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Consider the function $f(x)=\max(7-x,x+3).$ In which range does $f$ take its minimum value$?$

  1.  $-6\leq x<-2$
  2.  $-2\leq x<2$
  3.  $2\leq x<6$
  4. $6\leq x<10$
in Numerical Ability by Boss (47.5k points)
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3 Answers

+4 votes
Best answer

$ y = f(x) = \max\left( 7-x,x+3\right )\rightarrow(1)$

We can draw the graph



 To find the intersection point we can equate the two lines

$\implies7-x=x+3$
$\implies 2x = 4\implies x = 2$

Put $x = 2 $ in equation $(1)$ and get $y = 5$
$\implies$ Intersection point $(x,y)=(2,5)$

Among the given options only option C range includes $2.$

So, Correct answer is $(C).$

by Boss (47.5k points)
edited by
+1 vote

$\rightarrow $ If we try to decrease 7-x to a minimum then x+3 will increase and vice versa since x has opposite sign in both case.

$\rightarrow $ As a result the higher value among 7-x  and x+3  will be selected.

$\rightarrow $ So in order to cancel this effect both values should become same at some point and that should be the required answer.

$\rightarrow $  7-x = x+3 => 2x = 4 => x=2.

$\because$ x=2 comes in the range  2≤x<6

$\therefore$ Option C is the correct answer.

by Boss (18.3k points)
0 votes

$f(x)=\max(7-x,x+3)=\frac{|7-x+x+3|+|7-x-x-3|}{2}=\frac{|10|+|4-2x|}{2}=5+|2-x|\\ \therefore f(x)=5+|x-2|$

Now $f(x)$ will be minimum if $|x-2|$ is minimum.

For the option A, it is $-6\le x< -2 \Rightarrow -8\le x -2 < -4 \space \therefore |x-2| \in (4,8] $

 

For the option B, it is $-2\le x< 2 \Rightarrow -4\le x -2 < 0 \space \therefore |x-2| \in (0,4] $

 

For the option C, it is $2\le x< 6 \Rightarrow 0\le x -2 < 4 \space \therefore |x-2| \in [0,4) $

 

For the option D, it is $6\le x< 10 \Rightarrow 4\le x -2 < 8 \space \therefore |x-2| \in [4,8) $

 

Clearly, for the option C, the value of $|x-2|$ is minimum as $|x-2| \in [0,4)$. So the answer is C.

ago by (45 points)
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