Revised Datesheet 2021
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CBSE Class 12 Mathematics
Sample Paper 01 (202021)
Maximum Marks: 80
Time Allowed: 3 hours
General Instructions:
 This question paper contains two parts A and B. Each part is compulsory. Part A carries 24 marks and Part B carries 56 marks
 PartA has Objective Type Questions and Part B has Descriptive Type Questions
 Both Part A and Part B have choices.
Part – A:
 It consists of two sections I and II.
 Section I comprises of 16 very short answer type questions.
 Section II contains 2 case studies. Each case study comprises of 5 casebased MCQs. An examinee is to attempt any 4 out of 5 MCQs.
Part – B:
 It consists of three sections III, IV and V.
 Section III comprises of 10 questions of 2 marks each.
 Section IV comprises of 7 questions of 3 marks each.
 Section V comprises of 3 questions of 5 marks each.
 Internal choice is provided in 3 questions of Section –III, 2 questions of SectionIV and 3 questions of SectionV. You have to attempt only one of the alternatives in all such questions.
 Part – A Section – I
 Let P be the set of all subsets of a given set X. Show that ∪ : P × P ⟶ P given by (A, B) ⟶ A ∪ B and ∩ : P × P ⟶ P given by (A, B) ⟶ A ∩ B are binary operations on the set P.ORState with reason whether the function has inverse:
f : {1, 2, 3, 4}→{10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}  Define a function. What do you mean by the domain and range of a function? Give examples.ORDetermine whether the relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x and y live in the same locality}  If f : R → R is defined by f(x) = x2, write f1 (25).
 If A=[−23451−6]and B =[52−7364]find a matrix C such that A + B – C = 0.
 Consider the matrix A=[3−25691].
then write the values of a11, a12, a13, a21, a22 and a23.ORIf A = [54−26−17], Find 3A  If A = [2443], X = [n1], B = [811] and AX = B, then find n.
 Write a value of ∫(tan−1x)31+x2dxOREvaluate: ∫x(x4−x2+1)dx
 Find the value of c for which the area of figure bounded by the curve y = 3, the straight lines x=1 and x=c and the xaxis is equal to 163
 Find the general solution for differential equation: dydx=(1+x)(1+y2)ORFind the order and degree (if defined) of the differential equation d2ydx2+5x(dydx)2−6y=logx
 Is the measure of 5 seconds vector or scalar?
 Find a vector in the direction of vector 2ˆi−3ˆj+6ˆk which has magnitude 21 units.
 In fig. ABCD is a regular hexagon, which vector is Coinitial?
 Find a normal vector to the plane 2x – y + 2z = 5. Also, find a unit vector normal to the plane.
 Find the direction cosines of the line 4−x2=y6=1−z3.
 The probability that a student selected at random from a class will pass in Mathematics is 4/5, and the probability that he/she passes in Mathematics and Computer Science is 1/2. What is the probability that he/she will pass in Computer Science if it is known that he/she has passed in Mathematics?
 A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of getting at least one success.
 Section – II
 A man has an expensive square shape piece of golden board of size 24 cm is to be made into a box without top by cutting from each corner and folding the flaps to form a box.
 Volume of open box formed by folding up the flap:
 4(x3 – 24×2 + 144x)
 4(x3 34×2 + 244x)
 x3 – 24×2 + 144x
 4×3 – 24×2 + 144x
 In the first derivative test, if dydx changes its sign from positive to negative as x increases through c1, then function attains a:
 Local maxima at x = c1
 Local minima at x = c1
 Neither maxima nor minima at x = c1
 None of these
 What should be the side of the square piece to be cut from each corner of the board to be hold the maximum volume?
 14 cm
 12 cm
 4 cm
 5 cm
 What should be the maximum volume of open box?
 1034 cm3
 1024 cm3
 1204 cm3
 4021 cm3
 The smallest value of the polynomial x3 – 18×2 + 96x in [0, 9] is:
 126
 0
 135
 160
 Volume of open box formed by folding up the flap:
 A shopkeeper sells three types of flower seeds A1, A2, and A3. They are sold as a mixture where the proportions are 4:4:2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35%.
Based on the above information answer the following questions: The probability of a randomly chosen seed to germinate:
 0.69
 0.39
 0.49
 0.59
 The probability that the seed will not germinate given that the seed is of type A3:
 15100
 65100
 75100
 55100
 The probability that the seed is of the type A2 given that a randomly chosen seed does not germinate.
 2251
 5551
 5116
 1651
 Calculate the probability that it is of the type A1 given that a randomly chosen seed does not germinate.
 5122
 2251
 1651
 751
 The probability that it will not germinate given that the seed is of type A1:
 55100
 65100
 35100
 45100
 The probability of a randomly chosen seed to germinate:
 Part – B Section – III
 Prove that: cot−1{√1+sinx+√1−sinx√1+sinx−√1−sinx} =x2,0<x<π2
 Solve the matrix equation [5411]X=[1−213], where X is a 2 × 2 matrix.ORSolve [3−492][xy]=[102] for x and y.
 Find dydx if x=a(cosθ+θsinθ) and y=a(sinθ−θcosθ)
 Find the point at which the tangent to the curve y = √4x−3−1 has its slope 23.
 Evaluate: ∫ x log (1 + x) dxOREvaluate: ∫log(tanx2)sinxdx
 Find the area of the region common to the parabolas 4y2 = 9x and 3×2 = 16y.
