10 - Equations and Inequalities Questions Answers

If 9a+24b=1, a,b, +, prove that a2+b2 9( 4+93)3.

Asked By: JAMES GHOSH
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If a,b and c are positive real numbers, show that

ab+c+bc+a+ca+b32

Asked By: MROY
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How to do it without using cauchy schwartz inequality?
Asked By: LUFFY
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No. of integral roots of the eqn. x8 - 24 x7 - 18x5 + 39 x2 + 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 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

Asked By: AKASH
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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)

now get answer 

4[3t+17]=10

Asked By: GLADNESS
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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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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 2x88 + 3x87 − 13x2 + 5x + 9 = 0 is

Asked By: HIMANSHU MITTAL
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Joshi sir comment

 

2x88 + 3x87 − 13x2 + 5x + 9 = 0

implies that 2x88 + 3x87 = 13x2 - 5x - 9

or x87(2x+3) = 13x2 - 5x - 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,

x87(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

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