A) 374
B) 742
C) 472
D) 247
E) 274
Correct Answer: E
Solution :
\[y=(x+1)(x+2)(x+3)(x+4)(x+5)\] ...(i) Taking log on both sides, \[\log y=\log (x+1)+\log (x+2)+\log (x+3)\] \[+\log (x+4)+\log (x+5)\] Differentiating w.r.t.\[x\]on both sides, \[\frac{1}{y}.\frac{dy}{dx}=\frac{1}{(x+1)}+\frac{1}{(x+2)}+\frac{1}{(x+3)}\] \[+\frac{1}{(x+4)}+\frac{1}{(x+5)}\] \[\frac{dy}{dx}=y\left\{ \frac{1}{(x+1)}+\frac{1}{(x+2)}+\frac{1}{(x+3)} \right.\] \[\left. +\frac{1}{(x+4)}+\frac{1}{(x+5)} \right\}\]?. (ii) \[[\because {{y}_{at(x=0)}}=1.2.3.4.5=120]\] From Eq. (ii), \[{{\left( \frac{dy}{dx} \right)}_{at(x=1)}}={{y}_{(x=1)}}\left\{ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right\}\] \[=120\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5} \right)\] \[=120+60+40+30+24\] \[=274\]
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