Let `f:[-1/2,2] rarr R` and `g:[-1/2,2] rarr R` be functions defined by `f(x)=[x^2-3]` and `g(x)=|x|f(x)+|4x-7|f(x)`, where [y] denotes the greatest integer less than or equal to y for `yinR`. Then,A. f is discontinuous exactly at three points in [-1/2,2]B. f is discontinuous exactly at four points in [-1/2,2]C. g is not differentiable exactly at four points in [-1/2,2]D. g is not differentiable exactly at five points in [-1/2,2]

Let `f:[-1/2,2] rarr R` and `g:[-1/2,2] rarr R` be functions defined b

1.	Let `f:[-1/2,2] rarr R` and `g:[-1/2,2] rarr R` be functions defined by `f(x)=[x^2-3]` and `g(x)=\|x\|f(x)+\|4x-7\|f(x)`, where [y] denotes the greatest integer less than or equal to y for `yinR`. Then,A. f is discontinuous exactly at three points in [-1/2,2]B. f is discontinuous exactly at four points in [-1/2,2]C. g is not differentiable exactly at four points in [-1/2,2]D. g is not differentiable exactly at five points in [-1/2,2]
Answer» Correct Answer - B::C We have `f(x)=[x^(2)-3]=[x^(2)]-3` `and g(x)=(\|x\|+4x-7\|)f(x)` `Now, x in [-1//2,2] Rightarrow 0 le x^(2) le 4` So, `f(x)=[x^(2)-3` is discontinuous and hence non-differentiable at `x=1,sqrt2,sqrt3 and 2 "in "[-1//2,2]` Since, `g(x)=(\|x\|+4x-7\|)f(x)`. So, g(x) is not continuous and hence non-differentiable at `x=1,sqrt2,sqrt3 " in "(-1//2,2)` In the left neighbourhood of x=0, we find that `g(x)=(-5x+7)(-3)=15x-21` In the right neighbourhood of x=0, we have `g(x)=(-3x+7)(-3)=9x-21` Clearly, g is not differentiable at x=0 In the neighbourhood of x=7//4 `g(x)=(-3x+7)xx0=0` In the right neighbourhood f x=7/4. `g(x)=(-5x-7)xx0=0` So, g(x) is differentiable at x=0. Hence, g(s) is not differentiable at `x=0,1,sqrt2,7//4" in "(-1//2,2)`