Functions TG EAPCET Previous Year Chapters wise Questions

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To determine the range of the function $f(x)=\log_3 (5+4x-x^2)$, we need to analyze the quadratic expression $5+4x-x^2$.The expression $5+4x-x^2$ is a quadratic function of the form $ax^2+bx+c$, where $a=-1, b=4$, and $c=5$.Since $a=-1<0$, the parabola opens downwards, and the quadratic function achieves its maximum value at its vertex.The maximum value of a quadratic function $ax^2+bx+c$ is given by $-\frac{D}{4a}$, where $D=b^2-4ac$ is the discriminant.Calculate the discriminant:$D=4^2-4 \times (-1) \times 5=16+20=36$The maximum value of the quadratic $5+4x-x^2$ is:$-\frac{D}{4a}=-\frac{36}{4 \times (-1)}=9$This maximum is attained at $x=2$, and as $x \to -1^+$ or $x \to 5^-$ the expression $5+4x-x^2 \to 0^+$, so its range is $(0, 9]$.Considering the function $f(x)=\log_3 (5+4x-x^2)$, the possible values of $5+4x-x^2$ determine the input values for the logarithm.Since $0 < 5+4x-x^2 \leq 9$ and $\log_3$ is an increasing function, we get:$\log_3 (5+4x-x^2) \leq \log_3(9)$Given that $\log_3(9)=2$, we conclude:$f(x)=\log_3 (5+4x-x^2) \leq 2$Also, as $5+4x-x^2 \to 0^+$, we have $\log_3 (5+4x-x^2) \to -\infty$.Thus, the range of $f(x)$ is $(-\infty, 2]$.