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

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 $\left | adj\: A \right |=256,$ then the sum of all possible values of $\left | A \right |\; is$

(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

$\left | adj\: A \right |=\left | A \right |^{n-1}$

Where n is the order of the matrix

Here A is square matrix of order 3 ×3,so n =3 and we are given $\left | adj\: A \right |=256,$

$\left | A \right |^{3-1}=256$

$\left | A \right |^2=256$

$\left | A \right |=\sqrt{256}=16$

$Q6.If\: \begin{bmatrix}x-2 & 5+y \end{bmatrix}\begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix}=0,then\: x+y=$

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

Ans. (d) -3

The given equation is

$\begin{bmatrix}x-2 & 5+y \end{bmatrix}\begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix}=0$

$\begin{bmatrix}0+5+y & x-2+0 \end{bmatrix}=0$

$\begin{bmatrix}5+y & x-2 \end{bmatrix}=\begin{bmatrix} 0 & 0 \end{bmatrix}$

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

$A=\begin{bmatrix} a &0 & 0\\ 0 & b & 0\\ 0 &0 & c \end{bmatrix}$

We are given A² = A

$\begin{bmatrix} a &0 & 0\\ 0 & b & 0\\ 0 &0 & c \end{bmatrix}\begin{bmatrix} a & 0 & 0\\ 0& b &0 \\ 0& 0 & c \end{bmatrix}=\begin{bmatrix} a & 0 &0 \\ 0 &b & 0\\ 0 & 0 & c \end{bmatrix}$

$\begin{bmatrix} a^{2} &0 & 0\\ 0 & b^{2} & 0\\ 0 &0 & c^{2} \end{bmatrix}=\begin{bmatrix} a & 0 &0 \\ 0 &b & 0\\ 0 & 0 & c \end{bmatrix}$

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

$Q8.If\: \begin{bmatrix} 1 & -1\\ 2& 3 \end{bmatrix}+X=\begin{bmatrix} 3 & 4\\ 5 &6 \end{bmatrix},where\: X=\begin{bmatrix} a & b\\ c& d \end{bmatrix},then\: a+c-b-d=$

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

Ans. (d) -3

The given equation is

$\: \begin{bmatrix} 1 & -1\\ 2& 3 \end{bmatrix}+X=\begin{bmatrix} 3 & 4\\ 5 &6 \end{bmatrix}$

$Since\: X=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$

$\begin{bmatrix} 1 & -1\\ 2& 3 \end{bmatrix}+\begin{bmatrix} a & b\\ c& d \end{bmatrix}=\begin{bmatrix} 3 &4 \\ 5& 6 \end{bmatrix}$

$\begin{bmatrix} 1+a & -1+b\\ 2+c& 3+d \end{bmatrix}=\begin{bmatrix} 3 &4 \\ 5& 6 \end{bmatrix}$

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 + A^T(b) AA^T(C) A – A^T(d) A^T

Ans. (C) A – A^T

A matrix is symmetric when A^T= A

Since ( A – A^T ) ^T

=A^T-(A^T) ^T

= A^T– A,which is not possible

Q10. If A is a non-singular square matrix of order 3 such that $\left | A \right |$ = 3, then the value of $\left | 2A^{T} \right |$ is

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

Ans. (d) 24

We are given that

$\left | A \right |=3$

We have

$\left | A^{T} \right |=\left | A \right |=3$

$\left | 2A^{T} \right |$

$=2^{3}\left | A^{T} \right |\because A \: and\: A^{T}\: both \: are \: of\: same\: order$

= 8 ×3 =24

$Q11.If\: y=\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{c-b}+x^{a-b}}+\frac{1}{1+x^{a-c}+x^{b-c}},then\: \frac{dy}{dx}=$

(a) x^a+b+c(b) x^abc(c) 1/( x^a+ x^b+ x^c) (d) 0

Ans. (d) 0

The given function is

$y=\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{c-b}+x^{a-b}}+\frac{1}{1+x^{a-c}+x^{b-c}}$

$=\frac{1}{\left ( x^{a}+x^{b} +x^{c}\right )/x^{a}}+\frac{1}{\left ( x^{a}+x^{b} +x^{c}\right )/x^{b}}+\frac{1}{\left ( x^{a}+x^{b} +x^{c}\right )/x^{c}}$

$=\frac{x^{a}}{\left ( x^{a}+x^{b} +x^{c}\right )}+\frac{x^{b}}{\left ( x^{a}+x^{b} +x^{c}\right )}+\frac{x^{c}}{\left ( x^{a}+x^{b} +x^{c}\right )}$

$=\frac{x^{a}+x^{b}+x^{c}}{\left ( x^{a}+x^{b} +x^{c}\right )}=1$

$\frac{dy}{dx}=0$

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

$f\left ( x \right )=\left\{\begin{matrix} 2x+3,x\leq 4\\ x^{2}-5,x> 4 \end{matrix}\right.$

