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

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 ________________.

Answer (integer) 14

Solution

$$f(x) = \left\{ {\matrix{ {a\sin {\pi \over 2}(x - 1),} &amp; {for\,x \le 0} \cr {{{\tan 2x - \sin 2x} \over {b{x^3}}},} &amp; {for\,x &gt; 0} \cr } } \right.$$<br><br>For continuity at '0'<br><br>$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{\tan 2x - \sin 2x} \over {b{x^3}}} = - a$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{{{8{x^3}} \over 3} + {{8{x^3}} \over {3!}}} \over {b{x^3}}} = - a$$<br><br>$\Rightarrow 8\left( {{1 \over 3} + {1 \over {3!}}} \right) = - ab$<br><br>$\Rightarrow 4 = - ab$<br><br>$\Rightarrow 10 - ab = 14$

About this question

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

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