A) 5
B) -1
C) 4
D) -6
Correct Answer: B
Solution :
(b): \[p(-1)=0\Rightarrow {{\left( -1 \right)}^{4}}+3a(-1)+b=0\] \[\Rightarrow +1-3a+b=0\] \[\Rightarrow -3a+b=1\] ...,(1) \[p\left( 1 \right)=0\Rightarrow {{1}^{4}}+3a\left( 1 \right)+b=0\Rightarrow 3a+b=-1\] .....(2) Solving (1) and (2) we get \[3+2b=-1\]You need to login to perform this action.
You will be redirected in
3 sec