Question

If the equation \[{{\sin }^{4}}x-\left( k+2 \right){{\sin }^{2}}x-\left( k+3 \right)=0\] has a solution, then k must lie in a interval
A.$\left( -4,-2 \right)$
B.$\left[ -3,2 \right)$
C.$\left( -4,-3 \right)$
D.$\left[ -3,-2 \right)$

Answer 1

Hint: Put $t={{\sin }^{2}}x$ to the given equation to get a quadratic equation in ‘t’. Now get the roots of that quadratic equation using the quadratic formula, given as: -
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for $a{{x}^{2}}+bx+c=0$
Now, as $t={{\sin }^{2}}x$ so, put the roots from 0 to 1 as range of ${{\sin }^{2}}x$ is $\left[ 0,1 \right]$ . Get the value of k using this approach.

Complete step-by-step answer:
Given equation in the problem is \[{{\sin }^{4}}x-\left( k+2 \right){{\sin }^{2}}x-\left( k+3 \right)=0\] …………………………..(i)
Now, as we can rewrite the above equation as,
$\Rightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}-\left( k+2 \right){{\sin }^{2}}x-\left( k+3 \right)=0$
So, let us suppose $t={{\sin }^{2}}x$ to get the equation in simplified form as
$\Rightarrow {{t}^{2}}-\left( k+2 \right)t-\left( k+3 \right)=0$ …………………………………(ii)
Now, as we know roots of any quadratic equation $a{{x}^{2}}+bx+c=0$ is given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ ……………………………………(iii)
Hence, we can get roots of the quadratic equation given in equation (ii) are given with the help of above equation as
$\Rightarrow t=\dfrac{-\left( -\left( k+2 \right) \right)\pm \sqrt{{{\left( -\left( k+2 \right) \right)}^{2}}-4\times \left( -\left( k+2 \right) \right)\times \left( -\left( k+3 \right) \right)}}{2\times 1}$
$\Rightarrow t=\dfrac{\left( k+2 \right)\pm \sqrt{{{\left( k+2 \right)}^{2}}-4\times \left( k+2 \right)\left( k+3 \right)}}{2}$
$\Rightarrow t=\dfrac{\left( k+2 \right)\pm \sqrt{{{k}^{2}}+4k+4-4\left( {{k}^{2}}+3k+2k+6 \right)}}{2}$
$\Rightarrow t=\dfrac{\left( k+2 \right)\pm \sqrt{{{k}^{2}}+4k+4-4{{k}^{2}}-20k-24}}{2}$
$\Rightarrow t=\dfrac{\left( k+2 \right)\pm \sqrt{-3{{k}^{2}}-16k-20}}{2}$ ……………………………………………(iv)
Now, as the term $-3{{k}^{2}}-16k-20$ should be greater than 0 to get the real value of t $\left( t={{\sin }^{2}}x \right)$ because it is present in square root form and square root of any negative term will be imaginary. So, we get
$\Rightarrow -3{{k}^{2}}-16k-20\ge 0$
$\Rightarrow -\left( 3{{k}^{2}}+16k+20 \right)\ge 0$
On multiplying the whole equation with $'-'$ sign to both sides, we get
$\Rightarrow 3{{k}^{2}}+16k+20\le 0$
Now, we can factorize the above equation by splitting 16k as $10k+6k$ so that product of 10k and 6k is equal to the product of first and last term i.e. $3{{k}^{2}}$ and 20. Hence, we can rewrite the above equation as
$\Rightarrow 3{{k}^{2}}+10k+6k+20\le 0$
$\Rightarrow k\left( 3k+10 \right)+2\left( 3k+10 \right)\le 0$
$\Rightarrow \left( k+2 \right)\left( 3k+10 \right)\le 0$
$\Rightarrow 3\left( k+2 \right)\left( k+\dfrac{10}{3} \right)\le 0$
$\Rightarrow \left( k+2 \right)\left( k+\dfrac{10}{3} \right)\le 0$
$\Rightarrow \left( k-\left( -2 \right) \right)\left( k-\left( -\dfrac{10}{3} \right) \right)\le 0$ ……………………………………..(v)
As we know, if $\left( x-\alpha \right)\left( x-\beta \right)\le 0$ then $x\in \left[ \alpha ,\beta \right]$ , where $\alpha <\beta $
Hence, we get values of k from the equation (v) as
$\Rightarrow k\in \left[ -\dfrac{10}{3},-2 \right]$ ……………………………………(vi)
