Question

If \(k\in \mathbb{R}\) is such that the equation \[ 2\cosh^2x=3\sinh x+k \] has no real solution, then which of the following is correct?

• A. \(k<\frac{1}{2}\)
• B. \(k<\frac{3}{8}\)
• C. \(k<\frac{7}{8}\)
• D. \(k<\frac{5}{8}\)

Hint:

For equations involving \(\sinh x\) and \(\cosh x\), use \(\cosh^2x=1+\sinh^2x\) and then convert the equation into a quadratic.

Answer 1

The Correct Option is C

Step 1: Use the identity of hyperbolic functions.
We know that \[ \cosh^2x-\sinh^2x=1. \] Therefore, \[ \cosh^2x=1+\sinh^2x. \]

Step 2: Substitute \(\sinh x=t\).
Let \[ t=\sinh x. \] Since \(x\in \mathbb{R}\), \(\sinh x\) can take every real value, so \[ t\in \mathbb{R}. \] The equation becomes \[ 2(1+t^2)=3t+k. \] \[ 2t^2+2=3t+k. \] \[ 2t^2-3t+2-k=0. \]

Step 3: Apply the condition for no real solution.
For the quadratic equation \[ 2t^2-3t+2-k=0 \] to have no real solution, its discriminant must be negative.
So, \[ D<0. \] Here, \[ D=(-3)^2-4(2)(2-k). \] \[ D=9-8(2-k). \] \[ D=9-16+8k. \] \[ D=8k-7. \] For no real solution, \[ 8k-7<0. \] \[ 8k<7. \] \[ k<\frac{7}{8}. \]

Step 4: Final conclusion.
Therefore, \[ \boxed{k<\frac{7}{8}} \]