 Find the general solution of the differential equation: dydx + (sec x) y = tan x
 Find a vector of magnitude 49, which is perpendicular to both the vectors 2ˆi+3ˆj+6ˆk and 3ˆi−6ˆj+2ˆk
 The Cartesian equation of a line is x−53=y+47=z−62 Write its vector form.
 A die is rolled. If the outcome is an even number, what is the probability that it is a number greater than 2?ORTwo dice were thrown and it is known that the numbers which come up were different. Find the probability that the sum of the two numbers was 5.
 Section – IV
 Let f : N → N be defined by f(n)={n+12,ifnisoddn2,ifniseven for all n ∈ N. State whether the function f is bijective. Justify your answer.
 Differentiate the function with respect to x: sin−1{x+√1−x2√2},−1<x<1.
 Find the points of discontinuity, if any, of the function: f(x)={sinxx, if x<02x+3,x≥0ORFind dydx when y = (tan x)cot x + (cot x)tan x
 Find the intervals in which the function f(x) = (x – 1)3 (x – 2)2 is (i) increasing. (ii) decreasing.
 Evaluate: ∫(x2+1)(x4+x2+1)dx
 Find the area of the region bounded by the parabola y2 = 2x +1 and the line x – y – 1 = 0.ORFind the area enclosed by the curves 3×2 + 5y = 32 and y =x−2.
 In the differential equation show that it is homogeneous and solve it: (x – y)dydx = x + 3y.
 Section – V
 If A = 3275 and B = [6789], verify that (AB)1 = B1 A1.ORShow that A=[53−1−2] satisfies the equation A2 – 3A – 7I = 0 and hence find A1.
 Find the equation of the plane containing the line x+1−3=y−32=z+21 and the point (0, 7, 7) and show that the line x1=y−7−3=z+72 also lies in the same plane.ORFind the vector equation of the plane passing through the intersection of planes ¯r⋅(ˆi+ˆj+ˆk)=6 and →r⋅(2ˆi+3ˆj+4ˆk)=5 and the point (1,1,1).
 A company manufactures three kinds of calculators: A, Band C in its two factories I and II. The company has got an order for manufacturing at least 6400 calculators of kind A, 4000 of kind B and 4800 of kind C. The daily output of factory I is of 50 calculators of kind A, 50 calculators of kind B and 30 calculators of kind C. The daily output of factory II is of 40 calculators of kind A, 20 of kind B and 40 of kind C. The cost per day to run factory I is Rs 12000 and of factory II is Rs 15000. How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate this problem as an LPP and solve it graphically.ORAn oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps. D, E and F whose requirements are 45000 L, 3000 L and 3500 L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Distance in (km.) From/To A B D 7 3 E 6 4 F 3 2 Assuming that the transportation cost of 10 litres of oil is Rs.1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
CBSE Class 12 Mathematics
Sample Paper 01 (202021)
Solution
 Part – A Section – I
 Since, union operation ∪ carries each pair (A, B) in P × P to a unique element A ∪ B in P, ∪ is binary operation on P. Similarly,the intersection operation ∩ carries each pair (A, B) in P×P to a unique element A ∩ B in P, ∩ is a binary operation on P.ORf : {1, 2, 3, 4) → {10} given by
f{(1, 10), (2, 10), (3, 10), (4, 10)}
clearly f is manyone function
⇒ f is not bijective
⇒ f is not invertible  Definition: A relation R from a set A to a set B is called a function if each element of A has a unique image B.
It is denoted by the symbol f: A→B which reads ‘f’ is a function from A to B ‘f’ maps A to B.
Let f: A → B, then the set A is known as the domain of f & the set B is known as codomain of f.
The set of images of all the elements of A is known as the range of f.
Thus, Domain of f = {aa ∈ A,(a,f(a)) ∈ f )
Range of f = {f(a)  a ∈ A ,f(a) ∈ B }
Example: The domain of y = sin x is all values of x i.e. R, since there are no restrictions on the values for x.
The range of y is between −1 and 1. We could write this as −1 ≤ y ≤ 1.ORGiven that R = {(x, y) : x and y live in the same locality}
Clearly, (x,x) ∈ R as x and x live in the same locality.
⇒ R is reflexive.
Now, if (x, y) ∈ R, then x and y live in the same locality.
⇒ y and x live in the same locality.
⇒ (y,x) ∈ R
⇒ R is symmetric.
Further, let (x, y), (y, z) ∈ R
⇒ x and y live in the same locality and y and z live in the same locality.
⇒ x and z live in the same locality
⇒ (x, z) ∈ R
⇒ R is transitive.
Therefore, R is reflexive, symmetric and transitive.  Let f1(25) = x …(1)
Then, we have,
f(x) = 25
⇒ x2 = 25
⇒ x2 – 25 = 0
⇒ (x – 5)(x + 5) = 0
⇒ x = ±5 ⇒ f1(25) = {5, 5}  We have to find C,
Given A + B – C = 0
[−23451−6]+[52−7364]−C=0
C=[−23451−6]+[52−7364]
C=[35−387−2]  Clearly, the element in the 1st row and 2nd column is 2.
So, we write a12 = 2.
