Sets problem 30th October

Question

!!log_2(ax^2 + x + a) >= 1!! at !!x in R!!

Then !!a in!!

!!a)" "[1-(sqrt5)/2!! , infinity)

!!b)" "[1-(sqrt5)/2 , 1+(sqrt5)/2]!!

!!c)" "(0 , 1-(sqrt5)/2] U [1+(sqrt5)/2!!, infinity)

!!d)" "[1+(sqrt5)/2!! , infinity)

Solution

In the given equation !!log_2(ax^2 + x + a) >= 1!!

!!ax^2 + x + a >= 2^1!!

!!=> ax^2 + x + a-2 >= 0!!

Hence this condition satisfies for !!a>0 and D<=0!!

!!D = 1 - 4a(a-2) <= 0!!

!!=> 1 - 4a^2 + 8a <= 0!!

!!=> 4a^2 - 8a - 1 >= 0!!

Hence solving this , we will get

!!a = (8 +- sqrt(8^2 - 4(4)(-1)))/(2xx4)!!

!!a = 8/8 +- (sqrt(64+16))/8!!

!!a = 1 +- (sqrt(80))/8!!

!!a = 1 +- (sqrt5)/2!!

Since negative value of a is not possible.

Hence, !!a in [1+(sqrt5)/2!! , infinity)

Get it on Google Play