Let the function f(x)= (x2-1 ) |x2-ax+2 |+ |x| be not differentiable at the two points x= =2 and x= . Then the distance of the point ( , ) from the line 12x+5y+10=0 is equal to :
- 5
- 2
- 4
- 3
Answer
BONUS
Solution
|g(x)| has a corner where g=0; |x|= x is smooth
x=2 is a zero of x2-ax+2 ⇒ a=3 ⇒ (x-1)(x-2) ⇒ =1
Point (2,1): d= |12(2)+5(1)+10| 122+52= 3913=3
At x=1 the factor (x2-1) vanishes, so f is in fact differentiable there; the data are inconsistent ⇒ question dropped
Answer: Bonus (full marks to all who attempted)
Note: The official NTA answer key gives option (4). Our faculty team finds that no option is fully correct, so this question should be dropped (bonus: full marks to everyone who attempted it).