Mathematics: CUET Mock Test - 4 - Commerce MCQ

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/ dx) + dy/dx + dy/dx = ae^x + be^-x
From (1),
ae^x + be^-x = xy, so that we get
x(d²y/ dx) + 2(dy/dx) = xy
which is the required differential equation.

The correct answer is A - II, B - IV, C - I, D - III.

Key Points

A. Quantity index - II. Measures change in quantity of consumption of goods over a specific period of time: The changes in the volume or quantity of goods that are produced, consumed, and sold within a stipulated period are measured using a quantity index number. A relative change for particular quantities of goods is shown across a period by it.
B. Time series - IV. Statistical observation taken at different points of time for specific period of time: A collection of observations of well-defined data items obtained through repeated measurements over time is referred to as a time series. For instance, a time series could be made up of measurements of the value of retail sales each month of the year.
C. Price index - I. Measures relative price change over a period of time: The proportionate or percentage changes in a set of prices over time are measured by a price index. Changes in the prices of household goods and services are measured by a consumer price index (CPI). The real purchasing power of consumers' incomes and their welfare are affected by such changes.
D. Value index - III. Measures average value of goods for specific time period: A measure or ratio called a value index is used to describe the change in a nominal value in relation to its value in the base year. The figure for each point in time shows the percentage that a given value represents at that point in time, compared to its respective value at the base point in time. In active voice, we can say that the value index measures how much a nominal value has changed relative to its base year value. It also provides a point figure for each point in time, indicating the percentage of the given value compared to its base value.Hence, the correct match is A - II, B - IV, C - I, D - III.

The correct answer is 122.14
Key Points

Given that ∑p₀q₀ = 700, ∑p₀q₁ = 1450, ∑p₁q₀ = 855 and ∑p₁q₁ = 1300.
Laspeyer's Price Index number
= =
= 122.14 (approx)

Additional InformationDefinition: The Laspeyres price index is a method used to measure the changes in the cost of a fixed basket of goods and services over time. It is named after its creator, the German economist Étienne Laspeyres.
Calculation: The Laspeyres Index is calculated by dividing the cost of a group of commodities at current prices by the cost of the same group of commodities at base period prices, and then multiplying the result by 100. It should be noted that the base period index number is always set as 100 by the Laspeyres Index, which is used as a benchmark for comparing the changes in prices over time.
Formula: Laspeyer's Price Index number =

The correct answer is 20.3
Key Points
To find the values of a and b using the least square method, we need to minimize the sum of the squared errors between the predicted values and the actual values:

Therefore,

Thus, a + b = 16 + 4.3 = 20.3
Hence, the required answer is 20.3

A. Incorrect. If the discount rate (or required yield) is greater than the coupon rate, the present value of a bond is less than its face value, not more. This is because the fixed payments from the bond (the coupons and the face value at maturity) are being discounted at a higher rate, which reduces their present value.

B. Correct. An annuity that makes a periodic payment that begins on a certain date and continues forever is indeed called a perpetuity.

C. Correct. The issuer of a bond pays interest at fixed intervals at a fixed rate to the bondholder, and this is indeed referred to as the coupon payment.

D. Incorrect. A sinking fund is not a fixed payment made by a borrower to a lender every month to clear off the loan. A sinking fund is a means of repaying funds borrowed through a bond issue through periodic payments to a trustee who retires part of the issue by purchasing the bonds in the open market.

E. Correct. The issuer of a bond repays the principal (the face value) of the bond to the investor at a later date, which is indeed referred to as the maturity date.

Hence, option 3 is correct

Calculations:

Let's evaluate these statements one by one:

A. True. In the flat rate method of calculating EMI (Equated Monthly Installment), the EMI is calculated as (Principal + Total Interest) / Number of Payments. The principal and interest are summed and then divided by the term of the loan in months to obtain the EMI. This method is relatively simple but does not take into account that the principal amount outstanding reduces with each payment.

B. False. The formula for EMI in the reducing balance method, which takes into account that the principal outstanding reduces with each payment, is more complex than that stated. The correct formula is: EMI = P × , where P = Principal loan amount, r = monthly interest rate, n = number of monthly installments.

C. True. In a sinking fund, a fixed amount of money is set aside at regular intervals in a separate account. This is done to repay a debt or replace a future capital asset.

