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
This question is part of PrepWiser's free JEE Main question bank. 162 more solved questions on Limits, Continuity and Differentiability are available — start with the harder ones if your accuracy is >70%.