UPSC NDA II 2024 Mathematics Paper

Concept:The minimum of a logarithmic function with an increasing base is achieved at the minimum of its argument (a quadratic).Explanation:We have $f(x)=\log_{10}(x^{2}+2x+11)$.Since $\log_{10}$ is an increasing function, the minimum of $f$ occurs where the quadratic inside the log is smallest.The quadratic $x^{2}+2x+11$ can be written as $(x+1)^{2}+10$.This is always at least $10$, and the minimum $10$ is reached at $x=-1$.Alternatively, set the derivative $f'(x)=0$: $\frac{2x+2}{x^{2}+2x+11}=0$, giving $x=-1$.Then $f(-1)=\log_{10}\big((-1)^{2}+2(-1)+11\big)=\log_{10}(1-2+11)=\log_{10}10=1$.Thus the minimum value of $f(x)$ is $1$.Answer:B. 1