Question

For all ${}^{\prime }{x}^{\prime },x^2+2ax+(10-3a)>0$, then the interval in which ${}^{\prime }{a}^{\prime }$ lies is

• A. $a<-5$
• B. $-5<a<2$
• C. $a>5$
• D. $2<a<5$

Answer 1

Answer: (B)

$-5<a<2$

Given equation-

$x^2+2ax+(10-3a)>0$ for all $x\in R$.

Here, $A=1,B=2a,C=(10-3a)$

As we know that $A{x}^2+Bx+C>0$ for all $x\in R$ if-

$A>0$ and $D<0$

$A=1>0$

Now,

$D<0$

$B^2-4AC<0$

${\left(2a\right)}^2-4\left(1\right)(10-3a)<0$

$4a^2-40+12a<0$

$a^2+3a-10=0$

$(a+5)(a-2)<0$

$a\in (-5,2)$

$\therefore -5<a<2$