# 10 - Equations and Inequalities Questions Answers

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

**Asked By: JAMES GHOSH**

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**Joshi sir comment**

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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**Joshi sir comment**

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

4[3t+17]=10

**Asked By: GLADNESS**

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**Joshi sir comment**

let [ ] is greatest integer function

so 4[3t+17] = 10

or [3t+17] = 2.5

since greatest integer function can not be non integral so no t is possible.

now if [ ] is a normal bracket

then 3t+17 = 2.5 or 3t = -14.5 or t =-14.5/3 = -145/30 = -29/6

(α,β);(β,γ) &(γ,α) are respectively the roots of x²-2px+2=0, x²-2qx+3=0,x²-2rx+6=0. if α,β,γ are all positive , then the value of p+q+r is

**Asked By: AMIT DAS**

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**Joshi sir comment**

According to the given conditions

α+β=2p (1) α*β=2 (2)

β+γ=2q (3) β*γ=3 (4)

γ+α=2r (5) γ*α=6 (6)

so α+β+γ = p+q+r (7)

and α*β*γ=6 (8)

dividing eq. (8) by (2), (4) , (6) one by one we will get α, β, γ hence we will get p+q+r

If the eqaution x^3-3x+k, always has exactly one +ve real root , then find teh minimum value of [|k|].

**Asked By: AMIT DAS**

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**Joshi sir comment**

let f(x) = x^3-3x+k

so f^{'}(x) = 3 x^2 - 3

on comparing with 0, x = +1, -1

on double differentiating we get that function f(x) is minimum at +1 and maximum at -1

so for getting the only +ve real root plot a graph such that it could intersect positive x axis at any point and nowhere else. From the graph it is clear that k will be negative.

at x = +1, f(x) = k-2

at x = -1, f(x) = k+2

so on the basis of given conditions k-2<0 or k<2 and k+2<0 or k<-2

finally k<-2 so [|k|] has minimum value 2

x^7 - 3x^4+2x^3-k=0, k>0 has atleast m imaginary roots then m is ?

**Asked By: AMIT DAS**

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**Joshi sir comment**

x^7 - 3x^4+2x^3-k=0

according to sign change rule there are 3 pair of sign + -, - +, + - , see the sign before coefficient of different powers

so 3 positive real roots are confirmed

now take x = -x

we get -x^7 - 3x^4 - 2x^3 - k = 0, there is no sign change so no negative root

total real roots = 3

so imaginary roots = 4

The maximum number of real roots of the equation 2*x*^{88} + 3*x*^{87} − 13*x*^{2} + 5*x* + 9 = 0 is

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

2*x*^{88} + 3*x*^{87} − 13*x*^{2} + 5*x* + 9 = 0

implies that 2*x*^{88} + 3*x*^{87} = 13*x*^{2} - 5*x* - 9

or *x*^{87}(2x+3*)* = 13*x*^{2} - 5*x* - 9

now plot the left and right graphs for getting the intersecting points

Hint: Parabola will face in upward direction as a is positive, take x=0 we get -9 so 1 root will be positive and 1 negative,

*x*^{87}(2x+3*) *will be 0 for x = -3/2 and x = 0 besides it consider some values as x = 1 and -1

You will get one root between 0 and -1

now check both the graph at x= -2, you will get the idea about the second intersection