Question

sin^2 x+4sinx +5€[k,5k] => k=?​

Answer 1

ANSWER

f(x)=sin2x+4sinx+5

=(sinx+2)2+1

∴ max. value of f(x)=9+1=10sinx=1

min value of f(x)=2sinx=−1

∴ Range =[k,5k]=[2,10]

k=2

Answer 2

Step-by-step explanation:

$we \: will \: use \: completing \: the \: square \: method$

$let \sin(x) be \: t$

${t}^{2} + 4t + 5$

${t}^{2} + (2)(2)(t) + 4 + 1$

${(t + 2)}^{2} + 1$

$(\sin(x) + 2) + 1$

$range \: of \: sin \:x \: is \: from \: - 1to1$

$( - 1 \: to \: 1)$

$add \: 2$

$(1 \: to \: 3)$

$square \: both \: sides$

$(1 \: to \: \: 9)$

$add \: 1 \: on \: both \: sides$

$answer \: ranges \: from \: (2 \: to \: 10)$

$so \: the \: value \: of \: k \: is \: 2$