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.

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&lt;−4a≤e+1e or −12&gt;a≥−1+e22e Since −e4&gt;−1+e22e the answer is (A).

"e|sin⁡x|+e−|sin⁡x|+4a=0 will have exactly four different solution in..." was answered by our experts

High School
Algebra
Algebra I
Quadratic function and equation

Addison Gross

Answered question

2022-02-01

e^{| \sin x |} + e^{- | \sin x |} + 4 a = 0

will have exactly four different solution in

[0, 2 π]

.
A)

a \in [\frac{- e}{4}, \frac{- 1}{4}]

a \in [\frac{- 1 - e^{2}}{4 e}, \infty)

a \in R

D) None of these.

Answer & Explanation

Ydaxq

Beginner2022-02-02Added 12 answers

Let

e^{| \sin x |} = t \in [1, e]

So equation transformed into

t^{2} + 4 a t + 1 = 0

Above equation must have two distinct solution in [1, e]
For two distinct solution we must have

f (1) \geq 0, f (e) \geq 0, 4 a^{2} - 4 > 0 and 1 < - 2 a < e

where

f (t) = t^{2} + 4 a t + 1

Rosa Nicholson

Beginner2022-02-03Added 13 answers

Say

s = | \sin x |

, then

s \in [0, 1]

and we need to find for which s has

- 4 a = e^{s} + e^{- s}

two solutions. Since function

f (s) = e^{s} + e^{- s}

is even and increases for

s \geq 0

we see that

f_{max} = f (1) = e + \frac{1}{e} and f_{min} = f (0) = 2

. So

2 < - 4 a \leq e + \frac{1}{e}

- \frac{1}{2} > a \geq - \frac{1 + e^{2}}{2 e}

Since

- \frac{e}{4} > - \frac{1 + e^{2}}{2 e}

the answer is (A).

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