JEE Main & Advanced Mathematics Equations and Inequalities Question Bank Self Evaluation Test - Linear Inequalities

The least integer a, for which \[1+{{\log }_{5}}({{x}^{2}}+1)\le lo{{g}_{5}}(a{{x}^{2}}+4x+a)\] is true for all \[x\in R\] is

A) 6

B) 7

C) 10

D) 1

Correct Answer: B

Solution :

[b] For the validity of inequality \[a{{x}^{2}}+4x+a>0,\]

which is possible if a>0 and \[16-4{{a}^{2}}<0\]\[\Rightarrow a>2\] ?. (1)

Further, the inequality can be rewritten as

\[{{\log }_{5}}5+{{\log }_{5}}({{x}^{2}}+1)\le lo{{g}_{5}}(a{{x}^{2}}+4x+a)\]

\[\Rightarrow 5({{x}^{2}}+1)\le a{{x}^{2}}+4x+a\]

\[\Rightarrow (a-5){{x}^{2}}+4x+(a-5)\ge 0.\]

It holds if \[a-5>0\] and \[16-4{{(a-5)}^{2}}\le 0\]

\[\Rightarrow a>5\] And \[a\le 3\] or \[a\ge 7\Rightarrow a\ge 7\] ? (2)

Combining the results of (1) and (2) for common values, we get \[a\in [7,\infty )\]