Solutions of class 12 maths question paper 2021 preboard exam CBSE
Solutions of class 12 CBSE maths question paper 2021 preboard exam are created for helping the students of 12 class in CBSE board exam. The solutions of the question paper will give a complete idea about the type of questions and the ways of solving them. All solutions of class 12 CBSE maths question paper 2021 preboard exams are solved by an expert of maths subject. We hope these solutions will definitely help to boost the preparation of maths question papers in the board exam of CBSE.
This question paper contains two parts-A and B. Each part is compulsory. Part A carries 24 marks and part B has descriptive-type questions of 56 marks.
Part A
Section I -16 questions(1 marks each)
Section II -2 case studies. Each case study comprises of 5 case-based MCQ’s.questions (1 marks each)
Part B
Section III-10 questions of 2 marks each
Section IV-07 questions of 3 marks each
Section V-03 questions of 5 marks each
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Part -A
Section-I
Q1. Determine whether relation R in the set N of natural number defined as:
R = {(x,y) : y = x + 5 and x < 4} is symmetric.
Ans. The relation given us in the the set N of natural number
R = {(x,y) : y = x + 5 and x < 4}
x < 4 implies that x= 1,2,3
For x = 1,y = 1 +5 =6,for x = 2,y = 2+ 5 =7, for x =3,y = 3 +5=8
Therefore
R = {(1,6),(2,7),(3,8)}
Since (1,6) ∈ R but (6,1) ∉ R,so R is not symmetric
Q2.Let A = R – {3}, B = R – {1},consider the function f : A →B defined by
f(x) = (x -2)/(x -3). Check whether function is one-one or not.
Ans. We are given: A = R – {3}
B = R – {1}
F: A→B defied as
(x_{1 }∉ 3)
If f(x_{1})_{ = }f(x_{2}) then
x_{1}= x_{2}
Tis implies that x_{1}= x_{2} therefore is on-on
Q3.Let Set A = {1,2,3} define the relation R in the set A as follows:R ={(1,1),(2,2),(3,3),(1,3)} which order to be added to R make it smallest equivalence relation.
Ans. A relation is said to be an equivalence relation when it is transitive,reflexive and symmetric
For the Set A = {1,2,3}, R={(1,1),(2,2),(3,3),(1,3)}
For the Set A the relation is reflexive a ∈A, (a,a) ∈ R ,so minimum order added should be (1,1),(2,2),(3,3)
For the Set A the relation is symmetric , a ∈A,b∈A, (a,b) ∈ R and (b,a) ∈R ,so minimum order added should be {(1,3),(3,1}
For the Set A the relation is transitive , (a,b) ∈A,(b,c) ∈A then (a,c) ∈ R ,the order must be {(1,3),(3,3)}∈R,so (1,3) ∈ R
Therefore order to be added to R make it smallest equivalence relation is (3,1)
OR
Show that function f from z →z defined as f(x) = x + 2 is bijective.
Ans. For every value of x ∈z, f(x) = x +2: f(1) = 1+2 =3,f(2) =2+2=4 ,f(1) ≠f(2) for 1 ≠ 2, for every x ∈z,there is distintive image of z in z,so given function is one to one relation
For every value of x ∈ z,there is atleast one image of f in z: f(1) =2, 2∈z,f(o) =2 it indicates that for f(1) = f(2) ,0≠1,so function is onto relation
Since the function is one to one as well as onto, therefore it is bijective
Q4. Find the number of all possible matrices of order 3 ×3 with each entry 0 or 1.
Ans. The number of all possible entries in the matrix of order 3 ×3 =9
The number of choices to fill every entry is,2 = 0,9
So, the number of total possible matrices are =2^{9 }=512
Ans. We are given A, B and C matrix, such that 2A + 3B – 2C ia a null matrix where
According to question
OR
Q7.Find ∫ e^{x }sec x(1+tanx)dx
Ans.∫ e^{x }secx(1+tanx)dx
= ∫ e^{x }(secx + secx.tanx)dx
It is of the form
∫ e^{x}[f(x) + f′(x)]dx = e^{x }f(x) + C
where f(x) = secx
f′(x) = secx.tanx
so, our equations becomes
∫ e^{x }(secx + secx.tanx)dx = e^{x }secx + C
OR
Ans. We are given
Using the integral formula
2a = 1
a = 1/2
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Q8. Find the area of region bounded by curve y = cos x between x = 0 and x = π.
