Choose correct option(s) from the following.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

If $f (x)$ is a continuous function in $[0, 2]$ then there exist $x_{1}$ and $x_{2}$ where $x_{1}, x_{2} \in [0, 2]$ such that $f (x_{2}) - f (x_{1}) = \frac{1}{2} (f (2) - f (0))$ and $x_{2} - x_{1} = 1.$

There CANNOT be a continuous function $f : R \to R$ which attains each of the values in it’s range exactly three times

If $f (x)$ is a continuous function in $[0, 2]$ with $f (0) = f (2)$ then there exist $x_{1}$ and $x_{2}$ where $x_{1}, x_{2} \in [0, 2]$ such that $f (x_{2}) = f (x_{1})$ and $x_{2} - x_{1} = 1.$

There CANNOT be a continuous function $f : R \to R$ which attains each of the values in it’s range exactly two times

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

A) Consider $g (x) = f (x + 1) - f (x), x \in [0, 1] .$
Then $g (1) = f (2) - f (1) = - g (0) .$ Hence there exists $x_{0} \in [0, 1]$ such that $f (x_{0} + 1) = f (x_{0}) .$ So, we can take $x_{2} = x_{0} + 1$ and $x_{1} = x_{0} .$
B) Consider the function $g (x) = f (x + 1) - f (x) - \frac{1}{2} (f (2) - f (0)), x \in [0, 1],$ and apply the reasoning analogous to that used in the solution of the preceding option.
C) $g (x) = {\begin{matrix} x + 2 & i f & - 3 \leq x \leq - 1, \\ - x & i f & - 1 < x \leq 1, \\ x - 2 & i f & 1 < x \leq 3. \end{matrix}$ and define $f$ by the formula
$f (x) = g (x - 6 n) + 2 n f o r 6 n - 3 \leq x \leq 6 n + 3, n \in z .$
D) No such function possible