(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

$f\left ( x \right )=\left\{\begin{matrix} 2x+3,x\leq 4\\ x^{2}-5,x> 4 \end{matrix}\right.$

Checking the continuity of the function at x =4

Left-hand limit

$\lim_{x\rightarrow4^{-} }f\left ( x \right )=\lim_{x\rightarrow4^{-} }\left ( 2x+3 \right )=2\times 4+3=11$

LHL of f(x)= f(4) =2×4+3=11

Right-hand limit

$\lim_{x\rightarrow4^{+} }f\left ( x \right )=\lim_{x\rightarrow4^{+} }\left ( x^{2}-5 \right )=4^{2}-5=11$

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)

$Q19.If\: x=log_{e}y,then\: \frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}=$

(a) y (b) 2y (c) – 2y (d) -y

Ans.(d) -y

The given function is

Differentiating it w r t x

$\frac{1}{y}.\frac{dy}{dx}=1$

$\frac{dy}{dx}=y$

$\frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}$

$\therefore \frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}=\frac{dy}{dx}-2y=y-2y=-y$

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

$\frac{dx}{dt}=3sin^{2}t.\frac{d}{dt}sint$

$\frac{dx}{dt}=3sin^{2}t.cost...\left ( i \right )$

$\frac{dy}{dt}=3cos^{2}t.\frac{d}{dt}cost$

$\frac{dy}{dt}=3cos^{2}t\left ( -sint \right )$

$\frac{dy}{dt}=-3cos^{2}t.sint......\left ( ii\right )$

Dividing equation (ii) by equation (i)

$\frac{dy}{dt}=\frac{-3cos^{2}t.sint}{3sin^{2}t.cost}=\frac{-cost}{sint}=-cot.t$

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

$Q25.If\: A^{-1}=\begin{bmatrix} 3 & 1 &2 \\ 0 &1 &2 \\ 0& 2 & 1 \end{bmatrix},then\: \left | adj\: A \right |=$

(a) 1/9 (b) 1/81 (c) – 9 (d) -81

Ans. (b) 1/81

We are given

$A^{-1}=\begin{bmatrix} 3 & 1 &2 \\ 0 & 1 &2 \\ 0& 2 & 1 \end{bmatrix}$

We know

$adj.A=\left | A \right |.A^{-1}$

$\left | A \right |=\frac{1}{\left | A^{-1} \right |}$

$\left | A^{-1} \right |=3\left ( 1-4 \right )-1\left ( 0-0 \right )+2\left ( 0-0 \right )=-9$

$\left | A \right |=-\frac{1}{9}$

$adj.A=-\frac{1}{9}\begin{bmatrix} 3 &1 & 2\\ 0& 1 &2 \\ 0& 2 & 1 \end{bmatrix}$

$adj.A=\begin{bmatrix} 3/-9 &1/-9 & 2/-9\\ 0/-9& 1/-9 &2 /-9\\ 0/-9& 2/-9 & 1/-9 \end{bmatrix}$

$adj.A=\begin{bmatrix} -1/3 &-1/9 & -2/9\\ 0& -1/9 &-2 /9\\ 0& -2/9 & -1/9 \end{bmatrix}$

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)

Q27. If 5^x+ 5^y = 5^x+y, then dy/dx =

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

Ans.(d) -5^y-x

The given equation is

5^x+ 5^y = 5^x+y

Differentiating it both sides w r to x

$\frac{d}{dx}5^{x}+\frac{d}{dx}5^{y}=\frac{d}{dx}5^{x+y}$

$5^{x}log5+5^{y}log5.\frac{dy}{dx}=5^{x+y}.log5\frac{d}{dx}\left ( x+y \right )$

$5^{x}log5+5^{y}log5.\frac{dy}{dx}=5^{x+y}.log5\left ( 1+\frac{dy}{dx} \right )$

$5^{x}log5+5^{y}log5.\frac{dy}{dx}=5^{x+y}.log5 +5^{x+y}log5\frac{dy}{dx}$

$5^{x}+5^{y}.\frac{dy}{dx}=5^{x+y} +5^{x+y}\frac{dy}{dx}$

$5^{y}.\frac{dy}{dx}-5^{x+y}\frac{dy}{dx} =5^{x+y}-5^{x}$

$\frac{dy}{dx}\left ( 5^{y}-5^{x+y} \right ) =5^{x+y}-5^{x}$

$\frac{dy}{dx}=\frac{5^{x+y}-5^{x}}{5^{y}-5^{x+y}}$

$\because 5^{x+y}=5^{x}+5^{y}$

$=\frac{5^{x}+5^{y}-5^{x}}{5^{y}-5^{x}-5^{y}}$

$\frac{dy}{dx}=\frac{5^{y}}{-5^{x}}=-5^{y-x}$

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

$\frac{d}{dx}log\left ( x^{2} -y^{2}\right )-\frac{d}{dx}log\left ( x^{2}+y^{2} \right )=0$