Similarly we can find , a11 = 3; a12 = 2; a13 = 5; a21 = 6; a22 = 9 and a23 = 1ORWe have to find 3A, i,e;
3A=[3⋅53⋅43⋅(−2)3⋅63⋅(−1)3⋅7]=[1512−618−321]  Here,
[2443][n1]=[811]
In the LHS the first matrix is of order 2×2 and the second one is of order 2×1 which will result in the matrix of order 2×1.
⇒[2n+44n+3]=[811]
⇒2n+4=8
⇒2n=4
⇒ n = 2  I=∫(tan−1x)31+x2dx
Put tan1 x = t
⇒11+x2dx=dt
⇒∫t31dt
=t44+c
=(tan−1x)44 + cORI = ∫x(x4−x2+1)dx
Putting x2 = t and 2x dx = dt, we get
12⋅∫dt(t2−t+1)
= 12⋅∫dt(t−12)2+(√32)2.
=12⋅1(√32)tan−1(t−12)(√32)
=1√3⋅tan−1(2t−1√3)+C
= 1√3tan−1(2x2−1√3) +C  we have, ∫c03dx=163
3(x)c0=163
3c=163
c=169  The given differential equation can be rewritten as,
11+y2dy = (1+x)dx
Integrating on both sides
∫11+y2dy=∫(1+x)dx
⇒tan−1y=x+x22+cORIt is given that equation is d2ydx2+5x(dydx)2−6y=logx
⇒d2ydx2+5x(dydx)2−6y−logx=0
We can see that the highest order derivative present in the differential is d2ydx2.
Thus, its order is two.
The highest power raised to d2ydx2 is 1.
Therefore, its degree is one.  5 Seconds is a time period, it has only magnitude i.e; 5 and has no direction, So it is Scalar.
 We have to find a vector in the direction of vector 2ˆi−3ˆj+6ˆk which has magnitude 21 units.
Let →a=2ˆi−3ˆj+6ˆk
Then, →a=√(2)2+(−3)2+(6)2
=√4+9+36=√49= 7 units
Now unit vector in the direction of the given vectors →a is given as
ˆa=→a→a=17(2ˆi−3ˆj+6ˆk)
=27ˆi−37ˆj+67ˆk
Now, the vector of magnitude equal to 21 units and in the direction of →a is given by
21ˆa=21(27ˆi−37ˆj+67ˆk)=6ˆi−9ˆj+18ˆk  Coinitial vectors are vectors originated from same point.
Coinitial vectors are →a→y and →z  We know that the direction ratios of a vector normal to a plane are proportional to the coefficients of x, y, and z respectively, in the cartesian equation of a plane. Therefore, direction ratios of a vector →n normal to the given plane are proportional to 2, – 1, 2 and so →n=2ˆi−ˆj+2ˆk. Therefore, a unit vector normal to the plane is given by ˆn=→n→n=2ˆi−ˆj+2ˆk√4+1+4 =13(2ˆi−ˆj+2ˆk)
 According to the question, equation of line is
4−x2=y6=1−z3
It can be rewritten as
x−4−2=y6=z−1−3
Here, Direction ratios of the line are (2, 6, 3).
:. Direction cosines of the line are
−2√(−2)2+62+(−3)2,6√(−2)2+62+(−3)2
−3√(−2)2+62+(−3)2 i.e. −2√49,6√49 and −3√49
Thus, Direction cosines of line are(−27,67,−37)  Let M = Mathematics , C = Computer Science
By given data,
P(M) = 45 and P(M ∩ C) = 12
Required probability is given by,
P(CM)=P(C∩M)P(M)=1245=54×2=58  Let p denote the probability of getting an odd number in a single throw of the die
Then,
p=36=12 and q=1−p=1−12=12
Let X denote the number of successes in 6 trials. Then, X is a binomial variate with parameter n=6 and p=1/2
We know that, The probability of r successes in 6 trials is given by
P(X=r)=6Cr(1/2)6−r(1/2)r, where r = 0,1,2,…6
or, P(X=r)=6Cr(1/2)6, where r = 0, 1, 2,…6 …(i)
Probability of at least one success = P(X≥1)=1−P(X=0)
= =1−6C0(12)6 [Using (i)]
=1−164=6364  Section – II

 (a) 4(x3 – 24×2+144x)
 (a) Local maxima at x = c1
 (c) 4 cm
 (b) 1024 cm3
 (b) 0
 We have, A1: A2: A3 = 4: 4: 2
P(A1)=410,P(A2)=410 and P(A3)=210
Where A1, A2 and A3 denote the three types of flower seeds.
Let E be the event that seed germinates and ¯E be the event that a seed does not germinate.
