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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.
2nd PUC Maths Previous Year Question Paper March 2019 1

KSEEB Solutions

Question 4.
Find a value of x for which
2nd PUC Maths Previous Year Question Paper March 2019 2
Answer:
2nd PUC Maths Previous Year Question Paper March 2019 3
3 – x2 = 3 – 8 ⇒ x2 = 8 ⇒ x = ±2\(\sqrt{2}\)

Question 5.
If y = sin(x2 + 5) then find \(\frac{d y}{d x}\)
Answer:
y = sin(x2 + 5)
\(\frac{d y}{d x}\) = cos(x2 + 5)(2x)

Question 6.
Find \(\int(1-x) \sqrt{x} d x\)
Answer:
\(\int(1-x) \sqrt{x} d x\)
2nd PUC Maths Previous Year Question Paper March 2019 4

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

KSEEB Solutions

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(X1) = f(x2)
⇒ 2x1 = 2x2 ⇒ x1 = x2 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-1x = \(\frac{\pi}{2}\) – 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.
2nd PUC Maths Previous Year Question Paper March 2019 6

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
2nd PUC Maths Previous Year Question Paper March 2019 7
= \(\frac { 1 }{ 2 }\) |[-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)
2nd PUC Maths Previous Year Question Paper March 2019 8

Question 16.
Find ax + by2 = cosy then find \(\frac{dy}{d x}\)
Answer:
ax + by2 = cosy
⇒ \(\frac{d}{d x}\)ax + \(\frac{d}{d x}\)by2 = \(\frac{d}{d x}\)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 = x3
or dV = \(\left(\frac{d V}{d x}\right)\) ∆x = (3x2) ∆x
= (3x2) (0.02x) = 0.06x3m3
Thus, the approximate change in volume is 0.06 x3m3

KSEEB Solutions

Question 18.
Find \(\int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} d x.\)
Answer:
Put (1 – tan x) = t ⇒ – sec2 x dx = dt
2nd PUC Maths Previous Year Question Paper March 2019 9

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:
2nd PUC Maths Previous Year Question Paper March 2019 10

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
2nd PUC Maths Previous Year Question Paper March 2019 11

Question 23.
Find the distance of the point (3, -2, 1) from the plane 2x – y + 2Z + 3 = 0.
Answer:
2nd PUC Maths Previous Year Question Paper March 2019 12

KSEEB Solutions

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 < b3} is reflexive, symmetric or transitive.
Answer:
(i) For (a, a) ∈R, a < a3 which is false.
As for a = -2, a3 = -8
∴ a \(\nleq\) a3 as – 2 \(\nleq\) – 8
∴ R is not reflexive.

(ii) For (a, b), a < b3 and (b, a), b < a3 which is false
As for a = 1 and b = 2
a < b3 as 1 < 8
Now, b \(\nleq\) a3 as 2 \(\nleq\) 1
∴ R is not symmetric.

(iii) For (a, b), (b, c) and (a, c)
a < b3 and b < c3
Now a = 25, b = 3, c = 2
a < b3 and 25 < 27
b < c3 and 3 < 8
but a \(\nleq\) c3 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 (\(\frac { 33 }{ 65 }\))

KSEEB Solutions

Question 27.
By using elementary transformations, find the inverse of the matrix
2nd PUC Maths Previous Year Question Paper March 2019 13
Answer:
2nd PUC Maths Previous Year Question Paper March 2019 14

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)
2nd PUC Maths Previous Year Question Paper March 2019 15

KSEEB Solutions

Question 29.
Verify Rolle’s theorem for the function f(x) = x2 + 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) = 2X3 – 3x2 – 36x + 7 is strictly increasing.
Answer:
f(x) = 2x3 – 3x2 – 36x + 7
f ‘(x) = 6x2 – 6x – 36
f ‘(x) = 0 ⇒ 6x2 – 6x – 36 = 0
2nd PUC Maths Previous Year Question Paper March 2019 47
⇒ 6(x2 – x – 6) = 0
⇒ 6(x – 3)(x + 2) = 0
6 ≠ 0, x = 3, x = -2
2nd PUC Maths Previous Year Question Paper March 2019 16

Question 31.
Find \(\int x \log x d x\)
Answer:
2nd PUC Maths Previous Year Question Paper March 2019 17

KSEEB Solutions

Question 32.
Evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
Answer:
2nd PUC Maths Previous Year Question Paper March 2019 18

Question 33.
Find the area of the region bounded by the curve y2 = 4x and the line x = 3.
Answer:
Given curve is y2 = 4x and line is x = 3
y2 = 4x
2nd PUC Maths Previous Year Question Paper March 2019 19

Question 34.
Form the differential equation of the family of curves y = ae3x + be-2x by eliminating arbitrary constants a and b.
Answer:
Given, y = a e3x + b e-2x …. (1)
y’ = (a e3x) 3 – (be-2x) 2
y’ = 3 a e3x – 2 be-2x …. (2)
From (1) and (2)
y = a e3x + b e-2x × 3
y’ = 3a e3x – 2b e-2x × 1
2nd PUC Maths Previous Year Question Paper March 2019 20
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.

