2nd PUC Maths Model Question Paper 4 with Answers

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Karnataka 2nd PUC Maths Model Question Paper 4 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.
Operation * is defined by a * b = a. Is * is a binary operation on z+?
Answer:
Let a = 2 and b = 1
Then, a * b = 2 * 1 = 2 ∈ z+
⇒ Yes * is a binary operation.

Question 2.
Write the principal value branch of f(x) = sin-1x.
Answer:
\(\)

Question 3.
Define a diagonal matrix.
Answer:
A square matrix [Aij] is said to be a diagonal matrix if the non – diagonal elements are zero.
Ex: A = \(\left[\begin{array}{ll}
{1} & {0} \\
{0} & {2}
\end{array}\right]\)

2nd PUC Maths Model Question Paper 4 with Answers

Question 4.
If A = \(\left[\begin{array}{ll}
{4} & {7} \\
{3} & {5}
\end{array}\right]\) find |3A|
Answer:
3A = \(\left[\begin{array}{ll}
{12} & {21} \\
{9} & {15}
\end{array}\right]\) ⇒ |3 A| = 180 – 189 = -9.

Question 5.
Write the points of discontinuty for the function f [x] = [x], -3 < x < 3
Answer:
Point of discontinuty are x = -2, x = -1, x = 0, x = 1, x = 2.

Question 6.
Evaluate ∫ cos ecx (cos ecx – cot x) dx
Answer:
∫ cosec x (cosec x – cot x) dx
= ∫cosec2x – cosec x cot x )dx
= cosec2x dx – ∫ cosecx cot x dx
=-cotx – (-cosecx)
=-cotx + cosecx
=cosecx – cotx+c

Question 7.
Find the direction ratios of the vector, joining the points P(2, 3, 0) and Q( -1, -2, -3)., direction from P to Q.
Answer:
\(\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}\)
=((-1) î + (-2) ĵ + (-3)k̂) -(2î + 3ĵ + 0k̂)
= (-1 – 2) î+(-2 – 3)ĵ +(-3 – 0)k̂
\(\overrightarrow{\mathrm{PQ}}\) = -3î -5ĵ-3k̂.
Direction Ratios of \(\overrightarrow{\mathrm{PQ}}\) = (-3, -5, -3).

Question 8.
Find the Equation of a plane with the internet 2, 3 and 4 on x1 y1 and z axes respectively.
Answer:
we have a = 2, b = 3, c = 4.
Therefore \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ⇒ \(\Rightarrow \frac{x}{2}+\frac{y}{3}+\frac{z}{4}\)  = 1

Question 9.
Define optional solution in the linear programming problem.
Answer:
Any point in the feasible region that gives the optional value (maximum or minimum ) of the objective function is called an optional solution.

2nd PUC Maths Model Question Paper 4 with Answers

Question 10.
If A and B are independent events with P(A) = 0.3 P(B) = 0.4, find P(A n B). Since A and B are independent,
we have P(A∩B) = P(A). P(B)
= 0.3 × 0.4
P(A∩B) = 0.12

Part – B

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

Question 11.
Find the gof and fog iff (x) – 8x3 and
Answer:
g (x) = X1/3 ,
(gof) (x) = g (f(x) = g ( 8x3)
=(8x3)1/3=(23x3)1/3
(gof)(x) =2x
(fog)(x) = f(g(x) = f(x1/3) = 8(x1/3)3
(fog)(x)= 8x

Question 12.
Write the function \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\) o < x < π in the Simplest form.
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 1

Question 13.
Prove that 2sin -1\(\left(\frac{3}{5}\right)\) = tan -1 \(\left(\frac{25}{7}\right)\)
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 2
2nd PUC Maths Model Question Paper 4 with Answers - 3

Question 14.
Find the area of a triangle whose vertices are (1, 3), (2, 5) and (7, 5) using determinant.
Answer:
Area = \(\frac{1}{2}\left|\begin{array}{ll}
{x_{1}} & {y_{1}} \\
{x_{2}} & {y_{2}}
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
{1} & {3} & {1} \\
{2} & {5} & {1} \\
{7} & {2} & {1}
\end{array}\right|\)
= \(\frac{1}{2}\) {1[5 – 5] – 3[2 – 7] + 1[10 – 35]}
= \(\frac{1}{2}\) {0 – 3(- 5) + 1(-25)
= \(\frac{1}{2}\) {15 – 25}
= \(\frac{1}{2}\) {-10} = – 5
Thus |Area| – 5 sq. units.