Ans. The graph of the curve y = cos x between x = 0 and x = π is as shown below
The graph y = cosx is a periodic function,so the area from x =0 to x =π/2 = area from π/2 to π is same.
Area from x= 0 to π = 2× area of graph from 0 to π/2
Area from x= 0 to π = 2×Integration of cos x within the limit 0 to π/2
= 2(sinπ/2 – sin0)
= 2
Therefore area of region bounded by curve is 2 sq.unit
Q9. Find the degree of differential equation.
Ans. Highest order of derivative is (d²x/dx²) and its degree is 2
Hence the degree of differential equation is 2
OR
Find the integrating factor of differential equation
Ans. The differential equation given to us is
The integration factor (I) of linear equation dy/dx +py = Q is given as following
Rearranging the given equation in the standard form
dy/dx -y/x = 2x²
On comparing the standard equation and given equation ,we have p = -1/x
I = 1/x ( Since e^{logx} = x)
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Q10. Find the direction ratio of vector and hence calculate its direction cosines.
Ans. The given vector is = i +j -2k, its direction ratios are a =1,b = 1 and c = -2
The magnitude of = i +j -2k is = = √[1²+1²+(-2)²] =√(1 +1 +4) =√6
The direction cosines of vector are
Q11. Find the area of the parallelogram whose adjacent sides are determined by vectors.
Ans. Area of the parallelogram whose adjacent sides are (i-j +3k) and (2i -7j +k) is given by =
= [-1 – (-21) ]i – [1-6]j + [-7-(-2)]k
= 20i + 5j -5k
Area of the parallelogram is = 15√2 square unit.
Q12.If is a unit vector and
Ans. We are given is a unit vector such that
Q13.Find the equation of the line which is parallel to and passes through the point (5,2,-4).
Ans. Equation of the line passing throgh (5,2,-4) is
The given line is
Equation of a line passing through a point and parallel to another vector is given as
, where is the position vector of the required line and λ is a scaler
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Q14. Find the distance between two planes:2x +3y +4z=4 and 4x +6y +8z =12
Ans. The distance between two planes Ax +By +Cz = d_{1 }and Ax +By +Cz = d_{2}
is given by
The given planes are 2x +3y +4z=4 and 4x +6y +8z =12,rearranging the given second plains =2(2x+3y +4z)=12⇒2x +3y +4z=6
On comparing the equations of given plains with the equations standard plains, we have
A =2, B =3, C = 4, d_{1}= 4, d_{2}=6, Let D is the distance between two plains
Q15. Evaluate P(A∪B) if 2P(A) = P(B) = 5/13 and P(A/B) = 2/5.
Ans. Given, 2P(A) = P(B) = 5/13
So, P(A) = 1/2 × 5/13 = 5/26
And P(B) = 5/13
Also, P(A|B) = 2/5
We know, P(A|B)= P(A B)/P(B)
Also, P(A B) = P(A) + P(B) – P(A B)
Q16. Event A and B are such that P(A) =1/2, P(B) =7/12, and P(not A nor B)=1/4. State whether A and B are independent?
Ans. It is given that P(A) = 1/2, P(B) = 7/12, and P(not A nor B)=1/4
P(A′ B′) = 1/4
However, P(A) – P(B) = 1/2 × 7/12 = 7/24…(2)
Here, 2/4 ≠ 7/24
∴ P(A B) ≠ P(A).P(B)
Therefore, A and B are not independent events.
SECTION-II
Both case study-based questions are compulsory. Attempt any 4 sub-parts of each question 17 and 18. Each question carries 1 mark.
Q17.In a box manufacturing company during practical ability test of B.Tech for the post of an engineer,instructor of company instructed a participant to construct an open box with square base with a given quantity of cardboard of area C².
(i) If length of side of square base be x units and height is y units .Then area of material used is:
(a) x² + 4xy (b) x + 4x²y
(c) x + 4xy² (d) y² +4xy
Ans. Since the box is open,therefore the area of cardboard required
2×height(length + breadth)
2× y(x +x) + x×x
x² +4xy
Hence answer is (a)
(ii) The relation between x and y is :
(a) x = (c² -x²)/4y (b) y =(c²-x²)/4x
(c) x = (x²-c²)/4y (d) y =(x²-c²)/4x
Ans. Area of the cardboard = area of the box =c²
Area of the open box = x² +4xy
x² +4xy = c²⇒y = (c²-x²)/4x
Hence answer is (ii)
(iii) Volume of box expressed as function of x is :
(a) 1/4(c² x -x³) (b) 1/4(x³- c²x)
(c) 1/4(x²- cx) (d) 1/4(cx – x²)
Ans. Volume of the box,V = l×b×h = x × x×y= x²y
As it is calculated above in (i) y = (c²-x²)/4x
V = x²(c²-x²)/4x = x(c²-x²)/4= 1/4(c² x -x³)
Hence answer is (i)
(iv) At what value of ‘x’ volume is maximum ?
