## Polynomials

Question 1 |

1 | |

2 | |

3 | |

4 |

Question 1 Explanation:

P(x) should be like

Minimum degree of P(x) = 4

Maximum degree of P(x) = 5

Minimum degree of P(x) = 4

Maximum degree of P(x) = 5

Question 2 |

A polynomial p(x) is such that p(0) = 5, p(1) = 4, p(2) = 9 and p(3) = 20. The
minimum degree it can have is

1 | |

2 | |

3 | |

4 |

Question 2 Explanation:

Lets take p(x) = ax + b

p(0) = 5 ⇒ b = 5

p(1) = 4 ⇒ a+b = 4 ⇒ a = -1

p(2) = 9 ⇒ 40+b = 9 ⇒ -4+5 = 9, which is false.

So degree 1 is not possible.

Lets take p(x) = ax

p(0) = 5 ⇒ c = 5

p(1) = 4 ⇒ a+b+c = 4 ⇒ a+b = -1 -----(1)

p(2) = 9 ⇒ 4a+2b+c = 9 ⇒ 2a+b = 2 -----(2)

(2) - (1)

⇒ a = 3, b = -1-1 = -4

p(3) = 20 ⇒ 9a+3b+c = 20

⇒ 27-12+5 = 20

⇒ 20 = 20, True

Hence, minimum degree it can have.

p(0) = 5 ⇒ b = 5

p(1) = 4 ⇒ a+b = 4 ⇒ a = -1

p(2) = 9 ⇒ 40+b = 9 ⇒ -4+5 = 9, which is false.

So degree 1 is not possible.

Lets take p(x) = ax

^{2}+ bx +cp(0) = 5 ⇒ c = 5

p(1) = 4 ⇒ a+b+c = 4 ⇒ a+b = -1 -----(1)

p(2) = 9 ⇒ 4a+2b+c = 9 ⇒ 2a+b = 2 -----(2)

(2) - (1)

⇒ a = 3, b = -1-1 = -4

p(3) = 20 ⇒ 9a+3b+c = 20

⇒ 27-12+5 = 20

⇒ 20 = 20, True

Hence, minimum degree it can have.

Question 3 |

All complex roots | |

At least one real root | |

Four pairs of imaginary roots | |

None of the above |

Question 3 Explanation:

Since, the polynomial has highest degree 7. So there are 7 roots possible for it.

Now suppose if an imaginary number a+bi is also root of this polynomial. That means there must be even number of complex root possible because, they occur in pair.

A) All complex root.

This is not possible. The polynomial has 7 roots and as I mention a polynomial should have been number of complex root and 7 is not even. So this option is wrong.

B) At last one real root.

This is possible. Since polynomial has 7 roots and only even number of complex root is possible, that means this polynomial has max 6 complex roots and hence minimum one real root. So, this option is correct.

C) Four pairs of imaginary roots.

4 pair means 8 complex root. But this polynomial can have atmost 7 roots. So, this option is also wrong.

Now suppose if an imaginary number a+bi is also root of this polynomial. That means there must be even number of complex root possible because, they occur in pair.

A) All complex root.

This is not possible. The polynomial has 7 roots and as I mention a polynomial should have been number of complex root and 7 is not even. So this option is wrong.

B) At last one real root.

This is possible. Since polynomial has 7 roots and only even number of complex root is possible, that means this polynomial has max 6 complex roots and hence minimum one real root. So, this option is correct.

C) Four pairs of imaginary roots.

4 pair means 8 complex root. But this polynomial can have atmost 7 roots. So, this option is also wrong.

Question 4 |

If f(x

_{i}) ⋅ f(x_{i+1}) < 0 thenThere must be a root of f(x) between x _{i} and x_{i+1}. | |

There need not be a root of f(x) between x _{i} and x_{i+1}. | |

There fourth derivative of f(x) with respect to x vanishes at x _{i}. | |

The fourth derivative of f(x) with respect to x vanishes at x _{i+1}. |

Question 4 Explanation:

As f(x

means one of them is positive and one of them is negative, as their multiplication is negative.

So, when you draw the graph for f(x) where x

Definitely f(x) will cut the x-axis. So there will definitely be a root of f(x) between x

_{i}) ⋅ f(x_{i+1}) < 0means one of them is positive and one of them is negative, as their multiplication is negative.

So, when you draw the graph for f(x) where x

_{i}≤ x ≤ x_{i+1}.Definitely f(x) will cut the x-axis. So there will definitely be a root of f(x) between x

_{i}and x_{i+1}.
There are 4 questions to complete.