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

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(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

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$0$

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(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 -

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