Practicing Success
If $0<a, b <5$ and $\frac{x^2+5}{2}=x-2 \cos (a + bx)$ has at least one real root, then the greatest value of $a + b$ is |
$π/2$ $π$ $3π$ $4π$ |
$3π$ |
We have, $\frac{x^2+5}{2}=x-2 \cos (a + bx)$ $⇒x-\frac{x^2+5}{2}=2 \cos (a + bx)$ $⇒\frac{2x^2-x^2-5}{2}=2 \cos (a + bx)$ $⇒\frac{1}{2}[-(x^2 - 2x) −5] = 2 \cos (a + bx)$ $⇒\frac{1}{2}[-(x - 1)^2 −4] = 2 \cos (a + bx)$ $⇒-2-\frac{1}{2}(x - 1)^2 = 2 \cos (a + bx)$ Clearly, LHS = $-2-\frac{1}{2}(x - 1)^2 ≤- 2$ and, RHS ≥ -2. Thus, the two sides are equal if each is equal to -2. Now, RHS = -2 $⇒ \cos (a - bx) = -1$ $⇒ \cos (a + bx) = \cos π$ $⇒a + bx = 2n\, π±π$ ...(i) and, $LHS=-2 ⇒ x=1$ ...(ii) Putting $x = 1$ in (i), we get $a+b=2n\,π± π, n∈Z$ $⇒a+b=π, 3 π$ $[∵ 0<a+b<10]$ Hence, the greates value of $a + b$ is $3π$. |