# 134 - Mathematics Questions Answers

No. of integral roots of the eqn. x^{8} - 24 x^{7} - 18x^{5} + 39 x^{2} + 1155 = 0 .

**Asked By: VAIBHAV GUPTA**

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**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 ax^{2}+bx+c,bx^{2}+cx+a,cx^{2}+ax+b are positive for all x ε R and let

α=bc+ca+ab/a^{2}+b^{2}+c^{2} then

(A) α<4 (B) α<1 (C) α>1/4 (D) α>1

THIS IS MULTIPLE CHOICE QUESTION

**Asked By: AKASH**

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**Joshi sir comment**

if ax^{2}+bx+c is positive for real x

then b^{2}< 4ac similarly others

on adding all inequalities we get b^{2} + c^{2} + a^{2}< 4(ac+ab+bc)

now get answer

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

**Asked By: RESHU SINGH**

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**Joshi sir comment**

general term of the series = r(n+1-r)^{2 }= r(n+1)^{2} + r^{3} - 2(n+1)r^{2}

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.

**Asked By: SUSHOVAN HALDAR**

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**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]^{2 }+ y^{2}= 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 ^{n}C_{r}x^{r}y^{n-r}equals nx....How??

**Asked By: RESHU SINGH**

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**Joshi sir comment**

∑ r ^{n}C_{r}x^{r}y^{n-r}

= ∑ r ^{n}C_{r}x^{r}(1-x)^{n-r}

= ∑ r n!/r!(n-r)! x^{r}(1-x)^{n-r}

= n (n-1)!/(r-1)!(n-r)! x^{r}(1-x)^{n-r }

= n ^{(}^{n-1)}C_{(}_{r-1) }x^{r}(1-x)^{n-r}

= n ^{(}^{n-1)}C_{r }x^{r+1}(1-x)^{n-r-1 }let r = r+1

= nx ^{(}^{n-1)}C_{r }x^{r}(1-x)^{n-r-1 }

^{now solve}

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

**Asked By: PARTHASARATHY**

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**Joshi sir comment**

let lnt = x so dt/t = dx so dt = tdx = e^{x}dx

now ∫(ln t)dt/(t-1) = ∫xe^{x}dx/(e^{x}-1) = ∫x(e^{x}-1+1)dx/(e^{x}-1)

now solve

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

**Asked By: BONEY HAVELIWALA**

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**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?

**Asked By: BONEY HAVELIWALA**

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**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

∫√tanx

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

let √tanx = t

so sec^{2}xdx/2√tanx = dt

so (1+t^{4})dx/2t = dt

so dx = 2tdt/(1+t^{4})

now solve

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

^{ x −−>0}

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

exact 0 and lim tends to 0 are different