2nd PUC Maths Model Question Paper 3 with Answers

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Karnataka 2nd PUC Maths Model Question Paper 3 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.
Give an example of a relation which is symmetric only.
Answer:
R= {(1,2), (2,1),(1,3),(3,1),(2,3),(3,2)}

Question 2.
Find the principal value of cos^-1 \(\left(\frac{-1}{\sqrt{2}}\right)\)
Answer:
cos ^-1 (-x) = π – cos ^-1 x
∴ cos ^-1\(\left(\frac{-1}{\sqrt{2}}\right)\) = π – cos ^-1 \(\left(\frac{-1}{\sqrt{2}}\right)\)
= π – \(\frac{\pi}{4}\) = \(\frac{3\pi}{4}\)

Question 3.
Costruct a 2 x 2 matrix A = [aij] . whose elements are given by aij = 2i + j.
Answer:
A = \(\left[\begin{array}{ll}
{a_{11}} & {a_{12}} \\
{a_{21}} & {a_{22}}
\end{array}\right] \)
Now
a₁₁ 2(1) + 1 =3; a₁₂ = 2(1) + 2 = 4
a₂₁ 2(2) + 1 =5; a₂₂ = 2(2) + 2 = 6
∴ A = \(\left[\begin{array}{ll}
{3} & {4} \\
{5} & {6}
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{ll}
{2} & {3} \\
{-1} & {4}
\end{array}\right]\), find |2A|.
Answer:
Now 2A = \(\left[\begin{array}{cc}
{4} & {6} \\
{-2} & {4}
\end{array}\right]\)
∴ |2A| = \(\left|\begin{array}{cc}
{4} & {6} \\
{-2} & {4}
\end{array}\right|\) = 16+12 = 28

Question 5.
If y = e^3logx, then show that \(\frac{d y}{d x}=3 x^{2}\)
Answer:
y=e^3logx e^logx3=x³(as e^{log x} = x)
∴ y = x³
∴ \(\frac{d y}{d x}=3 x^{2}\)

Question 6.
Find the antiderivative of x²(1 – \(\frac{1}{x^{2}}\) ) with respect to x.
Answer:
Let I = ∫ x²(1 – \(\frac{1}{x^{2}}\) ) dx =∫(x² – 1)dx
I = \(\frac{x^{3}}{x} \) – x + c

Question 7.
Define the feasible region in LPP.
Answer:
The common region determined by all the constraints including non – negative constraints x , y ,≥ of a linear programming problem is called feasible region.

Question 8.
Find the unit vector in the direction of the vector \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Answer:

Question 9.
Find the direction Ratio of The line

Answer:

Question 10.
If P(E) = 0.6 and P(E∩F)=0.2 then find P(F/E)
Answer:
\(P(F / E)=\frac{P(F n E)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}\)

Part -B

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

Question 11.
Define Binary operation * on a set. Verify whether the operation * defined on z by a * b= ab + 1 is binary or not.
Answer:
A binary operation * on a set A is a function * : A x A → A. we denote * (a,b) by
Given, a * b = ab + 1.
Let a = 1, b = 2 Then 1 * 2 = (1) (2) + 1
= 2 + 1 + 3 ∉ z
∴ * is a binary operation on z.

Question 12.
Write \(\cot ^{-1}\left(\frac{1}{\sqrt{x^{2}-1}}\right)\) , x≥ in the simplest form.
Answer:

Question 13.
Find the Equation of a line passing through the points (3, 2) and (-1, -3) by using determinants.
Answer:
Let A(3,1), B(-1,-3) be given Pts.
C(x,y) beany point on the line AB
∴  A, B & C are collinear
∴ ∆ABC =0
\(\left|\begin{array}{ccc}
{x} & {y} & {1} \\
{3} & {2} & {1} \\
{-1} & {-3} & {1}
\end{array}\right|=0\)
⇒ x[2 + 3] -y [3 + 3] + 1[-9 + 2] = 0
⇒ 5x – 6y – 7 = 0

Question 14.
Show that

Answer:
Let Sin^-1 x = θ ⇒x = sinθ
we have sin^-1x (2sinθ \(\sqrt{1-\sin ^{2} \theta}\) )
Sin^-1 (2 sinθ cosθ ) = sin^-1 ( sin 2 θ ) = 2θ =2 sin^-1 x

