Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22 - Future Study Point

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

class 12 maths exam preboard term2

Solutions of Class 12 Maths Preboard -2 Exam Term-2 CBSE Board 2021-22

class 12 maths exam preboard term2

 

Solutions of Class 12 Maths Question Paper Preboard -2 Exam Term-2 CBSE Board 2021-22 are created here for helping class 12 maths students for the preparation of the term 2 CBSE board exam 2021-22. The maths pre-board question paper is taken from a reputed public school G.D Lancer Public School of Mohan Garden New Delhi. The question paper contains 50 MCQs among which students have to solve 40 questions each of one mark. The maths pre-board 2 question paper is divided into three sections A, B, and C. Section A has 20 questions among which students have to attempt any 16 questions. Section B has 20 MCQs, among which students have to attempt any 16 questions. Section C has 10 MCQs, among which students have to attempt any 8 questions.

Solutions of class 12  maths question paper 2021 preboard exam CBSE

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

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SECTION A

Q1. The range of the function f(x) = tan-1x + cot-1x  is

(a) [-π/2, π/2]       (b) [0, π]        (c) [0, π/2]      (d) {π/2}

Ans.  (d) {π/2}

The given function is f(x) = tan-1x + cot-1x

Let x = tan θ

tan-1(tan θ) + cot-1(tan θ)

θ + cot-1[cot (π/2 – θ)]

θ + π/2 – θ = π/2

For any value of x the value of f(x) is π/2,therefore its range is {π/2}

Q2. The value of the expression sec-1 (2) + sin-1 (1/2) + tan-1 (-3) is

(a) 5π/6      (b) π/3        (iii) -π/3         (d) π/6

Ans.(a) 5π/6

The given expression is

sec-1 (2) + sin-1 (1/2) + tan-1 (-√3)

sec-1 (2) =π/6, sin-1 (1/2) = π/6 and tan-1 (-√3) = -tan-1 (√3)=π/3 [tan-1 (-x)= -tan-1 (x)]

π/3 + π/6  + π/3

(2π +π +2π)/6 = 5π/6 = 5π/3

Q3. The relation R in the set  {a,b,c} given by R ={(a,a),(b,b),(a,b),(b,a)} is

(a) syemmetric and transitive, but not reflexive

(b) reflexive and syemmetric,but not transitive

(c)symmetric,but neither reflexive nor transitive

(d) an equivalence relation

Ans. (b) reflexive and syemmetric,but not transitive

The relation R in the set  {a,b,c} given by R ={(a,a),(b,b),(a,b),(b,a)} is

a ∈ {a,b,c}, (a,a) ∈ R,So this relation is reflexive

(a,b) ∈   {a,b,c} and  (b,a) ∈  R,So this relation is symmetric

(a,b) ∈ {a,b,c} and (b,c) ∈ {a,b,c} but (a,c) ∉ R,therefore this relationship is not transitive

Q4. A = {1,2,3,4}, A relation R in the set A is given by 

R = {(1,1),(2,3),(3,2),(4,3),(3,4)},then relation R is

(a) Reflexive     (b) symmetric      (c) Transitive    (d) Equivalence

Ans.  (b) symmetric

The relation R in the set A= {1,2,3,4} given by R ={(1,1),(2,3),(3,2),(4,3),(3,4)} is

2 ∈ A , (2,2) ∉ R , So this relation is not reflexive

(2,3),(3,4) ∈ R, (3,2) ,(4,3) ∈ R, So this relation is symmetric

Q5. If A is any square matrix of order 3 ×3   such that      then the sum of all possible values of     

(a) 256        (b) 16          (c) -16         (d) 0

Ans. (b) 16

The relationship between determinant of adjoint matrix and determinant of the matrix is as follows

