JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 7)
Let f : R $$ \to $$ R be a function defined as
$$f(x) = \left\{ \matrix{ {{\sin (a + 1)x + \sin 2x} \over {2x}},if\,x < 0 \hfill \cr b,\,if\,x\, = 0 \hfill \cr {{\sqrt {x + b{x^3}} - \sqrt x } \over {b{x^{5/2}}}},\,if\,x > 0 \hfill \cr} \right.$$
If f is continuous at x = 0, then the value of a + b is equal to :
$$f(x) = \left\{ \matrix{ {{\sin (a + 1)x + \sin 2x} \over {2x}},if\,x < 0 \hfill \cr b,\,if\,x\, = 0 \hfill \cr {{\sqrt {x + b{x^3}} - \sqrt x } \over {b{x^{5/2}}}},\,if\,x > 0 \hfill \cr} \right.$$
If f is continuous at x = 0, then the value of a + b is equal to :
$$-$$3
$$-$$2
$$ - {5 \over 2}$$
$$ - {3 \over 2}$$
Explanation
Given, $f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x}, & x<0 \\ b, & x=0 \\ \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}, & x>0\end{array}\right.$
$$ \begin{array}{ll} \because & f(x) \text { is continuous at } x=0 . \\\\ \therefore & \lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{+}} f(x)=f(0) \\\\ \because & f(0)=b \end{array} $$
Now, $\lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x+\sin 2 x}{2 x}\right)$
$$ \begin{aligned} \Rightarrow \quad \lim _\limits{x \rightarrow 0^{-}} f(x) & =\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{2 x}+\frac{\sin 2 x}{2 x}\right) \\\\ & =\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{(a+1) x} \times\left(\frac{a+1}{2}\right)+\frac{\sin 2 x}{2 x}\right) \\\\ & =\frac{a+1}{2}+1 \end{aligned} $$
Again, $\lim _\limits{x \rightarrow 0^{+}} f(x)=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}\right)$
$$ =\lim _\limits{x \rightarrow 0^{+}} \frac{\left(\sqrt{x+b x^3}-\sqrt{x}\right)\left(\sqrt{x+b x^3}+\sqrt{x}\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)} $$
$$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\left(x+b x^3-x\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)} \\\\ & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b x^2}+1\right)} \end{aligned} $$
$$ \Rightarrow \quad \lim _\limits{x \rightarrow 0^{+}} f(x)=1 / 2 $$
From Eq. (i), (ii), (iii) and (iv)
$$ \begin{aligned} &\frac{1}{2} =b=\frac{a+1}{2}+1 \Rightarrow b=\frac{1}{2}, a=-2 \\\\ &\therefore \quad a+b =\frac{-3}{2} \end{aligned} $$
$$ \begin{array}{ll} \because & f(x) \text { is continuous at } x=0 . \\\\ \therefore & \lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{+}} f(x)=f(0) \\\\ \because & f(0)=b \end{array} $$
Now, $\lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x+\sin 2 x}{2 x}\right)$
$$ \begin{aligned} \Rightarrow \quad \lim _\limits{x \rightarrow 0^{-}} f(x) & =\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{2 x}+\frac{\sin 2 x}{2 x}\right) \\\\ & =\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{(a+1) x} \times\left(\frac{a+1}{2}\right)+\frac{\sin 2 x}{2 x}\right) \\\\ & =\frac{a+1}{2}+1 \end{aligned} $$
Again, $\lim _\limits{x \rightarrow 0^{+}} f(x)=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}\right)$
$$ =\lim _\limits{x \rightarrow 0^{+}} \frac{\left(\sqrt{x+b x^3}-\sqrt{x}\right)\left(\sqrt{x+b x^3}+\sqrt{x}\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)} $$
$$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\left(x+b x^3-x\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)} \\\\ & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b x^2}+1\right)} \end{aligned} $$
$$ \Rightarrow \quad \lim _\limits{x \rightarrow 0^{+}} f(x)=1 / 2 $$
From Eq. (i), (ii), (iii) and (iv)
$$ \begin{aligned} &\frac{1}{2} =b=\frac{a+1}{2}+1 \Rightarrow b=\frac{1}{2}, a=-2 \\\\ &\therefore \quad a+b =\frac{-3}{2} \end{aligned} $$
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