# 5 - Applications of differentiation Questions Answers

**Asked By: KANDUKURI ASHISH**

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**Joshi sir comment**

$Thegeneralsolutionofthiskindof\phantom{\rule{0ex}{0ex}}differentialeq.isy={e}^{ax}\phantom{\rule{0ex}{0ex}}Letafunctiong\left(x\right)={e}^{ax}f\left(x\right)\left(1\right)\phantom{\rule{0ex}{0ex}}differentiateeq.\left(1\right)thriceforgetting{g}^{,,,}\left(x\right)\phantom{\rule{0ex}{0ex}}nowforgettingthesamefunctionasgiven\phantom{\rule{0ex}{0ex}}finda.yougetthesamefunctionfor\phantom{\rule{0ex}{0ex}}a=1/2.clearly,formin5zerosofg\left(x\right),\phantom{\rule{0ex}{0ex}}therearemin2zerosfor{g}^{,,,}\left(x\right).\phantom{\rule{0ex}{0ex}}solveandinformiffaceanyproblem.$

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**

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**Joshi sir comment**

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**

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**Joshi sir comment**

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**

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**Joshi sir comment**

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