Question

The sum of the all integral value(s) of $kϵ[0,15]$ for which the inequality $1+lo{g}_5(x^2+1)\le lo{g}_5(k{x}^2+4x+k)$ is true for all $xϵR$ , is

• A. $120$
• B. $110$
• C. $105$
• D. $99$

Answer 1

The correct option is A $120$

Hint: Simplify the inequality using the properties of logarithm.

Step 1: Simplifying the given inequality.

Rewrite 1 as ${\log }_{5}5$, then apply the algebraic property of logarithms $\log m+\log n=\log \left(mn\right)$ and take the antilogarithm of both sides.

$\begin{array}{cc}{\log }_{5}5+{\log }_5(x^2+1)& \le {\log }_5(k{x}^2+4x+k)\\ {\log }_5\left[5\right(x^2+1\left)\right]& \le {\log }_5(k{x}^2+4x+k)\\ 5(x^2+1)& \le k{x}^2+4x+k\\ 5x^2+5& \le k{x}^2+4x+k\\ x^2(5-k)-4x+5-k& \le 0\end{array}$

Step 2: Obtaining the case for the solution of $k$.

The given inequality will be true if the leading coefficient of $x^2(5-k)-4x+5-k$ is less than or equal to 0, that is $5-k\le 0$ and the discriminant is less than or equal to 0, that is, $16-4(5-k{)}^2\le 0$.

If $5-k\le 0$, then $k\ge 5$.

The solution set to $5-k\le 0$ is $k\in [5,\infty )$

Step 3: Factorization of the discriminant.

Factor the expression $16-4(5-k{)}^2$.

$\begin{array}{cc}16-4(5-k{)}^2& =4(4-(5-k{)}^2]\\ & =4[2-(5-k\left)\right][2+(5-k\left)\right]\\ & =4(-3+k)(7-k)\end{array}$

If $4(-3+k)(7-k)\le 0$, then $k\ge 7$ or $k\le 3$.

Hence the solution set to $4(-3+k)(7-k)\le 0$ is $k\in (-\infty ,3]\cup [7,\infty )$.

Step 4: Calculation of the value of $k$ from the argument of logarithm.

The expression ${\log }_5(k{x}^2+4x+k)$ will output real values if $k{x}^2+4x+k>0$.

The quadratic expression $k{x}^2+4x+k$ will be greater than 0 if $k>0$ and $16-4k^2<0$.

Solve $16-4k^2<0$ for $k$.

$\begin{array}{cc}16-4k^2& <0\\ 4(4-k^2)& <0\\ 4-k^2& <0\\ (2-k)(2+k)& <0\end{array}$

The expression $(2-k)(2+k)$ will be less than 0 when $k<-2$ or $k>2$.

Hence the solution set to $16-4k^2<0$ is $k\in (-\infty ,-2)\cup (2,\infty )$.

Step 5: Calculation of common value of $k$ and their sum.

The common solution to all the above inequalities is $[7,\infty )$.

Add all the numbers from 7 to 15 to obtain the sum of all the values of $k$ for which the given inequality is true all $x\in R$.

$7+8+9+10+11+12+13+14+15=99$

Final step: The sum of all the values of $k$ is 99.