e|sin⁡x|+e−|sin⁡x|+4a=0 will have exactly four different solution in [0,2π]. A) a∈[−e4,−14] B) a∈[−1−e24e,∞) C) a∈R D) None of these.

Question

e|sin⁡x|+e−|sin⁡x|+4a=0 will have exactly four different solution in [0,2π].  A) a∈[−e4,−14]  B) a∈[−1−e24e,∞)  C) a∈R  D) None of these.

Ydaxq · Accepted Answer

Let e|sin⁡x|=t∈[1,e]  So equation transformed into t2+4at+1=0  Above equation must have two distinct solution in [1, e]  For two distinct solution we must have f(1)≥0,f(e)≥0,4a2−4&amp;gt;0and 1&amp;lt;−2a&amp;lt;e  where f(t)=t2+4at+1

Rosa Nicholson · Accepted Answer

Say s=|sin⁡x|, then s∈[0,1] and we need to find for which s has −4a=es+e−s two solutions. Since function f(s)=es+e−s is even and increases for s≥0 we see that fmax=f(1)=e+1e and fmin=f(0)=2. So 2&amp;lt;−4a≤e+1e or −12&amp;gt;a≥−1+e22e Since −e4&amp;gt;−1+e22e the answer is (A).