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P(x)=ax^3+bx^2+cx+d. P(x) a. |a|+|b|+|c|+|d| c. |a|+|b|+|c|-|d|>=7 d. |a|+|b|+|c|+|d|>=7 Pls explain.

P(x)=ax^3+bx^2+cx+d. P(x)<=1 for all x, mod(x)<=1. Then


a. |a|+|b|+|c|+|d|<=7      b. |a|+|b|+|c|-|d|<=7     


c. |a|+|b|+|c|-|d|>=7       d. |a|+|b|+|c|+|d|>=7     


Pls explain.

Grade:12

2 Answers

ishwar lohar
18 Points
9 years ago

take x=1

then P(x) becomes

a+b+c+d-1<=0

Take

a=1

b=2

c=3

d=5

check the options which satisfy above values.

we get ans B.

Shyama P
35 Points
9 years ago

But when,

a=1

b=2

c=3

d=5

a+b+c+d-1<=0 is violated ?

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