If $$P\left( x \right)$$ is a polynomial such that $$P\left( {{x^2} + 1} \right) = {\left\{ {P\left( x \right)} \right\}^2} + 1$$ and $$P\left( 0 \right) = 0$$ then $$P'\left( 0 \right)$$ is equal to :
A.
1
B.
0
C.
$$-1$$
D.
none of these
Answer :
1
Solution :
$$\because P\left( {{x^2} + 1} \right) = {\left\{ {P\left( x \right)} \right\}^2} + 1$$ and $$P\left( x \right)$$ is a polynomial, we have $$P\left( x \right) = x$$ which also satisfies $$P\left( 0 \right) = 0.$$ Therefore, $$P'\left( x \right) = 1.$$ So, $$P'\left( 0 \right) = 1.$$
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-