Question

The least integral value of $a$ such that $\left( a-2 \right){{x}^{2}}+8x+a+4>0,\forall x\in R$ is.
(a) $3$
(b) $5$
(c) $4$
(d) $6$

Answer 1

Hint: For solving this question we will use the concept that for any quadratic function $f\left( x \right)=d{{x}^{2}}+bx+c$ if the coefficient of ${{x}^{2}}$ is greater than zero and its discriminant is less than zero then, for any real value of $x$ value of $f\left( x \right)>0$ always.

Complete step-by-step solution -
Given:
We have to find the least integral value of $a$ such that $\left( a-2 \right){{x}^{2}}+8x+a+4>0,\forall x\in R$ .
Now, before we proceed we should know that a quadratic function $f\left( x \right)=d{{x}^{2}}+bx+c$ will be always greater than zero for any real value of $x$ only if $d>0$ & ${{b}^{2}}-4dc>0$ .
Now, here it is given that, $\left( a-2 \right){{x}^{2}}+8x+a+4>0$ . So, here $d=a-2$ , $b=8$ and $c=a+4$ . And from the above discussion, we know that, $d>0$ & ${{b}^{2}}-4dc<0$ .
Now, we will solve each of the conditions separately and find the suitable values of $a$ .
Now, we solve for the condition $d>0$ where, $d=a-2$ . Then,
$\begin{aligned} & d>0 \\\ & \Rightarrow a-2>0 \\\ & \Rightarrow a>2 \\\ & \Rightarrow a\in \left( 2,\infty \right).................\left( 1 \right) \\\ \end{aligned}$
Now, we solve for the condition ${{b}^{2}}-4dc<0$ where, $d=a-2$ , $b=8$ and $c=a+4$ . Then,
$\begin{aligned} & {{b}^{2}}-4dc<0 \\\ & \Rightarrow {{8}^{2}}-4\left( a-2 \right)\left( a+4 \right)<0 \\\ & \Rightarrow 64-4\left( {{a}^{2}}+2a-8 \right)<0 \\\ & \Rightarrow 64-4{{a}^{2}}-8a+32<0 \\\ & \Rightarrow -4{{a}^{2}}-8a+96<0 \\\ & \Rightarrow 4{{a}^{2}}+8a-96>0 \\\ & \Rightarrow {{a}^{2}}+2a-24>0 \\\ & \Rightarrow {{a}^{2}}+6a-4a-24>0 \\\ & \Rightarrow a\left( a+6 \right)-4\left( a+6 \right)>0 \\\ & \Rightarrow \left( a+6 \right)\left( a-4 \right)>0 \\\ & \Rightarrow a<-6\text{ }\\!\\!\And\\!\\! \text{ }a>4 \\\ & \Rightarrow a\in \left( -\infty ,-6 \right)\bigcup \left( 4,\infty \right)...................\left( 2 \right) \\\ \end{aligned}$
Now, form (1) and (2) we have the following results:
$\begin{aligned} & a\in \left( 2,\infty \right) \\\ & a\in \left( -\infty ,-6 \right)\bigcup \left( 4,\infty \right) \\\ \end{aligned}$
Now, we will have to take the intersection of $a\in \left( 2,\infty \right)$ and $a\in \left( -\infty ,-6 \right)\bigcup \left( 4,\infty \right)$ . Then,
$\begin{aligned} & a\in \left[ \left( -\infty ,-6 \right)\bigcup \left( 4,\infty \right) \right]\bigcap \left( 2,\infty \right) \\\ & \Rightarrow a\in \left( 4,\infty \right) \\\ \end{aligned}$
Now, from the above result, we conclude that the value of $a$ should be greater than 4 so, that $a\in \left( 2,\infty \right)$ and $a\in \left( -\infty ,-6 \right)\bigcup \left( 4,\infty \right)$ . And ultimately $a\in \left( 4,\infty \right)$ if $\left( a-2 \right){{x}^{2}}+8x+a+4>0,\forall x\in R$ .
Now, it is obvious that the least integral value greater than 4 is 5 so, our final answer will be 5.
Thus, $a=5$ will be the least integral value of $a$ such that $\left( a-2 \right){{x}^{2}}+8x+a+4>0,\forall x\in R$.
Hence, option (b) will be the correct option.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to get the correct answer quickly. Moreover, we should be careful while solving inequalities and writing the suitable values of $a$ . And while giving the final answer we should be extra careful and don’t select the option (c) as it doesn’t lie in the set $a\in \left( 4,\infty \right)$.