Tim Vx+6 - sin(x-3)-3 273 (x-3) cos...
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Tim Vx+6 - sin(x-3)-3 273 (x-3) cos(x-3) is equal to : D

JEE/Engineering Exams
Maths
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( sqrt{x+6}-sin (x-3)-3 ) ( x rightarrow 3=(x-3) cos (x-3) ) Umit drectly we Applyng bra get So, go for L-Hopital Rule, ie, Differentiale both numerator & then apply bmity denonuinator, ( left.Rightarrow frac{1}{i}right)^{n} frac{d}{d x}(sqrt{n+6}-sin (x-3)-3] ) ( =frac{1}{2 sqrt{x+6}}-cos (x-3) ) ii) ( frac{d}{d x}[(x-3) cos (x-3)) ) ( =frac{(x-3)[-sin (x-1)]-cos (x-3)}{1} ) ( therefore frac{u}{x rightarrow 3} frac{sqrt{x+6}-sin (x-3)-3}{(x-3) cos (x-3)}=frac{4}{x rightarrow 3} frac{1}{2 sqrt{x+6}}-cos (x-3) ) ( frac{2 frac{1}{2 sqrt{9}}-frac{cos 0^{-}}{2.3}-1}{0-cos 0}=frac{1}{-1}=frac{1}{frac{5}{6}}=frac{1}{frac{1}{1}} )
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