D. False. The formula for yield to maturity (YTM) is more complex than the one stated. The YTM is the discount rate at which the sum of all future cash flows (both interest and principal payments) would equal the current market price of the bond. The precise calculation involves solving for the discount rate in a present value cash flow equation and typically requires numerical methods. The formula given is more of an approximation and is not correctly stated.

Given:
x² + y² = π²
y = sin x

Calculation:

We have,

⇒ x² + y² = π² is a circle of radius π and center at the origin.

Required area = Area of the circle (1st quadrant) -
⇒ Required Area =
⇒ Required Area =
⇒ Required Area =
⇒ Required Area =
⇒ Required Area = sq. units
∴ The required area is sq. units.

Concept:

Integration by Parts:
∫f(x)g(x)dx = f(x)∫g(x)dx − ∫[f′(x)∫g(x)dx]dx
Where f is first function and g is second function.

ILATE: (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent) rule is used to decide which function is first and second function.

Calculation:

Required area =
Using integration by parts, f(x) = x anf g(x) = sin x
⇒ A = [x∫ sin x dx]₀^π - |[∫(∫ sin x dx)dx]_π^2π|
⇒ A = [-x cos x + sin x]0π + [|-x cos x + sin x|]_π^2π
We know that,
sin 0 = sin π = sin 2π = 0 and cos 0 = cos 2π = 1, cos π = -1
Taking the limit of integration
⇒ A = [(π + 0) - (0 + 0) + |(-2π + 0) - (π + 0)|]
⇒ A = π + 3π
∴ Required area is 4π sq. unit.

Concept:
The angle between two planes is the angle between their normal.
If θ is the angle between the planes a₁x + b₁y + c₁z + d₁ = 0
And a₂x + b₂y + c₂z + d₂ = 0, then

Calculation:
On comparing the given equations of the planes with the standard equations
Here, a₁ = 3, b₁ = -6, c₁ = 2, d₁ = -7
a₂ = 2, b₂ = 2, c₂ = -2, d₂ = -5

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.

Concept:

If a plane is parallel to a line then, the sum of the product of direction ratios of the line and the plane is zero.

Calculation:

Given equation of a Line is

Direction Ratio's of line = 3, 4, 5.

By checking the options,

Option 1: Dr's of 2x + 2y + z - 1 = 0 are 2, 2, 1

Product of direction ratios = 3×2 + 4×2 + 5×1 = 6 + 8 + 5 = 19

Option 2: Dr's of 2x - y - 2z + 5 = 0 are 2, -1, -2

Product of direction ratios = 3×2 + 4×-1 + 5×-2 = 6 - 4 - 10 = -8

Option 3: Dr's of 2x + 2y - 2z + 1 = 0 are 2, 2, -2

Product of direction ratios = 3×2 + 4×2 + 5×-2 = 6 + 8 - 10 = 4

Option 4: Dr's of x - 2y + z - 1 = 0 are 1, -2, 1

Product of direction ratios = 3 × 1+ 4 × -2 + 5 × 1 = 3 - 8 + 5 = 0

Hence, option 4 is correct.

The equation of family of parabolas is Ax² + Bx + C = 0 where, A, B, C are arbitrary constant.
By differentiating the equation with respect to x till all the constants get eliminated,
Hence, d³y/dx³ = 0

d²y/dx² – 3dy/dx + 4y = 0 …..(1)
Let, y = e^mx be a trial solution of (1), then,
⇒ dy/dx = me^mx and d²y/dx² = m²e^mx
Clearly, y = e^mx will satisfy equation (1). Hence, we have,
m²e^mx – 3m * e^mx – 4e^mx = 0
⇒ m² – 3m – 4 = 0 (as e^mx ≠ 0) …….(2)
⇒ m² – 4m + m – 4 = 0
⇒ m(m – 4) + 1(m – 4) = 0
Or, (m – 4)(m + 1) = 0
Thus, m = 4 or m = -1
Clearly, the roots of the auxiliary equation (2) are real and unequal.
Therefore, the required general solution of (1) is
y = Ae^4x + Be^-x where A and B are constants.