2nd PUC Maths Model Question Paper 4 with Answers

Question 15.
Find \(\frac{d y}{d x}\) if 2x+3y=sin y
Answer:
2x + 3y = siny
diff w. r. t ‘x’, we get
2(1) + 3\(\frac{d y}{d x}\)=cosy\(\frac{d y}{d x}\)
⇒ (cosy – 3)\(\frac{d y}{d x}\)=2
\(\frac{d y}{d x}=\frac{2}{\cos y-3}\)

Question 16.
If x = at2 and y = 2 at slow that \(\frac{d y}{d x}=\frac{1}{t}\)
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 4

Question 17.
Find the approximate change in the volume v of a cube of a side x meters caused by increasing by 2%.
Answer:
Given, v = x3
diff sv. r. t .’x’.
sv = 3x2dx
= 3x(2)
sv = 6x2

Question 18.
Evaluate ∫sin3x dx
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 5

Question 19.
Evaluate \(\int_{0}^{\pi / 2} \cos 2 x d x\)
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 6

Question 20.
Find the orders and degree of different
Equation \(\left(\frac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0\)
Answer:
Order =2 degree = 1

2nd PUC Maths Model Question Paper 4 with Answers

Question 21.
Find a vector in the direction of the vector \(\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}\) that has magnitude 7 units.
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 7
2nd PUC Maths Model Question Paper 4 with Answers - 8

Question 22.
If \(\overrightarrow{\mathbf{a}}\) =5î – ĵ – 3k̂ and \(\overrightarrow{\mathbf{b}} \) = î + 3ĵ – 5k̂, then show that the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) are perpendicular.
Answer:
Given \(\overrightarrow{\mathbf{a}}\) =5î – ĵ – 3k̂
\(\overrightarrow{\mathbf{b}} \) = î + 3ĵ – 5k̂
\(\vec{a}+\vec{b}\) = 6î + 2ĵ – 8k̂
\(\vec{a}-\vec{b}\) = 4î – 4ĵ + 2k̂
\(\vec{a}+\vec{b}\)] . \(\vec{a}-\vec{b}\) =
=(6)(4) + (2(-4) + (-8)(2)
= 24 – 8 – 16 = 24 – 24=0 .
\(\vec{a}+\vec{b}\) & \(\vec{a}-\vec{b}\) are perpendicular.

Question 23.
Find the Vector equation of the line, passing through the points (-1, 0, 2) and (3,4,6)
Answer:
Let \(\overrightarrow{\mathbf{a}}\) = (-1) î + 0ĵ + (2)k̂
\(\overrightarrow{\mathbf{b}} \) =(3) î + (4)ĵ + (6)k̂
we have \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})\)
But, \(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \mathrm{k}\)
∴ r = [(-1) î+(0) ĵ+(2)k̂] + λ(4 î + 4ĵ +4k̂)

Question 24.
Two coins are topped once, find P (E/ F) where E: no tail appears, F : no head appears.
Answer:
s={HH, HT, TH, TT }
E={HH},F={TT} and E∩F={/}
∴ P(E) = \(\frac{1}{4}\) and P(E∩F)= {0}
Thus, P (E/F) = \(\frac{\mathrm{P}(\mathrm{EnF})}{\mathrm{P}(\mathrm{F})}=\frac{0}{(1 / 4)}=0\)

Part – C

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

Question 25.
Show that the relation R in the set of real number R defined as R = {(a,b):a≤b2} is number reflexive nor symmetric nor transitive
Answer:
It can be observed that \(\left(\frac{1}{2}, \frac{1}{2}\right)=\frac{1}{2} \not<\frac{1}{4}\)
∴ R is not reflexive.
Now (l,2)ERas 1 < 22
But, 2 is not less than 12
∴ (2, 1) ∉ R ∴ is not symmetric
Now,(5, 3) (3, 2) ∈ R
( as 5 > 32 = 9 and 3 < (2)2 = 4)
But,5 > (2)2 = 4
∴ (5,2) ∉ R ,
∴ R is not Transitive
∴ \(\overline{\mathrm{k}}\) is neither reflexive, nor symmetric nor Transitive.

