2nd PUC Maths Model Question Paper 1 with Answers

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Karnataka 2nd PUC Maths Model Question Paper 1 with Answers

Time: 3.15 Hours
Max Marks: 100

Instructions :

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. 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.
State with reason whether the function h:{2,3,4,5} → {7,9,11,13} has inverse with h={(2,7),(3,9),(4,11),(5,13)}
Answer:
Yes, since h is one-one and onto

Question 2.
Find |3A| If A = \(\left(\begin{array}{ll}
{4} & {-1} \\
{3} & {2}
\end{array}\right)\)
Answer:
|3A| = \(\left|\begin{array}{rr}
{12} & {-3} \\
{9} & {6}
\end{array}\right|\) = 75 +27 = 99

Question 3.
Find the value of sin ^-1(sin \(\frac{3 \pi}{5}\) )
Answer:

Question 4.
The greatest integer function a not differentiable at integral points give reason.
Answer:
Since Left hand limit ≠ Right hand limit
⇒ The greatest integer function is discontinuous & hence cannot be differentiable.

Question 5.
Construct a 2 x 2 matrix
Answer:
A = [a_ij] ,where a_ij = \(\frac{i-j}{2}\)
Answer:
A = \(\left[\begin{array}{ll}
{a_{11}} & {a_{12}} \\
{a_{21}} & {a_{22}}
\end{array}\right] \) = \(\left[\begin{array}{cc}
{0} & {-1} \\
{\frac{1}{2}} & {0}
\end{array}\right]\)

Question 6.
Evaluate ∫ sin(2+5x) dx
Answer:
Let I = ∫sin (2+5x) dx
I = \(\frac{-\cos (2+5 x)}{5}+C\)

Question 7.
If \(\overrightarrow{\mathrm{a}}\) is a non zero vector of magnitude a and λ \( \overrightarrow{\mathrm{a}}\)  is a unit vector, find the value of λ.
Answer:

Question 8.
Find the distance of the plane 2x – 3y + 4z – 6=0 from the orgin.
Answer:

Question 9.
Define the term constraints in the LPP.
Answer:
The linear Inequalities or Equations or restrictions on the variables of a Linear programming problems are called constraints.

Question 10.

Given is not a probability distribution why ?
Answer:
Since p(3) cannot be negative, hence it is not a probability distribution. Also probability lie b / w o and 1.

Part – B

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

Question 11.
A binary operation ^ on the set {1,2,3,4,5} defined by a ^b = min {a , b), write the operation table for operation ^.
Answer:

Question 12.
Prove that
2 tan ^-1 \(\frac{1}{2}\) tan ^-1 \(\frac{1}{7}\) = tan ^-1\(\frac{31}{17}\)
Answer:
We know that

Question 13.
Solve 2tan^-1 (cosx) =tan^-1(2cosecx)
Answer:
2tan^-1(cosx) = tan^-1(2cosecx)

Question 14.
Find the area of triangle whose vertices are (2, 0), (-1, 0), and (0, 3) by using determinant.
Answer:

Question 15.
Differentiate x^{sin x}, x>0 w.r.t x
Answer:
Lety= x^{sin x} ⇒ 1ogy = sinx.logx
differentiate w. r. t.’x’

Question 16.
If y = tan ^-1 \(\frac{\sin x}{1+\cos x}\) then prove that \(\frac{d y}{d x}=\frac{1}{2}\)
Answer:

Question 17.
Find the local maximum value of the function g(x) = x³-3x
Answer:
g(x) = x³ – 3x
g¹(x) = 3x – 3
∴ g¹(x) = 0 gives, 3x² – 3 = 0 (or) x² -1 = 0
(or) x = ±1
g¹¹(x) = 6x
At x = 1; g¹¹(1) = 6 > 0. ⇒ g is minimum,
At x=-1, g¹¹(-1) = -6 < 0 ⇒g is maximum
Thus g has local maximum at x = -1 and the local maximum value is
g(-1) = -1 + 3 = 2

Question 18.
Evaluate \(\int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x\)
Answer:

Question 19.
Evaluate \(\int \frac{d x}{x^{2}-6 x+13}\)
Answer:
Let I = \(\int \frac{d x}{x^{2}-6 x+13}\)
But, x² – 6x +13 = x² – 6x +9+4
= (x – 3)² + 2²
x² – 6x + 13 = x² – 6x + 9 + 4
=(x – 3)² + 2²
I = \(\int \frac{d x}{(x-3)^{2}+2^{2}}\)
= \(\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\)
Here a = 2 & S → X-3
i = \(\frac{1}{2} \tan ^{-1}\left(\frac{x-3}{2}\right)+C\)

Question 20.
Find the order and degree of differential equation:
\(\left(\frac{\mathbf{d}^{2} \mathbf{y}}{\mathbf{d x}^{2}}\right)^{2}+\cos \left(\frac{\mathbf{d} \mathbf{y}}{\mathbf{d x}}\right)=\mathbf{0}\)
Answer:
order= 2 Degree = 2

Question 21.
Show that

Answer:

Question 22.
Find \(|\overrightarrow{\mathrm{a}}|\) and \(|\overrightarrow{\mathrm{b}}|\) if \((\vec{a}+\vec{b})\)
Answer:

Question 23.
Find the equation of plane passing
through the line of intersection of the planes x + y + z= 6 and 2x + 3y + 4z – 5 = 0 and the point (1, 1, 1).
Answer:
x+y+z=6 ⇒ \(\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\) = -6
2x + 3y + 4z – 5= 0.
⇒ \(\hat{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \) = 5

where, x is some real number
= (x + y + z – 6) + λ (2 + 3 + 4 – 5)=0
Given that the plane passes through (1,1,1).
ie, (1 + 1 + 1 – 6) + λ(2 + 3 + 4 – 5)=0
⇒ -3 + (4) =0 .
⇒ 4λ=3
⇒ λ = \(\frac{3}{4}\)
∴ (2) = (x+y+z-6) +\(\frac{3}{4}\)
(2x+3y + 4z –5) = 0
⇒ 4x + 4y + 4z – 6 + 6x + 9y + 12 z – 15=0
= 10x + 13y + 16z – 21=0

Question 24.
Assume that each born child is equally likely to be a boyor girl. If a family has two children. What is the conditional probability of both are girls given that at least one is girl?
Answer:
Let b →boy, g →girl
∴ S = {(b,b),(g,b),(b,g),(g,g)}
Let E : both the children are girls = {g,g}
and F – ‘Atleast on children is girl
{(g,b),(b,g),(g,g)}
E n F = {g,g}

Part – C

Answer any Ten Quesitions : 10 x 3 = 30

Question 25.
Show that the relation R in the set A = {x/xez and 0 ≤ x ≤ 12} given by R = {(a,b)}/|a-b| is multiple of 4}. Is an equivalence relation?
Answer:
(1) Ris reflexive, as (1,1) = (1-1) =0 is multiple of 4
∴ R is reflexive – (1)
(2) Let (1,5) ∈ z l(5,1) ∈ z
Then, |1 – 5| = |-4| = 4 is a multiple of 4
and |1 – 5| =|5 – 1| =4 is amultiple of 4
∴ (1,5)∈ zI (5,1) ∈ z
⇒ R is Symmetric – (2)
(3) Let (1,5) ∈ zI (5,1) ∈ z. be multipole of 4
Then, (1,9) = |1 – 9| = |-8| = 8 : multiple of 4
∴ (1,9) ∈ z
∴ R is transitive – (3)
from (1),(2) and (3)
R is Equivalence Relation.

Question 26.
Write \(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)-\frac{3 \pi}{2}<x<\frac{\pi}{2}\) in the simplest form.
Answer:

Question 27.
Find the value of x and y in \(\left(\begin{array}{cc}
{x+2 y} & {2} \\
{4} & {x+y}
\end{array}\right)-\left(\begin{array}{ll}
{3} & {2} \\
{4} & {1}
\end{array}\right)=0\) where 0 is null matrix
Answer:

Question 28.
If x = \( \sqrt{a^{\sin ^{-1} t}}\) y = \( \sqrt{a^{\cos ^{-1} t}}\) Show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Answer:

Question 29.
Verify Rolle’s theorem for the funcitonf(x) = x² + 2x – 8,x ∈ [-4,2].
(1) f(x) = x² + 2x – 8 is continuous in [-4,2]
(2) f¹(x) = 2x + 2 exists in(-4,2)
(3) f(-4) = (-4)² + 2(-4)-8
= 16 – 8 – 8=0
f(2) = (2)²+2(2)-8 =4+4-8=0
∴ f (-4) = f(2)
∴ By Rolle’s theorem, f atleast one point in (-4,2)
∋ ; f¹(c)=0
⇒ 2c + 2 =0
⇒ 2c= — 2
∴ c = -1 ∈ (-4,2)
Hence, Rolle’s theorem is. verified.

Question 30.
Find the equation of tangent and normal to the curve
y = x⁴ – 6×3 +13x² – 10x + 5 at (0,5)
Answer:
y = x⁴ – 6×3 + 13x² -10x + 5 at (0,5)
\(\frac{d y}{d x}\) = 4x³– 18X² 26X – 10
at(0,5): \(\frac{d y}{d x}\) = -10 and \(\frac{d x}{d y}=\frac{1}{10}\)
∴  Slope of the tan gent =-10
Slope of the Normal = \( \frac{1}{10}\)
Equation of the tangent, dy, x
y – y₁=\(\frac{d y}{d x}\)(x – x₁)
⇒ y – 5 = (-10) (x – 0)= -10x
∴ 10x + y -5 = 0
Equation of the Normal
y – y₁=\(-\frac{d y}{d x}\)(x – x₁)
⇒ y-5 =\( \frac{1}{10}\) ⇒ 10y – 50 = x
∴ x – 10y + 50=0

Question 31.
Integrate : \(\frac{\sin x}{\sin (a+x)}\) with respect to x
Answer:
= \(\frac{\sin x}{\sin (a+x)}\) dx
Put a + x = t or x = t – a

Question 32.
Evaluate \(\int_{0}^{5}(x+1) d x\) as a limit of sum.
Answer:

Question 33.
Find the area of the region founded by y² =9x, x = 2, x= 4 and theX-axis in the I quadrant.
answeR:
Area of the Region, R = ∫y. dx

Question 34.
Find the equation of the through the point (-2,3) given that the slope of the tangent at any point (x>y) is \(\frac{2 x}{y^{2}}\)
Answer:
Given, slope of the tangent = \(\frac{2 x}{y^{2}}\)
ie, \(\frac{d y}{d x}=\frac{2 x}{y^{2}}\)
on separating the variable, we get

Question 35.
Find the area of triangle with vertices
A(1, 1, 2),B(2, 3, 5), C(l, 5, 5).
Answer:

Question 36.
Prove that \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}}]=2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]\)
Answer:

Question 37.
Two groups are competing for the position on the board of directors of a corporation. The probability of I and II groups will are 0.6 and 0.4 respectively. Further, if Im group wins, the probability of introducing a new product introduced was by the II group.
Answer:
A → probability of I group
B → probability of II group
∴ P(A) = 0.6,P(B)=0.4
Let C stands for winning among the two groups.

Question 38.
Find the angle between the line \(\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\) and the plane 10x + 2y – 11z = 3.
Answer:

Part – D

Answer any questions: ( 6 x 5 = 30 )

Question 39.
Consider f :R → Rdefined by f(x) = 4x+3. Show that f is invertible. Find the inverse of f.
Answer:
Let y ∈ R be any arbitraty Element
Then, y = 4x + 3 y-3
⇒ x = \(\frac{y-3}{4}\) for same x ∈ R
Define g:R→ Rby g(y) =\(\frac{y-3}{4}\)
Now,
(gof)(x) = g(f(x)) = g[4x +3 ]

This shows that gof = I_R and fog =I_R
f is invertible function
and hence \( f^{-1}=g=\sqrt{\frac{x-3}{4}}\)

