JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 14)
If $$f(x) = \left\{ {\matrix{
{{{\sin (a + 2)x + \sin x} \over x};} & {x < 0} \cr
{b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;} & {x = 0} \cr
{{{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{ {1 \over 3}}}} \over {{x^{{4 \over 3}}}}};} & {x > 0} \cr
} } \right.$$
is continuous at x = 0, then a + 2b is equal to :
is continuous at x = 0, then a + 2b is equal to :
0
-1
-2
1
Explanation
f(0-) = $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x + \sin x} \over x}$$
= $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x} \over {\left( {a + 2} \right)x}} \times \left( {a + 2} \right)$$ + $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin x} \over x}$$
= $$\left( {a + 2} \right)$$ + 1
= $$\left( {a + 3} \right)$$
f(0+) = $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{{1 \over 3}}}} \over {{x^{{4 \over 3}}}}}$$
= $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {1 + 3x} \right)}^{{1 \over 3}}} - 1} \over {{x^{{1 \over 3}}}}}$$
= $$\mathop {\lim }\limits_{x \to {0^ + }} {{1 + x - 1} \over x}$$
= 1
And f(0) = b
As f(x) is continuous at x = 0, then
f(0-) = f(0) = f(0+)
$$ \Rightarrow $$ $$a + 3$$ = b = 1
$$ \therefore $$ $$a$$ = -2 and b = 1
$$ \therefore $$ $$a$$ + 2b = -2 + 2 = 0
= $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x} \over {\left( {a + 2} \right)x}} \times \left( {a + 2} \right)$$ + $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin x} \over x}$$
= $$\left( {a + 2} \right)$$ + 1
= $$\left( {a + 3} \right)$$
f(0+) = $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{{1 \over 3}}}} \over {{x^{{4 \over 3}}}}}$$
= $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {1 + 3x} \right)}^{{1 \over 3}}} - 1} \over {{x^{{1 \over 3}}}}}$$
= $$\mathop {\lim }\limits_{x \to {0^ + }} {{1 + x - 1} \over x}$$
= 1
And f(0) = b
As f(x) is continuous at x = 0, then
f(0-) = f(0) = f(0+)
$$ \Rightarrow $$ $$a + 3$$ = b = 1
$$ \therefore $$ $$a$$ = -2 and b = 1
$$ \therefore $$ $$a$$ + 2b = -2 + 2 = 0
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