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

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(C) Given $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$.Factorizing the numerator and denominator: $f(x) = \frac{(x+5)(x-3)}{(2x+3)(x+5)}$.For $x 
e

Vedclass · Accepted Answer

$R - \left\{\frac{1}{2}, \frac{8}{7}\right\}$

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(C) Given $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$.Factorizing the numerator and denominator: $f(x) = \frac{(x+5)(x-3)}{(2x+3)(x+5)}$.For $x 
eq -5$,$f(x) = \frac{x-3}{2x+3}$.Let $y = \frac{x-3}{2x+3}$.$y(2x+3) = x-3$ $\Rightarrow 2xy + 3y = x - 3$ $\Rightarrow x(2y-1) = -3y - 3$.$x = \frac{-3y-3}{2y-1}$.Since $x$ is defined,$2y-1 
eq 0 \Rightarrow y 
eq \frac{1}{2}$.Also,$x 
eq -5 \Rightarrow \frac{-3y-3}{2y-1} 
eq -5$.$-3y-3 
eq -10y + 5$ $\Rightarrow 7y 
eq 8$ $\Rightarrow y 
eq \frac{8}{7}$.Thus,the range is $R - \left\{\frac{1}{2}, \frac{8}{7}ight\}$.

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 $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$ is

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