Mathematics: CUET Mock Test - 9 - CUET MCQ

Because , the inverse of an identity matrix is an identity matrix.

e^y =1/ (x+5)
Taking log both side we get
log e^y = log {1/(5+x)}
then, y = log {1/(5+x)}
Differentiating both side ,
dy/dx =( x+5) . {-1/(5+x)²}
dy/dx = -1/(5+x) ……..( 1)
Again Differentiating, d²y/dx²= 1/(5+x)²
d²y/dx²= {-1/(5+x)}²
From equation (1)
d²y/dx² = {dy/dx}²

Sol.[A]   The parametric form of the given equation is x = t, y = t². The equation of any tangent at
t is 2xt = y t². Differentiating, we get 2t = y₁. Putting this value in the equation of tangent, we have 2 x y₁/2 = y (y₁/2)² Þ 4xy₁ = 4y y₁²

                The order of this equation is one.

Calculations:
2x + 3y = 28
x + y = 10

At (0, 9.3) value of z = 4x + 2y is 18.6.
At (10, 0) value of z = 4x + 2y is 40.
At (2, 8) value of z = 4x + 2y is 24.
At (0, 0) value of z = 4x + 2y is 0.
Hence, the Correct Option is option no 2.

Concept:

Feasible Region:

The feasible region is the set of all feasible solutions that satisfy all the constraints of the linear programming problem. It is represented by shaded regions in the graph of the problem.

Convex set:

A convex set is a set of points in which any two points can be connected by a straight line that lies entirely within the set.

Calculation:

The feasible set in a linear programming problem is always a convex set.

Since, the constraints of the problem are linear, also the intersection of any two linear constraints is always a line or a plane, which is a convex set.

Also, the feasible set is always bounded because the objective function of a linear programming problem is always a linear function.

∴ The feasible region of LPP is always a convex set.

The correct answer is option 4.

To determine the range of f(x), we start by analyzing each term of the sum, which is of the form [1 + sin (kx)] for k = 1, 2, …, n.
Note that for any x in (0, π), the sine function satisfies -1 < sin (kx) ≤ 1.
Therefore, the expression 1 + sin (kx) takes values in the interval (0, 2].
The greatest integer function [1 + sin (kx)] returns:
• 0 if 1 + sin (kx) is in the interval (0, 1),
• 1 if 1 + sin (kx) is in the interval [1, 2), and
• 2 if 1 + sin (kx) is exactly 2.

For x in (0, π), it is possible to choose x such that for one selected value of k, sin (kx) reaches its maximum value of 1 (making 1 + sin (kx) = 2 and
hence [1 + sin (kx)] = 2), while for the remaining values of k, sin (kx) does not reach 1 so that [1 + sin (kx)] remains 1. In this situation, the sum f(x) becomes (n - 1) × 1 + 2 = n + 1.
On the other hand, if no term attains the value 2 and all terms yield the integer 1, then f(x) equals n.
Thus, the only possible values for f(x) are n and n + 1, which means that the range of f(x) is exactly {n, n + 1}.

Correct Answer :- b

Explanation:- A = {a, b, c} and R = {(a, a), (b, b), (c, c), (b, c)}

Any relation R is reflexive if fx Rx for all x ∈ R. Here ,(a, a), (b, b), (c, c) ∈ R. Therefore , R is reflexive.

For the transitive, in the relation R there should be (a,c)

Hence it is not transitive.

Hence , the given equation has only one solution.

Given:

A and B be two non zero square matrics

Calculations:

Matrix AB is defined means Columns is equal to the Rows of B

and BA is defined means Columns of B is equal to the Rows of A

Hence, Both matrices (A) and (B) have same order is Correct.

A. In the above matrix A = , we can see, the number of rows and columns are 2 respectively. Since the order of the matrix is 2 × 2, hence A is a square matrix.

B. The given 2 × 2 matrix

We first find the determinant of A.

Det A = (2 × 5) - (3 × -3) = 10 + 9 = 19

∴ |A| = 19

Since, |A| ≠ 0 ⇒ A-1 exists.

C. To know if a matrix is symmetric, find the transpose of that matrix. If the transpose of that matrix is equal to itself, it is a symmetric matrix. That is A = A^T

Here A = then A^T =

Here, A ≠ AT

Thus A is not a symmetric matrix.

D. We have already derived |A| = 19.

E. Null Matrix: If in a matrix all the elements are zero then it is called a null matrix. It is also called a zero matrix. Here we can see A is not null matrix.

Thus A, B, D is the correct answer.

The number of possible entries of 2 × 2 matrix is 4 Every entry has two choice, 0 or 1.

Thus, the total no. of choices is,

2 × 2 × 2 × 2 = 2⁴

= 16

Consider the element -3 in Δ= 
The cofactor of the element -3 is given by
A₂₂=(-1)²⁺² M₂₂
M₂₂=  =1(5)-(-6)(-7)=5-42=-37
A₂₂=(-1)²⁺² (-37)=-37.

Given that, A= 
|A| = 
|A|=-cos⁡θ (cos⁡θ )-cotθ(-tan⁡θ)
|A|=-cos²⁡θ+1=sin²⁡θ.

