IIT JEE , AIEEE Forum - Mathematics , Trignometry

neeraj_agarwal_1990

If th equation ${sin}^{-1}({x}^{2}+x+1) + {cos}^{-1}(ax+1) = \frac{\pi}{2}$ has exactly two solutions,then a can't have the integral value:

a)-1 b)0 c)1 d)2

ans-a,c,d

i m getting only (c)

x^2+x+1 = ax+1 should have exactly 2 roots

=> D>0
=> (1-a)^2 >0
=> a is not equal to 1.

do we also consider the domain of the inverse functions in this problem?
if yes...then how do we get 'a' from it??

hash_include

yes we do.. this is coz the sin inverse function is independent of a.
so it has a constraint on the values of x. this constraint on the value of x in turn causes a constraint on value of a.

kislay

well by putting constraint -1<= x^2+x+1<=1
we get x=[-1,0]
and from equation
x^2+x+1 = ax+1

we find minimum value of the equation...=(a-1)/2

for this t ohave roots in [-1,0]
f(-1)>=f((a-1)/2)
and
f(0)>=f((a-1)/2)
use this constraint to get the answer...

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