Where n is the order of the matrix

Here A is  square matrix of order 3 ×3,so n =3 and we are given

(a) 0         (b) -2         (c) -1         (d) -3

Ans. (d) -3

The given equation is

5 +y = 0 and x -2 =0

y = -5 and x =2

x +y = -5 +2 = -3

Q7. If A is a diagonal matrix of order 3 ×3 such that A² = A, then the number of possible matrices A are

(a) 4            (b) 8             (c) 16          (d) 32

Ans. (b) 8

Let the diagonal matrix A of order 3 is as follows

We are given A² = A

a² = a, b² = b,c² = c

a² -a=0,b² -b=0,c² -c=0

a(a -1) =0, b(b -1) =0, c(c -1) =0

a =0 or 1, b=0 or 1, c =0 or 1

There are 2 choices of matrix for a =0 or 1

There are 2 choices of matrix for b =0 or 1

There are 2 choices of matrix for c =0 or 1

According to multiplication rule there the number of possible matrices are 2×2×2=8

(a) 13          (b) 5            (c) -8           (d) -3

Ans.  (d) -3

The given equation is

Comparing the elements,LHS and RHS

1 +a = 3, -1 +b =4, 2 +c =5 and 3 +d =6

a = 2, b =5,c =3 and d=3

a +c -b -d = 2 + 3 – 5- 3=-3

Q9. If A is a symmetric matrix then which of the following is not Symmetric matrix,

(a) A + AT         (b) AAT         (C) A – AT           (d) AT

Ans. (C) A – A

A matrix is symmetric when AT = A

Since ( A – A )

=AT -(AT )

=  A– A,which is not possible

Q10. If A is a non-singular square matrix of order 3 such that     = 3,  then the value of    is

(a) 3           (b) 6          (c) 12          (d) 24

Ans. (d) 24

We are given that

We have

= 8 ×3 =24

(a) xa+b+c        (b) xabc     (c) 1/( xa +  x+ xc)    (d) 0

Ans.  (d) 0

The given function is

Q12.Suppose P,Q and R are different matrices of order 3× 5, a×b and c×d respectively, then value of ac + bd is, if matrix 2P + 3q – 4R is defined 

(a) 9           (b) 30          (c) 34          (d) 15

Ans. The sum of the P,Q, and R is defined 2P + 3q – 4R,so the order of all matrices will be same

3× 5= a×b = c×d

a =3,b=5,c=3,d=5

ac + bd =3×3 +5×5 =9 +25 =34

Q13. The function given below at x = 4 is

(a) Continuous but not differentiable

(b) Differentiable but not continuous

(c) Continuous as well as differentiable

(d) Neither continuous nor differentiable

Ans. The given function is

Checking the continuity of the function at x =4

Left-hand  limit

LHL of f(x)= f(4) =2×4+3=11

Right-hand  limit

RHL of f(x)= f(4) =4²-3=11

Hence the given functionx f(x) is continuos at x =4 and a continuos function is always differentiable

Q14.If x³ – 3x²y +y³ =2021 +xy then dy/dx =

(a) (3x² -6xy -y)/(3x² + 3y² -x)   

(b) (3x² -6xy -y)/(3y² – 3x² -x) 

 (c) (6xy +y -3x²)/(3y² – 3x² -x)

(d) (3x² -6xy -y)/(3x² + 3y² +x)

Ans.  (c) (6xy +y -3x²)/(3y² – 3x² -x)

The given equation is

x³ – 3x²y +y³ =2021 +xy

Differentiating it w r t  x

3x² – 3.d/dx(x²y) + 3y² dy/dx = d/dx(xy)

3x² -3[x² dy/dx +y  2x] + 3y² dy/dx = x.dy/dx +y

3x²-3x² dy/dx – 6y x + 3y² dy/dx = x.dy/dx +y

-3x² dy/dx + 3y² dy/dx – x.dy/dx = y – 3x² +6xy

dy/dx(-3x² + 3y²- x) = y – 3x² +6xy

dy/dx = (y – 3x² +6xy)/(-3x² + 3y²- x)=(6xy+y-3x²)/( 3y²- 3x²- x)

