First slide

Theory of equations

Question

If $\frac{x^{2} + 5}{2} = x - 2 \cos (m + nx)$ has at least one real root, then

Moderate

A

number of possible values of x is two
B

number of possible values of x is one
C

the value of $m + n is (2 n + 1) π$
D

the value of m + n is $2 nπ$

Solution

Given equation is
$\begin{aligned} x^{2} + 5 - 2 x = - 4 \cos (m + nx) \\ \Rightarrow (x - 1)^{2} + 4 = - 4 \cos (m + nx) \end{aligned}$
$LHS \geq 4 and RHS \leq 4$
So, solution is possible only if LHS = RHS = 4
$\begin{array}{l} ∴ x = 1 and \cos (m + n) = - 1 . \\ ∴ m + n = (2 n + 1) π, n \in I \end{array}$

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