∴P(EA1)=45100,P(EA2)=60100 and P(EA3)=35100 And P(¯EA1)=55100,P(¯EA2)=40100 and P(¯EA3)=65100 (c) ∴P(E)=P(A1)⋅P(EA1)+P(A2)⋅P(EA2)+P(A3)⋅P(EA3)
=410⋅45100+410⋅60100+210⋅35100
=1801000+2401000+701000=4901000=0.49  (b) P(¯EA3)=1−P(EA3)=1−35100=65100 [as given above]
 (d) P(A2¯E) =P(A2)⋅P(¯EA2)P(A1)⋅P(¯EA1)+P(A2)+P(¯EA2)+P(A3)⋅P(¯EA3)
=410⋅40100410⋅55100+410⋅40100+210⋅65100 =16010002201000+1601000+1301000
=16010005101000=1651  (b) P(A2¯E) =P(A1)⋅P(¯EA1)P(A1)⋅P(¯EA1)+P(A2)+P(¯EA2)+P(A3)⋅P(¯EA3)
=410⋅55100410⋅55100+410⋅40100+210⋅65100 =22010002201000+1601000+1301000
=22010005101000=2251  (a) P(¯EA1)=1−P(EA1)=1−45100=55100
 (c) ∴P(E)=P(A1)⋅P(EA1)+P(A2)⋅P(EA2)+P(A3)⋅P(EA3)
 Part – B Section – III
 We know that
1±sinx=cos2x2+sin2x2±2sinx2cosx2=(cosx2±sinx2)2
∴cot−1{√1+sinx+√1−sinx√1+sinx−√1−sinx}
=cot−1{√(cosx2+sinx2)2+√(cosx2−sinx2)2√(cosx2+sinx2)2−√(cosx2−sinx2)2}
=cot−1{cosx2+sinx2+cosx2−sinx2cosx2+sinx2−cosx2−sinx2} [∵√x2=x]
=cot−1{(cosx2+sinx2)+(cosx2−sinx2)(cosx2+sinx2)−(cosx2−sinx2)} [∵0<x2<π4∴cosx2>sinx2]
=cot−1(cotx2)=x2 [∵0<x2<π4]  Let A=[5411]B=[1−213]
⇒ AX = B
Or, X = A – 1B
A = 1
Now Cofactors of A are
C11 = 1 C12 = – 1
C21 = – 4 C22 = 5
⇒ adj A = [C11C12C21C22]T
(adj A) = [1−1−45]T
= [1−4−15]
Now, A – 1 = 1A adj A
A – 1 = 11[1−4−15]
So, X = [1−4−15][1−213]
Hence, X = [−3−14417]ORHere,
[3−492](xy)=(102)
⇒(3x−4y9x−2y)=(102)
⇒ 3x – 4y = 10 ……..(1)
9x + 2y = 2 ……….(2)
Solving both the equations, we get
x=1421
=23
Substituting the value of x in equation (1), we get
3×23−4y=10
⇒2−4y=10
⇒ 2 – 4y = 8
⇒ y = 2
∴ x=23 and y = 2.  x=a(cosθ+θ.sinθ)
dxdθ=a[−sinθ+θ.cosθ+sinθ.1]
dxdθ=aθ.cosθ …(i)
y=a(sinθ−θ.cosθ)
dydθ=a.[cosθ−(−θsinθ+cosθ.1)]
=a[cosθ+θ.sinθ−cosθ]
=aθ.sinθ …(ii)
dydx=aθ.sinθaθ.cosθ=tanθ  Slope of tangent to the given curve at (x, y) is
dydx=12(4x−3)−12×4=2√4x−3
Given that slope, dydx=23
So, 2√4x−3=23
or 4x – 3 = 9
or x = 3
Now y = √4x−3−1. So when x = 3, y = √4(3)−3−1=2
Therefore, the required point is (3, 2).  Let I = ∫xlog(1+x) dx. Then, we have
I = log (x + 1) x22−∫1x+1×x22dx
⇒I=x22 log (x + 1) 12∫x2x+1dx=x22 log (x + 1) −12∫x2−1+1x+1dx
⇒I=x22 log (x + 1) –12∫x2−1x+1+1x+1dx
⇒I=x22 log (x + 1) – 12{∫((x – 1) + 1x+1)dx}
⇒I=x22 log (x + 1) – 12(x22– x + log x + 1} + CORI =∫log(tanx2)sinxdx
Put log tan x2=t
⇒1tanx2sec2x2×12dx = dt
⇒cosx2sinx2×1cos2x2×12dx = dt
⇒12sinx2cosx2dx =dt
⇒1sinxdx=dt
I = ∫tdt
= t22 + c
= (logtanx2)22 + c
Area of the shaded region is given by=∫40[3√x2−3x216]dx
=[x3/2−x316]40
=[(4)3/2−(4)316]
=[8−6416]
= [8 – 4] = 4 sq.units The given differential equation is:
dydx + py = Q (where p = sec x and Q = tan x)
Now, I.F. =e∫pdt =e∫secxdx = elog(sec x + tan x) = sec x + tan x
The general solution of the given differential equation is given by the relation,
y(I.F.) = ∫(Q × I.F.) dx + C
⇒ y(sec x + tan x) = ∫tan x (sec x + tan x)dx + C
⇒ y(sec x + tan x) = ∫sec x tan xdx + ∫tan2xdx + C
⇒ y (sec x + than x) = sec x + ∫(sec2 x – 1)dx + C
⇒ y (sec x + tan x) = sec x + tan x – x + C.  Given:
→a=2ˆi+3ˆj+6ˆk
→b=3ˆi−6ˆj+2ˆk
→a×→b=ˆiˆjˆk2363−62
=(6+36)ˆi−(4−18)ˆj+(−12−9)ˆk
=42ˆi+14ˆj−21ˆk
⇒→a×→b=√422+142+(−212)
=√2401
= 49
Required vector =49×{→a×→b→a×→b}
=49×42ˆi+14ˆj−21ˆk49
=42ˆi+14ˆj−21ˆk  Given: The Cartesian equation of the line is x−53=y+47=z−62=λ (say)
⇒x−5=3λ,y+4=7λ,z−6=2λ
⇒x=5+3λ,y=−4+7λ,z=6+2λ
General equation for the required line is →r=xˆi+yˆj+zˆk
Putting the values of x, y, z in this equation,
→r=(5+3λ)ˆi+(−4+7λ)ˆj+(6+2λ)ˆk=5ˆi+3λˆj−4ˆj+7λˆj+6ˆk+2λˆk
⇒→r=(5ˆi−4ˆj+6ˆk)+λ(3ˆi+7ˆj+2ˆk)[since→r=→a+λ→b]  A die has 6 faces and its sample space S = {1, 2, 3, 4, 5, 6}
The total number of outcomes = 6
Let P(A) be the probability of getting an even number
The sample space of A = {2, 4, 6}
∴P(A)=36=12
Let P(B) be the probability of getting a number whose value is greater than 2
The sample space of B = {3, 4, 5, 6}
