If ${e^{\cos x}} - {e^{ - \cos x}} = 4,$ then the value of $\cos x$ is

${\log _e}\left( {2 + \sqrt 5 } \right)$

$ - {\log _e}\left( {2 + \sqrt 5 } \right)$

${\log _e}\left( { - 2 + \sqrt 5 } \right)$

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

{e^{\cos x}} - {e^{ - \cos x}} = 4,

then the value of

\cos x

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derived value of

\cos x

is greater than 1, which is not possible for the cosine function.

Practice All Chapter MCQs →

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)