Range of&nbsp;f(x)=sin2x+sin⁡x−1sin2x−sin⁡x+2Let&nbsp;sin2x=t⇒&nbsp;(y−1)t2−(y+1)t+2y+1=0Since&nbsp;t=sin⁡x&nbsp;is real,Discriminant≥0⇒(y+1)2−4(y−1)(2y+1)≥0⇒&nbsp;−7y2+6y+5≥0⇒&nbsp;7y2−6y−5≤0⇒&nbsp;y∈[3−2117,3+2117]But the correct answer is&nbsp;y∈[3−2117,12]

y−1=d2t−3t2−t+2Let z=3−2t⇒5≥z≥1d41−y+4=4+dz2−4z+11z=z+d11z≥2z⋅d11zwhich is attained if z2=11y≥?Now for 5≥a&gt;b≥1,a+11a−(b+11b)=d(a+b−11)(a−b)ab&lt;0,So, z+11z will be maximum for the the minimum value of z which is =1 here

Get the answer to "Range off(x)=sin2x+sin⁡x−1sin2x−sin⁡x+2Letsin2x=t⇒(y−1)t2−(y+1)t+2y+1=0Sincet=sin⁡xis real,Discriminant≥0⇒(y+1)2−4(y−1)(2y+1)≥0⇒−7y2+6y+5≥0⇒7y2−6y−5≤0⇒y∈[3−2117,3+2117]But the correct answer isy∈[3−2117,12]"

High School
Algebra
Algebra I
Quadratic function and equation

Oliver Carson

Answered question

2022-03-31

Range of $f (x) = \frac{\sin^{2} x + \sin x - 1}{\sin^{2} x - \sin x + 2}$
Let $\sin^{2} x = t$
$\Rightarrow (y - 1) t^{2} - (y + 1) t + 2 y + 1 = 0$
Since $t = \sin x$ is real,
$Discriminant \geq 0$
$\Rightarrow {(y + 1)}^{2} - 4 (y - 1) (2 y + 1) \geq 0$
$\Rightarrow - 7 y^{2} + 6 y + 5 \geq 0$
$\Rightarrow 7 y^{2} - 6 y - 5 \leq 0$
$\Rightarrow y \in [\frac{3 - 2 \sqrt{11}}{7}, \frac{3 + 2 \sqrt{11}}{7}]$
But the correct answer is $y \in [\frac{3 - 2 \sqrt{11}}{7}, \frac{1}{2}]$

Answer & Explanation

Yaritza Phillips

Beginner2022-04-01Added 12 answers

y - 1 = d \frac{2 t - 3}{t^{2} - t + 2}

Let

z = 3 - 2 t \Rightarrow 5 \geq z \geq 1

d \frac{4}{1 - y} + 4 = 4 + d \frac{z^{2} - 4 z + 11}{z} = z + d \frac{11}{z} \geq 2 \sqrt{z \cdot d \frac{11}{z}}

which is attained if

z^{2} = 11

y \geq ?

Now for

5 \geq a > b \geq 1

a + \frac{11}{a} - (b + \frac{11}{b}) = d \frac{(a + b - 11) (a - b)}{a b} < 0

,
So,

z + \frac{11}{z}

will be maximum for the the minimum value of z which is

= 1

here

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