If f: R rightarrow R defined by f(x) = begin{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the function $f(x)$ to be continuous at $x=0$, the left-hand limit, right-hand limit, and the value of the function at $x=0$ must be equal.$1$

Vedclass · Accepted Answer

$b=2 \log |a|$

Vedclass · Answer

(A) For the function $f(x)$ to be continuous at $x=0$, the left-hand limit, right-hand limit, and the value of the function at $x=0$ must be equal.$1$. The value of the function at $x=0$ is $f(0) = a^2 \cos^2(0) + b^2 \sin^2(0) = a^2(1) + b^2(0) = a^2$.$2$. The right-hand limit $(RHL)$ as $x \rightarrow 0^+$ is $\lim_{x \rightarrow 0^+} e^{ax+b} = e^{a(0)+b} = e^b$.$3$. The left-hand limit $(LHL)$ as $x \rightarrow 0^-$ is $\lim_{x \rightarrow 0^-} (a^2 \cos^2 x + b^2 \sin^2 x) = a^2(1) + b^2(0) = a^2$.For continuity, $LHL = RHL = f(0)$, so $e^b = a^2$.Taking the natural logarithm on both sides, we get $\ln(e^b) = \ln(a^2)$, which simplifies to $b = 2 \ln |a|$ or $b = 2 \log |a|$.

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} a^2 \cos ^2 x+b^2 \sin ^2 x, & x \leq 0 \\ e^{ax+b}, & x>0 \end{cases}$ is a continuous function, then:

Similar Questions

Show that every polynomial function is continuous.

Let $f(x) = \begin{cases} -2 \sin x, & \text{if } x \leq -\frac{\pi}{2} \\ A \sin x + B, & \text{if } -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & \text{if } x \geq \frac{\pi}{2} \end{cases}$. For what values of $A$ and $B$ is $f$ continuous?

Discuss the continuity of the function $f$ given by
$f(x) = \begin{cases} x, & \text{if } x \ge 0 \\ x^2, & \text{if } x < 0 \end{cases}$

For $x > 0$,let $h(x) = \begin{cases} \frac{1}{q} & \text{if } x = \frac{p}{q} \text{ (where } p, q \in \mathbb{N} \text{ are relatively prime)} \\ 0 & \text{if } x \text{ is irrational} \end{cases}$. Which of the following statements does not hold true?

The values of $a, b, c$ for which the function $f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & x < 0 \\ c, & x = 0 \\ \frac{(x+bx^2)^{1/2} - \sqrt{x}}{bx^{1/2}}, & x > 0 \end{cases}$ is continuous at $x = 0$, are

Vedclass Test Series

Exam Paper Generator

Online Exam Module