JEE ADVANCED MATHS QUESTION SEPTEMBER 2020
Answers
Let functions f : (-1, 1) → R and g : (-1, 1) → (-1, 1) be defined by,
f(x) = |2x - 1| + |2x + 1| and g(x) = x - [x].
where [x] denotes the greatest integer less than or equal to x. Let f o g : (-1, 1) → R be the composite function defined by (f o g)(x) = f(g(x)). suppose c is the number of points in the integral (-1, 1) at which f o g is NOT continuous and d is the number of points in the interval (-1, 1) at which f o g is NOT differentiable. The value of c + d is...
solution : here f(x) = |2x - 1| + |2x + |
g(x) = x - [x] = {x} where {.} is fraction part.
now f o g = f(g(x)) = |2{x} - 1| + |2{x} + 1|
take 2{x} -1 = 0 ⇒{x} = 1/2
case 1 : {x} ≤ 1/2
f o g = |2 × 0 - 1| + |2 × 0 + 1 | = 2
case 2 : {x} > 1/2
f o g = 2{x} -1 + 2{x} + 1 = 4{x}
Draw the graph using above conditions.
see the graph as shown in figure. it is clear that f o g is discontinuous at x = 0 so c = 1
not differentiable at x = 1/2, 0, -1/2 so d = 3 [ remember at sharp point function cannot be differentiable. ]
so c + d = 1 + 3 = 4
