JEE MATHEMATICS MCQ: THEORY OF EQUATION

1 Medium

1 of 1

Analyzing +4 | -1

f(x)

is a polynomial of degree

4, f^{\prime}(1)=f(2)=f(3)=f(4)=0

, then root of

f(x)=0

View answer and explanation

Correct option: A

\frac{5}{11}

Option C (

\frac{5}{11}

) is correct because it is derived from the condition

f'(1) = 0

using the given information about the polynomial

f(x)

More Questions from THEORY OF EQUATION

Practice All Chapter MCQs →

{x^{2n}} - 1 = 0

\frac{1}{x}+\frac{1}{\sqrt{\left(1-x^2\right)}}=\frac{35}{12}

x^7+14 x^5+16 x^3+30 x-560=0

x_1, x_2, x_3, \ldots, x_n

x^n+a x+b=0

\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right) \ldots\left(x_1-x_n\right)

P(x)=k x^3+3 x^2-3

Q(x)=2 x^3-5 x+k

(x-4)

k

{\left( {x - 1} \right)^3}

{x^4} + a{x^3} + b{x^2} + cx - 1,