Question 26.
Prove that tan-1 x + tan-1 \(\left(\frac{2 x}{1-x^{2}}\right)\) = tan -1\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\) , |x| < \(\frac{1}{3}\).
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 9

Question 27.
Find the values of x ,y and % in the following matrices
Answer:
\(\left(\begin{array}{ll}
{x+y} & {2} \\
{5+z} & {x y}
\end{array}\right)=\left(\begin{array}{ll}
{6} & {2} \\
{5} & {8}
\end{array}\right)\)
x + y = 6…(L1), 5 + 2 = 5 (L2), xy = 8….(L3)
From (2):
5 + 2=5 ⇒ 2 = 5-5=0
∴ 2 = o
From (1) :y = 6 – x
∴ (3) ⇒ x (6 – x) = 8 ⇒ 6x – x2 = 8
⇒ x2 – 6x + 8=0
⇒ (x – 4) (x – 2)=0
⇒ x – 4 or x = 2.
If x = 4, y = 6-4 =2
I x = 2, y = 6-2 = 4
Thus , x = 4, y = 2 and z = 0 or x = 2 ,y = 4 and z = 0.

2nd PUC Maths Model Question Paper 4 with Answers

Question 28.
Differentiate \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}}\) with respect to x.
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 10

Question 29.
If y = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) find \( \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}\)
Answer:
Put x = tanθ .’. θ = tan -1x
y = \(\sin ^{-1}\left[\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right]\)
y = sin -1[ sin 2θ] = 2θ
y =2tan -1x
diff w.r.t. ‘x’
\(\frac{d y}{d x}=\frac{2}{1+x^{2}}\)

Question 30.
Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is (a) Strictly increasing (b)Strictly decreasing.
Answer:
‘Given f (x) = 2x3 – 3x2 – 36x + 7
f1(X) = 6X2 – 6X – 36 = 6(x2 – x6)
f1(x) = 6(x + 2)(x – 3)
.’.f1 (x) = 0 gives x = -2andx = 3.The points x = -2 and x = 3 divides the real line into three disjoint intervals namely,
(-∞ , -2), (-2, 3) and (3, ∞).
2nd PUC Maths Model Question Paper 4 with Answers - 11

Question 31.
Evaluate : \(\int \frac{(1+\log x)^{2}}{x} d x\)
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 12

Question 32.
Evaluate : \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x d t\)
Answer:
Put sin -1 ∴ x = t x = sin t
Also \(\frac{1}{\sqrt{1-x^{2}}} d x=d t\)
∴ \(\frac{(x \sin -1 x)}{\sqrt{1-x^{2}}} d x=\int(\sin t) d t\)
= ∫ t sin t dt
Apply Integration by parts.
=t(-cost) – ∫(1)( -cost)dt
= t cos t + ∫ cos t dt
= -tcost + sint
= -sin -1x \(\sqrt{1-x^{2}}\) + sin (sin-1)
= -sin -1x \(\sqrt{1-x^{2}}\) + x + c
= x – sin-1x \(\sqrt{1-x^{2}}\) +c

2nd PUC Maths Model Question Paper 4 with Answers

Question 33.
Find the area between the curves y = x2 and y = x.
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 13
To find P:
y = x2 and y = x.
x = x2 => x2 – x = 0
⇒ x(x – 1)=0 ⇒ x = 0 or x-1
If x = 0,y = 0 ⇒ (0, 0) x = 1,y = 1=>(1, 1)
That points of Intersection are (0,0) and (1, 1).
∴ Area = \(\int[f(x)-g(x)] d x\)
But f(x) = x, Ig (x) = x2
2nd PUC Maths Model Question Paper 4 with Answers - 14
Thus, the required absolute area is = \(\frac{1}{6}\) Sq.units.

