Mathematics: CUET Mock Test - 9 - CUET MCQ

The given system of equations can be expressed in the form of AX=B,
⇒ X= A^-1 B

We know that, A^-1= (1/|A|) adj A
A^-1= (1/10) 
∴ X = A^-1 B= 

Calculations:

∫1/x((x⁵)+3)dx

To solve this integral, we can use the substitution method. Let:

u = x⁵+3

du = 5x⁴dx

​∫1/x((x5)+3)dx = ∫1/u((5x⁴))du = 1/5 [∫1/u((u-3))]du

By partial Fraction

1/5 ∫(A(u-3) + Bu)/u(u-3) du =1/5 ( -1/3 ∫1/u du + 1/3 ∫1/(u - 3))du

= 1/15 (Log u-3)/log u = =

is simplified as = q(x) − log|x + 1| + C

Then, q(x) = + x

Calculation:

The integral of the absolute value function requires splitting the integral into separate parts where the function inside the absolute value is positive and where it's negative.

This splits into three integrals because the absolute value functions changes at x=0 and x=2. However, since our upper limit is 1, we only consider x=0:

This simplifies to:

Each of these can be integrated separately:

Evaluating these at their limits gives:
= 5
So,

Order of a Differential Equation: The order of a differential equation is defined as the highest power of the derivative present in the equation. It is a non-negative integer that indicates the number of times the dependent variable has been differentiated with respect to the independent variable. In mathematical notation, if the highest derivative present in the equation is the n-th derivative, the order of the differential equation is n.

Degree of a Differential Equation: The degree of a differential equation is the highest power to which the highest-order derivative is raised, provided that the equation is a polynomial equation in derivatives. In other words, it is the exponent of the highest-order derivative term in the differential equation, assuming the equation can be expressed as a polynomial in its derivatives. Note that the degree is only well-defined for differential equations in which all terms are algebraic expressions involving the dependent variable, its derivatives, and the independent variable.

Hence, the value of a is 2 and b is 1, then value of 2a + 6b = 2(2) + 6(1) = 10 .

Concept:

Differential Equations by Variable Separable Method
If the coefficient of dx is the only function of x and coefficient of dy is only a function of y in the given differential equation then we can separate both dx and dy terms and integrate both separately.
∫f(x)dx = ∫g(y)dy

Calculation:
Given: xdy - ydx = 0
xdy = ydx
dy/y = dx/x
Integrating both sides, we get

ln y = ln x + ln c
Since ln x + ln y = ln (xy) will be:
⇒ ln(y) = ln cx
⇒ y = cx
Solution of the differential equation represents straight line passing through origin.

Concept:

Calculation:

Multiplying and dividing by sin x

Put cot x = t

⇒ dt = - coesc2 x

Hence the integral becomes

∴ The value of the integral is .

Given:

and

Concept:
Use concept of sum of matrix in which add elements of same places of both matrix.
determinant of matrix is expansion of matrix with respect to any one row or column.

Calculation:


A + B = 2(60 - 63) - 3(50 - 56) + 4(45 - 48)
A + B = - 6 + 18 - 12
A + B = 0
Hence the option (2) is correct.

Option 3 is correct answer.

It is an optimization method for a linear objective function and a system of linear inequalities or equations.

The linear inequalities or equations are known as constraints. The quantity which needs to be maximized or minimized (optimized) is reflected by the objective function.

The fundamental objective of the linear programming model is to look for the values of the variables that optimize (maximize or minimize) the objective function.

Multiple techniques can be used to solve a linear programming problem. These techniques include:

Simplex method
Solving the problem using R
Solving the problem by employing the graphical method
Solving the problem using an open solver

Concept Used:

Any matrix M can be written as the sum of symmetric and skew-symmetric matrix i.e.,

A = S + K

Where S is skew-symmetric which is given by and K is the symmetric part which is given as .

Explanation:

Given Matrix

⇒

⇒

skew-symmetric part of the matrix is

Thus, the skew-symmetric part of the matrix is

⇒

⇒

Hence the option (4) is correct.

Given:

and such that A2 = B

Concept:

Two matrices are equal when all positional elements are equal for both matrices.

Calculation:

and

Then

Given A2 = B then

⇒ α² = 4 and α - 1 = 1

⇒ α = ± 2 and α = 2

Then α = 2

Hence the option (3) is correct.

Consider y=2(tan⁡x)^cot⁡x
Applying log in both sides,
log⁡y=log⁡2(tan⁡x)^cot⁡x
log⁡y=log⁡2+log⁡(tan⁡x)^cot⁡x
log⁡y=log⁡2+cot⁡x log⁡(tan⁡x)
Differentiating both sides with respect to x, we get

dy/dx = 2(tan⁡x)^cot⁡x (−csc^2xlog(tanx)+cot^2x+1)
dy/dx = 2(tan⁡x)^cot⁡x (−csc^2xlog(tanx)+csc^2x)
dy/dx = 2(tan⁡x)^cot⁡x (csc²⁡x (1-log⁡(tan⁡x))
∴ dy/dx =2 csc^2⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))

Given that, y=4x⁴+2x
dy/dx = 16x³+2
d²y/dx² =48x²
48x²−96x³−12
=12(4x²-8x-1)

Consider y= 
y=5x³ (∴log⁡e^x=x)

∴ dy/dx = 5(3x²)=15x²

The marginal cost is given by the rate of change of revenue.
Hence, (dN(x)/dt) =0.18x²-0.02x+10.
∴ = 0.18(5)²-0.02(5)+10
= 4.5-0.1+10
= Rs. 14.4

Let the cost of apples per kg be x and the cost of oranges per kg be y

From the given information, we have 8x + 3y = 70 and 10x + 6y = 90

Solve the equations simultaneously to find x=3, y=2

Consider y=(3 cos⁡x)^x
Applying log on both sides, we get
log⁡y=log⁡(3 cos⁡x)^x
log⁡y=x log⁡(3 cos⁡x)
log⁡y=x(log⁡3+log⁡(cos⁡x))
Differentiating both sides with respect to x, we get

 log⁡3+log⁡(cos⁡x)-x tan⁡x
dy/dx =y(log⁡(3 cos⁡x)-x tan⁡x)
dy/dx =(3 cos⁡x)^x (log⁡(3 cos⁡x)-x tan⁡x)

Given that, y=e^2x+sin^-1⁡e^x

Consider y=7 

Let the length be l, width be b and the area be A.
The Area is given by A=lb

Given that, dl/dt =4cm/s and dA/dt =8 cm/s
Substituting in the above equation, we get

Given that, l=4 cm and b=1 cm

If A is a singular matrix, then |A|=0
In this case, if (adj A) B≠O, then solution does not exist and the system of equations is called inconsistent.

Consider y= 
Applying log to both sides, we get
log⁡y=log 
log⁡y= 
log⁡y= (1/2) (log⁡(x+1)-log⁡(3x-1))
Differentiating with respect to x, we get

Given that, y=3x²+log⁡(4x)

Consider  log⁡x

Differentiating w.r.t x by using chain rule, we get

Given that, 
y=3x²-2x+7

4=6x-2
6x=6
⇒ x=1

The given system of equations can be expressed in the form of AX=B,
⇒X=A^-1 B
We know that, A^-1=1/|A| adj A

  

Consider y= 
Applying log on both sides, we get
log⁡y=3e^3x log⁡x
Differentiating both sides with respect to x, we get

Given that, 

By using u.v rule, we get

Consider y= 
Differentiating w.r.t x by using chain rule, we get

Let the volume of cube be V.
V=x³