∴ (A ∩ B) = {4,6}
∴ P(A ∩ B) = 26=13
The probability of getting a number greater than 2 given that the outcome is even is given by: P(B/A)
=P(A∩B)P(A)
=1/31/2
=23
This is the required probabilityORTwo die having 6 faces each when tossed simultaneously will have a total outcome of 62 = 36
Let P(A) be the probability of getting a sum equal to 5
Let P(B) be the probability of getting 2 different numbers
Probability of getting 2 different numbers
= 1 – probability of getting same numbers
= 1 – 16
= 56
∴ P(B) = 56
Let P(A ∩ B) be the probability of getting a sum = 5 and two different numbers at the same time
The sample space of (A ∩ B) = {(1,4), (2,3), (3,2), (4,1)}
∴ P(A ∩ B) = 436=19
The probability that the sum = 5 given that two different numbers were thrown: P(A/B)
=P(A∩B)P(B)
=1/95/6
=215  Section – IV
 f : N → N be defined by f(n)={n+12,ifnisoddn2,ifniseven
 f(1) = 1+12 = 1 and f(2) = 22 = 1
The elements 1, 2, belonging to domain of f have the same image 1 in its codomain.
So, f is not oneone, therefore, f is not injective.  Every number of codomain has preimage in its domain e.g., 1 has two preimages 1 and 2.
So, f is onto, therefore, f is not bijective.
 f(1) = 1+12 = 1 and f(2) = 22 = 1
 Let, y=sin−1{x+√1−x2√2}
Put x = sinθ
∴y=sin−1(sinθ+√1−sin2θ√2)
⇒y=sin−1(sinθ+cosθ√2)
⇒y=sin−1{sinθ(1√2)+cosθ(1√2)}
⇒y=sin−1{sinθcosπ4+cosθsinπ4}
⇒y=sin−1{sin(θ+π4)}
Here, 1 < x < 1
⇒ 1 < sin θ < 1
⇒−π2<θ<π2
⇒(−π2+π4)<(π4+θ)<3π4
⇒−π4<(π4+θ)<3π4
So, from (i)
y=θ+π4 [ since ,sin−1(sinα)=α, if α∈[−π2,π2]]
⇒y=sin−1x+π4
Differentiating it with respect to x,
dydx=1√1−x2+0
∴dydx=1√1−x2  Given; f(x)={sinxx, if x<02x+3,x≥0
When x < 0, then
f(x)=sinxx
We know that sin x, as well as the identity function x, are everywhere continuous.
So, the quotient function sinxx is continuous at each x < 0
When x > 0, then
f(x) = 2x + 3, which is a polynomial function
Therefore, f(x) is continuous at each x > 0
Now, Let us consider the point x = 0
We have
(LHL at x = 0) = limx→0−f(x)=limh→0f(0−h)=limh→0f(−h)
=limh→0f(−h)=limh→0(sin(−h)−h)=limh→0(sin(h)h)=1
(RHL at x = 0) = =limx→0+f(x)=limh→0f(0+h)=limh→0f(h)=limh→0(2h+3)=3
∴limx→0−f(x)≠limx→0+f(x)
Thus, f(x) is discontinuous at x = 0
Hence, the only point of discontinuity for f(x) is x = 0ORWe have, y = (tan x)cot x + (cot x)tan x
y=elog(tanx)cotx+elog(cotx)tanx
⇒ y = ecot x log tan x + etan x log(cot x)
Differentiating with respect to x using chain rule and product rule,
dydx=ddx(ecotxlogtanx)+ddx(etanxlogcotx)
=ecotxlogtanxddx(cotxlogtanx)+etanxlogcotxddx(tanxlogcotx)
=elog(tanx)cotx[cotxddx(logtanx)+logtanxddx(cotx)]+elog(cotx)tanx[tanxddx(logcotx)+logcotxddx(tanx)]
=(tanx)cotx[cotx×(1tanx)ddx(tanx)+logtanx(−cosec2x)] +(cotx)tanx[tanx×(1cotx)ddx(cotx)+logcotx(sec2x)]
=(tanx)cotx[(cosec2xsec2x)(sec2x)−cosec2xlogtanx] +(cotx)tanx[(sec2xcosec2x)(−cosec2x)+sec2xlogcotx]
= (tan x)cot x[cosec2x – cosec2x log tan x] + (cot x) tan x [sec2x log cot x – sec2x]
= (tan x) cot x cosec2x[1 – log tan x] + (cot x) tan x sec2x[log cot x1]
The differentiation of the given function y is as above.  f(x) = (x – 1)3(x – 2)2
Therefore, on differentiating both sides w.r.t. x, we get,
f'(x) = (x – 1)3 ddx(x−2)2+(x−2)2⋅ddx(x−1)3
⇒ f'(x) = (x – 1)3.2(x – 2) + (x – 2)2 · 3(x 1)2
= (x – 1)2 (x – 2)[2(x – 1) + 3(x – 2)]
= (x – 1)2(x – 2)(2x – 2 + 3 x – 6)
⇒ f'(x) = (x 1)2(x – 2)(5x – 8)
Now, put f'(x) = 0
⇒ (x – 1)2(x – 2)(5x – 8) = 0
Either (x – 1)2 = 0 or x – 2 = 0 or 5x – 8 = 0
∴x=1,85,2
Now, we find intervals and check in which interval f(x) is strictly increasing and strictly decreasing.Interval f'(x) = (x – 1)2(x – 2)(5x – 8) Sign of f'(x) x < 1 (+)()() +ve 1 < x < 85 (+)()() +ve 85 < x < 2 (+)()(+) ve x > 2 (+)(+)(+) +ve So, the given function f(x) is increasing on the intervals (−∞,1)(1,85)and (2,∞) and decreasing on (85,2).
