# 159 - Mathematics Questions Answers

Find Sum to n terms

x/1-x² + x²/1-x⁴ + x⁴/1-x⁸ + ....

**Asked By: LUFFY**

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**Joshi sir comment**

$\frac{x}{1-{x}^{2}}=\frac{1}{1-x}-\frac{1}{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}breakalltermssimilarlyandsolve\phantom{\rule{0ex}{0ex}}$

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

**Asked By: LUFFY**

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**Solution by Joshi sir**

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

then a=?, b=?

**Asked By: RAJIV**

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**Joshi sir comment**

1+x^{2}+x^{4} = (1+x+x^{2})(1-x+x^{2})

now solve

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

**Asked By: VAIBHAV GUPTA**

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**Joshi sir comment**

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

Let f(x) = x^{135} + x^{125} - x^{115 }+ x^{5} +1. If f(x) is divided by x^{3}-x then the remainder is some function of x say g(x). Find the value of g(10)

**Asked By: VAIBHAV GUPTA**

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**Joshi sir comment**

for getting reminder put x^{3}= x so

x^{135} + x^{125} - x^{115 }+ x^{5} +1 will give

x^{45} + x^{41}*x^{2} - x^{38}*x +x*x^{2} +1

x^{15} + x^{13}*x^{2}*x^{2} - x^{12}*x^{2}*x +x*x^{2} +1

x^{17 }+ x^{3} +1

x^{5}*x^{2} + x + 1

x^{7 }+ x + 1

x^{2}*x + x +1

x^{3} + x +1

x+x+1

2x+1

now put x = 10

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}