Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The number of integers which $'a'$ can take so that $x^{2}-ax+1> 2a^{2}$ $\forall$ $x\; \epsilon \; R$ is

Option 1)
$0$

Option 2)
$1$

Option 3)
$2$

Option 4)
$3$

Answers (1)

$1\cdot x^{2}-ax+\left ( 1-2a^{2} \right )> 0\: \forall \: n\, \epsilon \, R$

$D< 0$ will be required condition

$a^{2}-4\left ( 1-2a^{2} \right )< 0$

$\Rightarrow \; 9a^{2}< 4\: \Rightarrow \: a\, \epsilon \, \left ( \frac{-2}{3}, \frac{2}{3} \right )$

$\therefore \: a=0$ is only integer.

$\therefore$ Option(B)

Sign of quadratic expression as positive. -

$ax^{2}+bx+c$ will be always positive for all $x\epsilon R$ , If $a> 0$ & $b^{2}-4ac< 0$ $\left ( Where\; a,b,c\; \epsilon\; R \right )$ .

- wherein

So, graph of $y= ax^{2}+bx+c$ will be always above x-axis so $ax^{2}+bx+c= 0$ has no real roots.

Option 1)

$0$

This is incorrect

Option 2)

$1$

This is correct

Option 3)

$2$

This is incorrect

Option 4)

$3$

This is incorrect

Posted by

subam

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions

anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question