Question 15.
If y = tan^-1 \(\left[\frac{3 x-x^{3}}{1-3 x^{2}}\right]\) , \(\frac{-1}{\sqrt{3}}\frac{1}{\sqrt{3}}\) then find \(\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} \mathbf{x}}\) ,
Answer:
y = tan^-1 \(\left[\frac{3 x-x^{3}}{1-3 x^{2}}\right]\)
Put x=tanθ (or) θ = tan^-1x
Then y = tan^-1 \(\left[\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \operatorname{tam}^{2} \theta}\right] \)
= tan^-1(tan 3θ ) = 3θ
y = 3tan ^-1 x
∴ \(\frac{d y}{d x}=\frac{3}{1+x^{2}}\)

Question 16.
Find \(\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} \mathbf{x}}\) , if Sin²x + cos²y = k , where k is constant.
Answer:
Sin²x + cos²y=k
2sinx cosx + 2cosy(-siny)\(\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} \mathbf{x}}\) = 0
⇒ 2cosysiny \(\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} \mathbf{x}}\) =2smxcosx
⇒ sin2y\(\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} \mathbf{x}}\) =sin2x
∴ \(\frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}\)

Question 17.
If the radius of the Sphere is measured as 7 cm with an error of 0. 02 cm, then find the approximate error in calculating its volume.
Answer:
Let r be the radius of the sphere, and ∆ be the error in measuring the radius. Then r = 7 cm and ∆ r = 0.02 cm
∴ volume of the sphere, v = \(\frac{4}{3} \pi r^{3}\)

Question 18.
Evaluation : \(\int \frac{\cos 2 x}{(\sin +\cos x)^{2}} d x\)
Answer:
Put (sinx + cosx)² =t
Then 2 (sinx+cosx) (cosx-sinx) dx=dt
=⇒ 2(cos² x-sin² x)dx = dt
⇒ 2cos2xdx =dt
⇒ cos2xdx = \(\frac{\mathrm{dt}}{2}\)

Question 19.
Evaluate :∫ tan ^-1 xdx
Answer:
Let I = ∫ tan ^-1 xdx =∫ tan ^-1 x (1)dx
Apply Integration by parts.

Question 20.
Find the projection of the vector \(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\) on the vector \(7 \hat{\mathrm{i}}-\hat{\mathrm{j}}+8 \hat{\mathrm{k}}\)
Answer:
\

Question 21.
Find the area of parallelogram whose adjacent sides are given by the vector
\(\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}\) and \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Answer:
Let \(\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}\) , \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
∴ Area of a paralleogram = \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\)
Now, \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{rrr}
{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\
{3} & {1} & {4} \\
{1} & {-1} & {1}
\end{array}\right|\)
= î[1+4] ĵ[+3-4] + k̂[-3-1]
\(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\) = 5î + ĵ -4k̂

Question 22.
Find the distance of a point (2, 5, -7) from the plane \(\overrightarrow{\mathbf{r}}\).( 6î – 3ĵ + 2k̂) = 4
Answer:
Let \(\vec{a}=2 \hat{j}+5 \hat{j}-7 \hat{k}\)
\(\vec{N}=6 \hat{i}-3 \hat{i}+2 \hat{k}\) and d = 4
There fore, the distance of the point (2,5,-7) from the given plane is

Question 23.
Find the order and Degree of the Different Euquation \(\frac{d^{3} y}{d x^{3}}+\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)
Answer:
Order=3 Degree = 1

Question 24.
Given that the event A and B are such that P(A)=\(\frac {1}{2}\) ,P(B) = K, find k
Answer:
if A and B are In dependent.
∵ A and B are Indepentent,
we have,
P(A∩B) =P(A).P(B)

Part – C

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

Question 25.
Show that the relation R in the set of all integers Z defined by R = {(a,b): 2 divides a-b} is an Equivalence relation.
Answer:
(1) R is Reflexive, as 2 divides (a – a) for all a ∈ z
(2) Further, If (a, b) ∈ R, then 2 divided a – b
∴  2 divides b -a
Hence, (b – b ) ∈ R
⇒ R is symmetric
(3) If (a – b) ∈ R and (b , c) ∈ R , then
a – b I b- c are divisible by 2
Now , a -c = (a – b) + ( b – c) is even
so, a – c is divisible by 2.
⇒ R is transitive. Thus, R is an Equivalence relation in z

Question 26.
Simplify :

Answer:

Question 27.
Express matrix A = \(\left[\begin{array}{cc}
{1} & {2} \\
{2} & {-1}
\end{array}\right]\) as the sum of a symmetrix and skew – symmetric matric matrix.
Answer:
We have

Thus P =\(\frac {1}{ 2 }\)(A + A¹) is a skew – symmetric matrix.
Also Let

Thus , A is represented as a Sum of Symmetric and skew – Symmetric matrix.

