5 - Applications of differentiation Questions Answers

sir pls provide a detailed solutions asap
Asked By: HARDIK
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Sir pls solve
Asked By: KANDUKURI ASHISH
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Joshi sir comment

The general solution of this kind of  differential eq. is y = eax Let a function g(x) = eaxf(x)    (1) differentiate eq. (1) thrice for getting g,,,(x) now for getting the same function as given  find a. you get the same function for  a = 1/2. clearly, for min 5 zeros of g(x), there are min 2 zeros for g,,,(x).  solve and inform if face any problem.

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 + πX2/8

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(xi)/3} < f -1({x1+x2+x3}/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< x2 < x3 ,  å{f -1(xi)/3} < f -1({x1+x2+x3}/3), 

The same result for f(x) will be 

  å{f (xi)/3} > f ({x1+x2+x3}/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) = x2 + cosα x + 1

its descreminant is negative and coefficient of x2 is positive so f(x) > 0

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