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...

Addison Gross

Addison Gross



e|sinx|+e|sinx|+4a=0 will have exactly four different solution in [0,2π].
A) a[e4,14]
B) a[1e24e,)
C) aR
D) None of these.

Answer & Explanation




2022-02-02Added 12 answers

Let e|sinx|=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)0,4a24>0and 1<2a<e
where f(t)=t2+4at+1
Rosa Nicholson

Rosa Nicholson


2022-02-03Added 13 answers

Say s=|sinx|, then s[0,1] and we need to find for which s has 4a=es+es two solutions. Since function f(s)=es+es is even and increases for s0 we see that fmax=f(1)=e+1e and fmin=f(0)=2. So 2<4ae+1e or 12>a1+e22e Since e4>1+e22e the answer is (A).

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