Question 28.
prove that the function is differentiable at a point c then it is also continuous at that point.
Answer:
Since f is differentiable at c,

Question 29.
Verify that mean value theorem for the function f (x) = x² – 4x – 3 in the interval [1,4].
Answer:
f (x) =x² – 4x – 3
is continuous in [ 1,4]
f¹ (x) = 2x -4 exists in (1,4).
Now f(a)=f(1)
=1² – 4 (1) -3 – 1 – 4 – 3 = -6
f(b)=f(4) = 4² – 4(4) – 3
= 16 – 16 – 3 =-3

Hence, Mean value Theorem is verified.

Question 30.
Find the equation of the targent to the curve given byx = asin³ t.y = bcos³t at point where t = \(\frac{\pi}{2}\)
Answer:
x =asin³t ; \(\frac{d x}{d t}\) =3a sin² cost
y = b cos³t \(\frac{d y}{d t}\) = – 3b cos² + sin t

Equation of targent at x = \(\frac{\pi}{2}\) is
y – y₁ = \(\frac{d y}{d t}\) (x – x₁)
y – 0 =(0)(x—a)
y = 0

Question 31.
Evaluate \(\int \frac{x+2}{2 x^{2}+6 x+5} d x\)
Answer:
Put 2x² + 6x + 5
(4x+6).dx = dt

Question 32.
Evaluate \(\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x\)
Answer:

Question 33.
Find area bounded by the parabola y² =4x and the line y = 2x

Answer:
To find P:
y² =4x and y = 2x
Now 4x² = 4x ⇒ x² = x
⇒ x (x – 1) = 0
⇒ x = 0, x = 1
If x = 0; y = 0 ⇒ (0,0)
x = 1 ; y =2 ⇒ (1,2)
Thus the points of Intersection are (0, 0) and (1,2)

Question 34.
Three vectors \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}} \& \overrightarrow{\mathbf{c}}\) satisfy the
condition \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0\) Evaluate the quantity
\(\mu=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}\) If \(|\overrightarrow{\mathrm{a}}|\) = 1,\(|\overrightarrow{\mathrm{b}}|\) = 4 and \(|\overrightarrow{\mathrm{c}}|\) = 2
Answer:

Question 35.
If \(\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}\) , \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \) and \(\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}\) then find \(\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})\) and \((\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{c}}\)
Answer:
\(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc}
{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\
{1} & {2} & {-3} \\
{3} & {4} & {-1}
\end{array}\right|\)
= î [-2 + 12] – ĵ[-l + 9] + k̂ [4 -6]
= î [10] -8 ĵ -2k̂ =10î – 8ĵ – 2k̂
\(\vec{a} \cdot(\vec{b} \times \vec{c})\) = 2(10) + (-3)(-8) + (4)(-2)
= 20 + 24 – 8
\(\vec{a} \cdot(\vec{b} \times \vec{c})\) = 36
Similary \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}
{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\
{1} & {2} & {-3} \\
{3} & {4} & {-1}
\end{array}\right|\)
= î [9 – 8] – ĵ[-6 – 4] + k̂[4 + 3]
= î + 10ĵ +7k̂
∴ \((\vec{a} \times \vec{b}) \cdot \vec{c}\) =(1) (3) +(10)(4) +(7) (-1)
= 3 + 40 – 7
\((\hat{a} \times \vec{b}) \cdot \vec{c}\) = 36

Question 36.
Find the Shortest distance between the lines
\(\mathbf{L}_{1}: \overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) – (1) and
\(\mathrm{I}_{2}: \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})-(2)\)
Answer:

Question 37.
Find the Differential Equation of the family of circle touching the x – axis at origin.
Answer:

