Question

The number of possible real solution(ss) of \(y\) in the equation \(y^2 - 2y\cos x + 1 = 0\) is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Divide the equation by y and recall that \(y+\dfrac{1}{y}\) never lies strictly between \(-2\) and \(2\) for a real, nonzero y.

Answer 1

The Correct Option is C

A quicker route uses the bound on $y+\frac{1}{y}$ for real, nonzero $y$, instead of computing a discriminant.

1. Rule out $y=0$: putting $y=0$ into $y^2-2y\cos x+1=0$ gives $1=0$, which is false, so $y=0$ is never a solution and we can safely divide the equation by $y$.
2. Divide the equation by $y$: this gives $y - 2\cos x + \frac{1}{y} = 0$, i.e. $y+\frac{1}{y} = 2\cos x$.
3. Bound the left side: for real, nonzero $y$, either $y>0$ and AM-GM gives $y+\frac{1}{y}\geq2$, or $y<0$, and applying AM-GM to $-y>0$ gives $y+\frac{1}{y}\leq-2$. So $y+\frac{1}{y}$ never falls strictly between $-2$ and $2$.
4. Bound the right side: $2\cos x$ always lies in $[-2,2]$ since $\cos x\in[-1,1]$.
5. Find where the two ranges meet: the only values common to 'at least $2$ or at most $-2$' and 'between $-2$ and $2$ inclusive' are exactly the two endpoints, $2$ and $-2$.

So $y+\frac{1}{y}=2$ forces $y=1$ (from $y^2-2y+1=(y-1)^2=0$), and $y+\frac{1}{y}=-2$ forces $y=-1$ (from $y^2+2y+1=(y+1)^2=0$). No other real value of $y$ can satisfy the original equation for any choice of $x$.

Let's summarize:

• Dividing by $y$ turns the equation into $y+\frac{1}{y}=2\cos x$.
• The left side only ever reaches $2$ or $-2$ among the values allowed on the right, giving exactly two possible real values of $y$: $1$ and $-1$.

So the number of possible real solutions of $y$ is $2$.