# 159 - Mathematics Questions Answers

Find Sum to n terms

x/1-x²  +  x²/1-x⁴  +  x⁴/1-x⁸ + ....

Joshi sir comment

1 . Iₙ = ∫ ( 1/ ( x² + a²)ⁿ  )dx

How to do this by substituting x = a tan Α

2.  Is this true

∫ f(x) d(kg(x)) = k ∫ f(x) d(g(x))

where k is a constant

Solution by Joshi sir

d/dx((1+x^2+x^4)/(1+x+x^2))=ax+b

then a=?, b=?

Joshi sir comment

1+x2+x4 = (1+x+x2)(1-x+x2)

now solve

Maths >> Calculus >> Functions IIT JEE

Find a formula for a function g(x) satisfying the following conditions

a) domain of g is (-∞ , ∞ )     b) range of g is [-2 , 8]     c)  g has a period π   d)  g(2) = 3

Joshi sir comment

g(x) = 3-5sin(2x-4)

Maths >> Calculus >> Functions IIT JEE

Let f(x) = x135 + x125 - x115 + x5 +1. If f(x) is divided by x3-x then the remainder is some function of x say g(x). Find the value of g(10)

Joshi sir comment

for getting reminder put x3= x so

x135 + x125 - x115 + x5 +1 will give

x45 + x41*x2 - x38*x +x*x2 +1

x15 + x13*x2*x2 - x12*x2*x +x*x2 +1

x17 + x3 +1

x5*x2 + x + 1

x+ x + 1

x2*x + x +1

x3 + x +1

x+x+1

2x+1

now put x = 10

No. of integral roots of the eqn. x8 - 24 x7 - 18x5 + 39 x2 + 1155 = 0 .

Joshi sir comment

two times sign change so 2 positive roots

let a,b,c be three distinct real numbers such that each of expression ax2+bx+c,bx2+cx+a,cx2+ax+b are positive for all x ε R and let

α=bc+ca+ab/a2+b2+c2 then

(A) α<4 (B) α<1 (C) α>1/4 (D) α>1

THIS IS MULTIPLE CHOICE QUESTION

Joshi sir comment

if ax2+bx+c is positive for real x

then b2< 4ac similarly others

on adding all inequalities we get b2 + c2 + a2< 4(ac+ab+bc)

Maths >> Algebra >> Binomial Theorem Engineering Exam

Find sum of series..1*(n)2  + 2*(n-1)+3*(n-2)2 +...+n .

Joshi sir comment

general term of the series = r(n+1-r)= r(n+1)2 + r3 - 2(n+1)r2

now put sigma then solve

### Tangents are drawn to the circle x^2 + y^2=12 at the points where it is met by the circle x^2 +y^2 - 5x + 3y - 2=0.Find the point of intersection of these tangents.

Joshi sir comment

sovle the two eqs for x and y,

method :

### x^2 + y^2=12

on putting this in 2nd eq. we get -5x+3y = -10 so x = (3y+10)/5

so [(3y+10)/5]+ y2= 12

now get y then x, then get the tangents in the first circle and then intersection point of the tangents.

If x+y = 1 ,then ∑ r  nCrxryn-requals nx....How??

Joshi sir comment

∑ r nCrxryn-r

∑ r nCrxr(1-x)n-r

∑ r n!/r!(n-r)xr(1-x)n-r

= n (n-1)!/(r-1)!(n-r)! xr(1-x)n-r

= n (n-1)C(r-1) xr(1-x)n-r

= n (n-1)Cxr+1(1-x)n-r-1             let r = r+1

= nx  (n-1)Cxr(1-x)n-r-1

now solve