The range of the real valued function f(x) = | vedclass

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(C) We know that for any real values of $x$,the expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.For $3 \

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$[1, 3]$

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(C) We know that for any real values of $x$,the expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.For $3 \sin x + 4 \cos x$,we have $a = 3$ and $b = 4$,so the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-5, 5]$.Adding $10$ to all parts,we get $-5 + 10 \leq 3 \sin x + 4 \cos x + 10 \leq 5 + 10$,which simplifies to $5 \leq 3 \sin x + 4 \cos x + 10 \leq 15$.Taking the reciprocal,the inequality reverses: $\frac{1}{15} \leq \frac{1}{3 \sin x + 4 \cos x + 10} \leq \frac{1}{5}$.Multiplying by $15$,we get $1 \leq \frac{15}{3 \sin x + 4 \cos x + 10} \leq 3$.Thus,the range of the function is $[1, 3]$.

Column $I$	Column $II$
$(A)$ If $-1 < x < 1$,then $f(x)$ satisfies	$(p)$ $0 < f(x) < 1$
$(B)$ If $1 < x < 2$,then $f(x)$ satisfies	$(q)$ $f(x) < 0$
$(C)$ If $3 < x < 5$,then $f(x)$ satisfies	$(r)$ $f(x) > 0$
$(D)$ If $x > 5$,then $f(x)$ satisfies	$(s)$ $f(x) < 1$

The range of the real valued function $f(x) = \frac{15}{3 \sin x + 4 \cos x + 10}$ is

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