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(A) For $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ to be defined for all $x \in R$, the expression inside the square root must be non-negative: $\ln(2\l

Vedclass · Accepted Answer

$5$

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(A) For $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ to be defined for all $x \in R$, the expression inside the square root must be non-negative: $\ln(2\lambda \cos x + 5) \geq 0$.This implies $2\lambda \cos x + 5 \geq e^0 = 1$, which simplifies to $2\lambda \cos x \geq -4$, or $\lambda \cos x \geq -2$.Since this must hold for all $x \in R$, we consider the range of $\cos x$, which is $[-1, 1]$.If $\lambda > 0$, the minimum value of $\lambda \cos x$ is $-\lambda$. Thus, $-\lambda \geq -2 \implies \lambda \leq 2$. So, $\lambda \in \{1, 2\}$.If $\lambda If $\lambda = 0$, the expression becomes $\ln(5) \geq 0$, which is true.Thus, the possible integral values for $\lambda$ are $\{-2, -1, 0, 1, 2\}$.The total number of such values is $5$.

The number of integral values of $\lambda$ for which the function $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ is defined for all $x \in R$ is:

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