Number of integral values of lambda for which | vedclass

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

Vedclass · Accepted Answer

$5$

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(A) For the function $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ to be defined for all $x \in R$, the expression inside the square root must be non-negative, and the argument of the logarithm must be positive.$1$. $\ln(2\lambda \cos x + 5) \geq 0 \implies 2\lambda \cos x + 5 \geq e^0 = 1$.$2$. This simplifies to $2\lambda \cos x \geq -4$, or $\lambda \cos x \geq -2$.$3$. Since this must hold for all $x \in R$, we consider the range of $\cos x$, which is $[-1, 1]$.$4$. If $\lambda > 0$, the minimum value of $\lambda \cos x$ is $-\lambda$. Thus, $-\lambda \geq -2 \implies \lambda \leq 2$. So, $\lambda \in \{1, 2\}$.$5$. If $\lambda $6$. If $\lambda = 0$, the expression becomes $\ln(5) \geq 0$, which is true. So, $\lambda = 0$ is a solution.$7$. The set of integral values is $\{-2, -1, 0, 1, 2\}$.$8$. The total number of such integral values is $5$.

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

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