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Q1- [8+6+6 marks] Solve for :
a)
b) .
c) .
Ans-
.
Let 2x = t
So 4t2- 17t+4 = 0
4t2-(16+1)t+4 =0
4t2-16t-t+4 =0
4t(t-4)-1(t-4) = 0
So we get factors (4t-1)(t-4) = 0
Then 4t-1 = 0
4*2x -1 =0
4*2x = 1
2x = ¼
2x = 4-1 => 2x = 2-2
On comparing with bases we get x= -2
Or t-4 = 0
2x = 4 => 2x = 22
on comparing bases we get x = 2 ans
.
ex1/2 – (ex)1/2 = 0
let x1/2 = t ,then we have et – (et2)1/2 = 0
on squaring both the sides ,we get - e2t – et2 = 0
e2t = et2
on comparing with common base e
we get 2t = t2
=> t2 – 2t =0
=> t(t-2) = 0 => t = 0, 2
If t = 0 or t = 2
=> X1/2 = 0 or x1/2 = 2
On squaring x = 2 or 4 ans
.
Using ln property we have-
ln(x+1) = 0 ,x >-1
= 0 or ln(x+1) = 0
x+1 = e0
x+1 = 1 => x = 0
ans
Q2-[10+10 marks]
a) Using logarithmic differentiation, find the derivative of:
.
b) Find an equation of the tangent line to graph of at the point whose -coordinate is .
Ans-
a)Using logarithmic differentiation, find the derivative of:
.
Let y = f(x)
y =
taking log both sides-
ln y = ln(x+1)(2x+1)(3x+1) - ln
ln y = ln(x+1) + ln(2x+1)+ ln(3x+1) – 1/2ln(4x+1)
on differentiating –
y’/y = 1/(x+1) + 2/(2x+1) + 3/(3x+1)- 2/(4x+1)
y’ = y (1/(x+1) + 2/(2x+1) + 3/(3x+1)- 2/(4x+1))
y’ = f(x) (1/(x+1) + 2/(2x+1) + 3/(3x+1)- 2/(4x+1)) ans
b) Find an equation of the tangent line to graph of at the point whose -coordinate is .
Sol-
dy/dx = 3cos x - 2sin 2x
dy/dx at x = = m
so we have m = 3 cos - 2sin2*
m = 3*0 – 2*0 , here sin pi = 0
m= 0
so the equation of tangent line is y = mx+c
=> y = 0 +c
=> y = c ans
Q3-[14+6 marks]
a) Let . Find the maximum and minimum values for , if any.
b) Find the average value of the function on the interval .
Ans-
a)Let . Find the maximum and minimum values for , if any.
Sol-
On differentiating both sides-
P’(t) = - 120 sin 6t
For .
We have P’(t) = 0
-120sin6t = 0
=> sin6t = 0
=> 6t = sin-1(0)
=> 6t = 0 => t = 0
Now P’’(t) = -720cos 6t
At t = 0 we have P’’(t) = -720 = (-ve) ,So it is maximum point at t=0.
Hence maximum value is P(0) = 120 ans
b)Find the average value of the function on the interval .
Average value is defined in [a,b] as-
¯f=1/b−a ∫ab f(x)dx
Here f(x) =
So we have f = 1/9-1 ∫19 dx
F= 1/8 (2x1/2)91 = 1/8 *(6-2) = ½ ans
Q4- Let be the region that is bounded by the graphs of , and .
a) Sketch the region .
b) Find the area of the region .
Ans-
a)
y1 =>4x2 – 4x = y and the line y2=>y = 3/4
Ans-b)
and .
x = ¾ and ¼
Area = ꭍ 4x-4x2 – ¾ dx ------------------------(1)
= 4x2/2 – 4x3/3-3x/4
= At x= ¾ we get Area = 0
= Now we put x= ¼ in the eqn (1) we get
= ½
Hence Area = ½ square unit Ans
Q5- Let be a density function.
a) Find .
Sol- The CDF of X is found by integrating the PDF from 0 to x.
F (x) = ꭍx0 f(x) x ′ dx′ =
b) Find .
Sol- P(X>5)= 1- P(5)= 1- 5/18 = 13/18
c) Find mean of .
Sol- E [X] = ꭍ63 dx = 1
=
d) Find variance of .
Sol- E[x2] = ꭍ63 dx = 4/3
Var[X] = E[x2] - E [X] = 4/3 – 1 = 1/3 ans
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