(a) x = c/3 (b) x = c²/√3
(c) x = c/√3 (d) x =√3 c
Ans.
For getting maximum value of volume ,the slope of the V-x curve should be 0.
V = 1/4(c²x – x³)
dV/dx = 1/4 d/dx(c²x – x³) = 1/4(c² – 3x²)
1/4(c² – 3x²) = 0
c² – 3x² = 0
x = c/√3
Hence answer is (iii)
(v) Maximum value of V is :
(a) c²/6√3 (b) c/3√6
(c) c²/6√3 (d) c/6√3
Ans. As we have determined in (iv) the value of x for which volume is maximum
So putting the value of x = c/√3 in volume equation
V = 1/4(c²x – x³)
V(max.) = 1/4[c² ×c/√3 – (c/√3)³]
= 1/4[c³/√3 -c³/3√3]
= c³/4√3[1-1/3] = 2c³/12√3 = c³/6√3
Hence answer is (iii)
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Q18. A car manufacturing factory has two plants X and Y.Plant X manufactures 70 % of cars and plant Y manufactures 30 %. 80% of cars at plant X and 90 % of cars at plant Y are rated of standard quality. A car is chosen at random and is found to be of standard quality.
Let A = Event car of standard quality
B_{1 }=Event car is manufactured in X
B_{2} = Event car is manufactured in Y
Based on the information given above answer the following (any four)
(i) Value of P(B_{1 }) will be:
(a) 70 % (b) 7/10
(c) 70/10 (d) 3/10
Ans.B_{1 }=Event car is manufactured in X
P(B_{1}) →Probability of the car that is manufactured in X plant
P(B_{1}) = 70 %
Hence answer is (a)
(ii) Value of P(B_{2 }) will be:
(a) 10/3 (b) 30%
(c) 3/10 (d) 7/10
B_{2 }=Event car is manufactured in Y
P(B_{2}) →Probability of the car that is manufactured in Y plant
P(B_{2}) = 30 %
Hence answer is (b)
(iii) Probability that standard quality car is manufactured in plant X will be:
(a) 9/10 (b) 8/10
(c) 7/10 (d) 3/10
Ans. The probability that a standard quality car is manufactured in plant X
It means if the selected car is manufactured in plant X then it is of standard quality , its probability is given by = P(E/B_{1}) =80% = 8/10
Hence answer is (ii)
(iv) Probability that standard quality car is manufactured in plant Y will be:
(a) 9/10 (b) 8/10
(c) 7/10 (d) 3/10
Ans. The probability that a standard quality car is manufactured in the plant Y
It means if the selected car is manufactured in plant Y then it is of standard quality, its probability is given by = P(E/B_{2}) =90% = 9/10
(v) If a car is selected at random is of standard quality. What is the probability standard quality car has manufactured from plant X.
(a) 83/56 (b) 56/100
(c) 56/83 (d) 27/100
Ans. The probability that a standard quality car is manufactured in plant X
It means if the selected car is of standard quality then it is manufactured in plant X ,its probability is given by = P(B_{1}/E)
Applying Bays theorem
Where P(B_{1}) = 70 % =0.70, P(B_{2}) = 30 %=0.30
P(E/B_{1}) = 80% =0.80, P(E/B_{2}) = 90%=0.90
Part -B
SECTION-III
Q19. Write in the simplest form.
Ans. We are given the expression
Let x = secθ, θ = Sec^{-1}x
Q20.Find the value of x, y, and z if the matrix
satisfying equation AA´= I
Ans. The matrix(A) given to us
A’ → Transpose of the matrix A
I→ Unit matrix of the order 3× 3
According to question
AA’ = I
Since matrices are equal so corresponding elements will be equal
2x² =1⇒ x =±1/√2, 6y² = 1⇒y =±1/√6,3z²=1⇒z=±1/√3
OR
Find the non-zero values of x satisfying, Matrix equation:
Ans.The matrix equation is given to us
On comparing the corresponding elements
3x +8 = 20 ⇒ x = 4, this value of x also satisfy other equations 2x+12x = 48,x² +8x = 12x
Q21.Find the relationship between a and b. So that function ‘f’ defined.
is continuous at x = 3
Ans. As we know a function f(x) is continuous at x= a, when its left-hand limit , right-hand limit at x =a is equal to the value of the function at x =a.
