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Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationMaximum & minimum valuesEasy2 minQB

MathematicsEasysingle choice

(L-1)If $0 < a < 5, 0 < b < 5$ and $\frac{x ^{2} + 5}{2} = x - 2 cos (a + b x)$ is satisfied for atleast one real $x$ , then value of $a + b$ may be equal to

Options:

Answer:

Solution:

$cos (a + b x) = - 1 - \frac{( x - 1 ) ^{2}}{4}$ exists only when $x = 1$

at $x = 1; a + b = π$

$cos (a + bx) = \frac{- ( x ^{2} - 2 x + 5 )}{4} = - 1 - \frac{( x - 1 ) ^{2}}{4}$

$\Rightarrow x = 1$

$\Rightarrow a + b = 5$

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Maximum & minimum values

⏱ 2mℹ️ Source: QB

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