Continuity and Differentiability
Mean Value Theorem and fixed point applications
GRB_1000_MCQ
Grade Class 12
Question:
Let $f'(x)$ be a continuous function which maps from $[0,1] \to [p(a), p(b)]$. If $p(x)$ is a differentiable function on $[a, b]$ such that $p(g(x)) = x$, $g(0) = a$ and $g(1) = b$, then which of the following is/are true?
(c) $\dfrac{\displaystyle\int_0^1 f'(x)\, dx}{\displaystyle\int_0^1 g'(x)\, dx} \leq p'(c)$ for some $c \in (a, b)$
$f(0) + 2 < f(1)$
$f(1) \leq 1 + f(0)$
$\dfrac{\displaystyle\int_0^1 f'(x)\, dx}{\displaystyle\int_0^1 g'(x)\, dx} \leq p'(c)$ for some $c \in (a, b)$
There exists $k \in [0,1]$ such that $f'(k) = k$
Step-by-Step Solution
Key Concept: The key idea is to first determine the range of the function $f'(x)$. By using the given conditions $p(g(x))=x$, $g(0)=a$, and $g(1)=b$, it follows that $p(a)=0$ and $p(b)=1$. Since $f'(x)$ maps $[0,1]$ to $[p(a), p(b)]$, we deduce that $f'(x) \in [0,1]$ for all $x \in [0,1]$. This crucial bound for $f'(x)$ is then used in conjunction with the Mean Value Theorem, Cauchy Mean Value Theorem, and the Intermediate Value Theorem to evaluate each option.
Step 1: Analyze option (b): $f(1) \leq 1 + f(0)$. Since $f'(x)$ maps $[0,1] \to [p(a), p(b)]$ and $p(g(x)) = x$, differentiating gives $p'(g(x))\cdot g'(x) = 1$. By the mean value theorem applied to $f$ on $[0,1]$:
$$f(1) - f(0) = f'(c)$$
for some $c \in (0,1)$. Since $f'(x) \leq p(b) \leq 1$ (from the mapping condition), $f(1) - f(0) \leq 1$, so $f(1) \leq 1 + f(0)$. Option 2 is TRUE.
Step 2: Analyze option (a): $f(0) + 2 < f(1)$. This would require $f(1) - f(0) > 2$, but from Step 1 we showed $f(1) - f(0) \leq 1$. So option 1 is FALSE.
Step 3: Analyze option (c). By the Cauchy Mean Value Theorem applied to $\int_0^1 f'(x)dx = f(1)-f(0)$ and $\int_0^1 g'(x)dx = g(1)-g(0) = b-a$:
$$\frac{\int_0^1 f'(x)dx}{\int_0^1 g'(x)dx} = \frac{f(1)-f(0)}{b-a}$$
By MVT, there exists $c \in (a,b)$ such that this ratio equals $p'(c)$ (or is bounded by it). Option 3 is TRUE.
Step 4: Analyze option (d). Since $f'(x)$ is continuous on $[0,1]$ and maps into $[p(a), p(b)]$, and $p(g(0)) = 0$ implies $p(a) = 0$, $p(g(1)) = 1$ implies $p(b) = 1$. So $f'(x)$ maps $[0,1] \to [0,1]$. By the Intermediate Value Theorem (or Brouwer's fixed point theorem for continuous maps $[0,1]\to[0,1]$), there exists $k \in [0,1]$ such that $f'(k) = k$. Option 4 is TRUE.
Correct Answer: 2, 3, 4