Q15. The slope of the tangent to the curve y = x³, at the point (2,8) is

(a) 2           (b) 6          (c) 11          (d) 12

Ans.(d) 12

The given curve is

y = x³

Differentiating the given function with r t x

dy/dx = d/dx(x³) =3x²

(dy/dx)(2,8)= 3×2²= 12

Q16. Corner point of the feasible region determined by the system of linear constraints are (0,3),(1,1) and (3,0). Let Z=px + qy, where p,q >0. Condition on p and q so that the minimum of z occurs at (3,0) and (1,1) is.

(a) p = 2q         (b) q = 2p        (c) p = 3q      (d) p = q

Ans.  (b) q = 2p

The points of linear constraints are given (0,3),(1,1) and (3,0)

Max. Z =px + qy, where p,q >0

The value of Z at (0,3)  = 3q

The value of Z at (1,1)  = p+q

The value of Z at (3,0)  = 3p

Since minimum of z occurs at (3,0) and (1,1)

Max. Z at (3,0) =Max. Z at (1,1)

3p = p+q

2p = q

Q17. The points on the curve 4x² + 9y² = 36 at which tangent to the curve is parallel to x – axis, is

(a) (±2, 0)        (b) (0, ±2)       (c) (0, ±3)       (d) (±3, 0)

Ans. The given curve is

4x² + 9y² = 36

Differentiating the given equation with r t x

8x + 18y.dy/dx =0

8x = -18y.dy/dx

Since the tangent to the curve is parallel to x -axis so,slope of the tangent is equal to the 0

dy/dx = 0

8x = 0

x = 0

Tangent and the curve intersect at the point of contact ,therefore putting the value of x =0 in the equation of the curve

9y² = 36

y = ±2

Therefore the point on the given curve  at which tangent to the curve is parallel is (0, ±2)

Q18. The interval in which y = -x3    + 3x² + 2021 is increasing is 

(a) (-∞, ∞)        (b) (0, 2)       (c) (2, ∞)       (d) (-2, 0)

Ans. (b) (0, 2)

The given function is

y = -x3    + 3x² + 2021

Differentiating it w r t x

dy/dx = -3x² + 6x

For the function to be increasing,we must have

dy/dx >0

-3x² + 6x > 0

-x² + 2x > 0

x( -x + 2) > 0

⇒o < x < 2

The f(x) is increasing in the interval (0,2)

 

(a) y         (b) 2y        (c) – 2y       (d) -y

Ans.(d) -y

The given function is

Differentiating it w r t x

Q20. If x=sin³ t,y = cos³ t then dy/dx =

(a) tan t        (b) cot t      (c) -tan t      (d) -cot t 

Ans. (d) -cot t

The given functions are

x=sin³ t,y = cos³ t

Differentiating them w r to t

Dividing equation (ii) by equation (i)

SECTION B

Q21. If   sin-1x +sin-1y = 2π/3, cos-1x + cos-1y =

(a) -π/3          (b) π/3          (c) π           (d) π/2

Ans.(b) π/3

The given function is

sin-1x +sin-1y = 2π/3

π/2 – cos-1x +π/2 –  cos-1y =2π/3

π -( cos-1x  + cos-1y)   =2π/3

-(cos-1x  + cos-1y ) = 2π/3 – π = (2π -3π)/3 = -π/3

cos-1x  + cos-1y = π/3

Q22. Let f : R → R be defined as f(x) = 7x – 5, then

(a) f is one-one onto 

(b) f is many-one onto 

(a) f is one-one but not onto 

(a) f is neither one-one nor onto 

Ans.(a) f is one-one onto

f : R → R be defined as f(x) = 7x – 5

Let x,y ∈ R such that f(x) = f(y)