Question 34.
Form the differential equation representing the family of curves y = a sin(x+b) where a and b are arbitrary constants.
Answer:
y=asin(x+b)-(1)
diff (1) w . r. t‘x’
\(\frac{d y}{d x}\)= acos(x+b)-(2)
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = -asm(x + b) -(3)
Eliminating a and b from (1), (2) and (3)
we get \(\frac{d^{2} y}{d x^{2}}+y=0\)
This is the required differential Equation.

Question 35.
Find the area of a triangle having the points A(l,l,l,) and C (2,3,1) and as its vertices using vector method.
Answer:
We have
2nd PUC Maths Model Question Paper 4 with Answers - 15
2nd PUC Maths Model Question Paper 4 with Answers - 16

Question 36.
Prove that \([\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{d}}]=[\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{e}}]+[\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{d}}]\)
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 17

2nd PUC Maths Model Question Paper 4 with Answers

Question 37.
Find the distance between parallel lines \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}+\mathbf{m}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\) and \(\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+n(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
Answer:
We have \(\vec{a}_{1}=\hat{i}+2 \hat{j}-4 \hat{k}\) , [/latex]\vec{a}_{2}=3 \hat{i}+3 \hat{j}-5 \hat{k}[/latex] and \(\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\)

Question 38.
Let f : N → R be a function defined by f(x) = 4x2 + 12x + 5 .Show that f : N → S, where S is the range of function f, is invertible. Find the inverse of f.
Answer:
Let y ∉ s be any arbitrary element
Then, y = 4x2+12x + 5
2nd PUC Maths Model Question Paper 4 with Answers - 18
2nd PUC Maths Model Question Paper 4 with Answers - 19

Part – D

Answer any six questions. ( 6 x 5 = 30 )

Question 39.
Two cards are drawn successively with Replacement from a well shuffled deck of 52 cards. Find the probability distribution of the number of aces.
Answer:
Let x={0,1,2} denote the number of aces. ∴ p(x = 0) = p (non – ace) x p (non – ace)
= \(\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}\)
P (X = 1)=p (ace I non – ace) + p (non – ace) = p(ace).p(non – ace) + p(non – ace).p(ace)
= \(\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52}=\frac{24}{169}\)
p(x=2)= p(ace& ace) = \(\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}\)
∴ Thus, required probability distribution, is
2nd PUC Maths Model Question Paper 4 with Answers - 20
2nd PUC Maths Model Question Paper 4 with Answers - 21
2nd PUC Maths Model Question Paper 4 with Answers - 22
This shows that gof=IN and fog = IS
⇒ is invertible function
and hence, \(f^{-1}(x)=g=\sqrt{\frac{x+4}{4}}-\frac{3}{2}\)

Question 40.
Verify that (A + B) (C) = AC + BC,
If A = \(\left[\begin{array}{ccc}
{0} & {6} & {7} \\
{-6} & {0} & {8} \\
{7} & {-8} & {0}
\end{array}\right]\), B = \(\left[\begin{array}{lll}
{0} & {1} & {1} \\
{1} & {0} & {2} \\
{1} & {2} & {0}
\end{array}\right]\) and C = \(\left[\begin{array}{c}
{2} \\
{-2} \\
{3}
\end{array}\right]\)
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 23
2nd PUC Maths Model Question Paper 4 with Answers - 24

2nd PUC Maths Model Question Paper 4 with Answers

Question 41.
Solve the following system of equations by matrix method x+y+z=6, x-y-z = -4 and x+ 2y = -1.
Answer:
Consider
AX=B-(1) => X =A-1 B – (2)
Here
A = \(\left[\begin{array}{ccc}
{1} & {1} & {1} \\
{1-1} & {-1} \\
{1} & {2} & {-2}
\end{array}\right]\) , X = \(\left[\begin{array}{l}
{\mathrm{X}} \\
{\mathrm{Y}} \\
{\mathrm{Z}}
\end{array}\right]\) , B = \(\left[\begin{array}{r}
{6} \\
{-4} \\
{-1}
\end{array}\right]\)
To find A-1:
= 1 (2 + 2)- 1 (-2 + 1) + 1(2 + 1)
=1(4) – 1(-1) + 1(3)
=4 + 1 + 3
|A| = 8 ≠ 0
2nd PUC Maths Model Question Paper 4 with Answers - 25
2nd PUC Maths Model Question Paper 4 with Answers - 26

