Mathematics: CUET Mock Test - 2 - CUET MCQ

We have, x = a + bt + ct²  ……….(1)
Let, v and f be the velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct²)/dt = b + ct   ……….(2)
And f = dv/dt = d(b + ct)/dt = c  ……….(3)
Clearly, when c = 0, then f = 0 that is, acceleration of the particle is zero.
Hence in this case the particle moves with an uniform velocity.

We have, t = x²/2 + x
Therefore, dt/dx = 2x/2 + 1 = x + 1
Thus, if v be the velocity of the particle at time t, then
v = dx/dt = 1/(dt/dx)
= 1/(x + 1) = (x + 1)^-1
Thus dv/dt = d((x + 1)^-1)/dt
= (-1)(x + 1)^-2 d(x + 1)/dt
= -1/(x + 1)² * dx/dt
As, 1/(x + 1) = dx/dt,
So, -(dx/dt)²(dx/dt)
Or dv/dt = -v²*v  [as, dx/dt = v]
= -v³
We know, dv/dt = acceleration of a particle.
As, dv/dt is negative, so there is a retardation of the particle.
Thus, the retardation of the particle = -dv/dt = v³ = cube of the particle.

We assume that the particle moves with uniform acceleration 2f m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b   ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft² + bt + a   ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21   ……….(3)
16f + 4b + a = 43   ……….(4)
49f + 7b + a = 91  ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t² + 5t + 7
Putting t = 3, f = 1 and b = 5 in (1),
We get, the velocity of the particle in 3 seconds,
= [v]_{t = 3} = (2*1*3 + 5)m/sec = 11m/sec.

We have, t = ax² + bx + c  ……….(1)
Differentiating both sides of (1) with respect to x we get,
dt/dx = d(ax² + bx + c)/dx = 2ax + b
Thus, v = velocity of the particle at time t
= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)^-1  ……….(2)
Initially, when t = 0 and v = u, let x = x₀; hence, from (1) we get,
ax₀² + bx₀ + c = 0
Or ax₀² + bx₀ = -c   ……….(3)
And from (2) we get, u = 1/(2ax₀ + b)
Thus, 1/v² – 1/u² = (2ax + b)² – (2ax₀ + b)²
= 4a²x² + 4abx – 4a²x₀² – 4abx₀
= 4a²x² + 4abx – 4a(ax₀² – bx₀)
= 4a²x² + 4abx – 4a(-c)   [using (3)]
= 4a(ax² + bx + c)
Or 1/v² – 1/u² = 4at

Let, x be the distance travelled by the particle in time t seconds.
Then, v = dx/dt = 3t² – 4t + 5
Or ∫dx = ∫ (3t² – 4t + 5)dt
So, on integrating the above equation, we get,
x = t³ – 2t² + 5t + c where, c is a constant.  ……….(1)
Therefore, the distance travelled by the particle at the end of 3 seconds,
= [x]_{t = 3} – [x]_{t = 0}
= (3³ – 2*3² + 5*3 + c) – c [using (1)]
= 24 cm.

We assume that the particle moves with uniform acceleration 2f m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b  ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft² + bt + a   ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21  ……….(3)
16f + 4b + a = 43  ……….(4)
49f + 7b + a = 91   ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t² + 5t + 7
Therefore, the distance described by the particle in 3 seconds,
= [x]_{t = 3} = (3² + 5*3 + 7)m = 31m

Since the particle is moving in a straight line under a retardation kv³, hence, we have,
dv/dt = -kv³   ……….(1)
Or dv/v³ = -k dt
Or ∫v^-3 dv = -k∫dt
Or v^-3+1/(-3 + 1) = -kt – c [c = constant of integration]
Or 1/2v² = kt + c   ……….(2)
Given, u = v when, t = 0; hence, from (2) we get,
1/2u² = c
Thus, putting c = 1/2u² in (2) we get,
1/2v² = kt + 1/2u²
Or 1/v² = 1/u² + 2kt

Let v be the velocity of the particle at time t seconds after start (that is at a distance s from O). Then,
v = ds/dt = d[(t – 1)²(t – 2)²]/dt
Or v = (t – 2)(3t – 4)
Clearly, v = 0, when (t – 2)(3t – 4) = 0
That is, when t = 2
Or 3t – 4 = 0 i.e., t = 4/3
Now, s = (t – 1)(t – 2)²
Therefore, when t = 4/3, then s = (4/3 – 1)(4/3 – 2)² = 4/27
And when t = 2, then s =(2 – 1)(2 – 2)² = 0
Therefore, the velocity of the particle is zero, when its distance from O is 4/27 units and when it is at O.

Let, y = sin(x + α)/sin(x + β)
Then,
dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin²(x + β)
= sin(x+β – x-α)/sin²(x + β)
Or sin(β – α)/sin²(x + β)
So, for minimum or maximum value of x we have,
dy/dx = 0
Or sin(β – α)/sin²(x + β) = 0
Or sin(β – α) = 0 ……….(1)
Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).
Therefore, y has neither a maximum or minimum value.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x² – 24x + 45
3x² – 24x + 45 = 0
Or x² – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.
Putting, x = 3 in (1)
Thus, its maximum value is,
f(3) = 3³ – 12*3² + 45*3 + 8 = 62.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x² – 24x + 45
3x² – 24x + 45 = 0
Or x² – 8x + 15 = 0
Or(x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.
Putting, x = 5 in (1)
Thus, its maximum value is,
f(3) = 5³ – 12*5² + 45*5 + 8 = 58.

Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t³/12 – 6t²/3 + 6t/2 + 1
So, v = dx/dt = t³/3 – 2t²/ + 3t + 1
Thus, dv/dt = t² – 4t + 3
And d²v/dt² = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t² – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d²v/dt²]_{t = 3} = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x² – 24x + 45
3x² – 24x + 45 = 0
Or x² – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x² – 24x + 45
3x² – 24x + 45 = 0
Or x² – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.

The given function is f(x) = ∣x − 1∣, x ∈ R.
It is known that a function f is differentiable at point x = c in its domain if both
 − hf(c + h) – f(c)
And
l + hf(c + h) – f(c) are finite and equal.
To check the differentiability of the function at x = 1,
LHS,
Consider the left hand limit of f at x=1

= 
= 
= −1
RHS,
Consider the right hand limit of f at x − 1

= 
= 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.
As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.
Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.
Therefore, f(x) has a local minima at x = 1.

Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t³/12 – 6t²/3 + 6t/2 + 1
So, v = dx/dt = t³/3 – 2t²/ + 3t + 1
Thus, dv/dt = t² – 4t + 3
And d²v/dt² = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t² – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d²v/dt²]_{t = 3} = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Putting t = 3 in (1) we get,
3³/3 – 2(3)²/ + 3(3) + 1
= 1 cm/sec.

The equation of the given parabola is, y² = 5x ……….(1)
Differentiating both sides of (1) with respect to y, we get,
2y = 5(dx/dy)
Or dx/dy = 2y/5
Take any point P((5/4)t², (5/2)t). Then, the normal to the curve (1) at P is,
-[dx/dy]_P = -(2*5t/2)/5 = -t
By the question, slope of the normal to the curve (1) at P is tan45°.
Thus, -t = 1
Or t = -1
So, the required equation of normal is,
y – 5t/2 = -t(x – 5t²/4)
Simplifying further we get,
4(x – y) = 15

Given, y² = 3x ……….(1) and y = 2x + 4 ……….(2)
Differentiating both sides of (1) with respect to y we get,
2y = 3(dx/dy)
Or dx/dy = 2y/3
Let P (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is
-[dx/dy]_P = -2y₁/3
If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,
-2y₁/3*2 = -1
Since the slope of the line (2) is 2
Or y₁ = 3/4
Since the point P(x₁, y₁) lies on (1) hence,
y₁² = 3x₁
As, y₁ = 3/4, so, x₁ = 3/16
Therefore, the required equation of the normal is
y – y₁ = -(2y₁)/3*(x – x₁)
Putting the value of x₁ and y₁ in the above equation we get,
16x + 32y = 27.

The given equation of the circle is,
2x² + 2y² + 5 x + y – 15 = 0
Or x² + y² + 5/2 x + y/2 – 15/2 = 0 ………..(1)
The required length of the tangent drawn from the point (7, 2) to the circle (1) is,
√(7² + 2² + 5/2 (7) + 1/2 – 15/2)
= √(49 + 4 + 35/2 + 1 – 15/2)
= √64
= 8 units.

The equation of any circle through the points of intersection of the given circle is,
x² + y² + 2x + 4y + 1 + k(x² + y² – 1) = 0
x² + y² + 2x(1/(k + 1)) + 2*2y/(k + 1) + (1 – k)/(1 + k) = 0
Clearly, the co-ordinates of the center of the circle (1) are, (-1/(1 + k), -2/(1 + k)) and its radius,
= √[(1/(1 + k))² + (2/(1 + k))² – ((1 – k)]/(1 + k))
= √(4 + k²)/(1 + k)
Clearly, the line x + 2y + 5 = 0 is tangent to the circle (1), hence, the perpendicular distance of the line from the center of the circle = radius of the circle
± (-1/(1 + k))– 2(2/(1 + k)) + 5/ √(1² + 2²) = √(4 + k²)/(1 + k)
Or ±(5k/√5) = √(4 + k²)
Or 5k² = 4 + k²
Or 4k² = 4
Or k = 1 [as, k ≠ -1]
Putting k = 1 in (1), equation of the given circle is,
x² + y² + x + 2y = 0

We have, x²/a + y²/b = 1 ……….(1)
and
x²/c + y²/d = 1 ……….(2)
Let, us assume curves (1) and (2) intersect at (x₁, y₁). Then
x₁²/a + y₁²/b = 1 ……….(3)
and
x₁²/c + y₁²/d = 1 ……….(4)
Differentiating both side of (1) and (2) with respect to x we get,
2x/a + 2y/b(dy/dx) = 0
Or dy/dx = -xb/ya
Let, m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x₁, y₁); then,
m₁ = [dy/dx]_{(x1, y1)} = -(bx₁/ay₁) and m₂ = [dy/dx]_{(x1, y1)} = -(dx₁/cy₁)
By question as the curves (1) and (2) intersects at right angle, so, m₁m₂ = -1
Or -(bx₁/ay₁)*-(dx₁/cy₁) = -1
Or bdx₁² = -acy₁² ……….(5)
Now, (3) – (4) gives,
bdx₁²(c – a) = acy₁²(d – b) ……….(6)
Dividing (6) by (5) we get,
c – a = d – b
Or a – b = c – d.