7x – 5 = 7y – 5

x = y

∴ f is one one

Let y = f(x)

y = 7x – 5

x =(y+5)/7

f will be onto if for any real number (y ) in co-domain R ,there exist (y+5)/7 in R

f[(y+5)/7]  =7(y+5)/7 – 5 =y +5 -5 =y

∴ f is onto

Q23. A relation R in the set of real numbers R is given by

R ={(a,b) : a > b, a,b ∈ R}, the relation R is

(a) Reflexive       (b) Symmetryric     (c) Transitive    (d) Equivalence

Ans.The relation R in the set of real numbers R is defined as

R ={(a,b) : a > b, a,b ∈ R}

(a,b) ∈ R but (a,a) ∉R, since a ≠b .So R is not reflexive

(a,b) ∈ R but (b,a) ∉R since b<a, So R is not symmetric

Let there exist (a,b) ∈R where a>b and (b,c) ∈ R where b >c ⇒ a>c that implies (a,c) ∈R,therefore relation R is transitive

Q24. If a ≤  2 sin-1x +cos-1x ≤ b, then

(a) a = 0, b =π        (b) a = π, b = 2π       (c) a = -π/2, b =π/2      (d) a =0, b = π/2

Ans.(a) a = 0, b =π

The given inequality is

a ≤  2 sin-1x +cos-1x ≤ b

We have

-π/2 ≤ sin-1x  ≤ π/2

-π/2 +π/2 ≤sin-1x +π/2 ≤ π/2+ π/2

0 ≤ sin-1x + (sin-1x + cos-1x) ≤ π

0 ≤ 2 sin-1x  + cos-1x ≤ π

Hence a = 0 and b = π

(a) 1/9           (b) 1/81            (c) – 9          (d) -81

Ans. (b) 1/81

We are given

We know

 

adj A = -1/3(1/81 -4/81) = 1/81

Q26. If A and B are two square matrices of the same order such that, AB = A and BA = B, then (A +B)(A – B) =

(a) A ² – B²             (b) 2A – 2B            (c) 2A + 2B        (d) 0

Ans.(d) 0

We are given that

AB = A and BA = B

Multiplying both sides by B ,BA = B

BAB= B ²

B(AB) = B²

BA = B² (since A B= A)

Similarly AB = A²

(A +B)(A – B)

=A (A – B) + B(A – B)

= AB – AB + BA – BA (Matrix product is didtributive but not commutative)

=0

Q27. If 5x+ 5y = 5x+y, then dy/dx =

(a) 5x-y         (b)5y-x      (c) -5x-y      (d) -5y-x

Ans.(d) -5y-x

The given equation is

5x+ 5y = 5x+y

Differentiating it both sides w r to x

Q28. The interval on which the function f(x) = 2x³ – 3x² – 36x + 10 is decreasing is

(a) (-∞,-2)      (b) (-2,3)      (c) (2,3)     (d) (3, ∞)

Ans.(b) (-2,3)

The given function is

Let y =f(x)

y= 2x³ – 3x² – 36x + 10

Differentiating it both sides w r to x

dy/dx = 6x² -6x -36

For the function to be increasing,we must have

dy/dx < 0

6x² -6x -36 < 0

x² -x -6 < 0

x² -3x +2x-6 < 0

x(x -3) +2(x -3) <0

(x -3)(x +2) <0

x <3 and x > -2

(-2,3)

Q29. If the curve ay + x² =7 andcx³ = y,cut orthogonaally at (1,1), then the value of a is :

(a) 1      (b) 3      (c) -6      (d) 6

Ans. (d) 6

he curves are

ay + x² =7 andcx³ = y

Differentiating them w r t x

a dy/dx + 2x   =0 and 3x² =dy/dx

dy/dx =-2x/a and dy/dx = 3x²

(dy/dx)(1,1)=(dy/dx)(1,1)

-2 ×1/a = 3×1

a = 6

Since both of the curvrs cut orthogonaally at (1,1)

