JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 20)
Let a, b $$\in$$ R, b $$\in$$ 0, Define a function
$$f(x) = \left\{ {\matrix{ {a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr {{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr } } \right.$$.
If f is continuous at x = 0, then 10 $$-$$ ab is equal to ________________.
$$f(x) = \left\{ {\matrix{ {a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr {{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr } } \right.$$.
If f is continuous at x = 0, then 10 $$-$$ ab is equal to ________________.
Answer
14
Explanation
$$f(x) = \left\{ {\matrix{
{a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr
{{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr
} } \right.$$
For continuity at '0'
$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{\tan 2x - \sin 2x} \over {b{x^3}}} = - a$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{{{8{x^3}} \over 3} + {{8{x^3}} \over {3!}}} \over {b{x^3}}} = - a$$
$$ \Rightarrow 8\left( {{1 \over 3} + {1 \over {3!}}} \right) = - ab$$
$$ \Rightarrow 4 = - ab$$
$$ \Rightarrow 10 - ab = 14$$
For continuity at '0'
$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{\tan 2x - \sin 2x} \over {b{x^3}}} = - a$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{{{8{x^3}} \over 3} + {{8{x^3}} \over {3!}}} \over {b{x^3}}} = - a$$
$$ \Rightarrow 8\left( {{1 \over 3} + {1 \over {3!}}} \right) = - ab$$
$$ \Rightarrow 4 = - ab$$
$$ \Rightarrow 10 - ab = 14$$
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