13Yup done...!
Guess what, i took d/dx(xn)= nxn while solving in the test [16]....khair chodo ab kya kar sakte hain....thnx karna
If t1, t2, t3,...tn are the roots of the equation -
xn +px + q =0 , then
(t1- t2)(t1- t3)(t1- t4) ... (t1- tn) = ?
(a) n.p (b) n.t1n +p
(c) n.t1n-1 +p (d) n.t1 +p
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2 Answers
1P(x)=(x-t_{1})(x-t_{2})..............(x-t_{n})
\frac{P(x)}{(x-t_{1})}=(x-t_{2})..............(x-t_{n})\lim_{x\rightarrow t_{1}}\frac{P(x)}{(x-t_{1})}=(t_{1}-t_{2})(t_{1}-t_{3})..............(t_{1}-t_{n})
applying l'hospital's rule
\lim_{x\rightarrow t_{1}}\frac{P'(x)}{(1)}=(t_{1}-t_{2})(t_{1}-t_{3})..............(t_{1}-t_{n})
\frac{P'(t)}{(1)}=(t_{1}-t_{2})(t_{1}-t_{3})..............(t_{1}-t_{n})
hence asnswer is :
nt1n-1+p
13