24Q3 plzzz
Q1 Let S denote the set of all values of a for which 4^{x^2}+2(2a+1)2^{x^2}+4a^2-3>0 is satisfied for all reals..
I am getting ans as (-∞,-1) but ans is diff
Q2 The number of integral solns of x(\frac{5-x}{x+1})(x+\frac{5-x}{x+1})=6
I ma getting 1 integral sols where as ans is 2 integral solns
Q3 The number of subsets of real solns of x^4=\frac{11x-6}{6x-11}
whats the significance of subsets here ??
Q4 Let x_1>x_2>x_3>x_4>x_5>x_6>0
A=x_1+x_2+x_3+x_4;B=x_1x_3+x_1x_5+x_3x_5+x_2x_4+x_2x_6+x_4x_6;C=x_1x_3x_5+x_2x_4x_6
then calculate the number of negative roots of eqn 2x^3-Ax^2+Bx-C=0
Ans given is 3 ..but how is it possible becoz there is no sign change for f(-x)
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18 Answers
106haan... the solution set is {-1,2,0.5} so the no. of subsets of this solution set = 23=8
66Yes x=-1 is a root and so are x=2, x=12.
These are the only real roots.
341??!!
x=-1 is a root. If you draw rough graphs of LHS and RHS you will see two other intersection points, reciprocals of each other, one lying between 0 and 6/11 and the other >11/6.
4I somehow didn't understand Q 4
x1 , x2 ... x6 R roots of the given eqn ???
1second one --- simplify to get x 5-xx+1 x2+5x+1 = 6
clearly x cannot be negative ,as then all the terms will be negative.
also x < 5 as again l.h.s will be -ve.
x is also > 0
it only remains to check 1,2,3,4 -- from which we get 1 and 2
24@nishant sir
in psot#5
didnt understand this "first use the condition that roots are imaginary
if they are real then both roots should be -ve"
341Q1: A small correction to Nishant Sir's post #4. If the roots are real, the greater root must be greater than 1.
4@eureka is the eqn given in qs 4 a 6th degree one or 3rd degree???????
62Q4)
a direct outcome of descarte's rule of sign change is that f(x) has no -ve roots!
PS: oh i din see eureka's comment... Yes the answer is wrong
621) you need to find the condition for no +ve root of t2+2(2a+1)t+4a2-3 = 0
first use the condition that roots are imaginary
if they are real then both roots should be -ve
so f(0)>0 and -b/2a<0
so you have 4a2-3>0 and 2(2a+1)>0
so a>√3/2 and a>-1/2
I think the answer should be (-∞, -1) U (-1/2, √3/2)
Check for open and closed brackets...
24do we need to do all that ???
btw i solved Q2....
6=1.2.3
put x=1 => LHS=6=RHS
put x=2 => LHS=6=RHS
put x=3 => LHS≠RHS
so two solns i.e x=1,2