Boost your Understanding -1

Take $2^{x}=y$ where $x \epsilon R, y >0$

This question gets transformed as follows :

How many integral values of $a$ lie in the interval $[-1000,1000]$ such that :

$y^{2}-ay + 3-a \le 0$

is satisfied by at least one $y>0$

For this inequality to be satisfied there must exist at least one positive root of the quadratic.

Hence $Discriminant \ge 0$

${a}^{2}+4a-12 \ge 0$

$\Rightarrow a \le -6$ or $a\ge 2$

Now since we want at least one positive root hence we can do it in two ways either make two cases as one positive and one negative root or both positive roots Or we can find the condition of getting both non-positive roots and exclude all those possible values of $a$ .

I will do it the second way.

Condition for both non-negative roots :

the value of quadratic at $y=0$ is greater than or equal to $0$ and the vertex of the parabola made by the quadratic in co-ordinate plane must have negative abscissa.

$\Rightarrow 3-a \ge 0$

$\Rightarrow \frac{a}{2} \le 0$

Finally $a \le 0$

Excluding these values of $a$ and taking the intersection of the rest of the values with the values of $a$ obtained in the condition $Discriminant >0$ we have :

$a \ge 2$

Finally giving us $999$ values.

The inequality can be rewritten as

$(4^x+3) \leq a(2^x+1)$

Noting that $2^x+1>0, \forall x$ ,

$\frac{4^x+3}{2^x+1} \leq a$

Since we are interested in solutions for $a$ , we need

$a \geq \min_x \frac{4^x+3}{2^x+1}$

Differentiating the function on the right and equating to zero

$\frac{d}{dx}\frac{4^x+3}{2^x+1}=\frac{4^x\ln(4)(2^x+1)-2^x\ln(2)(4^x+3)}{(2^x+1)^2}=0$

Solving,

$2^x\ln(2)(4^x+2\times 2^x-3)=0$

which implies

$((2^x)^2+2\times 2^x-3)=(2^x+3)(2^x-1)=0$

Since, $2^x>0$ , the minimizing solution is $2^x=1$ . Substituting we have

$a\geq \frac{1+3}{1+1} = 2$

Hence, the number of integer values of $a$ satisfying thw inequality = $1000-2+1=\boxed{999}$

$4^{ x }-a.2^{ x }-a+3\leq 0\\ Let\quad t=2^{ x }\\ Since\quad x\epsilon \left( { -\infty },{ \infty } \right) \rightarrow t\epsilon \left( { 0 },{ \infty } \right) \\ Rewriting\quad the\quad equation,\\ t^{ 2 }-at-a+3\leq 0\\ t=\frac { a\pm \sqrt { { (a+2) }^{ 2 }-16 } }{ 2 } \\ Condition\quad of\quad the\quad discriminant:\\ { (a+2) }^{ 2 }-16\ge 0\\ { (a+2) }^{ 2 }\ge 16\\ a\ge 2\quad or\quad a\le -6\\ Condition\quad for\quad t\quad to\quad satisfy\quad its\quad bounds\quad t\epsilon \left( { 0 },{ \infty } \right) \\ t\ge 0\\ a\pm \sqrt { { (a+2) }^{ 2 }-16 } \ge 0\\ \underline { Case\quad 1: } \\ a\quad is\quad positive\quad and\quad greater\quad than\quad 1\quad \\ There\quad is\quad at\quad least\quad one\quad root.\quad i.e.\\ \\ a+\sqrt { { (a+2) }^{ 2 }-16 } \ge 0\quad is\quad always\quad satisfied.\\ \underline { Case\quad 2: } \\ a\quad is\quad negative\quad and\quad lesser\quad than\quad 5.\\ a-\sqrt { { (a+2) }^{ 2 }-16 } \ge 0\quad will\quad never\quad hold.\\ a+\sqrt { { (a+2) }^{ 2 }-16 } \ge 0\quad have\quad to\quad check\quad if\quad this\quad will\quad hold.\\ For\quad the\quad above\quad condition\quad to\quad hold,\quad \\ \sqrt { { (a+2) }^{ 2 }-16 } \ge -a\\ \\ { (a+2) }^{ 2 }-16\ge { a }^{ 2 }\\ { a }^{ 2 }+4a-12\ge { a }^{ 2 }\\ a\ge 3\\ But,\quad our\quad intial\quad assumption\quad was\quad a\quad is\quad negative.\\ There\quad are\quad no\quad negative\quad values\quad of\quad a\quad \\ that\quad will\quad satisfy\quad the\quad conditions.\\ Hence,\quad a\epsilon \left[ 2,1000 \right] \Rightarrow \boxed { \boxed { 999\quad values } }$