Let C denote the family of circles touching x – axis at origin.
Let (0, a) be the coordinates of the centre of any member of the family.
∴ Equation of familycis
x² + (y-a)² = a²
(or) x² + y² =2ay-(1)
where, a is arbitrary constant.
Diff (1) w. r. t ‘x’


This is the required Differential Equation of the given family of Θ^les

Question 38.
An Insurance company insured 2,000 scooter drivers, 4,000 car drivers and 6,000 truck drivers, The probability . of an accident are 0. 01, 0.03 and 0.15 respectively, one of the insured person meets with an accident. what is the probability that he is a scooter driver.
Answer:
Let A → Scooter drivers
B → car drivers
C → struck drivers
probability of selecting insured person
P(A)= \(\frac{1}{3}\) p(B) = \(\frac{1}{3}\),P(C) = \(\frac{1}{3}\)
Let D be the change of selecting accident person
P(D/A) = 0.01, P(D/B) = 0.03
P(D/C) = 0.05
P(D) = P(A).P(D/A)P(D/B) + P(C).P(D/C)
= (\(\frac{1}{3}\))(0.01) + (\(\frac{1}{3}\))(0.03) + (\(\frac{1}{3}\))(0.15)
= (0.33) (0.01) + (0.33) + (0.03) + (0.03) + (0.33) (0.15)
=0.0033+0.0099 + 0.0495
P(D) = 0.0627
we have to find P(A/D)

Part – D

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

Question 39.
Let f: N → S, where S is the range of the function is invertible. Also, find the invers off.
Answer:
Let y∉s be any arbitrary element Then, y = 4x² +12x + 15

Question 40.
Verify (B + C)A =BA + CA, if A =\(\left[\begin{array}{cc}
{2} & {3} \\
{4} & {-5}
\end{array}\right]\) B = \(\left[\begin{array}{cc}
{3} & {8} \\
{11} & {21}
\end{array}\right]\) and C \(\left[\begin{array}{cc}
{7} & {13} \\
{5} & {19}
\end{array}\right]\)
Answer:

Question 41.
Solve the Equations x-y + 3z = 10, x-y-z = -2, x-y-z = -2 and 2x + 3y + 4z = 4 by matrix method.
Answer:
Consider
Ax = B (1) x = A^-1B…….(2)
Have
A = \(\left[\begin{array}{ccc}
{1} & {-1} & {3} \\
{1} & {-1} & {-1} \\
{2} & {3} & {4}
\end{array}\right]\)
x = \(\left[\begin{array}{l}
{x} \\
{y} \\
{z}
\end{array}\right]\)
B = \(\left[\begin{array}{c}
{10} \\
{-2} \\
{4}
\end{array}\right]\)
To find A^-1 :|A| = \(\left[\begin{array}{ccc}
{1} & {-1} & {3} \\
{1} & {-1} & {-1} \\
{2} & {3} & {4}
\end{array}\right]\)
= 1 [-4 + 3] +1[4 + 2] + 3[3 + 2]
= 1(-1)+ 1(6) + 3(5) = -1 + 6+ 15
|A|≠ 20 ≠ 0

Question 42.
If y = Ae^mx + Be^nx then prove that
\(\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0\)
Answer:
y = Ae^mx + Be^nx (1)
\(\frac{d y}{d x}\) = Ame^mx + Bne^nx …. (2)
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\) = Am²e^mx + Bn²e^nx …. (3)
Now , \(\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y\)
Am²e^mx + Bn²e^mx – (m + n )[Ame^mx + Bne^nx ] + mn[Ae^mx + Be^nx]
= Am²e^mx + Bn²e^nx – Am²e^mx – Bme^nxn – Amne^mx – Bn²e^nx + Amne^mx +Bmne^nx
= 0

Question 43.
A ladder 5m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall at the rate of 2cm / sec. How fast is its heiht on the wall decreasing when the foot of the ladder is 4cm away from the wall ?
Answer:

Let AC be the ladder.
Let AB = x I BC = y
By data: \(\frac{d x}{d t}\) =2cm/sec
we have to find \(\frac{d y}{d t}\)
From ∆^le ABC, x² +y² = 5²
x² + y² = 25 .
=> \(2x\frac{d x}{d t} + 2y\frac{d y}{d t}/[latex] = 0
[latex]\frac{d y}{d t}\) = \(\frac{-x}{y}\) (2)
\(\frac{d y}{d t}\) = \(\frac{-2x}{y}\) …… (1)
Again, when x = 4m , we have
(4)² +y² = (5)²
=> y² = 25 -16 =>y² = 9
y = 3m
Thus when x = 4, y = 3 from (1):
\(\frac{d y}{d t}=\frac{-2(4)}{3}=\frac{-8}{3}\) ft/sec
Thus, the upper end is moving down wards at the rate of \(\frac{-8}{3}\) m /sec.

