Mathematics: CUET Mock Test - 5

The zero-length interval property is one of the properties used in definite integrals and they are always positive. The zero-length interval property is .

= 3(4) – 4
= 8

= – 2(-3)
= 6

= 2(4) – 2
= 6

= 7(e⁶ – e²)

= – 4

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

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 .

In the reverse integral property the upper limits and lower limits are interchanged. The reverse integral 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 c ∈ (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.

The matrix in which the number of columns is greater than the number of rows is called a vertical matrix. There the matrix which follows the condition m>n is a vertical 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

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 c ∈ (0,1) such that f’(a + θh) = 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 c ∈ (0,1) such that f’(a + θh) = 0

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

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

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

Given function is f(x) = tan x on 
F(x) = tan x is not defined at x on 
So, f(x) is not continuous on .
Hence, Rolle’s theorem is not applicable.

Given f(x) = ex, a = 0, b = 1

e^c = e - 1

Differentiation of the function f(x) = sin(x²) is done with chain rule. First we differentiate sin function which becomes cos and then differentiate the inner (x²) which becomes 2x, hence it comes out to be 2xcos(x²).

We know that derivative of tanx is sec²(x), now in the above question we get tan(x + 4), hence its derivative comes out to be sec²(x + 4), as the inside expression (x + 4) is differentiated into 1.
Therefore the answer is sec²(x + 4).

We differentiate the given function with the help of chain rule so we first differentiate the outer function which becomes –sin and then we differentiate the inner function sinx which is differentiated and comes out to be cosx, hence the differentiated function comes out to be -sin (sinx).cosx.

Differentiating on both sides we get 2 + 3dy/dx = cosx.

Given function is y = 
Now taking RHS and substituting x = tang in it and then we get,
y = , Now it becomes the expansion of the function tan3g,
Hence the given function becomes y= tan-1(tan3g). Which is equal to 3g, now substituting the value of g= tan-1x, now after differentiating both sides we get the answer 

It is a standard rule for derivative of a function of this form in which the original power comes in front and the value in the power is decreased by one. Therefore the only option of this form is nx^n-1 from the given options.

We know x-y = 1, hence we differentiate it on both sides:
We get 1- dy/dx = 0, dy/dx = 1, hence the value of dy/dx comes out to be 1.

We differentiate the given function with help of chain rule and hence the outer function becomes –sin and the inner function is differentiated into 2x, therefore the answer comes out to be -sin (x² + 5).2x.

The derivative of cotx is –cosec²x, as this function has a fixed derivative like sinx has its derivative cosx. Therefore the answer to the above question is –cosec²x.

We differentiate the given function with the help of chain rule and hence
the outer function is differentiated into cos, and the inner function comes out to be a and the constant b becomes 0, which is multiplied to the whole function and the answer comes out to be
⇒ a.cos (ax + b).

 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).

 A function is said to be continuous when it is both left continuous and right continuous. Mathematical expression for a function f is right continuous on (a,b) is lim_x→a+⁡f(x)=f(a).

Minor matrix is not a type of matrix. Scalar, diagonal, symmetric are various type of matrices.

The three conditions required for a function f is said to be continuous on (a,b) if f is continuous at each point of (a,b), f is right continuous at x = a, f is left continuous at x = b.