JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 14)
If the function $$f(x) = \left\{ {\matrix{
{a|\pi - x| + 1,x \le 5} \cr
{b|x - \pi | + 3,x > 5} \cr
} } \right.$$
is continuous at x = 5, then the value of a – b is :-
is continuous at x = 5, then the value of a – b is :-
$${2 \over {\pi - 5 }}$$
$${2 \over {5 - \pi }}$$
$${-2 \over {\pi + 5 }}$$
$${2 \over {\pi + 5 }}$$
Explanation
As f(x) is continuous at x = 5 then
$$\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} f\left( {5 - h} \right) = f\left( 5 \right) = \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right)$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} \left( {a\left| {\pi - \left( {5 - h} \right)} \right| + 1} \right) = a\left| {\pi - 5} \right| + 1$$
= $$\mathop {\lim }\limits_{h \to 0} \left( {b\left| {5 + h - \pi } \right| + 3} \right)$$
$$ \Rightarrow $$ $${a\left| {\pi - 5} \right| + 1}$$ = $$a\left| {\pi - 5} \right| + 1$$
= $${b\left| {5 - \pi } \right| + 3}$$
Note : As $$\pi $$ = 3.14, then $$\pi $$ - 5 = 3.14 - 5 = -1.84 < 0
$$ \therefore $$ $${\left| {\pi - 5} \right|}$$ = - ($$\pi $$ - 5) and $${\left| {5 - \pi } \right|}$$ = (5 - $$\pi $$)
$$ \Rightarrow $$ $$ - a\left( {\pi - 5} \right) + 1$$ = $$ - a\left( {\pi - 5} \right) + 1$$ = $${b\left( {5 - \pi } \right) + 3}$$
$$ \Rightarrow $$ $$ - a\left( {\pi - 5} \right) + 1$$ = $${b\left( {5 - \pi } \right) + 3}$$
$$ \Rightarrow $$ $$\left( {a - b} \right)\left( {5 - \pi } \right) = 2$$
$$ \Rightarrow $$ $$\left( {a - b} \right) = {2 \over {\left( {5 - \pi } \right)}}$$
$$\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} f\left( {5 - h} \right) = f\left( 5 \right) = \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right)$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} \left( {a\left| {\pi - \left( {5 - h} \right)} \right| + 1} \right) = a\left| {\pi - 5} \right| + 1$$
= $$\mathop {\lim }\limits_{h \to 0} \left( {b\left| {5 + h - \pi } \right| + 3} \right)$$
$$ \Rightarrow $$ $${a\left| {\pi - 5} \right| + 1}$$ = $$a\left| {\pi - 5} \right| + 1$$
= $${b\left| {5 - \pi } \right| + 3}$$
Note : As $$\pi $$ = 3.14, then $$\pi $$ - 5 = 3.14 - 5 = -1.84 < 0
$$ \therefore $$ $${\left| {\pi - 5} \right|}$$ = - ($$\pi $$ - 5) and $${\left| {5 - \pi } \right|}$$ = (5 - $$\pi $$)
$$ \Rightarrow $$ $$ - a\left( {\pi - 5} \right) + 1$$ = $$ - a\left( {\pi - 5} \right) + 1$$ = $${b\left( {5 - \pi } \right) + 3}$$
$$ \Rightarrow $$ $$ - a\left( {\pi - 5} \right) + 1$$ = $${b\left( {5 - \pi } \right) + 3}$$
$$ \Rightarrow $$ $$\left( {a - b} \right)\left( {5 - \pi } \right) = 2$$
$$ \Rightarrow $$ $$\left( {a - b} \right) = {2 \over {\left( {5 - \pi } \right)}}$$
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