# 42 - Algebra Questions Answers

There are 8 distinct boxes and each box can hold any number of balls.A child having 4 identical balls tandomly choosen four boxes.Then another child having four balls,identical to the previous mentioned,again puts one ball in each of the arbitary choosen four boxes.The probability that there are balls in atleast 6 boxes is

Sir pls solve

**Asked By: KANDUKURI ASHISH**

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Between two numbers, whose sum is 13/6, an even number of arithmetic means are inserted, the sum of these means exceed their number by unity. How many means are there?

**Asked By: PREM**

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**Joshi sir comment**

$If\frac{9}{a}+\frac{24}{b}=1,a,b,\in {\mathrm{\mathbb{R}}}^{+},provethat{a}^{2}+{b}^{2}\ge 9{(4+\sqrt[3]{9})}^{3}.$

**Asked By: JAMES GHOSH**

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**Joshi sir comment**

**Asked By: PREM**

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**Joshi sir comment**

If a,b and c are positive real numbers, show that

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$

**Asked By: MROY**

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**Asked By: LUFFY**

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**Joshi sir comment**

$Probabilitytogetfollowingdifferences\phantom{\rule{0ex}{0ex}}differenceprobability\phantom{\rule{0ex}{0ex}}06/36\phantom{\rule{0ex}{0ex}}110/36\phantom{\rule{0ex}{0ex}}28/36\phantom{\rule{0ex}{0ex}}36/36\phantom{\rule{0ex}{0ex}}44/36\phantom{\rule{0ex}{0ex}}52/36\phantom{\rule{0ex}{0ex}}Sorequiredprobability=1-2\left[\frac{6}{36}\left(\frac{4}{36}+\frac{2}{36}\right)+\frac{10}{36}\left(\frac{2}{36}\right)\right]\phantom{\rule{0ex}{0ex}}nowsolve\phantom{\rule{0ex}{0ex}}problemisbasedonintegralvaluessoareaisnotveryuseful$

Find Sum to n terms

x/1-x² + x²/1-x⁴ + x⁴/1-x⁸ + ....

**Asked By: LUFFY**

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**Joshi sir comment**

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

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