1 Crore+ students have signed up on EduRev. Have you? 
A particle moves in a straightline OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct^{2} (a, b > 0). What will be the nature of motion of the particle when c = 0?
We have, x = a + bt + ct^{2} ……….(1)
Let, v and f be the velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct^{2})/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.
A particle moving in a straight line traverses a distance x in time t. If t = x^{2}/2 + x, then which one is correct?
We have, t = x^{2}/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 = v^{2}*v [as, dx/dt = v]
= v^{3}
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 = v^{3} = cube of the particle.
A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds, respectively.What is the velocity of the particle in 3 seconds?
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 = ft^{2} + 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 = t^{2} + 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.
The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax^{2} + bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?
We have, t = ax^{2} + bx + c ……….(1)
Differentiating both sides of (1) with respect to x we get,
dt/dx = d(ax^{2} + 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 = x_{0}; hence, from (1) we get,
ax_{0}^{2} + bx_{0} + c = 0
Or ax_{0}^{2} + bx_{0} = c ……….(3)
And from (2) we get, u = 1/(2ax_{0} + b)
Thus, 1/v^{2} – 1/u^{2} = (2ax + b)^{2} – (2ax_{0} + b)^{2}
= 4a^{2}x^{2} + 4abx – 4a^{2}x_{0}^{2} – 4abx_{0}
= 4a^{2}x^{2} + 4abx – 4a(ax_{0}^{2} – bx_{0})
= 4a^{2}x^{2} + 4abx – 4a(c) [using (3)]
= 4a(ax^{2} + bx + c)
Or 1/v^{2} – 1/u^{2} = 4at
A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t^{2} – 4t + 5 cm/second. What will be the distance travelled by the particle during first 3 seconds after the start?
Let, x be the distance travelled by the particle in time t seconds.
Then, v = dx/dt = 3t^{2} – 4t + 5
Or ∫dx = ∫ (3t^{2} – 4t + 5)dt
So, on integrating the above equation, we get,
x = t^{3} – 2t^{2} + 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}
= (3^{3} – 2*3^{2} + 5*3 + c) – c [using (1)]
= 24 cm.
A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?
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 = ft^{2} + 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 = t^{2} + 5t + 7
Therefore, the distance described by the particle in 3 seconds,
= [x]_{t = 3} = (3^{2} + 5*3 + 7)m = 31m
A particle moves in a horizontal straight line under retardation kv^{3}, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?
Since the particle is moving in a straight line under a retardation kv^{3}, hence, we have,
dv/dt = kv^{3} ……….(1)
Or dv/v^{3} = k dt
Or ∫v^{3} dv = k∫dt
Or v^{3+1}/(3 + 1) = kt – c [c = constant of integration]
Or 1/2v^{2} = kt + c ……….(2)
Given, u = v when, t = 0; hence, from (2) we get,
1/2u^{2} = c
Thus, putting c = 1/2u^{2} in (2) we get,
1/2v^{2} = kt + 1/2u^{2}
Or 1/v^{2} = 1/u^{2} + 2kt
The distance s of a particle moving along a straight line from a fixedpointO on the line at time t seconds after start is given by x = (t – 1)^{2}(t – 2)^{2}. What will be the distance of the particle from O when its velocity is zero?
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.
What will be the nature of the equation sin(x + α)/sin(x + β)?
Let, y = sin(x + α)/sin(x + β)
Then,
dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin^{2}(x + β)
= sin(x+β – xα)/sin^{2}(x + β)
Or sin(β – α)/sin^{2}(x + β)
So, for minimum or maximum value of x we have,
dy/dx = 0
Or sin(β – α)/sin^{2}(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.
Given, f(x) = x^{3} – 12x^{2} + 45x + 8. What is the maximum value of f(x)?
We have, f(x) = x^{3} – 12x^{2} + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x^{2} – 24x + 45
3x^{2} – 24x + 45 = 0
Or x^{2} – 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) = 3^{3} – 12*3^{2} + 45*3 + 8 = 62.
Given, f(x) = x^{3} – 12x^{2} + 45x + 8. What is the minimum value of f(x)?
We have, f(x) = x^{3} – 12x^{2} + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x^{2} – 24x + 45
3x^{2} – 24x + 45 = 0
Or x^{2} – 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) = 5^{3} – 12*5^{2} + 45*5 + 8 = 58.
A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t^{4}/12 – 2t^{3}/3 + 3t^{2}/2 + t + 15. At what time is the velocity minimum?
Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t^{3}/12 – 6t^{2}/3 + 6t/2 + 1
So, v = dx/dt = t^{3}/3 – 2t^{2}/ + 3t + 1
Thus, dv/dt = t^{2} – 4t + 3
And d^{2}v/dt^{2} = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t^{2} – 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, [d^{2}v/dt^{2}]_{t = 3} = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Given, f(x) = x^{3} – 12x^{2} + 45x + 8. At which point does f(x) has its maximum?
We have, f(x) = x^{3} – 12x^{2} + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x^{2} – 24x + 45
3x^{2} – 24x + 45 = 0
Or x^{2} – 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.
Given, f(x) = x^{3} – 12x^{2} + 45x + 8. At which point does f(x) has its minimum?
We have, f(x) = x^{3} – 12x^{2} + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x^{2} – 24x + 45
3x^{2} – 24x + 45 = 0
Or x^{2} – 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.
At which point does f(x) = x – 1 has itslocal minimum?
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.
A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t^{4}/12 – 2t^{3}/3 + 3t^{2}/2 + t + 15. What is the minimum velocity?
Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t^{3}/12 – 6t^{2}/3 + 6t/2 + 1
So, v = dx/dt = t^{3}/3 – 2t^{2}/ + 3t + 1
Thus, dv/dt = t^{2} – 4t + 3
And d^{2}v/dt^{2} = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t^{2} – 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, [d^{2}v/dt^{2}]_{t = 3} = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Putting t = 3 in (1) we get,
3^{3}/3 – 2(3)^{2}/ + 3(3) + 1
= 1 cm/sec.
What will be the equation of the normal to the parabola y^{2} = 5x that makes an angle 45° with the x axis?
The equation of the given parabola is, y^{2} = 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)t^{2}, (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 – 5t^{2}/4)
Simplifying further we get,
4(x – y) = 15
What will be the equation of the normal to the parabola y^{2} = 3x which is perpendicular to the line y = 2x + 4?
Given, y^{2} = 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 (x_{1}, y_{1}) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is
[dx/dy]_{P} = 2y_{1}/3
If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,
2y_{1}/3*2 = 1
Since the slope of the line (2) is 2
Or y_{1} = 3/4
Since the point P(x_{1}, y_{1}) lies on (1) hence,
y_{1}^{2} = 3x_{1}
As, y_{1} = 3/4, so, x_{1} = 3/16
Therefore, the required equation of the normal is
y – y_{1} = (2y_{1})/3*(x – x_{1})
Putting the value of x_{1} and y_{1} in the above equation we get,
16x + 32y = 27.
What will be the length of a tangent from the point (7, 2) to the circle 2x^{2} + 2y^{2} + 5 x + y = 15?
The given equation of the circle is,
2x^{2} + 2y^{2} + 5 x + y – 15 = 0
Or x^{2} + y^{2} + 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,
√(7^{2} + 2^{2} + 5/2 (7) + 1/2 – 15/2)
= √(49 + 4 + 35/2 + 1 – 15/2)
= √64
= 8 units.
What will be the equation of the circle which touches the line x + 2y + 5 = 0 and passes through the point of intersection of the circle x^{2} + y^{2} = 1 and x^{2} + y2 + 2x + 4y + 1 = 0?
The equation of any circle through the points of intersection of the given circle is,
x^{2} + y^{2} + 2x + 4y + 1 + k(x^{2} + y^{2} – 1) = 0
x^{2} + y^{2} + 2x(1/(k + 1)) + 2*2y/(k + 1) + (1 – k)/(1 + k) = 0
Clearly, the coordinates 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 + k^{2})/(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/ √(1^{2} + 2^{2}) = √(4 + k^{2})/(1 + k)
Or ±(5k/√5) = √(4 + k^{2})
Or 5k^{2} = 4 + k^{2}
Or 4k^{2} = 4
Or k = 1 [as, k ≠ 1]
Putting k = 1 in (1), equation of the given circle is,
x^{2} + y^{2} + x + 2y = 0
If the curves x2/a + y2/b = 1 and x2/c + y2/d = 1 intersect at right angles, then which one is the correct relation?
We have, x^{2}/a + y^{2}/b = 1 ……….(1)
and
x^{2}/c + y^{2}/d = 1 ……….(2)
Let, us assume curves (1) and (2) intersect at (x_{1}, y_{1}). Then
x_{1}^{2}/a + y_{1}^{2}/b = 1 ……….(3)
and
x_{1}^{2}/c + y_{1}^{2}/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, m_{1} and m_{2} be the slopes of the tangents to the curves (1) and (2) respectively at the point (x_{1}, y_{1}); then,
m_{1} = [dy/dx]_{(x1, y1)} = (bx_{1}/ay_{1}) and m_{2} = [dy/dx]_{(x1, y1)} = (dx_{1}/cy_{1})
By question as the curves (1) and (2) intersects at right angle, so, m_{1}m_{2} = 1
Or (bx_{1}/ay_{1})*(dx_{1}/cy_{1}) = 1
Or bdx_{1}^{2} = acy_{1}^{2} ……….(5)
Now, (3) – (4) gives,
bdx_{1}^{2}(c – a) = acy_{1}^{2}(d – b) ……….(6)
Dividing (6) by (5) we get,
c – a = d – b
Or a – b = c – d.
