Number of solution(s) of the equation ln(1 + | vedclass

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

Question

(A) The given equation is $\ln(1 + \sin^2 x) = 1 - \ln(5 + x^2)$.Rearranging the terms,we get $\ln(1 + \sin^2 x) + \ln(5 + x^2) = 1$.Using the propert

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The given equation is $\ln(1 + \sin^2 x) = 1 - \ln(5 + x^2)$.Rearranging the terms,we get $\ln(1 + \sin^2 x) + \ln(5 + x^2) = 1$.Using the property $\ln(a) + \ln(b) = \ln(ab)$,we have $\ln((1 + \sin^2 x)(5 + x^2)) = 1$.Taking the exponential of both sides,we get $(1 + \sin^2 x)(5 + x^2) = e^1 = e \approx 2.718$.Since $0 \le \sin^2 x \le 1$,the term $(1 + \sin^2 x)$ ranges from $1$ to $2$.Since $x^2 \ge 0$,the term $(5 + x^2)$ is at least $5$.Therefore,the product $(1 + \sin^2 x)(5 + x^2) \ge 1 	imes 5 = 5$.Since $5 > e$,there is no value of $x$ that satisfies the equation.Thus,the number of solutions is $0$.

Number of solution$(s)$ of the equation $\ln(1 + \sin^2 x) = 1 - \ln(5 + x^2)$ is -

Similar Questions

$\tanh^{-1}\left(\frac{1}{3}\right) + \coth^{-1}(2) = $

The value of $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \infty\right)}$ is equal to

The real root of the equation ${\log _4}\{ {\log _2}(\sqrt {x + 8} - \sqrt x )\} = 0$ is..........

If ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$,then in which interval does $x$ lie?

If $x = \log_{b}a$,$y = \log_{c}b$,and $z = \log_{a}c$,then the value of $xyz$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module