Medium INTEGER +4 / -1 PYQ · JEE Mains 2021

Let f : R $\to$ R be a function defined as $$f(x) = \left\{ {\matrix{ {3\left( {1 - {{|x|} \over 2}} \right)} & {if} & {|x|\, \le 2} \cr 0 & {if} & {|x|\, > 2} \cr } } \right.$$

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 ______________.

Answer (integer) 4

Solution

<p>$$f(x) = \left\{ {\matrix{ {3\left( {{{1 - \left| x \right|} \over 2}} \right)} & {if\,\left| x \right| \le 2} \cr 0 & {if\,\left| x \right| > 2} \cr } } \right.$$</p> <p>$g(x) = f(x + 2) - f(x - 2)$</p> <p>$$f(x) = \left\{ {\matrix{ {0,} & {x < - 2} \cr {{3 \over 2}(1 + x),} & { - 2 \le x < 0} \cr {{3 \over 2}(1 - x),} & {0 \le x < 2} \cr {0,} & {x > 2} \cr } } \right.$$</p> <p>$$f(x + 2) = \left\{ {\matrix{ {0,} & {x < - 4} \cr {{3 \over 2}( 3 + x),} & { - 4 \le x < - 2} \cr {{3 \over 2}( - 1 - x),} & { - 2 \le x < 0} \cr {0,} & {x > 4} \cr } } \right.$$</p> <p>$$f(x - 2) = \left\{ {\matrix{ {0,} & {x < 0} \cr {{3 \over 2}(x - 1),} & {0 \le x < 2} \cr {{3 \over 2}( - 1 - x),} & {2 \le x < 4} \cr {0,} & {x > 4} \cr } } \right.$$</p> <p>$g(x) = f(x + 2) + f(x - 2)$</p> <p>$$ = \left\{ {\matrix{ {{{3x} \over 2} + 6,} & { - 4 \le x \le 2} \cr { - {{3x} \over 2},} & { - 2 < x < 2} \cr {{{3x} \over 2} - 6,} & {2 \le x \le 4} \cr {0,} & {\left| x \right| > 4} \cr } } \right.$$</p> <p>So, n = 0 and m = 4</p> <p>$\therefore$ m + n = 4</p>

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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