If $x^2+2 a x+10-3 a>0$ for each $x \in R$ then

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

x^2+2 a x+10-3 a>0

for each

x \in R

then

View answer and explanation

Correct option: B

-5<\mathrm{a}<2

Option D (

-1 \leq x \leq 1

) is correct because the given equation simplifies to 0 for all x within the domain of the inverse trigonometric functions.

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\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)