Question 42.
If ey(x+1) = 1 Prove that \(\frac{d y}{d x}=-e^{y}\) and hence prove that \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)
Answer:
ey(X + 1) = 1 ⇒ ey = \(\frac{1}{x+1}\) ….. (1)
Differentiating with respect to ‘x’
2nd PUC Maths Model Question Paper 4 with Answers - 27
2nd PUC Maths Model Question Paper 4 with Answers - 28

Question 43.
The length of a rectangle is decreasing at the rate of 3cm / minute and width – y is increasing at the rate of 2 cm / minute, when x= 10 cm and y = 6 cm, find the rate of change of (i) the perimeter and (ii) the area of the rectangle.
Let → length and y → width.
Data: x = 10cm, y = 6cm
\(\frac{d x}{d t} = -3cm/min,\frac{\mathrm{d} y}{\mathrm{dt}}\) + 2cm / min

(i) perimeter, p=2(x + y)
\(\frac{d p}{d t}=2\left[\frac{d x}{d t}\right]\) = 2[-3+2] = 2(-)
\(\frac{d p}{d t}\) = -2 cm/min

(ii) Araea , A = xy
\(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{y} \frac{\mathrm{dx}}{\mathrm{dt}}\)
= (10)(2) + (6)(-3) = 20-18
\(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\) = 2cm2/min

2nd PUC Maths Model Question Paper 4 with Answers

Question 44.
Find the integral of \(\sqrt{x^{2}+a^{2}}\) w . r .t
‘x’ and hence evaluate \(\int \sqrt{x^{2}+4^{x}+6} d x\)
Answer:
Let I = \(\sqrt{x^{2}+a^{2}}\)
= \(\sqrt{x^{2}+a^{2}}(1) dx\)
Apply Integration by parts
2nd PUC Maths Model Question Paper 4 with Answers - 29
2nd PUC Maths Model Question Paper 4 with Answers - 30
2nd PUC Maths Model Question Paper 4 with Answers - 31
2nd PUC Maths Model Question Paper 4 with Answers - 32

Question 45.
Using Integration find the area of triangle bounded by the verticed A (2, 0), B ( 4, 5) and C (6, 3).
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 33
Let A (2,0), B = (4, 5), C= (6, 3) be the vertical of a triangle ABC.
Area of ∆ ABC = Area of ∆ ABD + Area of trapezium BDEC – Area of ∆ ACE (i) Equation of the side AB,
2nd PUC Maths Model Question Paper 4 with Answers - 34

(ii) Equation of the side BC,
\(\frac{y-3}{x-6}=\frac{5-3}{4-6}=\frac{2}{-2}\) = -1
y – 3 = – (x – 6)
y-3 = -3 – x + 6
y = 3 – x+6

(iii) Equation of side AC,
\(\frac{y-0}{x-2}=\frac{3-0}{6-2}=\frac{3}{4}\)
y = \(\frac{3}{4}(x-2)\)
Hence, Area of ∆ ABC =
2nd PUC Maths Model Question Paper 4 with Answers - 35
.’. Area of ∆ ABC| = 9 s.q. units Thus, the required area will be 9 sq. units.

Question 46.
Solve the differential Equation ydx -(x+2y2) dy
Answer:
y dx – (x + 2y2) dy = 0
( x + 2y2)dy = ydx
2nd PUC Maths Model Question Paper 4 with Answers - 36
This is a Linear Differential Equation in ‘x’.with p = \(\frac{1}{y}\) , Q = 2y?
2nd PUC Maths Model Question Paper 4 with Answers - 37
2nd PUC Maths Model Question Paper 4 with Answers - 38
This is the required General Solution of the ordinary Diffrential Equation.

