# 134 - Mathematics Questions Answers

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

How to inegrate ( In t) dt /(t-1)

Joshi sir comment

let lnt = x so dt/t = dx so dt = tdx = exdx

now ∫(ln t)dt/(t-1) = ∫xexdx/(ex-1) = ∫x(ex-1+1)dx/(ex-1)

now solve

ƒ(x)=|x-2| and g=ƒoƒ so g'(x)=.......? ; 2<x<4

Joshi sir comment

for 2<x<4,   f(x) = x-2

so g(x) = f(x-2) = x-2-2 = x-4

so g'(x) = 1

ƒ(x)=3|2+x| so ƒ'(-3)=? ,if its -3 , how?

Joshi sir comment

ƒ(x)=3|2+x|

for x>-2,  f(x) = 3(2+x)  so  f'(x) = 3

for x<-2, f(x) = -3(2+x) so f'(x) = -3

Maths >> Calculus >> Integrations IIT JEE

∫√tanx

Joshi sir comment

let  √tanx = t

so sec2xdx/2√tanx = dt

so (1+t4)dx/2t = dt

so dx = 2tdt/(1+t4)

now solve

What is the value of 0⁄0? if its otherthan 1, then how this can be true:      lim      (ex-1)/x = 1

x −−>0