[85,2]  To find: ∫(x2+1)(x4+x2+1)dx
Formula Used: ∫1a2+x2dx=1atan−1xa+C
On dividing by x2 in the numerator and denominator of the given equation,
⇒∫1+1x2x2+1+1x2dx
⇒∫(1+1x2)(x−1x)2+3dx
Let y=x−1x
Differentiating write x,
dy=(1+1x2)dx
substituiting in the original equation,
⇒∫dyy2+(√3)2
⇒1√3tan−1y√3+c
Substituting for y=x−1x
⇒1√3tan−1(x−1x√3)+C
⇒1√3tan−1(x2−1√3x)+c
Therefore, we have the value of given integral as.
∫(x2+1)(x4+x2+1)dx=1√3tan−1(x2−1√3x)+C  To find area bounded by
y2 = 2x +1 …(i)
and x – y = 1 …(ii)
Equation (i) is a parabola with vertex (−12,0) and passes through (0, 1) (0, 1).
Equation (ii) is a line passing through (1, 0) and (0, 1), points of intersection of parabola and line are (3, 2) and (0, 1).
A rough sketch of the curve is given as:
Shaded region represents the required area. It is sliced in rectangles of area (x1 – x2) △y.
It slides from y = 1 to y = 3, so
Required area of the shaded region =Area of the Region ABCDA
=∫3−1(x1−x2)dy
=∫3−1(1+y−y2−12)dy
=12∫3−1(2+2y−y2+1)dy
=12∫3−1(3+2y−y2)dy
=12[3y+y2−y33]3−1
=12[(9+9−9)−(−3+1+13)]
=12[9+53]
=326
Required area =163 sq. unitsORTo find area enclosed by
3×2 + 5y = 32
3×2 = 5 (y−325) …(i)
And
y=x−2
⇒y={−(x−2), if x−2<1(x−2), if x−2≥1
⇒y={2−x, if x<2x−2, if x≥2...(2)
Equation (i) represents a downward parabola with vertex (0,325) and equation (ii) represents lines.
A rough sketch of curves is given as:
Thus the Required area of the Region = Area of Region ABECDA
A = Region ABEA + Region AECDA
=∫32(y3−y4)dx+∫2−2(y1−y2)dx
=∫32(32−3x25−x+2)dx+∫2−2(32−3x25−2+x)dx
=∫32(32−3x2−5x+105)dx+∫2−2(32−3x2−10+5x5)dx
=15[∫32(42−3x2−5x)dx+∫2−2(22−3x2+5x)dx]
A=15[(42x−x3−5x22)32+(22x−x3+5x22)2−2]
=15[{(126−27−452)−(84−8−10)}+{(44−8+10)−(−44+8+10)}
=15[{1532−66}+{46+26}]
=15[212+72]
A=332 sq. units.  (x – y)dydx = x + 3y
⇒dydx=x+3yx−y
⇒dydx=1+3yx1−yx
⇒dydx=f(yx)
⇒ the given differential equation is a homogenous equation.
The solution of the given differential equation is:
Put x = vy
⇒dxdy=v+ydvdy
v+xdvdx=1+3vxx1−vxx
⇒xdvdx=1+3v1−v−v=1+3v−v+v21−v=1+2v+v21−v
⇒1−v1+2v+v2dv=dxx
Integrating both the sides we get:
⇒∫1−v1+2v+v2dv=∫dxx+c
⇒∫v−11+2v+v2dv=−∫dxx+c
⇒ln1+2v+v22 = lnx + ln c
Resubstituting the value of y = vx we get
⇒ln1+2yx+(yx)22 = lnx + lnc
⇒ log x + y + 2x(x+y) = c  Section – V
 Clearly, A= 3275 = 15 – 14 = 1 ≠ 0. So, A is invertible
Let Aij be the cofactor of element aij in A = [aij]. Then,
A11 = (1)1+1 5 = 5, A12 = (1)1+2 7 = 7, A21 = (1)2+1 2 = 2 and A22 = (1)2+2 3 = 3
∴adjA=[A11A12A21A22]T=[5−7−23]T=[5−2−73]
Hence, A1 = 1A adj A = [5−2−73]
We have, B = [6789]
∴B=6789 = 54 – 56 = – 2 ≠ 0
So, B is invertible
Let Bij be the cofactors of bij in B = [bij]. Then,
B11 = (1)1+1 9 = 9, B12 = (1)1+2 8 = 8, B21 = (1)2+1 7 = 7 and B22 = (1)2+2 6 = 6
∴adjB=[B11B12B21B22]T=[9−8−76]T=[9−7−86]
Hence, B1 = 1B adj B = −12[9−7−86]
We know that adj AB = adj B. adj A.