Formula Used:

2 sin x cos x =sin 2x

Calculation:

Let . . .(1)

Now, Put sin x - cos x = t

⇒​ ( sin x + cos x ) dx = dt

and (sin x - cos x)² = t²

⇒ 1 - 2 sin x cos x = t2

⇒ 1 - sin 2x = t2

⇒ 1 - t2 = sin 2x

Substituting all the values in (1)

Solution

⇒

Dividing and multiplying the term by its conjugate.

⇒ × dx

⇒ ...(1 - sin²x = cos²x)

⇒ =

Since, (1/cos²x = sec²x and sin/cos²x = tan x × sec x)

⇒ ∫(sec²x - tan x sec x) dx

⇒ ∫sec²x dx - ∫tan x sec x dx

⇒ tan x - sec x + c

The correct option is 2.

Let −cot⁻¹x=t
Differentiating w.r.t x, we get

=e^t
Replacing t with -cot^-1x, we get

Let y=x^1/4. Let x=81 and Δx=1
Then, Δy=(x+Δx)^1/4-x^1/4
Δy=82^1/4-81^1/4
82^1/4=Δy+3
dy is approximately equal to Δy is equal to
dy = (dy/dx)Δx

∴ The approximate value of 82^1/4 is 3+0.00925=3.00925

As it represents the identity (b+a)² it satisfies the identity (b+a)² = (a² + b² +2ab) and is not linear, cubic or an imaginary equation so the correct option is Identity Equation.

Let, AB be the lamp-post whose foot is A, and B is the source of light, and given (AB)’ = 12(½) ft.
Let MN denote the position of the man at time t where (MN)’ = 5ft.
Join BN and produce it to meet AM(produced) at P.
Then the length of man’s shadow= (MP)’
Assume, (AM)’ = x and (MP)’ = y. Then, (PA)’ = (AM)’ + (MP)’ = x + y
And dx/dt = velocity of the man = 3
Clearly, triangles APB and MPN are similar.
Thus, (PM)’/(MN)’ = (PA)’/(AB)’
Or, y/5 = (x + y)/12(½)
Or, (25/2)y = 5x + 5y
Or, 3y = 2x
Or, y = (2/3)x
Thus, dy/dt = (2/3)(dx/dt)
As, dx/dt = 2,
= 2/3*3 = 2mph

Let x be the radius of the sphere.
Then, x=6cm and Δx=0.07cm
The volume of a sphere is given by V= (4/3)πx³

dV=4×π×6²×0.07
dV=10.08π cm³

Let cos^-1⁡x=t
Differentiating w.r.t x, we get

= t²/2
Replacing t with cos^-1x,we get

We have, x = t³ + 6t² – 15t + 18
Let, v be the velocity of the particle at the end of t seconds. Then, v = dx/dt = d/dt(t³ + 6t² – 15t + 18)
So, v = 3t² + 12t – 15
Thus, velocity of the particle at the end of 2 seconds is, [dx/dt]_{t = 2} = 3(2)² + 12(2) – 15 = 21cm/sec.

If L₁ and L₂ have the direction ratios
a₁,b₁,c₁ and a₂,b₂,c₂  respectively then the angle between the lines is given by

• (A) Graphical Method is a method to solve linear programming problems with two variables using graphs, so it matches with (III).
• (B) Objective Function is a function to be maximized or minimized in linear programming, so it matches with (II).
• (C) Constraints are the set of inequalities that limit the decision variables, so it matches with (IV).
• (D) Feasible Region is the area that satisfies all the constraints in a graph, so it matches with (I).

• (A) Feasible Solution: A feasible solution satisfies all constraints of the problem (which is (III)).
• (B) Infeasible Solution: An infeasible solution does not satisfy one or more constraints of the problem (which is (II)).
• (C) Optimal Solution: The optimal solution is the one that maximizes or minimizes the objective function and satisfies all constraints (which is (IV)).
• (D) Linear Inequalities: Linear inequalities are constraints that limit the values of the decision variables (which is (I)).

• (A) Linear Programming Problem refers to a problem where the objective function and constraints are linear, so it matches with (II).
• (B) Objective Function is a function to be either maximized or minimized in the problem, so it matches with (I).
• (C) Constraints are inequalities or equations that restrict the values of decision variables, so it matches with (III).
• (D) Feasible Region is the region in the graph that satisfies all the constraints, so it matches with (IV).

 x -(xy)^½ dy = ydx
[x-(xy)^½] - y = dx/dy
dx/dy = x/y - [(xy)^½]/y……………….(1)
Let V = x/y
x = Vy
dx/dy = V + ydv/dy……..(2)
V + ydv/dy = V - (V)^½
ydv/dy = -(V)^½
dv/(V)^½ = -dy/y
Integrating both the sides, we get
(V^(-½+1))/(-½ + 1) = -log y + c
2(V)^½ = -log y + c
2(y/x)^½ = -log y + c
2(x)^½ = (y)^½(-log y + c)
2(x)^½ = (y)¹/2log y + c(y)^½ 
(x)^½ = [(y)^½]/2 log y + [c(y)^½]/2

α,β,γ are the angles which the position vector  makes with the positive x-axis ,y-axis and z-axis respectively are called direction angles.