If the polynomial equation $${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ..... + {a_2}{x^2} + {a_1}x + {a_0} = 0,\,n$$ positive integer, has two different real roots $$\alpha $$ and $$\beta ,$$ then between $$\alpha $$ and $$\beta ,$$ the equation $$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ..... + {a_1} = 0$$ has :
A.
exactly one root
B.
at most one root
C.
at least one root
D.
no root
Answer :
at least one root
Solution :
Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ..... + {a_2}{x^2} + {a_1}x + {a_0}$$ which is a polynomial function in $$x$$ of degree $$n.$$
Hence, $$f\left( x \right)$$ is continuous and differentiable for all $$x.$$
Let $$\alpha < \beta $$
We are given, $$f\left( \alpha \right) = 0 = f\left( \beta \right)$$
By Rolle's theorem, $$f'\left( c \right) = 0$$ for some value $$c,\,\alpha < c < \beta .$$
Hence, the equation $$f'\left( x \right) = n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ..... + {a_1} = 0$$ has at least one root between $$\alpha $$ and $$\beta .$$
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-