Q30. If log(x² -y²)/(x² +y²) =a, then dy/dx =

(a) y/x        (b) -y/x         (c) x/y      (d) -x/y

Ans.(c) x/y

The given function is

log(x² -y²)/(x² +y²) =a

log(x² -y²) -log(x² +y²)=a

Q31. The area of a triangle with vertices (-3,0),(3,0) & (0,k) is 9 sq.units. The value of k is (k>0)

(a) 3        (b) 6        (c) 9       (d) 12

Ans.(a) 3

Area of triangle formed by joining the given points must be zero because points given to us are collinear

Area of triangle =1/2[x1(y2-y3) +x2(y3-y1) + x3(y1-y2) ]=9

x1=-3,y1=0,x2=3,y2=0,x3=0,y3=k

1/2[-3(0-k) +3(k-0) + 0(0-0) ] = 9

3k +3k  = 18

6k = 18

k = 18/6 =3

(a) 2       (b) 6        (c) 9         (d) 5

Ans. The given matrix equation is

Comparing LHS and RHS

x + y +z = 6…..(i)

y + z =3……..(ii)

z = 2………(iii)

Puuting the value of z from equation (iii) in equation (ii)

y + 2 = 3⇒y =1

Putting the value of y and z in equation (i)

x +1 +2 =6⇒ x=3

Hence 2x +y – z = 2×3 +1 -2=5

(a) 2         (b) -2        (c) 1        (d) -1

Ans.(b) -2

The given function is

tan y = x

y=tan-1x

Differentiating it w r t x

Differentiating again it w r t x

Q34. If a non-singular Matrix A satisfy 2A² + A – I = 0, then A-1=

(a) 2A        (b) 2A +1        (c) 4A +2I        (d) 2A – 4I

Ans.(b) 2A +1

The given equation is

2A² + A – I = 0

I = 2A² + A

Multiplying the equation by A-1

I A-1= 2A²A-1+ AA-1

IA-1= 2A + I

A-1= =2A +I

Q35. The maximum value of the function f(x) = 4 sin x.cos x is

(a) 2        (b) 4        (c) 1        (d) 8

Ans.(a) 2

The given function is

f(x) =4 sin x.cos x is

Let y = 4 sin x.cos x

Since 2 sin x.cos x = sin 2x

y = 2sin 2x

dy/dx = 2cos 2x.2 =4 cos 2x

For getting critical points putting dy/dx =0

4 cos 2x =0

cos 2x =0 = cos π/2

2x = π/2

x = π/4

d2x/dx2=4(-sin 2x).2=-8sin 2x

Since the rate of change of the slope (-8 sin 2π/4 =-8) <0

The value of the function is maximum at x= π/4

Hence the maximum value of the f(x) is

4 sin x.cos x = 4 sin π/4. cos π/4 =4. 1/√2 .1/√2 =4/2 =2

Q36.If the objective function z = ax + y is minimum at (1,4) and its minimum value is 13,then value of a is 

(a) 1       (b) 4         (c) 9       (d) 13

Ans.(c) 9

The given function is

z = ax + y

The minimum value of the function is  at (1,4) which is 13

Putting the value x =1 and z = 13

a×1 + 4 = 13

a = 13 – 4 = 9

Q37. Let L be the set of all lines in a plane . A relation R in L is given by R = {(L1,L2) :L1,and L2   intersect at exactly one point ,L1,L2∈},then the relation R is 

(a) Reflexive       (b) Symmetric      (c) Transitive      (d) Equivalence

Ans.(b) Symmetric

The given relation is  R = {(L1,L2) :L1,and L2   intersect at exactly one point ,L1,L2∈}

L1 and L2 ∈ set of lines in a plane L1,L1 or L2,L∉ R because a line can not intersect to itself,therefore R is not reflexive

L1 and L2 ∈ set of lines in a plane (L1,L2 ) or (L2,L1 ) ∈R because  line L1  intersect L2,then line L2 ralso intersect line L1 threfore R is symmetric

