9but i feel rupals q is wrong as it doesnt make sense when b = ac instead of b2=ac
ques 1. Number of points where y=log(ax2+2b|x|+c) x2 is not defined where b2=ac and b<0 and c>1 are? ques 2. If f(x)=2x+|x|, g(x)=1/3(2x-|x|) and h(x)=f(g(x)), then the domain of sin-1(h(h(h(........h(h(x).....))) is? (n times)
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11 Answers
62ax2+2bx+c
discriminant is given by b2-ac
so D=0
so the roots are |x|=-b/a
(|x|+b/a)2
which is positive adn not defined only when |x|=- b/a = -c
ax2+2bx+c is never equal to 1 . (why/???) Not a very tough proof.. just think it out!
Also now find the points where log(x2) is nto defined.. (only at x=0)
So the answer I think should be 1 point namely x=0
62If f(x)=2x+|x|, g(x)=1/3(2x-|x|) and h(x)=f(g(x)),
the domain of sin-1(h(h(h(........h(h(x).....))) is? (n times)
when x>0
g(x) = 1/3 x (which is again greater than zero)
Thus f(g(x)) = 2/3x+1/3 x =x
also if x<0, then g(x) = x
f(g(x)) = 2x-x = x
h(x)=x for all x
domain of the above function is given by domain of
sin -1x
9byah
ax2+2bx+c is never equal to 1 . (why/???) Not a very tough proof.. just think it out!
the abv fact is wrong :(
a is obviously <0 and f(0) = c>1 and its a part inverted parabola facing down y axis
also D is not 0 as b=ac and not b2 =ac
962sorry celestine...
i messsed it a bit..
even i solved it assuming b2=ac
i guess i have made more mistakes in the last 2 months than my entire JEE preparations :(
11sir 2nd one is correct of urs
and 4 the 1st question to make ny sense it has to be b^2=ac
1ax2+2b|x|+c this will never have real roots..
but we are given the condition that b2=ac
11look closely Rupal now
y=log(ax2+2b|x|+c) x2
now we know that the base of log >1 and the x can never be equal to 0 or negative
as x2 is given so only possibility is x=0
POINT 1 : x=0
now we look at (ax2+2b|x|+c)>1
as u said earlier that
b2=ac , b<0 , c>1
we know that the x coordinate of the min value is -b/2a
ax2+2bx +c-1>0
let us take a possibility when a<0
so the min value of the function (-D/4a) will be positive and also the x value
so 2 roots will be here as the the parabola is opening downwards and c>1
now let us take a>0
the min value of the function could be positive or negative depending on the sign of D
if D>0 then we have 2 roots with the parabola opening upwards
now we will get 2 more roots because of the mod sign!!!!!!!!
Hence we get a total of 7 roots