Mathematics: CUET Mock Test - 2

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.

Poisson distribution shows the number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

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.

Bernoulli trials is termed after swiss mathematician Jacob Bernoulli. 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.

Bernoulli trial has only two possible outcomes and is mutually exclusive. Those two outcomes are ‘success’ and ‘failure’. So, it is also called as a ‘yes’ or ‘no’ question.

Poisson distribution shows the number of times an event is likely to occur within a specified time. It is used only for independent events that occur at a constant rate within a given interval of time.

Poisson distribution shows the number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

The equation is cos^-1 x + cos^-1 y + cos^-1 z = 3π
This means cos^-1 x = π, cos^-1 y = π and cos^-1 z = π
This will be only possible when it is in maxima.
As, cos^-1 x = π so, x = cos^-1 π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.

Bayes theorem is the method in which the calculated probabilities are revised with values of new probabilities, whereas Updation theorem, Revision theorem and Dependent theorem are not related to the concept of probability.

Conditional probability P(A | B) indicates the probability of event ‘A’ happening given that event B has happened.
Which in formula can be written as P(A|B) = P(A∩B)/P(B).
Whereas formula’s P(A | B) = P(A∩B)/P(A), P(A | B) = P(A)/P(B), P(A|B) = P(B)/P(A) doesn’t satisfies the specified conditions.

Let E = event that the man reports that six in the throwing of the die and let, S1 = event that six occurs and S2 = event that six does not occur.
P(S1) = Probability that six occurs = 1/6.
P(S2) = Probability that six does not occur = 5/6.
Also, P(E|S1) = P(Probability that man reports six occurs when six actually has occurred on the die) = 3/4.
P(E|S2) = P(Probability that man reports six occurs when six not actually occurred on the die) =
1 – 3/4 = 1/4.
By using Bayes’ theorem,
P(S2 | E) = P(S1)P(E | S1)/(P(S1)P(E│S1)+P(S2)P(E | S2))
= (1/6 × 3/4) / ((1/6 × 3/4)) + (5/6 × 1/4)) = 3/8.