

Question 37.

Find the Equation of the line which passes through the point (1,2,3) and is paralled to the vector \(3\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) both in vector and form and cartesian form.

Vector form :

Answer:

Question 38.

Bag I contains 3 red and 4 black balls and while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bag and it is found to be red. Find one of the bag and it is was drawn from bag II?

Answer:

Let A be the event of choosing the bag I B be the event of choosing the bag II and C be the event of drawing a red ball

Part – D

Answer any Six Questions : ( 6 x 5 = 30 )

Question 39.

Consider f:R_{+} → [5,∞] given by f (x) = 9x^{2} + 6x – 5, Show that f is invertible with f^{-1}(y) = \(\frac{\sqrt{y+6}-1}{3}\)

Answer:

Question 40.

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

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

{0} & {-3} & {0} \\

{0} & {0} & {4}

\end{array}\right|\), verify that A^{3} – 3A^{2} -10 A+ 24I =0, where o is zero matrix of order 3 x 3.

Answer:

Question 41.

Solve by Matrix method : X + Y + Z = 6; x – 2y + 3z = 6 and x – y + z = 2

Answer:

Consider AX = B – (1) => X = A^{-1} B – (2)

Question 42.

IF y = Sin^{-1}x, show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=0\)

Answer:

Question 43.

A man of height 2 meters walks at a uniform speed of 5 km / hour, away from a lamp post which is 6 meters high. Find the rate at which the length of his shadow increases.

Answer:

Let AB be the lamp -post, the lamp being at the position B and Let MN be the Man at a particular time t and Let AM = 1 meters. Then MS is the shadow of the man.

Let MS. = S meters

Note the ∆ MSN ∼ ∆ ASM

(or) \(\frac{M S}{A S}=\frac{M N}{A B}\)

(or) AS = 3S(as MN =2 andAB=6)

Thus, AM = 3S – S= 2S B

ut AM =1

So, 1 = 2S

∴ \(\frac{d l}{d t}=2 \frac{d s}{d t}\)

Since \(\frac{d l}{d t}\) = 5 km /h. Hence the lenght of the shadow increases at the rate of \(\frac{5}{2}\) km/h

Question 44.

Find the Integral of \(\frac{1}{\sqrt{x^{2}}+a^{2}}\) w- r -t. x I hence evaluate \( \int \frac{1}{\sqrt{x^{2}+2 x+4}} d x\)

Answer:

Question 45.

Find the area of the region enclosed by the parabola x^{2} = 4y and the line x = 4y -2 and the y – axis.

Answer:

we have , x^{2} = 4y(parabola) – (1)

x = 4y – 2 ⇒ x – 4y + z = 0 -(2)

(2) is a line passing through (o,1/2)& (- 2, 0) and the

x -axis ⇒ y =0 -(3)

Question 46.

Derive the equation of the plane in normal form both in the cartesian and vector form

Answer:

Consider a plane whose perpendicular distance from the origin is d. where d ≠ 0.

This is the Required Equation of a plane in normal form.

To find P and Q

x^{2}=4y and x = 4y – 2

⇒(4y – 2)^{2} = 4y

⇒16y^{2} + 4-1 6y = 4y

⇒ 16y^{2} – 20y + 4 = 0

⇒ (y – 1) (16y – 4) = 0

y = 1 or y = 1/4

If y = 1; x = -1 (-1,1/4)

Thus , the points of Intersection are (2,1)I (-1,1/4)

∴ Area of the Region O B Q C = Area of B C Q+ Area of QCO.

Question 47.

Find the particular solution of the differential Equation \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=1\) when y = 0 andx = 1

when y = 0 and x = 1.

Answer:

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

This is the Required particular solution of the given differential Equation.

Question 48.

A die is thrown 6 times, if ‘getting an odd number is success’ what is the probability of (i) 5 success (ii) at least 5 success, (iii) at most 5 success.

Answer:

Let P = probability of getting an odd number = \(\frac{3}{6}=\frac{1}{2}\)

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

& n = 6

we have, p (x) = n C_{x} p^{x} q ^{n-x}

p(x) = 6C_{x} \(\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{2}\) …….(i)

(ii) p(x ≥ 5): p(x ≥ 5) = p(x = 5) +p (x = 6)

(iii) p(x ≤ 5): p(x ≤ 5) = 1 – p (x > 5)

= 1 – P(x = 6)

Part – E

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

Question 49.

(a) A furniture dealer deals in only two items – tables and chairs. He has Rs . 50, 000 to invest and has storage space of at most 60 pieces. A table costs Rs 2500 and a chair Rs 500. He estimates that from the sale of one table, he can make a profit of Rs 250 and that from the sale of one chair a profit of Rs 75. How many tables and chairs he should buy from the available money so as to maximize his total profit assuming that he can sell all the items which he buys.

Answer:

Let x → Tables y → chairs

Maximise : Z = 250 x 75 y

SIT : C_{1}: 2500 x + 500y ≤ 50, 000

c_{2}: (1)x +(1)y ≤ 60

where , x, y ≥ 0

Now, 2500x + 500y =50,000

25x + 5y = 500 – (1)

x + y = 30 – (2)

Eq(1) passes through A(0,100) and B(20,0)

Eq(2) passes through A(0,60) and B(60,0)

Here O C E B represents feasible Region I it is bounded

The optimal solution

(z) max =Rs.650 at x = 10 & y = 50

Thus, the dealer gets maximum profit of Rs. 6250 on selling 10 tables and 50 chairs.

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

{x+y+2 z} & {x} & {y} \\

{z} & {y+z+2 x} & {y} \\

{z} & {x} & {z+x+2 y}

\end{array}\right|=2(x+y+z)^{3}\)

Answer:

Question 50.

(a) Prove that \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-c) d x \)and hence evaluate \(\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x\)

Answer:

Consider \(\int_{0}^{a} f(x) d x\)

put x = a – t ⇒ dx = -dt

where x = 0;t=a.

& x = a; t =0

(b) Find all points of discontinuty of f(x), where f is defined by

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

{x^{3}-3, \text { if } x \geq 2} \\

{x^{2}+1, \text { if } x<2}

\end{array}\right\}\) ……(4)

Answer:

f(2) = 2^{3} – 3 = 8 – 3 = 5

Lt f(x) = Lt (x^{2} + 1) = 2^{2} + 1 = 4 + 1 = 5

x → 2 x → 2

Lt f(x) = Lt (x^{3} – 3) = 2^{2} – 3 = 8 – 3 = 5

x → 2^{+} x → 2

Since, Lt f(X) = Lt f(x) = Lt f(x) = f(2) = 5

x → 5 x → 2^{+}

∴ f is continuous at x = 2

There is no point of discontinuty.