Question

Let $f\left(x\right)=|kx+5|$, domain of $f\left(x\right)$ is $[-5,7]$, domain of $f\left(h\right(x\left)\right)$ is $[-6,1]$ & range of $h\left(x\right)$ is the same as the domain of $f\left(x\right)$, then value of $k$ is

Answer 1

Domain of $f\left(x\right)$ i.e. $x$ lies between $-5\le x\le 7$
Domain of $f\left(h\left(x\right)\right)$ i.e., $h\left(x\right)$ lies $-6\le h\left(x\right)\le 1$
Range of $h\left(x\right)\Rightarrow f\left(h\left(x\right)\right)$
$-5\le |kh\left(x\right)+5|\le 7$
$-5\le kh\left(x\right)+5\le 7⟶1$
$-7\le kh\left(x\right)+5\le 5⟶2$
Now, $-6k\le kh\left(x\right)\le k$
if $k>0$
$-6k+5\le kh\left(x\right)+5\le k+5$
Which can be compared to
$-5\le kh\left(x\right)+5\le 7$
$-6k+5=-5\&k+5=7$
$\frac{10}{6}=k=\frac{5}{3}<k=2$
If k is negative then $-k\le kh\left(x\right)\le 6k$
$-k+5\le kh\left(x\right)+5\le 6k+5$
comparing with equation 2
$-k+5=-7,k=+12$
$6k+5=5$ then, $k=0$
So, if $k>0,$ then $k=2,\frac{5}{3}$
$k<0,$ then $k=0,k=+12$ (which cant be true)