Search any question & find its solution
Question:
Answered & Verified by Expert
The range of the function $f(x)=-\sqrt{5-6 x-x^2}$ is
Options:
Solution:
2945 Upvotes
Verified Answer
The correct answer is:
$[-\sqrt{14}, 0]$
$f(x)=-\sqrt{5-6 x-x^2}$
Let $f(x)=y$
$\begin{aligned} & \Rightarrow \quad y=-\sqrt{-\left(x^2+6 x-5\right)} \\ & \Rightarrow-y=\sqrt{-\left(x^2+6 x+9-9-5\right)} \\ & \Rightarrow-y=\sqrt{-\left\{(x+3)^2-14\right\}}\end{aligned}$
On squaring both sides,
$y^2=14-(x+3)^2$
$\Rightarrow \quad(x+3)^2=14-y^2$
$\therefore \quad x=\sqrt{14-y^2}-3$
which is defined only when $y \in[-\sqrt{14}, \sqrt{14}$
But $f(x)$ will give only negative values.
Hence, range of $f(x)=[-\sqrt{14}, 0]$
Let $f(x)=y$
$\begin{aligned} & \Rightarrow \quad y=-\sqrt{-\left(x^2+6 x-5\right)} \\ & \Rightarrow-y=\sqrt{-\left(x^2+6 x+9-9-5\right)} \\ & \Rightarrow-y=\sqrt{-\left\{(x+3)^2-14\right\}}\end{aligned}$
On squaring both sides,
$y^2=14-(x+3)^2$
$\Rightarrow \quad(x+3)^2=14-y^2$
$\therefore \quad x=\sqrt{14-y^2}-3$
which is defined only when $y \in[-\sqrt{14}, \sqrt{14}$
But $f(x)$ will give only negative values.
Hence, range of $f(x)=[-\sqrt{14}, 0]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.