60 - Calculus Questions Answers

how to find differential eqn. of all conics whose axes coincide with coordinate axes??   tell me the eqn. of tht. conic

Asked By: SHALINI SINGH
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Joshi sir comment

Parabola   eq.  y2 = 4ax ,

on differentiating this eq. with respect to x we get 2y(dy/dx) = 4a                        

on taking this value of 4a in eq. y2= 4ax we will get the differential eq. of parabola.

similarly for ellipse    x2/a2  +   y2/b2  =   1

on differentiaing this eq. twice we will get two additional eq. 

by solving these eq. get an eq. free from a and b, this will be the differential eq. of the ellipse.

similarly for hyperbola

 

 

 

Maths >> Calculus >> Integrations IIT JEE

0π [cot-1x]dx

Asked By: NIKHIL VARSHNEY
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Joshi sir comment

i think the question will be 0π [cotx]dx because limits are in angular terms.

by graph given below we get answer as -π/2

here in the graph same coloured shaded region are dx*(-1) always, because for every similar pair [+k]+[-k] = -1 , here k is a real number, so complete area = -1(π/2)

Maths >> Calculus >> Integrations Others
  • what is the integration of cosx 2
Asked By: VIKRANT KUMAR
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Joshi sir comment

cosx2 = (cos2)x

so ∫ (cos2)x dx = (cos2) x / logecos2

Solution by Joshi sir

 

cosx2 = (cos2)x

so ∫ (cos2)x dx = (cos2) x / logecos2 + C            formula ∫axdx = ax/logea

 

lim ∑ nr=1 1/n er/n n tends to infinity.

Asked By: NIKHIL VARSHNEY
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Joshi sir comment

limn->∞ r=1er/n/n

This question is related to definite integration, consider 1 and divide it into n parts, upto nth part total value = r/n is equivalent to x and 1/n is dx

so question will be

01 exdx 

now solve it

ABC is an isosceles triangle inscribed in a circle of radius r. If  AB=AC and h is the altitude from A to BC. and P and φ denote the perimeter and area of the triangle respectively, then  lim n−» 0   φ/p³ is equal to??

Asked By: AMIT DAS
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Joshi sir comment

 

first draw a cirle and triange inside it , consider angle B and angle C asα so angle A = 180-2α, let centre of the circle is O and D is the point in the line BC where altitude meets in the line BC, given that altitude = h and radius = r

so AB = AC = h cosecα and BC = 2 h cotα so 

p = 2 h (cosecα + cotα)

and φ = 1/2  2 h cotα h = h2 cotαa

and in triangle OBD angle OBD = 2α - 90    so cos(2α - 90) = h cotα /r   implies that h = 2 r sin2α 

now i think limit will be based on h not n

 lim h−» 0   φ/p³  = 1/128r

for getting solution put p and φ first then put h in terms of r, you will get an equation based on α and r,

convert limh->0  to  limα->0     because when h will be 0, α will also be 0 

 

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), 

Maths >> Calculus >> Integrations IIT JEE

Evaluate: 0ò11/{ (5+2x-2x2)(1+e(2-4x)) } dx

Asked By: KAMAL
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Joshi sir comment

let I = 0ò11/{ (5+2x-2x2)(1+e(2-4x)) } dx

on using the property 0a f(x) dx = 0af(a-x)dx

we get 2I = 0ò11/(5+2x-2x2) dx

now solve this 

Maths >> Calculus >> Functions IIT JEE

Let f(x) be a real valued function not identically equal to zero such that f(x+yn)=f(x)+(f(y))n; y is real, n is odd and n >1 and f'(0) ³ 0. Find out the value of f '(10) and f(5).

Asked By: KAMAL
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Joshi sir comment

take x = 0 and y = 0 and n = 3

we get f(0) = f(0) + f(0)n or f(0) = 0

now take x = 0,  y = 1 and n = 3

so f(1) = f(0) + f(1)3

so get f(1) = 1, other values are not excetable because f(x) be a real valued function not identically equal to zero and f'(0) ³ 0

similarly get the other values 

you will get f(5) = 5

so f(x) = x generally then find f (x) , i think it will be 1 

Maths >> Calculus >> Integrations IIT JEE

Evaluate 0òx [x] dx .

Asked By: KAMAL
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Joshi sir comment

 

 

integer nearst to x and less than x will be [x]

so 0òx [x] dx = 0ò1 0 dx + 1ò2 1 dx + 2ò3 2 dx + ..................... + [x]-1[x] [x]-1 dx + [x][x] dx

now solve it

 

A function f : R® R satisfies f(x+y) = f(x) + f(y) for all x,y Î R and is continuous throughout the domain. If I1 + I2 + I3 + I+ I5 = 450, where In = n.0òn f(x) dx. Find f(x).

Asked By: KAMAL
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Joshi sir comment

according to the given condition f(nx) = n f(x)

0òn f(x) dx

consider x = ny so this integration will become 0ò1 f(ny) ndy = n0ò1 f(y) dy = n2 0ò1 f(x) dx

now by using these conditions I1 = 13 0ò1 f(x) dx similarly others 

put these values and get the answer

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