KSEEB Solutions

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:
2nd PUC Maths Previous Year Question Paper March 2019 21

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
2nd PUC Maths Previous Year Question Paper March 2019 22
= -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:
2nd PUC Maths Previous Year Question Paper March 2019 23
Theorem: Derive intercept form of the equation of a plane
Let the equation of a plane be
Ax + By + Cz + 0 = 0 (1)
2nd PUC Maths Previous Year Question Paper March 2019 24
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

KSEEB Solutions

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”.
2nd PUC Maths Previous Year Question Paper March 2019 25

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(x1) = f(x2)
4x1 + 3 = 4x2 + 3
4x1 = 4x2
∴ x1 = x2.
∴ 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 A2 – 23A – 40I = 0
Answer:
We have
2nd PUC Maths Previous Year Question Paper March 2019 26
2nd PUC Maths Previous Year Question Paper March 2019 27

KSEEB Solutions

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
2nd PUC Maths Previous Year Question Paper March 2019 28
|A| = 3(2 – 3) + 2(4 + 4) + 3(-6 -4) = -17
Hence, A is non singular matrix and so its inverse exists. Now
A11 = -1 A12 = -8 A13 = -10
A21 = -5 A22 = -6 A23 = +1
A31 = -1 A32 = +9 A33 = +7
2nd PUC Maths Previous Year Question Paper March 2019 29

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
y1 = \(\frac{1}{\sqrt{1-x^{2}}}\) i.e., (1 – x2)y12 = 1
so (1 – x2). 2y1y2 + y12(0 – 2x) = 0
Hence (1 – x2) y2 – xy1 = 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 cm2/min.

KSEEB Solutions

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:
2nd PUC Maths Previous Year Question Paper March 2019 30
2nd PUC Maths Previous Year Question Paper March 2019 31

Question 45.
Using the method of integration, find the smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2.
Answer:
Given circle x2 + y2 = 4 and the line is x + y = 2
x2 + (2 – x)2 = 4
[∴ y = 2 – x]
2nd PUC Maths Previous Year Question Paper March 2019 32
2x2 – 4x = 0 ⇒ 2x(x – 2) = 0 ⇒ x = 0, 2
Now, x2 + y2 = 4 and x + y = 2
⇒ y2 = 4 – x2 y = 2 – x,
⇒ y = \(y=\sqrt{2^{2}-x^{2}}\)
Required Area
2nd PUC Maths Previous Year Question Paper March 2019 33

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 x2 + y2 = 4 and the line is x + y = 2
x2 + (2 – x)2 = 4 [∴ y = 2 – x]
2nd PUC Maths Previous Year Question Paper March 2019 34
2x2 – 4x = 0 ⇒ 2x(x – 2) = 0 ⇒ x = 0, 2
Now, x2 + y2 = 4 and x + y = 2
⇒ y2 = 4 – x y = 2 – x,
⇒ y = \(\sqrt{2^{2}-x^{2}}\)
Required Area
2nd PUC Maths Previous Year Question Paper March 2019 35

KSEEB Solutions

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:
2nd PUC Maths Previous Year Question Paper March 2019 36
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
2nd PUC Maths Previous Year Question Paper March 2019 37
2nd PUC Maths Previous Year Question Paper March 2019 38
2nd PUC Maths Previous Year Question Paper March 2019 39

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:
2nd PUC Maths Previous Year Question Paper March 2019 40

KSEEB Solutions

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
2nd PUC Maths Previous Year Question Paper March 2019 41

(b) Evaluate
2nd PUC Maths Previous Year Question Paper March 2019 42
Answer:
Taking a, b, c commn from R1, R2, R3
2nd PUC Maths Previous Year Question Paper March 2019 43
2nd PUC Maths Previous Year Question Paper March 2019 44

KSEEB Solutions

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
2nd PUC Maths Previous Year Question Paper March 2019 45
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)
2nd PUC Maths Previous Year Question Paper March 2019 46
⇒ K(5) + 1 = 3(5) – 5
⇒ 5K+ 1 = 10 ⇒ 5K = 9
⇒ K = \(\frac { 9 }{ 5 }\)