The equation of the given parabola is, y² = 5x ……….(1)
Differentiating both sides of (1) with respect to y, we get,
2y = 5(dx/dy)
Or dx/dy = 2y/5
Take any point P((5/4)t², (5/2)t). Then, the normal to the curve (1) at P is,
-[dx/dy]_P = -(2*5t/2)/5 = -t
By the question, slope of the normal to the curve (1) at P is tan45°.
Thus, -t = 1
Or t = -1
So, the required equation of normal is,
y – 5t/2 = -t(x – 5t²/4)
Simplifying further we get,
4(x – y) = 15
The co-ordinates of the foot of the normal are, P((5/4)t², (5/2)t).
As t = 1, so putting the value of t = 1, we get,
P = (5/4, 5/2).

Given, y² = 3x ……….(1) and y = 2x + 4 ……….(2)
Differentiating both sides of (1) with respect to y we get,
2y = 3(dx/dy)
Or dx/dy = 2y/3
Let P (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is
-[dx/dy]_P = -2y₁/3
If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,
-2y₁/3*2 = -1
Since the slope of the line (2) is 2
Or y₁ = 3/4
Since the point P(x₁, y₁) lies on (1) hence,
y₁² = 3x₁
As, y₁ = 3/4, so, x₁ = 3/16
Therefore, the required equation of the normal is
y – y₁ = -(2y₁)/3*(x – x₁)
Putting the value of x₁ and y₁ in the above equation we get,
16x + 32y = 27
And the coordinates of the foot of the normal are (x₁, y₁) = (3/16, 3/4)

Given, x² + 3y² = 12 Or x²/12 + y²/4 = 1
Differentiating both sides of (1) with respect to y we get,
2x*(dx/dy) + 3*2y = 0
Or dx/dy = -3y/x
Suppose the normal to the ellipse (1) at the point P(√12cosθ, 2sinθ) makes an angle 60° with the major axis. Then, the slope of the normal at P is tan60°
Or -[dx/dy]_P = tan60°
Or -(-(3*2sinθ)/√12cosθ) = √3
Or √3tanθ = √3
Or tanθ = 1
Now the centre of the ellipse (1) is C(0, 0)
Therefore, the slope of the line CP is,
(2sinθ – 0)/(√12cosθ – 0) = (1/√3)tanθ = 1/√3 [as, tanθ = 1]
Therefore, the line CP is inclined at 30° to the major axis.

x² – y² = 2a² ……….(1) and x² + y² = 4a² ……….(2)
Adding (1) and (2) we get, 2x² = 6a²
Again, (2) – (1) gives,
2y² = 2a²
Therefore, 2x² * 2y² = 6a² * 2a²
4x²y² = 12a²
Or x²y² = 3a⁴
Or 2xy = ±2√3
Differentiating both side of (1) and (2) with respect to x we get,
2x – 2y(dy/dx) = 0
Or dy/dx = x/y
And 2x + 2y(dy/dx) = 0
Ordy/dx = -x/y
Let (x, y) be the point of intersection of the curves(1) and (2) and m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x, y); then,
m₁ = x/y and m₂ = -x/y
Now the angle between the curves (1) and (2) means the angle between the tangents to the curve at their point of intersection.
Therefore, if θ is the required angle between the curves (1) and (2), then
tanθ = |(m₁ – m₂)/(1 + m₁m₂)|
Putting the value of m₁, m₂ in the above equation we get,
tanθ = |2xy/(y² – x²)|
As, 2xy = ±2√3a² and x² – y² = 2a²
tanθ = |±2√3a²/-2a²|
Or tanθ = √3
Thus, θ = π/3.

Bernoulli trials is also called a Dichotomous experiment and is repeated n times. If in each trial the probability of success is constant, then such trials are called Bernoulli trails.

The above is a condition for a symmetric relation.
For example, a relation R on set A = {1,2,3,4} is given by R={(a,b):a+b=3, a>0, b>0}
1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.
Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each other and is a part of the relation. Hence, they are symmetric.

Bernoulli trials is also called a Dichotomous experiment and is repeated n times. Bernoulli trials is also called as a ‘yes’ or ‘no’ questions because it has only two outcomes, those are ‘success’ and ‘failure’.

We know that sin(x) = sin(2A * π + x) where A can be positive or negative integer.
If A is -1, then sin(6) = sin(-2π + 6);
If A is 1, then sin(6) = sin(2π + 6);

Formula for binomial distribution is P [X = x] = ⁿC_x p^x q^n-x
Where n is the number of trials, x is the number of successes and (n-x) is failures.