What will be the coordinates of the foot of the normal to the parabola y^{2} = 5x that makes an angle 45° with the x axis?
The equation of the given parabola is, y^{2} = 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)t^{2}, (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 – 5t^{2}/4)
Simplifying further we get,
4(x – y) = 15
The coordinates of the foot of the normal are, P((5/4)t^{2}, (5/2)t).
As t = 1, so putting the value of t = 1, we get,
P = (5/4, 5/2).
What will be the coordinates of the foot of the normal to the parabola y^{2} = 3x which is perpendicular to the line y = 2x + 4?
Given, y^{2} = 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 (x_{1}, y_{1}) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is
[dx/dy]_{P} = 2y_{1}/3
If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,
2y_{1}/3*2 = 1
Since the slope of the line (2) is 2
Or y_{1} = 3/4
Since the point P(x_{1}, y_{1}) lies on (1) hence,
y_{1}^{2} = 3x_{1}
As, y_{1} = 3/4, so, x_{1} = 3/16
Therefore, the required equation of the normal is
y – y_{1} = (2y_{1})/3*(x – x_{1})
Putting the value of x_{1} and y_{1} in the above equation we get,
16x + 32y = 27
And the coordinates of the foot of the normal are (x_{1}, y_{1}) = (3/16, 3/4)
If the normal to the ellipse x^{2} + 3y^{2} = 12 at the point be inclined at 60° to the major axis, then at what angle does the line joining the curve to the point is inclined to the same axis?
Given, x^{2} + 3y^{2} = 12 Or x^{2}/12 + y^{2}/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.
What will be the value of angle between the curves x^{2}  y2 = 2a^{2} and xv + y^{2} = 4a^{2}?
x^{2} – y^{2} = 2a^{2} ……….(1) and x^{2} + y^{2} = 4a^{2} ……….(2)
Adding (1) and (2) we get, 2x^{2} = 6a^{2}
Again, (2) – (1) gives,
2y^{2} = 2a^{2}
Therefore, 2x^{2} * 2y^{2} = 6a^{2} * 2a^{2}
4x^{2}y^{2} = 12a^{2}
Or x^{2}y^{2} = 3a^{4}
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 m_{1} and m_{2} be the slopes of the tangents to the curves (1) and (2) respectively at the point (x, y); then,
m_{1} = x/y and m_{2} = 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 = (m_{1} – m_{2})/(1 + m_{1}m_{2})
Putting the value of m_{1}, m_{2} in the above equation we get,
tanθ = 2xy/(y^{2} – x^{2})
As, 2xy = ±2√3a^{2} and x^{2} – y^{2} = 2a^{2}
tanθ = ±2√3a^{2}/2a^{2}
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.
(a_{1}, a_{2}) ∈R implies that (a_{2}, a_{1}) ∈ R, for all a_{1}, a_{2}∈A. This condition is for which of the following relations?
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 are also called as _____ or _____ questions.
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] = ^{n}C_{x} p^{x} q^{nx}
Where n is the number of trials, x is the number of successes and (nx) is failures.
P(x; μ) = (e^{μ}) (μ^{x}) / x! is the formula for _____
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!
P [X = x] = ^{n}C_{x} p^{x }q^{nx} is the formula for _____
Formula for binomial distribution is P [X = x] = ^{n}C_{x} p^{x} q^{nx}
Where n is the number of trials, x is the number of successes and (nx) is failures.
The term Bernoulli trials is termed after which swiss mathematician?
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.
The Poisson distribution comes under which probability distribution?
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.
What is the formula for the Poisson distribution probability?
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!
What will be the value of x + y + z if cos^{1} x + cos^{1} y + cos^{1} z = 3π?
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.
Method in which the previously calculated probabilities are revised with values of new probability is called ____
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.
Formula for conditional probability P(A  B) is _______
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(AB) = P(A∩B)/P(B).
Whereas formula’s P(A  B) = P(A∩B)/P(A), P(A  B) = P(A)/P(B), P(AB) = P(B)/P(A) doesn’t satisfies the specified conditions.
A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
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(ES1) = P(Probability that man reports six occurs when six actually has occurred on the die) = 3/4.
P(ES2) = 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.
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 








