JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 21)
Let g : R $$\to$$ R be given by $$g(x) = f(x + 2) - f(x - 2)$$. If n and m denote the number of points in R where g is not continuous and not differentiable, respectively, then n + m is equal to ______________.
Explanation
$$f(x) = \left\{ {\matrix{ {3\left( {{{1 - \left| x \right|} \over 2}} \right)} & {if\,\left| x \right| \le 2} \cr 0 & {if\,\left| x \right| > 2} \cr } } \right.$$
$$g(x) = f(x + 2) - f(x - 2)$$
$$f(x) = \left\{ {\matrix{ {0,} & {x < - 2} \cr {{3 \over 2}(1 + x),} & { - 2 \le x < 0} \cr {{3 \over 2}(1 - x),} & {0 \le x < 2} \cr {0,} & {x > 2} \cr } } \right.$$
$$f(x + 2) = \left\{ {\matrix{ {0,} & {x < - 4} \cr {{3 \over 2}( 3 + x),} & { - 4 \le x < - 2} \cr {{3 \over 2}( - 1 - x),} & { - 2 \le x < 0} \cr {0,} & {x > 4} \cr } } \right.$$
$$f(x - 2) = \left\{ {\matrix{ {0,} & {x < 0} \cr {{3 \over 2}(x - 1),} & {0 \le x < 2} \cr {{3 \over 2}( - 1 - x),} & {2 \le x < 4} \cr {0,} & {x > 4} \cr } } \right.$$
$$g(x) = f(x + 2) + f(x - 2)$$
$$ = \left\{ {\matrix{ {{{3x} \over 2} + 6,} & { - 4 \le x \le 2} \cr { - {{3x} \over 2},} & { - 2 < x < 2} \cr {{{3x} \over 2} - 6,} & {2 \le x \le 4} \cr {0,} & {\left| x \right| > 4} \cr } } \right.$$
So, n = 0 and m = 4
$$\therefore$$ m + n = 4
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