Let (L1,L2 ) ∈ R and  (L2,L3 ) ∈R ⇒(L1,L3 ) ∉ R because (L2,L3 )  inersect each other and (L1,L2 ) intersect each other doesn’t mean (L1,L3 ) intersect each other

Q38. If X → Y is defined, then f is

Q38. Preboard class 12 maths term 1

(a) Bijective function     (b) Many-one and onto      (c)  Many-one and into  function     (d)  One-one but not onto

Ans.The given function is defined as f: X → Y

Let function is f(x)

Here f(3) = c and f(4) = c but 3 ≠ 4, the function is many one because two elements have the same image

The range ,R of the f(x) is ={a,b,c) and codomain is ={a,b,c,d},therefore f(x) is a into function (since range ≠codomain)

Hence f(x) many-one and into function

Q39. The feasible region for an LPP is always a …….polygon

(a) Convex      (b) Concave       (c) either (a) or (b)     (d) neither (a) nor (b)  

Ans.(a) Convex

The feasible region for an LPP is always a convex polygon

Each angle of a convex polygon is less than 180°

At least one of the angles of a concave polygon is greater than 180°

Q40.The tangent to the curve y = ex at the point (0,1) meets x -axis at

(a) (1,0)    (b) (-1,0)      (c) (0,0)      (d) (2,0)  

Ans. (c) (0,0)

The given curve is

y = ex

Differentiating it w r t x

dy/dx = d/dx( ex)= ex

Slope of the tangent to the curve is ex

The value of the slope

(dy/dx)(0,1)=e0= 1 (i.e m=1)

The equation of a line passing through  (x1, y1 ) is given as

y – y1=m(x- x1)

Since tangent is passing through (0,1)

y – 1=1(x- 1)

y -1 = x – 1

y -x =0

The tangent meets x-axis,so putting y= 0

x=0

Hence tangent to the given cueve meets x -axis at (0,0)

SECTION C

Q41. The feasible region, for the inequalities x + 2y ≤ 6, y ≥6, 0 ≤ x lies in

(a) First Quadrant      (b) Second Quadrant      (c) Third Quadrant    (d) Fourth Quadrant

Ans.(a) First Quadrant

The given inequalities are x + 2y ≤ 6, y ≥0, 0 ≤ x

Solving the equations and drawing their graphs x + 2y = 6

Putting x =0, we get y= 3 and y =0,we get x =6

y =0 (X axis),  x=0 (Y axis)

Q41 Class 12 maths term 1 2021-22

Q42.Which of the following function is decreasing on (0,π/2)

(a) sin x         (b) cos x        (c) tan x         (d) sin 2x

Ans. (b) cos x

Let y = cos x

dy/dx = -sin x

Since sin x > 0 for (0,π/2)

∴ -sin x < 0 for (0,π/2)

Hence cos x  is decreasing on (0,π/2)

Q43. If the function f(x) = sin x – ax + b, is decreasing on x ∈ R,then a belongs to

(a) (1,∞)        (b) [0,∞)         (c) (0,∞)        (d) [1,∞)

Ans.(a) (1,∞)

The given function f(x) = sin x – ax + b, is decreasing on x ∈ R

Differentiating f(x) with repect to x

f'(x) = cos x -a

If a function is decreasing then

f'(x) ≤ 0

cos x -a ≤0

cos x ≤ a

a ≥ cos x

Since cos x ≥ 1

Therefore a is greater and equal to 1

Hence for a the interval is [1,∞)

Q44.In a linear programming problem ,if the feasible region is bounded then objective function Z = px + qy has

(a) Maximum  value only     

 (b) Minimum value only   

 (c) Maximum and minimum value both   

(d) Neither maximum nor minimum value

Ans.(c) Maximum and minimum value both

In linear programming problem two types of feasible regions are there

Bounded : Has maximum and minimum value both

Unbounded : Has either maximum or minimum value

is singular matrix, then the value of x is

(a) 3          (b) -2         (c) 0       (d) 2

Ans.(d) 2

The given matrix is

Since it is given that A is singular

Therefore

6x × 2 – 8×3 =0

12 x =24

x = 2

CASE STUDY

The fuel cost per hour for running a train is proportioal to the square of the the speed it generates in km per hour. If the fuel costs Rs 48 per hour at speed 16 km per hour and the fixed charges to run the train amount to Rs 1200 per hour. Assume the speed of the train as v km/h.

case study class 10 maths preboard

Based on the given information, answer the following questions.

