# 7 - Applications of differentiation Questions Answers

**Asked By: SHAZIYA ANSARI**3 year ago

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1. A particle moves according to a law of motion

s(t) = t^{3 }- 12t^{2} + 36t , t ≥ 0

where, *t* is measured in seconds and *s* in meters.

a) Find the acceleration at time t and after 3 seconds.

b) Graph the position, velocity and acceleration function for 0≤t≤8

c) When is the particle speeding up? When is it slowing down?

**Asked By: NORNAZIHAH BINTI ZAKARIA**5 year ago

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SCRA 2013

f(x) = |x^{2}-2| ; -3<x<3

QUES.- Consider The Following Statements:

1. The Absolute Maxmimum Value of The Function is 2.

2. The Absolute Minimum Value of The Fuction is 0.

WHICH OF THE ABOVE STATEMNTS IS/ARE CORRECT?

**Asked By: GAUTAM SAGAR**5 year ago

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Q1. Let f be a twice differentiable function on R.Given that f''(x)>0 for all x element of R.Then which one is true and why?

a.f(x)=0 has exactly two solutions on R.

b.f(x) =0 has a positive solution if f(o)=0 and f'(0)=0

c.f(x)=0 has no positive solution if f(o)=0 and f'(o)>0.

4. f(x)=0 has no positive solution if f(0)=0 and f'(0)<0

**Asked By: KALPANA SAI**6 year ago

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A window frame is shaped like a rectangle with an arch forming the top ( ie; a box with a semicircle on the top)

The vertical straight side is Y, the width at the base and at the widest part of the arch (Diam) is X.

I know that the perimeter is 4.

Find X if the area of the frame is at a maximum?

**Asked By: ROBERT DEPANGHER**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

perimeter = 4 so 2Y + X + πX/2 = 4

and area = XY + [π(X/2)^{2}]/2

put Y from 1st equation to 2nd we get A = X[4-X{1+(π/2)}]/2 + πX^{2}/8

now calculate dA/dX and then compare it to zero

X will be 8/(4+π)

If f(x) is a monotonically increasing function " x Î R, f "(x) > 0 and f ^{-1}(x) exists, then prove that å{f ^{-1}(x_{i})/3} < f ^{-1}({x_{1}+x_{2}+x_{3}}/3), i=1,2,3

**Asked By: KAMAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

f(x) is monotonically increasing so f '(x)>0 and f '' (x) >0 implies that increment of function increases rapidly with increase in x

These informations provide the following informations about the nature of inverse of f(x)

1) f ^{-1} (x) will also be an increasing function but its rate of increase decreases with increasing x

2) for x_{1 }< x_{2} < x_{3} , å{f ^{-1}(x_{i})/3} < f ^{-1}({x_{1}+x_{2}+x_{3}}/3),

The same result for f(x) will be

å{f (x_{i})/3} > f ({x_{1}+x_{2}+x_{3}}/3),

f(x+y) = f(x) + f(y) + 2xy - 1 " x,y. f is differentiable and f'(0) = cos a. Prove that f(x) > 0 " x Î R.

**Asked By: KAMAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

f(x+y) = f(x) + f(y) + 2xy - 1

so f(0) = 1 obtain this by putting x = 0 and y = 0

now f ^{' }(x+y) = f ^{'} (x) + 2y on differentiating with respect to x

take x = 0, we get f ^{'} (y) = f ^{'}(0) + 2y

or f ' (x) = cosα + 2x

now integrate this equation within the limits 0 to x

we get f(x) = x^{2} + cosα x + 1

its descreminant is negative and coefficient of x^{2} is positive so f(x) > 0