Left-hand limit of the function( the function defined for x<3 is f(x) = ax +1)
Right-hand limit of the function( the function defined for x>3 is f(x) = bx +3)
Since it is given to us that function is continuous at x =3
∴ LHL = RHL
3a + 1 = 3b +3
3a – 3b =2
Therefore required relation is 3a – 3b =2
Q22.Find the equations of all lines having slope zero which are tangent to the curve
Ans. We are given the curve
We are given,the slope of the curve is 0, so dy/dx = 0
x = 1
Value of y when x = 1
The slope is passing through the coordinates (1,1/2)
The equations of the lines through the point (x_{1},y_{1}) is given as
(y-y_{1}) = m(x-x_{1})
(y-1/2) = 0(x -1)
y-1/2 = 0
Therefore the equation of the tangent is y-1/2 = 0
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Q23.Integrate: ∫ cos³ x e^{log sinx }dx
Ans. The integral function given to us is
∫ cos³ x e^{log sinx }dx
I = ∫ cos³ x sinx dx [e^{log x }= x]
Let t = sinx
dt/dx = cos x
Putting the value of dt and t in the function
Putting the value sinx =t
Substituting the value of t= sin x
Therefore the integration of the given function is
OR
Ans. The given function is
f(x)=x^{17}.cos^{4} x
Putting the value x = -x
f(-x) =(-x)^{17}.cos^{4} (-x)
f(-x) = -x^{17}.cos^{4} x = -f(x)
∴ f(x) is an odd function
Using the property
Q24. Find the area of the region bounded by line x = 2 and parabola y² = 8x.
Ans. The equations of parabola y² = 8x and of the line is x = 2
Drawing the graph of the parabola and the line
The area bounded by the parabola and the line is = 2× area of AOC
Area of AOC = Integration of the function y with respect to x within the limits x =0 and x =2.
y² = 8x
y =±2√2√x
Since y is in the first quadrant
∴ y = 2√2√x
The area bounded by the parabola and the line is (AOBC)
Q25.For given differential equations find the particular solutions satisfying giving condition:
Ans. The differential equation given is
y = c sec x
Putting values of y =1, x = 0
1 =c sec 0⇒ c = 1`
Putting value of c = 1,we get
y = 1 ×sec x
y = sec x
Therefore required particular solution of the given differential equation is y= sec x.
Q26. If such that
is perpendicular to Find the value of λ.
Ans. We are given that
is perpendicular to
Since is perpendicular to
(2-λ)3 + (2 + 2λ) = 0
6- 3λ + 2 + 2λ = 0
8 – λ = 0
λ = 8
Therefore the required value of λ is = 8
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Q27. Show that line joining the origin to the point (2,1,1) is perpendicular to the line determined by points (3,5,-1) and (4,3,-1).
Ans. Let the origin is O(0,0,0) and the point A(2,1,1) , the points of another line B(3,5,-1) and C(4,3,-1)
A line passing through (x_{1,} y_{1, }z_{1}) and (x_{2,} y_{2, }z_{2}) has the direction ratios
(x_{2 }– x_{1}), (y_{2 }– y_{1}), (z_{2 }– z_{1})
Direction ratios of OA =(2-0),(1-0),(1-0)= 2,1,1
Direction ratios of BC =(4-3),(3-5),(-1+1)= 1,-2,0
Two lines having direction ratios a_{1,} b_{1, }c_{1 }and a_{2,} b_{2, }c_{2 }are perpendicular to each other if
a_{1 }a_{2 }+ b_{1} b_{2} + c_{1} c_{2 }= 0
a_{1 }= 2, b_{1 }= 1, c_{1 }= 1, a_{2 }= 1, b_{2 }= -2, c_{2 }= 0
2×1+1×-2+1×0 =2-2+0 = 0
Therefore given two lines OA and BC are perpendicular to each other
Q28. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?
Ans.We are given that an urn contains ,10-black balls,5-white balls
Total balls in the urn are = 10+5=15 and number of black balls=10
The probability of first ball drawn is black
= (number of black ball)/(Total ball) = 10/15= 2/3
Since two balls are drawn from the urn one after the other without replacement
∴ The probability of second ball drawn is black if first ball drawn is black = (number of remaining black ball)/(Total remaining ball)= 9/14
According to multiplication rule of probability = (2/3)× (9/14) =3/7
The probability that both drawn balls are black is 3/7
OR
If P(A) = 2/5, P(B) = 3/10, P(A∩B) = 1/5. Find the value of P(A’/B’)
Ans. We have to find out P(A’/B’) =?
