If the domain of the function f(x) = sqrt{ln( | vedclass

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

Question

(D) For the function $f(x) = \sqrt{\ln(m\sin x + 4)}$ to be defined for all $x \in R$,the expression inside the square root must be non-negative,i.e.,

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) For the function $f(x) = \sqrt{\ln(m\sin x + 4)}$ to be defined for all $x \in R$,the expression inside the square root must be non-negative,i.e.,$\ln(m\sin x + 4) \ge 0$.This implies $m\sin x + 4 \ge e^0$,which simplifies to $m\sin x + 4 \ge 1$.Thus,$m\sin x \ge -3$.Since the domain is $R$,this inequality must hold for all $x \in R$. The range of $\sin x$ is $[-1, 1]$.If $m > 0$,the minimum value of $m\sin x$ is $-m$,so $-m \ge -3 \Rightarrow m \le 3$.If $m If $m = 0$,$0 \ge -3$,which is true.Combining these,$m \in [-3, 3]$.The integral values of $m$ are $\{-3, -2, -1, 0, 1, 2, 3\}$.Counting these,there are $7$ possible integral values.

If the domain of the function $f(x) = \sqrt{\ln(m\sin x + 4)}$ is $R$,then the number of possible integral values of $m$ is:

Similar Questions

If the domain of the function $f(x) = \frac{[x]}{1+x^2}$,where $[x]$ is the greatest integer $\leq x$,is $(2, 6)$,then its range is

If $e^x + e^{f(x)} = e$, then the domain of $f(x)$ is

The range of the function $f(x) = -\sqrt{-x^2-6x-5}$ is

The domain of the function $f(x) = \sqrt{\frac{4-x^2}{[x]+2}}$,where $[x]$ denotes the greatest integer not more than $x$,is

If the domain of the function $f(x) = \frac{1}{\sqrt{10+3x-x^2}} + \frac{1}{\sqrt{x+|x|}}$ is $(a, b)$,then $(1+a)^2 + b^2$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module