Question

Suppose the domain of the function $y=f\left(x\right)$ is $-1\le x\le 4$ and the range is $1\le y\le 10$. Let $g\left(x\right)=4-3f(x-2)$. If the domain of $g\left(x\right)$ is $a\le x\le b$ and range of $g\left(x\right)$ is $c\le x\le d$ then which of the following relation hold good?

• A. $2a+4b+c+d=0$
• B. $a+b+d=8$
• C. $5b+c+d=5$
• D. $a+b+c+d+18=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $a+b+d=8$
C $a+b+c+d+18=0$
D $5b+c+d=5$

Domain of f(x) is $-1\le x\le 4$
Given , $g\left(x\right)=4-3f(x-2)$
Here , $-1\le x-2\le 4$
$\Rightarrow 1\le x\le 6$
So, domain of g(x) is [1,6]. a=1,b=6
Range of f(x) is $1\le f\left(x\right)\le 10$
$\Rightarrow 1\le f(x-2)\le 10$
$\Rightarrow 3\le 3f(x-2)\le 30$
$\Rightarrow -30\le -3f(x-2)\le -3$
$\Rightarrow -26\le 4-3f(x-2)\le 1$
So,$c=-26,d=1$