Question 44.
Find the Integral of \(\frac{1}{\sqrt{x^{2}-a^{2}}}\) w.r.t xI hence evaluate \(\int \frac{1}{\sqrt{x^{2}-2 x}} d x\)
Answer:

Question 45.
Using integration find the area bounded by the circle x² +y² = 16 and the parabola y² = 6x
Answer:
To Find P

x²+y² = 16 & y²=6x
x² + 6x =16 x² + 6x-16 = 0
(x + 8)(x – 2)=0
x = – 8, x = 2
But x ≠ 8, x = 2
If x = 2,then 4+y² =16 ⇒ y² =12
⇒ y = ±2y√3
Thus the points of Intersection are
p(2,2√3) and Q(2,-2√3). Thus , the Required Area
-4n

Question 46.
Derive the Equation of the line in space passing through a point and parallel to a vector both in the 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, \(\overrightarrow{\mathrm{AP}} \| \overrightarrow{\mathrm{b}}\)
ie, \(\overrightarrow{\mathrm{AP}}=\lambda \overrightarrow{\mathrm{b}}\)
where λ – some Re al number

Equating the corresponding components of î ,ĵ and k̂, respectively,
x – x₁ = aλ , y-y₁= bλ,z-z₁ = cλ
⇒ \(\frac{x-x_{1}}{a}=\lambda, \frac{y-y_{1}}{b}=\lambda, \frac{z-z_{1}}{c}=\lambda\)
Thus, we have,
\(\frac{\mathbf{x}-\mathbf{x}_{1}}{\mathbf{a}}=\frac{\mathbf{y}-\mathbf{y}_{1}}{\mathbf{b}}=\frac{\mathbf{z}-\mathbf{z}_{1}}{\mathbf{c}}\)

Question 47.
Solve the Differential Equation
\(x\log x \frac{d y}{d x}+y=\frac{2}{x} \log x\)
Answer:

Question 48.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Answer:
For this Random Experiment,the no:of sample points = 36
E = ={(1,1) (2,2) (3,3)(4,4)(5,5)(6,6) }
Let P = probability of getting doublet = \(\frac{6}{36}=\frac{1}{6}\)
q= probability of not getting doublet = \(1-\frac{1}{6}=\frac{5}{6}\)
and n = 4
we have

Part – E

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

Question 49.
a)One kind of cake requires 200g of flour and 25g of fat and another kind of cake requires lOOg of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredient, used in making the cakes. ( 6 marks)
Answer:
Let x → cake 1 and y → cake 2.

Minimize : Z = X + Y
SIT: 200x + 100y ≥ 5000
25x + 50y ≥ 1000
where , x , y, ≥ 0
Now, 200 x + 100y =5000 or
2x +y = 50 – (1)

Equation (l) passes through A(0,50)lB(25,0).
Also, 25x + 50 y =1000……(2)

Equation (2) passes through C(0,20)ID(40,0)

Here DEA represents feasible Region and (1) it is unbounded. The feasible solution exists all these points. .
The optimal solution is obtained as follows

(Z)_min = 30atx = 20 and y=10
Thus, minimum cakes obtained is 30 at 20g flour and 10g of fat

(b). Find the value of k if \(f(x)=\left\{\begin{array}{l}
{k x+1 ; \text { if } x \leq 5} \\
{3 x-5 ; \text { if } x>5}
\end{array}\right\}\) is continuous at x = 5(4 marks)
Answer:
f(-x) n= [tan(-x)]⁹ =(-tanx)⁹ =-tan⁹ x
f(-x) = -f(x)
=> f is an odd function.
Thus, I = 0

(b). Prove that \(\left|\begin{array}{ccc}
{1+a} & {1} & {1} \\
{1} & {1+b} & {1} \\
{1} & {1} & {1+c}
\end{array}\right|\) = ab + bc + ca + abc (4 marks)
Answer: