Question

ation 51 : Find the value(s) of'a' for which the inequality tan x + (a + 1)tanx-(2-3)<0, is true for
at least one xe, 0,
2
siglo​

Answer 1

Answer:

Set tanx=y. Note that y can take only positive values since x∈(0,π2).

Then

y2+(a+1)y−(a−3)<0

⟺(y+a+12)2−(a+1)24−(a−3)<0(completing the square)

⟺(y+a+12)2<a2+6a−114=(a+3)2−204

So,

−(a+3)2−204−−−−−−−−−−−√<y+a+12<(a+3)2−204−−−−−−−−−−−√

⟺−(a+3)2−204−−−−−−−−−−−√−a+12<y<(a+3)2−204−−−−−−−−−−−√−a+12

We just need to make sure that there exists a positive y that satisfies this. So we need to make sure that the upper bound is positive.

Thus, (a+3)2−204−−−−−−−−−−−√−a+12>0⟺(a+3)2−20−−−−−−−−−−−√>a+1

If a+1 negative then the inequality is true, and if a+1≥0, we can square both sides and simplify to obtain a>3 which is consistent with a+1≥0. Also, we need (a+3)2−20 to be non-negative, so (a+3)2≥20⟺a≥20−−√−3 or a≤−20−−√−3.

Thus, a>3 or a≤−20−−√−3.

Step-by-step explanation:

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Answer 2

a=√2...

hope it helps......... ...........