Mathematics: CUET Mock Test - 4 - CUET MCQ

The product of any matrix with itself can be found only when it is a square matrix.i.e. m = n.

y = Acos(αx) + Bsin(αx)

dy/dx = -Aαsin(αx) + Bαcos(αx)

d²y/dx² = -Aα²cos(αx) - Bα²sin(αx)

= -α²(Acos(αx) + Bsin(αx))

= -α² * y

d²y/dx² + α²*y = 0

Concept:

The probability density function for any f(x) is defined as:

Calculation:

Explanation:

If X is a random variable with mean μ, then the variance of X, denoted by Var (X) is given by:

Var (X) = E[(X - μ)]², where μ = E(X)

For a discrete random variable X, the variance of X is obtained as follows:

where the sum is taken all over the values of x for which pX(x) > 0. So the variance of X is the weighted average of the squared deviations from the mean μ, where the weights are given by the probability function pX(x) of X.

Important Points

• The standard deviation of X is defined as the square root of the variance.
• The variance cannot be negative, because it is an average of squared quantities.

• Var (X) is often denoted as σ².

Hence, If X is a random variable with mean μ, then the variance of X, denoted by Var (X) is given by Var (X) = E[(X - μ)]²

The number of rows (m) and the number of columns (n) in the given matrix A=  is 2. Therefore, the order of the matrix is 2×2(m×n)

Given that,  
Separating the variables, we get

log⁡(y - 3) = log⁡(x - 3) + log⁡C₁
log⁡(y - 3) - log⁡(x - 3) = log⁡C₁
 = log⁡C₁
y-3 = C₁(x-3)
This is the general solution for the given differential equation where C₁ is a constant

Consider the function y = 2x
Differentiating w.r.t x, we get
y’= dy/dx = 2
Substituting in the equation xy’-y, we get
xy’- y = x(2) - 2x = 2x - 2x = 0
Therefore, the function y = 2x is a solution for the differential equation xy’ - y = 0.

Consider the function y = log⁡x
Differentiating w.r.t x, we get

Differentiating (1) w.r.t x, we get

∴ 
= 

Consider the function y = 7x²
Differentiating w.r.t x, we get
dy/dx = 14x
⇒ dy/dx -14x = 0
Hence, the function y = 7x² is a solution for the differential equation dy/dx - 14x = 0

• (A) Derivative of Composite Functions → (I) A method used to differentiate a function formed by composing two or more functions, such as using the chain rule.
• (B) Exponential Functions → (II) The derivative of the exponential function e^x is e^x.
• (C) Second-Order Derivative → (III) A derivative that provides information about the concavity of the function’s graph.
• (D) Implicit Differentiation → (IV) A method used to differentiate equations where variables are not explicitly defined, often involving the use of the chain rule.

• (A) ∫0 to 1 x dx: The integral of x from 0 to 1 gives 1/2 (which is (III)).
• (B) ∫0 to 2 (x + 2) dx: The integral of (x + 2) from 0 to 2 gives 6 (which is (IV)).
• (C) ∫0 to 2 x² dx: The integral of x² from 0 to 2 gives 8/3 (which is (I)).
• (D) ∫1 to 2 (x³ + x) dx: The integral of (x³ + x) from 1 to 2 gives 9 (which is (II)).

Consider the function y = e^-2x
Differentiating both sides w.r.t x, we get

dy/dx = -2y
⇒ 

We have xy = ae^x + be^-x ……(1)
Differentiating (1) with respect to x, we get
x(dy/dx) + y = ae^x + be^-x …..(2)
Differentiating (2) now, with respect to x, we get
x(d²y/ dx²) + dy/dx + dy/dx = ae^x + be^-x
From (1),
ae^x + be^-x = xy, so that we get
x(d²y/ dx²) + 2(dy/dx) = xy
which is the required differential equation.

Concept:

Matrix A of dimension n x n is called invertible if and only if

• There exists another matrix B of the same dimension, such that AB = BA = I, where I is the identity matrix of the same order.
• Matrix A is invertible if its inverse exists or |A| ≠ 0 for which AA-1 = I
• Matrix B is known as the inverse of matrix A.
• Inverse of matrix A is symbolically represented by A^-1.
• If the determinant of the matrix is zero, then it will not have an inverse; the matrix is then said to be singular.

Calculation:

Let A be a 2 × 2 symmetric matrix with integer entries is given by

Now checking the statements,

the first column of A is the transpose of the second row of A.

and are transpose

So, =

⇒ a = b = c

⇒

⇒ |A| = 0

Hence, statement 1 is wrong.

Second row of A is the transpose of the first column of A

and and are transpose

So, =

⇒ a = b = c

⇒

⇒ |A| = 0

Hence, statement 2 is wrong.

A is a diagonal matrix with nonzero entries in the main diagonal

​​

⇒

Hence, statement 3 is correct.

Equation of the normal at a point P(x, y) is given by
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x₁ , 0).
From (1), we get
y(dy/dx) = x₁ – x ….(2)
Now, giving that PQ² = k²
Or, x₁ – x + y² = k²
=>y(dy/dx) = ± √(k² – y²) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k² – y²) = ∫-dx
Or, x² + y² = k² passes through (0, k)
Thus, it is a circle.