$\frac{1}{x^{2}-y^{2}}\left ( 2x-2y\frac{dy}{dx} \right )-\frac{1}{x^{2}+y^{2}}\left ( 2x+2y\frac{dy}{dx} \right )=0$

$-\frac{dy}{dx}\left (\frac{2y}{x^{2}-y^{2}} +\frac{2y}{x^{2}+y^{2}}\right )=\frac{2x}{x^{2}+y^{2}}-\frac{2x}{x^{2}-y^{2}}$

$-\frac{dy}{dx}\left ( \frac{2yx^{2}+2y^{3}+2yx^{2}-2y^{3}}{\left ( x^{2}-y^{2} \right )\left ( x^{2}+y^{2} \right )} \right )=\frac{2x^{3}-2xy^{2}-2x^{3}-2xy^{2}}{\left ( x^{2}-y^{2} \right )\left ( x^{2}-y^{2} \right )}$

$-\frac{dy}{dx}=\frac{2x^{3}-2xy^{2}-2x^{3}-2xy^{2}}{2yx^{2}+2y^{3}+2yx^{2}-2y^{3}}$

$-\frac{dy}{dx}=\frac{-4xy^{2}}{4yx^{2}}=-\frac{y}{x}$

$\frac{dy}{dx}=\frac{y}{x}$

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

$Q32.If\: \begin{bmatrix} 1 & 1 &1 \\ 0 & 1 & 1\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x\\ y \\ z \end{bmatrix}=\begin{bmatrix} 6\\ 3 \\2 \end{bmatrix},then\: 2x+y+z=$

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

Ans. The given matrix equation is

$\: \begin{bmatrix} 1 & 1 &1 \\ 0 & 1 & 1\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x\\ y \\ z \end{bmatrix}=\begin{bmatrix} 6\\ 3 \\2 \end{bmatrix}$

$\begin{bmatrix} x+y+z\\ y+z \\z \end{bmatrix}=\begin{bmatrix} 6\\ 3 \\2 \end{bmatrix}$

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

$Q33.If\: tany=x,then\: when\: x =1,value\: of\: 4\frac{d^{2}y}{dx^{2}}$

(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

$\frac{dy}{dx}=\frac{1}{1+x^{2}}$

Differentiating again it w r t x

$\left ( \frac{d^{2}y}{dx^{2}} \right )=\frac{-2x}{\left ( 1+x^{2} \right )^{2}}$

$\left ( \frac{d^{2}y}{dx^{2}} \right )_{x=1}=\frac{-2\times 1}{\left ( 1+1^{2} \right )^{2}}=\frac{-2}{4}=-\frac{1}{2}$

$4\left ( \frac{d^{2}y}{dx^{2}} \right )_{x=1}==-\frac{4}{2}=-2$

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 L₂intersect 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 L₂intersect at exactly one point ,L₁,L₂∈}

L₁and L₂∈ set of lines in a plane L₁,L₁or L₂,L₂∉ R because a line can not intersect to itself,therefore R is not reflexive

L₁and 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

(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 = e^x 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 = e^x

Differentiating it w r t x

dy/dx = d/dx( e^x)= e^x

Slope of the tangent to the curve is e^x

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)

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

$Q45.If\: A=\begin{bmatrix} 6x &8 \\ 3& 2 \end{bmatrix}$

is singular matrix, then the value of x is

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

Ans.(d) 2

The given matrix is

$A=\begin{bmatrix} 6x &8 \\ 3& 2 \end{bmatrix}$

Since it is given that A is singular

Therefore

$\left | A \right |=0$

$\begin{vmatrix} 6x &8 \\ 3& 2 \end{vmatrix}=0$

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.

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

$C=\frac{375}{2}v+\frac{1200000}{v}$

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

Differentiating the function with respect to v

$\frac{dC}{dv}=\frac{d}{dv}\left ( \frac{375}{2}v+\frac{1200000}{v} \right )$

$\frac{dC}{dv}=\frac{375}{2}+\frac{1200000}{v^{2}}$

$\frac{dC}{dv}=\frac{375}{2}-\frac{1200000}{v^{2}}$

$\frac{375}{2}-\frac{1200000}{v^{2}}=0$

$\frac{375}{2}v^{2}=1200000$

$v^{2}=\frac{2400000}{375}=6400$

$v=80\: km/h$

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

$C_{min}=\frac{375v}{2}=\frac{375\times 80}{2}=375\times 40=15000$

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

$C=\frac{375}{2}v+\frac{1200000}{v}$

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

$C_{total}=\frac{375\times 80}{2}+\frac{1200000}{80}$

$C_{total}=\frac{375\times 80}{2}+\frac{1200000}{80}=15000+15000=30000$

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

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