If domain of $f(x)=\sqrt{x^2+b x+4}$ is $\mathbf{R}$, then maximum possible integral value of $b$ is

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

If domain of

f(x)=\sqrt{x^2+b x+4}

\mathbf{R}

, then maximum possible integral value of

b

View answer and explanation

Correct option: C

Option C (4) is correct because it is the maximum integer value of b for which the discriminant of the quadratic inside the square root is non-positive, ensuring the expression is defined for all real numbers.

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The solution of the equation

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is If

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and

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then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

and

{\alpha _1},{\beta _1}

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

then the equation whose roots are

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

is If

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

for all values of

x

, then

a

lies in the interval If

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

and

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

and

\;a \ne 0,

then Root(s) of the equation

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

belonging to the domain of definition of the function

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

is (are)