Number of integral values of lambda for which | vedclass

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

Question

(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

$5$

Vedclass · Answer

(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

Similar Questions

If $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right)$,then:

If $90^{\circ} < \alpha < 180^{\circ}$,$\sin \alpha = \frac{\sqrt{3}}{2}$ and $180^{\circ} < \beta < 270^{\circ}$,$\sin \beta = -\frac{\sqrt{3}}{2}$,then $\frac{4 \sin \alpha - 3 \tan \beta}{\tan \alpha + \sin \beta} = $

The value of $6(\sin^6 \theta + \cos^6 \theta) - 9(\sin^4 \theta + \cos^4 \theta) + 4$ is

If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is

Which of the following relations is correct for $0^{\circ} < \theta < 90^{\circ}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module