Mathematics: CUET Mock Test - 5 - CUET MCQ

Concept:

Objective function: Linear function Z = ax + by, where a, b are constants, which has to be maximized or minimized is called a linear objective function.
In the above example, Z = ax + by is a linear objective function. Variables x and y are called decision variables.

By putting values of variables (coordinates of the point) in linear objective function we get the value of the point.

Calculations:

Given, Objective function for all LPP is z = 2x + 3y

Putting coordinates of points in the equation we get value of the point

e.g for corner point (0, 2)

z = 2x + 3y = 2 × 0 + 3 × 2 = 6

The difference of the maximum and minimum values of z is = 36 - 6 = 30

Concept:

The solution of the linear differential equation is given by

y × I.F =

Where I.F =

Explanation -

we have ⇒

Now integrating factor is I.F. =

Now the solution of the differential equation is -

⇒

⇒

⇒

⇒

Hence the option (i) is true.

The answer is C. We eliminate constants.

We have

y=ax+x²

Differentiating with respect to x,

(a² + b²)×1 - (2a) × b
= a² +b² - 2ab
= (a - b)²

Determinant = [(a² + b²).1 - (-2ab)]
= (a+b)²

The modulus function has V-shaped graph,which means that it has only one minima.

Concept:

We know, for valid PDF :

Calculation:

The graph of the pdf is given in triangular form with the base length as 3 and the height as λ.

We know, for valid PDF :

The above is representing the area under the curve between x ϵ [-2, 1].

⇒ 1/2 (base) × Height = 1 

⇒ 1/2 (1- (-2)) × λ = 1 

⇒ λ = 2/3

We have , 

Therefore, range of f(x) is {-1}.

For even function: f(-x) = f(x) , 
therefore, f(− x)
 = sin (− x)² = sin x² = f(x).

Acceleration is the derivative of velocity, so we integrate a(t) to get v(t):

v(t)=∫a(t)dt=∫sin(t)dt=−cos(t)+C

Now, we are given that the velocity at t=0 is 10 km/hr. We can use this information to find the constant C:

v(0)=−cos(0)+C=−1+C=10

Solving for C, we get C=11.

Now, we have the velocity function:

v(t)=−cos(t)+11

Finally, we integrate v(t) to get the displacement function s(t):

s(t)=∫v(t)dt=∫(−cos(t)+11)dt

s(t)=−sin(t)+11t+D

Now, we need to find the constant D. We are given that the car moves from t=0 to t=π/2, and we know that s(0)=0 (starting position). Plugging in these values, we can solve for D:

s(0)=−sin(0)+11(0)+D=0

D=0

So, the displacement function is:

s(t)=−sin(t)+11t

Now, to find the distance traveled, we evaluate s(t) over the given time interval:

Distance=s(π/2​)−s(0)

Distance=(−sin(π/2​)+11(π/2​))−(−sin(0)+11(0))

Distance=−1+11π​/2 = 16.27887 kilometers

Therefore, the distance traveled by the car from t=0 to t=π/2​ is 16.27887 kilometers​.

Order of the D.E. is 3
Order of a differential equation is the order of the highest derivative present in the equation.

Equation of ellipse :

 x²/a² + y²/b² = 1

Differentiation by x,

2x/a² + (dy/dx)*(2y/b²) = 0

dy/dx = -(b²/a²)(x/y)

-(b²/a^2) = (dy/dx)*(y/x) ----- eqn 1

Again differentiating by x,

d²y/dx² = -(b²/a²)*((y-x(dy/dx))/y²)

Substituting value of -b²/a² from eqn 1

d²y/dx² = (dy/dx)*(y/x)*((y-x(dy/dx))/y²)

d²y/dx² = (dy/dx)*((y-x*(dy/dx))/xy)

(xy)*(d²y/dx²) = y*(dy/dx) - x*(dy/dx)²

(xy)*(d²y/dx²) + x*(dy/dx)²- y*(dy/dx) = 0

Given:

Line and point (3, 4, 5)

Concept:

If two vectors are perpendicular to each other then dot product of both is zero.

Calculation:

Let point A = (3, 4, 5)

and

Then point B = (k, 2k + 1, 3k + 2) on the line

Now, the line AB = B - A = (k - 3, 2k - 3, 3k - 3) .

DRs of given line (1, 2, 3)

We know that the Line AB is perpendicular to the given line

Then

(k - 3, 2k - 3, 3k - 3) ⋅ (1, 2, 3) = 0

⇒ k - 3 +4k - 6 + 9k - 9 = 0

⇒ 14k = 18 ⇒

Then line

The perpendicular length from point A on given line is the magnitude of AB.

Hence option (2) is correct.

l2 + m2 + n2 = 1
The cosines l, m, and n of the angles made by a vector with the coordinate axes are also known as the direction cosines of a vector.
For any vector, the sum of the squares of these direction cosines always equals one i.e l² + m² + n² = 1
This is a consequence of the generalization of the Pythagorean theorem known as the squared Euclidean norm (for 3D vectors).
Hence, Option 2 is Correct.

• (A) Cartesian Equation of a Line → (II) Given by the form
• (x - x₁) / a = (y - y₁) / b = (z - z₁) / c, which represents a line passing through (x₁, y₁, z₁) in direction (a, b, c).
• (B) Skew Lines → (I) Two lines that do not intersect and are not parallel, meaning they exist in different planes.
• (C) Distance of a Point from a Plane → (III) The shortest perpendicular distance from a point to a plane, calculated using the equation: d = |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²).
• (D) Angle Between Two Planes → (IV) Determined by the normal vectors of the planes using the dot product formula:
• cos(θ) = (n₁ • n₂) / |n₁| |n₂|.

= – 4

In the reverse integral property the upper limits and lower limits are interchanged. The reverse integral property of definite integrals is .

The adding intervals property of definite integrals is.

The number of elements for a matrix with the order m × n is equal to mn, where m is the number of rows and n is the number of columns in the matrix.

The given matrix  has 3 rows and 2 columns. Therefore, the order of the matrix is 3×2.

According to Rolle’s theorem, if f : [a,b] → R is a function such that

• f is continuous on [a,b]
• f is differentiable on (a,b)
• f(a) = f(b) then there exists at least one point c ∈ (a,b) such that f’(c) = 0

According to Rolle’s theorem, if f : [a,a+h] → R is a function such that

• f is continuous on [a,a+h]
• f is differentiable on (a,a+h)
• f(a) = f(a+h) then there exists at least one θ ∈ (0,1) such that f’(a+θh) = 0

A square matrix is a matrix in which the number of rows(m) is equal to the number of columns(n). Therefore, the matrix which follows the condition m=n is a square matrix.

Rolle’s theorem is just a special case of Lagrange’s mean value theorem when f(a) = f(b) and Lagrange’s mean value theorem is also called the mean value theorem.

According to Lagrange’s mean value theorem, if f : [a,b] → R is a function such that f is differentiable on (a,b) then the formula for Lagrange’s theorem is f’(c) = .

According to Rolle’s theorem, if f : [a,b] → R is a function such that

• f is continuous on [a,b]
• f is differentiable on (a,b)
• f(a) = f(b) then there exists at least one point c ∈ (a,b) such that f’(c) = 0

 function f defined on (a,b) is said to be continuous on (a,b) if it is continuous at every point of (a,b) i.e., if lim_x→c⁡f(x) = f(c) ∀ c ∈ (a,b).