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Statement -1 ax^(3)+bx+c=0 where a,b,c e...

Statement -1 `ax^(3)+bx+c=0` where `a,b,c epsilonR` cannot have 3 non-negative real roots.
Statement 2 Sum of roots is equal to zero.

A

Statement -1 is true, Statement -2 is true, Statement -2 is a correct explanation for Statement-1

B

Statement -1 is true, Statement -2 is true, Statement -2 is not a correct explanation for Statement -1

C

Statement -1 is true, Statement -2 is false

D

Statement -1 is false, Statement -2 is true

Text Solution

Verified by Experts

The correct Answer is:
A
Let `y=ax^(3)+bx+c`
`:.(dy)/(dx)=3ax^(2)+b`
For maximum or minimum `(dy)/(dx)=0` we get
`x=+-sqrt(-b/(3a))`
Case If `a gt0, bgt0` then `(dy)/(dx)gt0`
In this case function is increasing so it has exactly one root
Case II If `alt0, blt0` ten `(dy)/(dx)lt0`
In this case function is decreasing so it has exactly one root.
Case III `agt0, blt0` or `alo0, bgt0` then `y=ax^(3)+bx+c` is maximum at one point and minimum at other point.
Hence all roots can never be non -netative.
`:.` Statement -1 is false. But ltgtbrgt Sum of roots `=-("Coefficient of" x^(2))/("Coefficient of"x^(3))=0`
i.e. Statement -2 is true.
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