Question 47.
Derive equation perpendicular to a given vector and passing through a given point both in the vector and cartesian form.
2nd PUC Maths Model Question Paper 4 with Answers - 39
Let \(\overrightarrow{\mathrm{r}}\) be the position vector of any point p (x, y, z) in the plane.
Then the point p lies in the plane if
2nd PUC Maths Model Question Paper 4 with Answers - 40
This is the required equation of a plane in vector form
Cartisian form
2nd PUC Maths Model Question Paper 4 with Answers - 41
2nd PUC Maths Model Question Paper 4 with Answers - 42
This is the required equation of a plane in cartesian form.

2nd PUC Maths Model Question Paper 4 with Answers

Question 48.
There are 5 % defective itemas in a large bulk of items, what is the probability that sample of 10 items will include not more than one defective item?
Answer:
Let p= probability of getting defective item = 0.05
q = probability of getting non – defective
item = 0. 95
and n=10
we have, by Binomial distribution.
p(x) = nCx px qn-x
p(x) = 10Cx (0.05)x (0.95)10 – x
To find p =(x ≤ 1): p = p(x = 0) + p( x = 1)
=10 Co (0.05)0 (0.95)10 +10C1 (0.95)9 +10C1(o.o5)1 (0.95)9
= (0.95)9 [0.95 + 0.5]
p(x ≤ 1) = (0.95)9 [1.95]

Part – E

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

Question 49.
(a) Prove that \(\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a}(x) d x\) when f (2a – x) = f(x) I Hence evaluate \(\int_{0}^{\pi}(\cos x) d x\) (6 Marks)
on using p2 we have
2nd PUC Maths Model Question Paper 4 with Answers - 43
Let t = 2a-x ∴ dt = -dx
when x = a; t = a
I x = 2a; t = 0
∴ 2nd Integral in RHS becomes
2nd PUC Maths Model Question Paper 4 with Answers - 44
2nd PUC Maths Model Question Paper 4 with Answers - 45
2nd PUC Maths Model Question Paper 4 with Answers - 46

(b) Prove that \(\left|\begin{array}{ccc}
{\mathbf{1}} & {\mathbf{x}} & {\mathbf{x}^{2}} \\
{\mathbf{x}^{2}} & {\mathbf{1}} & {\mathbf{x}} \\
{\mathbf{x}} & {\mathbf{x}^{2}} & {\mathbf{1}}
\end{array}\right|=\left(\mathbf{x}^{3}-\mathbf{1}\right)^{2}\)
Answer:
2nd PUC Maths Model Question Paper 4 with Answers - 47
Expanding along
= (x3-1)2 {[l-0]-0[x-0]+0[x2-x2] }
= (X3-1)2(1)
A = (x3 -1)2

2nd PUC Maths Model Question Paper 4 with Answers

Question 50.
(a) Solve the following linear programming problem graphically
Maximise , z =3x+2y S.Tx + 2y ≤ 10,
3x + y ≤ 15,x ≥ 0, y ≥ 0. (6 Marks)
Answer:
consider, x + 2y =10 ………. (1)
2nd PUC Maths Model Question Paper 4 with Answers - 48
∴Equation (l) passes through A(0,5) and B (10,0)
3x + y = 15 …………. (2)
2nd PUC Maths Model Question Paper 4 with Answers - 49
∴ Equation (2) passes through c(0,15) and D (5 , 0)
2nd PUC Maths Model Question Paper 4 with Answers - 50
Here ODEA represents feasible region and it is bounded. The feasible solutions exists at all these points. The optiomal solution is obtained as follows.
2nd PUC Maths Model Question Paper 4 with Answers - 51
Thus, (z)max = 18at x = 4 and y = 3

2nd PUC Maths Model Question Paper 4 with Answers

(b) Find the relation between a and b so that the function f defined by f (x) = {ax + 1,if x≤ 3 is continuous bx + 3, if x > 3} at x = 3. (4 Marks)
Answer:
we have , f(x) = 3a + 1
2nd PUC Maths Model Question Paper 4 with Answers - 52
Also, f is continous at x = 3
2nd PUC Maths Model Question Paper 4 with Answers - 53
⇒ 3a +1 = 3b + 3 = 3a +1
consider, 3a +1 = 3b + 3
3a = 3b + 2
a = \(\frac{3 b+2}{3}\)