(D + 1)²y = 0
Or, (D² + 2D+ 1)y = 0
⇒ d²y/dx² + 2dy/dx + y = 0 ……….(1)
Let y = e^mx be a trial solution of equation (1). Then,
⇒ dy/dx = me^mx and d²y/dx² = m²e^mx
Clearly, y = e^mx will satisfy equation (1). Hence, we have
⇒ m².e^mx + 2m.e^mx + e^mx = 0
Or, m² + 2m +1 = 0 (as, e^mx ≠ 0) ………..(2)
Or, (m + 1)² = 0
⇒ m = -1, -1
So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is
y = (A + Bx)e^-x where A and B are two independent arbitrary constants ……….(3)
Given, y = 2 log_e 2 when x = log_e 2
Therefore, from (3) we get,
2 log_e 2 = (A + B log_e2)e^-x
Or, 1/2(A + B log_e2) = 2 log _e2
Or, A + B log_e2 = 4 log_e2 ……….(4)
Again y = (4/3) log_e3 when x = log_e3
So, from (3) we get,
4/3 log_e3 = (A + Blog_e3)
Or, A + Blog_e3 = 4log_e3 ……….(5)
Now, (5) – (4) gives,
B(log_e3 – log_e2) = 4(log_e3 – log_e2)
⇒ B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe^-x

• Correct the solution to state: “The equation of the family of parabolas with directrices parallel to the x-axis is y = Ax² + Bx + C, where A, B, C are arbitrary constants.”
• Show differentiation steps:
  • y = Ax² + Bx + C.
  • dy/dx = 2Ax + B.
  • d²y/dx² = 2A.
  • d³y/dx³ = 0.

Since, x = a cos(αt + β)
Therefore, dx/dt = -a α sin(αt + β)
And, d²x/dt² = -a α² cos(αt + β)
= -α² a cos(αt + β)
Or, d²x/dt² = -α²x [as a cos(αt + β) = x]
So, d²x/dt² + α²x = 0

The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))² – 4dy/dx = 0
Or, x²(dy/dx)² – 2(xy – 2)dy/dx + y² = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x²
Let, 1 – xy = t²
⇒ x(dy/dx) + y = -2t(dt/dx)
⇒ x²(dy/dx) = t² – 1 – 2tx(dt/dx), so that (3) gives
t(x(dt/dx) – (t ± 1)) = 0
Hence, either t = 0
⇒ xy = 1 which is satisfied by (1, 1)
Or, x dt/dx = t ± 1
⇒ dx/x = dt/t ± 1
⇒ t ± 1 = cx
For x = 1, y = 1 and t = 0
⇒ c = ± 1, so the solution is
t = ± (x – 1) => t² = (x – 1)²
Or, 1 – xy = x² – 2x + 1
Or, x + y = 2
Thus, the two curves that satisfies are xy = 1 and x + y = 2

dy/dx = (6x + 9y – 7)/(2x + 3y – 6)
So, dy/dx = (3(2x + 3y) – 7)/(2x + 3x – 6) ……….(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx – 2)
Therefore, from (1) we get,
1/3(dz/dx – 2) = (3z – 7)/(z – 6)
Or, dz/dx = 2 + (3(3z – 7))/(z – 6)
= 11(z – 3)/(z – 6)
Or, (z – 6)/(z – 3) dz = 11 dx
Or, ∫(z – 6)/(z – 3) dz = ∫11 dx
Or, ∫(1 – 3/(z – 3)) dz = 11x + c
Or, z – log |z – 3| = 11x + c
Or, 2x + 3y – 11x – 3log|2x + 3y -3| = c
Or, 3y – 9x – 3log|2x + 3y – 3| = c
Or, 3x – y + log|2x + 3y – 3| = -c/3

Given, d²y/dx² = e^2x(12 cos3x – 5 sin3x) ………(1)
Integrating (1) we get,
dy/dx = 12∫ e^2xcos3x dx – 5∫ e^2x sin3x dx
= 12 * (e^2x/2² + 3²)[2cos3x + 3sin3x] – 5 * (e^2x/2² + 3²)[2sin3x – 3cos3x] + A (A is integration constant)
So dy/dx = e^2x/13 [24 cos3x + 36 sin3x – 10 sin3x + 15 cos3x] + A
= e^2x/13(39 cos3x + 26 sin3x) + A
⇒ dy/dx = e^2x(3 cos3x + 2 sin3x) + A ……….(2)
Again integrating (2) we get,
y = 3*∫ e^2x cos3xdx + 2∫ e^2x sin3xdx + A ∫dx
y = 3*(e^2x/2² + 3²)[2cos3x + 3sin3x] + 2*(e^2x/2² + 3²)[2sin3x – 3cos3x] + Ax + B (B is integration constant)
y = e^2x/13(6 cos3x + 9 sin3x + 4 sin3x – 6 cos3x) + Ax + B
or, y = e^2x sin3x + Ax + B

The original given equation is,
y = ae^3x + be^x ……….(1)
Differentiating the above equation, we get
dy/dx = 3ae^3x + be^x ……….(2)
d²y/dx² = 9ae^3x + be^x ……….(3)
Now, to obtain the final equation we have to make the RHS = 0,
So, to get RHS = 0, we multiply, (1) with 3 and (2) with -4,
Thus , 3y = 3ae^3x + 3be^x ……….(4)
And -4(dy/dx) = -12ae^3x – 4 be^x ……….(5)
Now, adding the above three equations,(i.e. (3) + (4) + (5)) we get,
d²y/dx² – 4dy/dx + 3y = 0

The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))² – 4dy/dx = 0
Or, x²(dy/dx)² – 2(xy – 2)dy/dx + y² = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x²

(dy/dx)= (dx/dy)-1
So, d²y/dx² = – (dx/dy)^-2 d/dx(dx/dy)
= – (dy/dx)²(d²x/dy²)(dy/dx)
⇒ d²y/dx² + (dy/dx)³ d²x/dy² = 0

 Let 2 tan^-1⁡x = y