Let a, b, c 1, 2, 3, 4.

Question

Let a, b, c \in  {1, 2, 3, 4}. If the probability, that $a{x}^{2}+2\sqrt{2}bx+c>0$ for all $x\in R$, is $\frac{m}{n}$, $gcd(m,n)=1$, then $m+n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 81

Accepted Answer

For the quadratic expression

$a{x}^{2}+2\sqrt{2}\, bx+c$

to be positive for all $x\in R$, we use the NCERT condition for a quadratic $A{x}^{2}+Bx+C$ to satisfy

$A{x}^{2}+Bx+C>0\, x\in R,$

which is:

$A>0$

Discriminant $

Here,

$A=a,\, B=2\sqrt{2}\, b,\, C=c$

Since $a\in {1,2,3,4}$, we always have

$a>0$

So we only need

$D={B}\^{2}-4ACNow,

$D=(2\sqrt{2}\, b{)}^{2}-4ac=8{b}^{2}-4ac$

Thus,

$8{b}^{2}-4acDividing by $4$:

$2{b}\^{2}-acSo the required condition is

$ac>2{b}^{2}$

Now $a,b,c\in {1,2,3,4}$, so total number of choices is

$4	imes 4	imes 4=64$

We count favourable triples $(a,b,c)$.

Case 1: $b=1$
Then

$ac>2$

Now $a,c\in {1,2,3,4}$. Total pairs $(a,c)$ are $16$.

We exclude pairs with $ac\le 2$:

$(1,1)$ gives $ac=1$

$(1,2)$ gives $ac=2$

$(2,1)$ gives $ac=2$

So favourable pairs:

$16-3=13$

Case 2: $b=2$
Then

$ac>8$

Possible products greater than $8$ are obtained from:

$(3,3)$ gives $9$

$(3,4)$ gives $12$

$(4,3)$ gives $12$

$(4,4)$ gives $16$

So favourable pairs = $4$.

Case 3: $b=3$
Then

$ac>18$

But maximum possible value of $ac$ is

$4\cdot 4=16$

So no favourable pair.

Case 4: $b=4$
Then

$ac>32$

Again impossible, since maximum $ac=16$.

So no favourable pair.

Hence total favourable triples are

$13+4+0+0=17$

Therefore, the required probability is

$\frac{17}{64}$

Thus,

$m=17,\, n=64$

and

$m+n=17+64=81$

So the answer is

$81$

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