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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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