Question 40.
I A = \(\left(\begin{array}{ccc}
{2} & {3} & {4} \\
{0} & {-2} & {1} \\
{3} & {4} & {5}
\end{array}\right)\) , B= \(\left(\begin{array}{ccc}
{2} & {0} & {-3} \\
{4} & {0} & {-1} \\
{3} & {4} & {5}
\end{array}\right)\), C = \(\left(\begin{array}{ccc}
{5} & {6} & {7} \\
{-1} & {2} & {3} \\
{4} & {-5} & {4}
\end{array}\right)\)
Prove that A(BC) = (AB)C
Answer:

Question 41.
Solve by matrix method
2x+y + 2z = 5, x – y – z = 0, x + 2y + 3z = 5
Consider AX = B → (1) => X = A^-1/ B -(2)

= 2 (-3 + 2) – 1(3 + 1) + 2(2 + 1)
= 2(-1) – 1(4) + 2(3)
=-2-4+6
|A| = 0
A^-1 does not exist because |A| =0 & hence cannot determine the value of x, y and z.

Question 42.
If = y = 5cos (logx) + 7 sin(logx), show that x²y₂ + xy₁= 0
Answer:
y = 5 cos (logx) + 7 sin (logx)

Question 43.
Find the integral of \(\frac{1}{x^{2}-a^{2}}\) with respect x and evaluate \(\int \frac{d x}{x^{2}-8 x+5}\)
Answer:


missing

Question 44.
Sand is pouring from a pipe at the rate of 12cm³/s. The falling sand form a cone on the ground in such a way that the height of the cone is always one – sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4cm?
Answer:

Question 45.
Find the area of the region bounded by the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) by the method of integration.
Answer:
Area of ABCD =4 [Area of sector OAB Y

Question 46.
Find the area of the region bounded by the differential equation \(\frac{\mathrm{d} y}{\mathrm{d} x}\) + ycot = 4xcosec x ,x ≠ 0 given that y=0 when x = \(\frac{\pi}{2}[latex]
Answer:

Question 47.
Derive the equation of a line in a space passing through two given points both in the vector and Cartesian form.
Answer:

Let [latex]\overrightarrow{\mathrm{r}}\) be any position vector on the Line L
Let \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) be two given points on the line L
clearly p is on the line L

Question 48.
Five cards are drawn successively with replacement from well shuffled deck of 52 cards. What is the probability that (i) all the five cards are spade, (ii) none is spade.
Answer:
Let p = probability of drawing spade =\(\frac{13}{52}=\frac{1}{4}\)
q = probability of drawing cards = \(1-\frac{1}{4}=\frac{3}{4}\)
n = 5

Part – E

Answer any one question ( 1 x 10 = 10 )

Question 49.
(a). Maximize and Minimize
z = 3x + 9y – (6)
x + 3y ≤60
S/T . x + y ≥ 10 x ≤ y
and x ≥ 0, y ≥ 0. graphically
Answer:
x + 3y = 60 – (1)

A(0,20) I B(60,0)
x + y = 10-(2)

c(0,10) and D(10,0) x = y-(3)

(b) Show that

Answer:

Question 50.
(a) Prove that

and evaluate \(\int_{-1}^{1} \sin ^{5} x \cos ^{4} x d x\)
Answer:

\(\int_{-a}^{a} f(x) d x=0\)
Further , Let I = \(\int_{-1}^{1} \sin ^{5} x \cos ^{4} x d x\)
Let f(x)=sin⁵ x cos⁴ x
Then f(-x) = sin⁵(-x) cos⁴(-x)
= – sin⁵ x cos⁴
f(-x) = – f(x)
⇒ f is odd fimction.
∴ from the above property, I = 0

(b) Find all points of discontinuty of f(x),
where f is defined by
\(f(x)=\left\{\begin{array}{l}
{x^{3}-3, \text { if } x \geq 2} \\
{x^{2}+1, \text { if } x<2}
\end{array}\right\}\)
Answer:
f(2)= 2³ – 3 = 8 – 3 = 5
Lt f(x) = Lt (x² + 1) = 2² + 1 = 4 + 1 = 5
x → 2 x → 2
Lt f(x) = Lt (x³ – 3) = 2² – 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⁺
If f is an odd function, then
∴f is continuous at x = 2
f(-x) = -f(x) and so (2)becomes,
There is no point of discontinuty.