Continuity Condition
For the function $f(x)$ to be continuous at $x = 0$, the limit of $f(x)$ as $x$ approaches 0 must exist and be equal to $f(0)$.
$ \lim_{x \to 0} f(x) = f(0) $
Given $f(0) = b$, we require:
$ \lim_{x \to 0} \frac{a|x| + x^2 - 2(\sin|x|)(\cos|x|)}{x} = b $
Simplify the trigonometric term: $2(\sin|x|)(\cos|x|) = \sin(2|x|)$.
$ \lim_{x \to 0} \frac{a|x| + x^2 - \sin(2|x|)}{x} = b $
Limit Calculation
We evaluate the limit using the left-hand limit (LHL) and the right-hand limit (RHL).
Right-Hand Limit (RHL)
For $x \to 0^+$, $|x| = x$. The limit becomes:
$ \lim_{x \to 0^+} \frac{ax + x^2 - \sin(2x)}{x} = \lim_{x \to 0^+} \left( \frac{ax}{x} + \frac{x^2}{x} - \frac{\sin(2x)}{x} \right) $
$ = \lim_{x \to 0^+} \left( a + x - \frac{\sin(2x)}{x} \right) $
Using the standard limit $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$, we get:
$ \text{RHL} = a + 0 - 2 = a - 2 $
Left-Hand Limit (LHL)
For $x \to 0^-$, $|x| = -x$. The limit becomes:
$ \lim_{x \to 0^-} \frac{a(-x) + x^2 - \sin(2(-x))}{x} = \lim_{x \to 0^-} \frac{-ax + x^2 + \sin(2x)}{x} $
$ = \lim_{x \to 0^-} \left( \frac{-ax}{x} + \frac{x^2}{x} + \frac{\sin(2x)}{x} \right) $
$ = \lim_{x \to 0^-} \left( -a + x + \frac{\sin(2x)}{x} \right) $
Using the standard limit, we get:
$ \text{LHL} = -a + 0 + 2 = -a + 2 $
Determining Constants a and b
For continuity, LHL = RHL:
$ a - 2 = -a + 2 $
Solving for $a$:
$ 2a = 4 \implies a = 2 $
The value of the limit is found by substituting $a=2$ into either LHL or RHL:
$ \text{Limit} = a - 2 = 2 - 2 = 0 $
Since the limit must equal $f(0)$, we have $b = 0$.
Final Calculation
We found $a = 2$ and $b = 0$. The question asks for $a + b$.
$ a + b = 2 + 0 = 2 $
Similar Questions
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is continuous at $x = 0$, then $a^2 + b^2$ is equal to
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
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Important Questions from Differential Calculus
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then. - If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
Let $[t]$ denote the greatest integer less than or equal to $t$. If the function
$f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$
is continuous at $x = 0$, then $a^2 + b^2$ is equal to
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