∴adjAB=[9−7−86][5−2−73]=[94−39−8234]
We also know that AB = A B
∴ AB = 1 × 2 = 2 ≠ 0
So, AB is invertible
Hence, (AB)1 = 1AB adj (AB) = 1−2[94−39−8234]=−12[94−39−8234]…(i)
Also, B1 A1 = −12[9−7−86][5−2−73]=−12[94−39−8234] …(ii)
From (i) and (ii), we get
(AB)1 = B1 A1ORWe have, A=[53−1−2]
3A=3[53−1−2]=[159−3−6]
And 7I=7[1001]=[7007]
∴A2−3A−7I=[229−31]−[159−3−6]−[7007]
=[22−15−79−9−0−3+3−01+6−7]=[0000]
=0 Hence proved.
Since, A2 – 3A – 7I = 0
⇒ A1[(A2) – 3A – 7I] = A10
⇒ A1A.A – 3A1A – 7 A1 I = 0 [∵ A10 = 0]
⇒ IA – 3I – 7A1 = 0 [∵ A1A = I]
⇒ A – 3I – 7A1 = 0 [∵ A1I = A1]
⇒ 7A1 = A + 3I
=[−5−312]+[3003]=[−2−315]
∴A−1=−17[−2−315]  We know that the plane passing through (x1, y1, z1) is given by
a(x – x1) + b(y – y1) + c(z – z1) = 0 …(i)
Required plane is passing through (0, 7, 7), so
a(x – 0) + b(y – 7) + c(z + 7) = 0
ax + b(y – 7) + c(z + 7) = 0 …(ii)
Plane (ii) also contains line x+1−3=y−32=z+21 so, it passes through point (1, 3, 2),
a(1) +b(3 – 7) + c(2 + 7) = 0
a – 4b + 5c = 0 …(iii)
Also, plane (ii) will be parallel to line, so, a1a2 + b1b2 + c1c2 = 0
(a) (3) + (b) (2) + (c) (1) = 0
3a + 2b + c = 0 …..(iv)
Solving (iii) and (iv) by crossmulitplication,
a(−4)(1)−(5)(2)=b(−3)(5)−(−1)(1)=c(−1)(2)−(−4)(−3)
a−4−10=b−15+1=c−2−12
a−14=b−14=c−14 = λ (say)
⇒ a = 14λ, b = 14λ, c = 14λ
Put a, b, c in equation (ii);
ax + b(y – 7) + c(z + 7) = 0
(14λ) x + (14λ) (y – 7) + (14λ) (z + 7) = 0
Dividing by (14λ), we get,
x + y – 7 + z + 7 = 0
x + y + z = 0
So, equation of plane containing the given point and line is x + y + z = 0
The other line is x1=y−7−3=z+72
So, a1a2 + b1b2 + c1c2 = 0
(1) (1) + (1) (3) + (1) (2) = 0
1 – 3 + 2 = 0
0 = 0
LHS = RHS
So, x1=y−7−3=z+72 lie on plane x + y + z = 0OR→n1=ˆi+ˆj+ˆk,→n2=2ˆi+3ˆj+4ˆk
d1=6, d2 = 5
Using the relation
→r⋅(→n1+λ→n2)=d1+λd2
→r⋅[(1+2λ)ˆi+(1+3λ)ˆj+(1+4λ)ˆk]=6−5λ…… (1)
taking →r=xˆi+yˆj+zˆk
(xˆi+yˆj+zˆk).[(1+2λ)i+(1+3λ)j+(1+4λ)ˆk]=6−5λ
(1+2λ)x+(1+3λ)y+(1+4λ)z=6−5λ
(x + y + z – 6) + λ(2x + 3y + 4y + 4z + 5) = 0 ….. (2)
plane passes through the point (1, 1, 1)
λ=314
Put λ in eq (1),
→r⋅[(1+37)ˆi+(1+914)ˆj+(1+67)ˆk]=6−1514
→r⋅(107ˆi+2314ˆj+137ˆk)=6914
→r⋅(20ˆi+23ˆj+26ˆk)=69  Let factory (I) run x days and factory (II) run y days respectively to produce the three kinds of calculators. The LPP is to minimize the cost of the production of the three kinds of calculators.Hence let the
equation representing the total cost (in Rs) be 12000x + 15000y. Let z be the objective function which represents the total cost.Hence Z = 12000x + 15000y, which is to be minimised
Subject to the constraints
50x+40y≥6400 or 5x+4y≥640 ( by dividing throughout by 10 )
50x+20y≥4000 or 5x+2y≥400 (by dividing throughout by 10 )
30x+40y≥4800or3x+4y≥480 (by dividing throughout by 10 )
x≥0 and y≥0 ( non negative constraints which will restrict the solution in the first quadrant only.)
