and Let fix) = ax + b for real a, b...
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and Let fix) = ax + b for real a, b and c with a + 0. If the vertical asymptote of y = f(x) is x = 4x+c the vertical asymptote of y = f'(x) is x = - find the value(s) that b can take on. AW

JEE/Engineering Exams
Maths
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Vertical asymptote as ( x rightarrow frac{-5}{4} ; y rightarrow pm infty ) vertical asymptote of ( f-1 ) = Move" a ental asymptote of ( f(x) ) ( therefore ) as ( quad y rightarrow frac{3}{4} ; x rightarrow pm infty ) (i) ( x rightarrow-frac{5}{4} quad y=frac{a x+b}{4 x+c} ) be 15 ( begin{aligned} c+4 x &=0 Rightarrow c-5 & Rightarrow c=5 end{aligned} quadleft[begin{array}{l}b cos ^{prime} t+a text { pest all it } c_{0}end{array}right. ) ( y rightarrow frac{3}{4} quad x rightarrow pm infty ) ( because quad y=frac{a x+b}{4 x+5}=frac{a+b / x}{4+5 / x} ) ( operatorname{as} x rightarrow pm infty ) ( y=frac{a}{4}=frac{3}{4} cdot 1, a=3 ) ( y=frac{3 x+6}{4 x+5} quad ) At ( x=-frac{5}{4} quad begin{array}{l}3 x+6 eq 0 Rightarrow 12+quad-3 times 65end{array} ) ( =frac{15}{4} )
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