As we know, P(A’/B’) = P(A’∩B’)/P(B’)
Since P(A’∩B’) = P(A∪B)
P(A’/B’) = P(A∪B)/P(B’)…….(i)
P(B’) = 1- P(B) = 1 – 3/10 = 7/10
Let’s find out P(A∪B),for this applying the following formula
P(A∪B) = P(A) + P(B) – P(∩B)
From the question,we have P(A) = 2/5, P(B) = 3/10, P(A∩B) = 1/5
P(A∪B) = 2/5 + 3/10 – 1/5 = 5/10 =1/2
Putting the value of P(B’) and P(A∪B) in (i)
P(A’/B’) = (1/2)/(7/10)= 10/14 = 5/7
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Part -B
SECTION-IV
Q29. Let A = {1,2,3……9} and R be the relation in A × A defined by (a,b)R(c,d) if a +d = b+c for (a,b),(c,d) in A ×A. Prove that R is equivalence relation.Also obtain equivalence class [2,5].
Ans. We are given that A = {1,2,3……9}and R be the relation in A × A defined by (a,b)R(c,d) if a +d = b+c for (a,b),(c,d) in A ×A
R in A × A
(a,b)R(c,d) if (a,b),(c,d) ∈ A ×A
a +d = b+c
Considering (a,b) R (a,b)
a +b = b+a which is true for every value of (a,b) ∈ R(A ×A)
Hence R is reflexive
(a,b)R(c,d) if (a,b),(c,d) ∈ A ×A
a +d = b+c
Checking it for (c,d)R(a,b)
c + b = d +a, which is true for every value of (c,d) ,(a,b)∈ R(A ×A)
(c,d) R(a,b) is true
Hence R is symmetric
Let (a,b)R(c,d)and(c,d)R(e,f)
a+d = b+c and c+f = d+e
⇒ a+d = b+c and d+e = c+f
⇒ (a+d) – (d+e) = (b+c) – (c+f)
⇒ a-e = b-f
⇒ a+f = b+e
(a,b)R(e,f)
So,R is transitive
Hence, R is an equivalence relation
Equivalence class [2,5] in A = {1,2,3,4,5,6,7,8,9}
a +d = b+c ⇒ a -b = c-d
2 – 5 = -3⇒ a-b =c-d =-3 for every (a,b),(c,d) ∈ A [2,5]
∴ Forming the ordered pair (c,d) satisfying the condition d= c+3
(2,5) R (c,d)
(2,5) R (1,4),(2,5),(3,6),(4,7),(5,8),(6,9)
It is symmetric also,so the following relations also true
(1,4)R(2,5),(3,6)R(2,5),(4,7)R(2,5),(5,8)R(2,5),(6,9)R(2,5)
Therefore there 11 equivalent class of [2,5]
Q30. Differentiate the function:
w.r.t. ‘x’
Ans.
Taking logarithms of both sides
Now, differentiating both sides
Q31. Find dy/dx, if x^{2/3} + y^{2/3} = a ^{2/3}
Ans.
Differentiating w.r.t x
OR
If y = sin^{-1}x show that:
(1 – x²) d²y/dx² – x(dy/dx) = 0
Ans. We are given y = sin^{-1}x
Differentiating both sides
Squaring both sides
Differentiating both sides
Hence Proved
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Q32. Prove that y = (4sinθ/2+cosθ)-θ is an increasing function of θ in [0, π/2]
Ans. The given function is
Differentiating both sides w.r.t θ
Observing right hand side cosθ > 0 for θ ∈[0,π/2],(4-cosθ)>0 since -1≤cosθ≤1
(2+cosθ)² is always + ve.
Hence y = (4sinθ/2+cosθ)-θ is an increasing function of θ in [0, π/2]
The integral function is given to us
Let u = cos x
du = sin x.dx
Substituting the value u = cos x and du =sin x.dx
Substituting back the value u = cos x
= 1/3 -1 = -2/3
OR
Evaluate:
Ans. Evaluating the given integral function
t = x^{4} + 1
dt/dx = 4x³
dt = 4x³ dx ⇒ dx = dt/4x³
Substituting x^{4} + 1 = t and value of dx in the function
Since x^{4} + 1 is always positive
Q34. Find the area of the region bounded by y² = 9x. x = 2, x = 4 and x-axis in the first quadrant.