Now, as we know $t={{\sin }^{2}}x$ and we know range of $\sin x$ is given as $\sin x\in \left[ -1,1 \right]$ $\Rightarrow -1\le \sin x\le 1$
So, we get $0\le {{\sin }^{2}}x\le 1$
Hence, value of t will lie in $\left[ 0,1 \right]$ .So, from equation (iv) we get
$0\le t\le 1$
$0\le \dfrac{\left( k+2 \right)\pm \sqrt{-3{{k}^{2}}-16k-20}}{2}\le 1$
Multiplying while equation by 2, we get
$0\le \left( k+2 \right)\pm \sqrt{-3{{k}^{2}}-16k-20}\le 2$
Adding $-k-2$ to all the sides of equation, we get
$\Rightarrow -k-2\le k+2-k-2\pm \sqrt{-3{{k}^{2}}-16k-20}\le 2-k-2$
$\Rightarrow -k-2\le \pm \sqrt{-3{{k}^{2}}-16k-20}\le -k$
On squaring the whole equation, we get the above equation as
$\Rightarrow {{\left( -k-2 \right)}^{2}}\le {{\left( \pm \sqrt{-3{{k}^{2}}-16k-20} \right)}^{2}}\le {{\left( -k \right)}^{2}}$
$\Rightarrow {{k}^{2}}+4k+4\le -3{{k}^{2}}-16k-20\le {{k}^{2}}$ …………………………………………………..(vii)
Now, taking the first two terms of the above equation, we get
${{k}^{2}}+4k+4\le -3{{k}^{2}}-16k-20$
${{k}^{2}}+4k+4+3{{k}^{2}}+16k+20\le 0$
$4{{k}^{2}}+20k+24\le 0$
On dividing the whole equation by 4, we get
${{k}^{2}}+5k+6\le 0$
Factorizing the above equation, we get
$\Rightarrow {{k}^{2}}+3k+2k+6\le 0$
$\Rightarrow k\left( k+3 \right)+2\left( k+3 \right)\le 0$
$\Rightarrow \left( k+2 \right)\left( k+3 \right)\le 0$
$\Rightarrow \left( k-\left( -2 \right) \right)\left( k-\left( -3 \right) \right)\le 0$ …………………………………..(viii)
As, we know if $\left( x-\alpha \right)\left( x-\beta \right)\le 0$ then $x\in \left[ \alpha ,\beta \right]$ , where $\alpha <\beta $
Hence, we get value of k from equation (viii), we get
$k\in \left[ -3,-2 \right]$
Now, taking last two terms of equation (vii), we get
$-3{{k}^{2}}-16k-20\le {{k}^{2}}$
$0\le {{k}^{2}}+3{{k}^{2}}+16k+20$
$4{{k}^{2}}+16k+20\ge 0$
On dividing the whole equation by 4, we get
${{k}^{2}}+4k+5\ge 0$ …………………………………….(ix)
Now, as we know discriminant of any quadratic $a{{x}^{2}}+bx+c=0$ is given as
$D={{b}^{2}}-4ac$
If roots are real, then $D\ge 0$ and if roots are imaginary, the $D<0$ .
Now, discriminant of equation (ix) is given as
$\Rightarrow D={{\left( 4 \right)}^{2}}-4\times 5=16-20=-4$
$\Rightarrow D=-4$ i.e. $D<0$
Hence, roots will not be real so, we cannot factorize equation (ix). So, we can write equation (ix) as
$\Rightarrow {{k}^{2}}+4k+4+1\ge 0$
$\Rightarrow {{\left( k+2 \right)}^{2}}+1\ge 0$
Now, as ${{\left( k+2 \right)}^{2}}$ will always be positive for any $k\in R$ So, equation (ix) will be true for $k\in R$.
Hence, we need to take intersection of $k\in \left[ \dfrac{-10}{3},-2 \right]$ and $k\in \left[ -3,-2 \right]$ and $k\in R$ .
Hence, we get the intersection of them as $k\in \left[ -3,-2 \right]$ .
Hence, option (D) is the correct answer.

Note: One may use only $D={{b}^{2}}-4ac>0$ to get the range of k for having real roots of the given equation, which is wrong. As ${{\sin }^{2}}x$ has a fixed range from 0 to 1. So, finding roots of the given expression will only help to get possible values of k. So, be careful with it.
As ${{\sin }^{2}}x$ has two possible values in the solution. So, one may think there are two possible inequalities, one when we take positive sign of $\dfrac{\left( k+2 \right)\pm \sqrt{-3{{k}^{2}}-16k-20}}{2}$ and another using negative sign of the above term. It will be a lengthier approach, so put $\dfrac{\left( k+2 \right)\pm \sqrt{-3{{k}^{2}}-16k-20}}{2}$ from 0 to 1. It will be less time taking. So, be careful with this step.