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Jo)=x-ak-3x3dx x- a x (x-ax- 3x + 3a) x-a 3 -a o 3 3 o 3 x x(x-a) (x-3) (y3) )= x-d) x (x-3)+3) xaf) = how to solvejurther
jo)=x-Ak-3x3dX X- A X (x-Ax- 3x + 3A) X-A 3 -A O 3 3 O 3 X X(X-A) (X-3) (Y3) )= X-d) x (x-3)+3) XAf) = How to Solvejurther
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Deepak singh 1![best-answer]()
Refer attached option d is correct
if x= lamda, you can't cancel then what?
why we cant cancel.? If x =lembda , then we can solve in this way .
doubt
please answer
since f(x) is polynomial function and hence continuous on R . So f(lembda) = lim(x tends to lembda) f(x)
inverse of (x-lembda) is 1/(x-lembda) . So (x-lembda)/(x-lembda)= 1 . We can easily cancel as field is R .