Q46. Given that the fuel cost per hour is k times the square of the speed the train generates in km/h, the value of 16 k is:

(a) 1          (b) 2          (c) 3          (d) 4

Ans. (c) 3

It is given that

Fuel cost per hour =k( speed the train)²

48 = k×16²=256k

k = 48/256 = 3/16

Hence the value of 16 k= 16 × 3/16 = 3

Q47. If the train has travelled a distance of 1000 km,then the total cost of running the train is given by function:

(a) (375/4)v + 60000/v

(b) (375/8)v + 60000/v

(c) (375/2)v + 60000/v

(d) (375/2)v + 1200000/v

Ans.(d) (375/2)v + 1200000/v

Speed of the train = v km/h

Distance covered by the train is = 1000 km

The time taken to cover the 1000 km = distance/time = 1000/v

The total cost of fuel to cover a distance of 1000 km =Fixed charges + Charges based on per km =

Fixed charges for 1 hour =Rs 1200

Fixed charges for 1000/v h is = 1200 ×(1000)/v =1200000/v

Fuel cost per hour =k( speed the train)²=(3/16)v²

Fuel cost for 1000/v hours = (3/16)v² ×1000/v=(375/2)v

The total cost of fuel to cover a distance of 1000 km =1200000/v + (375/2)v

Q48.The most economical speed to run the train (in km/h) is :

(a) 50        (b) 80         (c) 400      (d) 800

Ans.(b) 80

Total cost (C) of running the train is given by

For acheiving the minimum cost, dc/dv must be equal to 0

Differentiating the function with respect to v

Q49. The fuel cost (in Rs) for the train to travel 1000 km at the most economical speed is:

(a) 15000        (b) 75000       (c) 100000      (d) 150000

Ans.(a) 15000

The total cost of fuel to cover a distance of 1000 km =Fixed charges + charges based on per km

The function of total cost contains fixed charge and the fuel cost per hour,therefore for cost of the fuel at the most economical speed(80 km/h) the fixed charge must be equal to 0

Q50. The total cost of the train to travel 1000 km at the most economical speed is:

(a) 15000       (b) 30000        (c) 100000      (d) 150000

Ans.(b) 30000

Total cost (C) of running the train for 1000 km is given by

Total cost of fuel at most economical speed (i.e 80 km/h) is

The total cost of fuel to cover a distance of 1000 km =Fixed charges + charges based on per km

 

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Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles
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Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula
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Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion
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Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ?
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Chapter 8- Motion Last years question papers & sample papers

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Chapter 1-Real number Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial Chapter 10-Circles
Chapter 3-Linear equations Chapter 11- Construction
Chapter 4- Quadratic equations Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume
Chapter 6-Triangle Chapter 14-Statistics
Chapter 7- Co-ordinate geometry Chapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds Chapter 12- Electricity
Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process Chapter 14-Sources of Energy
Chapter 7-Control and Coordination Chapter 15-Environment
Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

Solutions of class 10 last years Science question papers

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NCERT solutions of class 11 maths

Chapter 1-Sets Chapter 9-Sequences and Series
Chapter 2- Relations and functions Chapter 10- Straight Lines
Chapter 3- Trigonometry Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations Chapter 15- Statistics
Chapter 8- Binomial Theorem  Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

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NCERT solutions of class 12 maths

Chapter 1-Relations and Functions Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra
Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability Chapter 13-Probability
Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions
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