Ans. The graph of the curve y² = 9x and the lines x = 2, x = 4 is shown below.
We have to find out the area bounded by x -axis, y² = 9x, y = 2 and y = 4 which is ABCD
The area of ABCD = Integration of y w.r.t x within the limits x =2 and x =4
y² = 9x ⇒ y = 3√x
= 16 -4√2
Hence area bounded by x -axis, y² = 9x, y = 2 and y = 4 is (16 -4√2) sq.unit
Q35. Find the general solution of the following differential equation
Cos²x (dy/dx) + y = tan x, 0 ≤ x ≤ π/2
Ans. The given equation is
Writing the given equation in the form of the general equation dy/dx +Py = Q
Where P = sec²x and Q = tanx.sec²x
We know the general solution of the equation dy/dx +Py = Q
y ×(I.F) = ∫(Q × I.F) dx +C
It is known to us that the integrating factor(I.F) is calculated as follows
I.F = e^{∫Pdx}=e^{∫sec²dx} =e^{tanx}
Putting the value of Q = tanx.sec²x and of I.F in the general solution
y e^{tanx} =∫(tanx.sec²x) e^{tanx} dx +C
Let t = tanx ⇒ dt/dx = sec²x ⇒ dt = dx.sec²x
Substituting tanx = t and dx.sec²x = dt
ye^{t}=∫t.dt .e^{t}+ C
ye^{t}= t.e^{t}– e^{t}+ C
y = t -1 + C/e^{t}
Putting the value of t = tanx
y = tanx + C’ is the required solution where -1+ C/e^{t }= C’
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Part -B
SECTION-V
Q36. Use product:
to solve the system of equations:
3x – 2y + 3z = 8, 2x +y – z = 1, 4x – 3y + 2z = 4
OR
Solve the following system of equations by matrix method:
3x + 2y – 2z = 3
x + 2y + 3z = 6
2x – y + z = 2
Ans. Given system of equations is:
3x + 2y – 2z = 3
x + 2y + 3z = 6
and 2x – y + z = 2
or AX = B
X = A^{-1 }B
∴ For A^{-1}, Cofactors are
A_{11} = 5, A_{12}= 5, A_{13} = -5,
A_{21} = 0, A_{22}= 7, A_{23} = 7,
A_{31} = 10, A_{32}= -11, A_{33} = 4
|A| = 3(5) + 2(5) + (-2)(-5) = 35
∴ x = 1, y =1 and z = 1
Q37. Find the shortest distance between lines whose vector equations are:
Ans. The shortest distance between two lines is
We are given here the vector equations
On comparing these equations with standard vector equations
We have
Therefore the shortest distance between both of the lines is
OR
Find the equation of the plane which contains the line of intersection of the planes and which is perpendicular to the plane
Q38.Minimize z = 13x – 15y
Such that x + y ≤ 7
2x – 3y + 6 ≥ 0, x ≥0, y ≥0
Ans. We have to minimize z=13x – 15y such that x + y ≤ 7,2x – 3y + 6 ≥ 0, x ≥0, y ≥0
Solutions of the equation x + y = 7 , for x = 0,y = 7 and for y=0,x = 7,ploting the line between (0,7) and (7,0)
Solutions of the equation 2x – 3y + 6 = 0, for x = 0, y= 2, for y=0, x= -3,plotting the lines between (0,2) and (-3,0)
The solutions of the equation x = 0,is y axis and of y= 0 is x axis
Locating the co-ordinates (0,7),(7,0),(0,2),(-3,0) and (0,0)
The corner points of the graph of inequalities are (0,0),(3,4),(0,2),(7,0)
Evaluating z=13x – 15y on every points (0,0),(3,4),(0,2),(7,0)
On (0,0), z = 0, on (3,4) , z = 13×3 -15×4 =39-60 =-21, on (0,2), z =-30,on (7,0), z =91
The minimum value of z is -30
OR
The corner points of feasible region of LPP are (0,8), (4,10), (6,8), (6,5), (0,0) and (5,0). Let the objective function be z = 3x – 4y. Answer the following:
(a) Find the point at which the minimum value of z occurs.
(b) Find the point at which the maximum value of z occurs.
(c) Find the sum of the minimum value of z and the maximum value of z.
(d) Let z = px + qy, where p, q > 0
Ans.Find the relation between p and q, so that minimum of z occurs at (0,8) and (4,10).
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