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 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

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
 = 0
y - 3 = 0 is the general solution for the given differential equation.

Given that, 

Separating the variables, we get
2 cos⁡y dy = 3 sin⁡x dx
Integrating both sides, we get
∫ 2 cos⁡y dy = ∫ 3 sin⁡x dx
2 sin⁡y = 3(-cos⁡x) + C
3 cos⁡x + 2 sin⁡y = C

Given that, 
Separating the variables, we get
dy = (3e^x + 2)dx
Integrating both sides, we get
 -----(1)
y = 3e^x + 2x + C which is the general solution of the given differential equation.

Given that, 


Separating the variables, we get
dy = (4x - 2)² dx
Integrating both sides, we get



Given that, y = 1/3 when x = 1
Therefore, equation (1) becomes,


Hence, the particular solution for the given differential solution is y = 

Separating the variables, we get

Integrating both sides, we get

First, for integrating logy/y
Let log⁡y = t
Differentiating w.r.t y, we get
1/y dy=dt
∴ 
= 
Hence, equation (1), becomes

Given y = 2, we get x = 2
Substituting the values in equation (2), we get

C = 0
Therefore, the equation becomes

∴ (log⁡y)²  - (log⁡x)² = 0.

Given that, 
Separating the variables, we get
7y dy = 3x² dx
Integrating both sides, we get



Given that y = 1, when x = 1
Substituting the values in equation (1), we get


Hence, the particular solution of the given differential equation is:

⇒ 7y² = 2x³ + 5

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

dy/dx = -2y
⇒ 

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 = 5 cos⁡3x
Differentiating w.r.t x, we get
y’ = dy/dx = -15 sin⁡3x
Differentiating again w.r.t x, we get
y” = d²y/dx² = -30 cos⁡3x
⇒ y” + 6y=0.
Hence, the function y = 5 cos⁡3x is a solution for the differential equation y” + 6y = 0.

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

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

∴ 
=