(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

Given, y = Ax + B/x
⇒ xy = Ax² + B ……….(1)
Differentiating (1) with respect to x, we get,
d(xy)/dx = d/dx(Ax² + B)
or, xdy/dx + y = A * 2x ……….(2)
Differentiating again with respect to x, we get,
x*d²y/dx² + dy/dx + dy/dx = A*2 ……….(3)
Eliminating A from (2) and (3) we get,
x² d² y/dx² + 2xdy/dx = 2Ax  [multiplying (3) by x]
or, x² d² y/dx² + 2xdy/dx = xdy/dx + y [using (2)]
or, x² d² y/dx² + xdy/dx – y = 0

(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
So, C = 4

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

(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

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 integrationconstant)
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

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

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.

Here, y²(dy/dx)² + 4x² + 4xy(dy/dx) = (y² + 2x²)[1 + (dy/dx)²]
⇒ dy/dx = y/x ± √(1/2(y/x)²) + 1
Let, y = vx
⇒ v + x dv/dx = v ± √(1/2(v)²) + 1
Integrating both sides,
±√dv/(√(1/2(v)²) + 1) = ∫dx/x
cx^±1/√2 = y/x + √(y² + 2x²)/x²

Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t² – 5t
Or, dv = 3t² dt – 5tdt
Or, ∫dv = 3∫t² dt – 5∫t dt
Or, v = t³ – (5/2)t² + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t³ – (5/2)t² + 5
Or, dx/dt = t³ – (5/2)t² + 5 ………..(2)
Thus, the velocity of the particle at the end of 4 seconds,
= [v]_{t = 4} = (4³ – (5/2)4² + 5 ) cm/sec [putting t = 4 in (2)]
= 29 cm/sec

The given form of equation can be written as,
dy/dx + 1/√(a² + x²) * y = (√(a² + x²) – x)/√(a² + x²) ……(1)
We have, ∫1/√(a² + x²)dx = log(x + √(a² + x²))
Therefore, integrating factor is,
e^{∫1/√(a2 + x2)} = e^{log(x + √(a2 + x2))}
= x + √(a² + x²)
Therefore, multiplying both sides of (1) by x + √(a² + x²) we get,
x + √(a² + x²dy/dx + (x + √(a² + x²))/ √(a² + x²)*y = (x + √(a² + x²))(√(a² + x²) – x)/√(a² + x²)
or, d/dx[x + √(a² + x²)*y] = (a² + x²) ………..(2)
Integrating both sides of (2) we get,
(x + √(a² + x²) * y = a²∫dx/√(a² + x²)
= a² log (x + √(a² + x²)) + c

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

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 – a/x * y = (x + 1)/x …….(1)
Multiplying both sides of equation (1) by
e^∫-a/xdx
= e^{-a log x}
= e^{log x-a}
= x^-a
We get, x^-ady/dx – x^-a (a/x)y = x^-a (x + 1)/x
Or, d/dx(y . x^-a) = x^-a + x^{-a – 1} …….(2)
Integrating both sides of (2) we get,
y. x^-a = x^{-a + 1}/(-a + 1) + x^{-a – 1 + 1}/(-a -1 + 1) + c
= x^-a.x/(1 – a) + x^-a/-a + c
Or, y = x/(1 – a) – 1/a + cx^a

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

Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t² – 5t
Or, dv = 3t² dt – 5t dt
Or, ∫dv = 3∫t² dt – 5∫t dt
Or, v = t³ – (5/2)t² + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t³ – (5/2)t² + 5
Or, dx/dt = t³ – (5/2)t² + 5 ………..(2)
Or, dx = t³ dt – (5/2)t² dt + 5 dt
Integrating this we get,
x = (1/4)t⁴ – (5/2)t³/3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t⁴ – (5/6)t³ + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]_{t = 4} = (1/4)4⁴ – (5/6)4³ + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm

Here, y²(dy/dx)² + 4x² + 4xy(dy/dx) = (y² + 2x²)[1 + (dy/dx)²]
⇒ dy/dx = y/x ± √(1/2(y/x)²) + 1
Let, y = vx
⇒ v + x dv/dx = v ± √(1/2(v)²) + 1
Integrating both sides,
±∫dv/(√(1/2(v)²) + 1) = ∫dx/x
cx^±1/√2 = y/x + √(y² + 2x²)/x² (put v/√2 = tan t)
putting x = 1, y = 0, we get c = √2
So, the curve is given by,
√2x^±1/√2 = y/x + √(y² + 2x²)/x²