Now, considering the inequations as equations, we get
5x + 4y = 640 …(i)
5x + 2y = 400 …(ii)
3x + 4y = 480 …(iii)
Table of values for line 5x + 4y = 640 is given below.x 128 0 y 0 160 So, the line (i) passes through the points with coordinates ( 128, 0) and (0, 160)
On replacing the coordinates of the origin O (0, 0) in the inequality 5x+4y≥640 ,
we get
0+0≥640 [which is false]
So, the half plane of the inequality of the line ( i) is away from the origin, means that the point ( 0,0) which is the origin is not in the feasible region of the inequality of the line ( i).
Table of values for the line ( ii ) 5x + 2y = 400 is given below.x 80 0 y 0 200 So, the line (ii) passes through the points with coordinates (80, 0) and (0, 200).
On replacing the coordinates of the origin O (0, 0) in the inequality,5x+2y≥400
we get
0+0≥400 [which is false]
So, the half plane for the inequality of the line (ii) is away from the origin, which means that the point O(0, 0) is not a point in the feasible region of the inequality of line (ii).
Table of values for line (iii) 3x + 4y = 480 is given below.x 160 0 y 0 120 So, the line (iii) passes through the points with coordinates (160, 0) and (0, 120).
On replacing the coordinates of the origin O (0, 0) in the inequality 3x+4y≥480,
we get
0+0≥480 [which is false]
So, the half plane for the inequality of the line (iii) is away from the origin, which means that the origin O(0, 0) is not in the feasible region for the inequality of the line (iii) .
Also, x≥0 andy≥0 so the feasible region lies in the first quadrant.
The point of intersection of lines (i) and (iii) is B (80, 60) and lines (i) and (ii) is C (32, 120).
The graphical representations of the system of inequations as given below
Clearly, feasible region is ABCD is an unbounded feasible region, where the coordinates of the corner points are A(160,0),B (80, 60), C (32, 120) and D(0, 200).
The values of Z at corner points are as followsComer Points Z = 12000x + 15000y A(160,0) Z=12000×160+0=1920000 B(80, 60) Z=12000×80+15000×60= 1860000 (minimum) C(32,120) Z=12000×32+15000×120= 2184000 D(0, 200) Z=0+15000×200=3000000 In the table above , we find that minimum value of Z is 1860000 occur at the point B(80, 60). But we can’t say that it is a minimum value of Z as the region is unbounded.
Therefore, we have to draw the graph of the inequality 12000x + 15000y < 1860000 or 12x + 15y < 1860 ( when dividing throughout by 1000)
From the figure, we see that the open half plane represented by 12x + 15y < 1860 has no point in common with the feasible region. Thus, the minimum value of Z is Rs 1860000 attained at the point with coordinates (80, 60). Hence, factory (I) should run for 80 days and factory (II) should run for 60 days to get a minimum cost of Rs. 1860000.ORIn this problem, we note that the total quantity of oil available
= (7000 + 4000) = 11000 L
And total requirement of oil
= (4500 + 3000 + 3500) = 11000L
⇒ Total availability = Total requirement
Let depot A supply xlitre of oil to petrol pump D and y litre to E so that supplies to F will be (7000 – x – y)L
All given information can be represented diagrammatically as:
Since petrol pump D requires 4500 L and it has already received xlitres from depot A, it must received (4500 – x)L from depot B, Similarly, E receives (3000 – y)L from depot B and F receives 3500 – (7000 – x – y)L from the depot B.
has already received xlitres from depot A, it must received (4500 – x) L from depot B, Similarly, E receives (3000 – y) L from depot B and F receives 3500 – (7000 xy) L from the depot B.
Now, total transportation cost (in Rs)
= 7x + 6y + 3(7000 – x – y) + 3(4500 – x) + 4(3000 – y) + 2 (x + y – 3500)
= 3x + y + 39500
Hence, the given problem can be formulated as an L.P.P. as follows
Minimize Z = 3x + y + 39500
Subject to constraints
x+y≤7000
x≤4500
y≤3000
x+y≥3500
x≥0,y≥0
Now, reducing the all inequalities into equations, we have
x + y = 7000 ……. (i)
x = 4500 ……. (ii)
y = 3000 ……. (iii)
x + y = 3500 ……. (iv)
x = 0, y = 0 …… (v)
Now, tracing all the given equation of lines on a graph and shade the region satisfied by all the inequalities.
Here, the feasible region is ABCDEA, which is bounded and comer points are A (4500, 0), B(3500, 0), C(500, 3000), D(4000, 3000) and E (4500, 2500)
Now, evaluating the Z for each comer point i.e.Corner Z = 3x + y + 39500 A(4500, 0) Z = 53,000 B(3500, 0) Z = 50,000 C(500, 3000) Z = 44,000 → Minimum D(4000, 3000) Z = 54500 E(4500, 2500) Z = 55500 ⇒ Transportation cost will be minimum when x = 500 and y = 3000
Hence, 500, 3000, 3500 litres are supplied from depot A and 4000,0,0 litres are supplied from depot B to petrol pump D, E and F respectively with minimum transportation cost.