Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

Let f : [0, 1] → [0, 1] be a continuous function. Then,

Options:

f(x) = x for at least one x ∈ (0, 1)

f(x) will be differentiable in [0, 1]

f(x) + x = 0 for at least one x such that 0 ≤ x ≤ 1

none of these

Correct Answer:

f(x) = x for at least one x ∈ (0, 1)

Explanation:

Since, $f:[0,1] \rightarrow[0,1]$ is a continuous function.

Therefore, Range $(f) \subset[0,1]$ and f(x) attains every value between f(0) and f(1)

Let $g(x)=f(x)-x$ for all $x \in[0,1]$.

Clearly, g(x) is also continuous on [0, 1]

Also, $g(0)=f(0)-0=f(0)>0$

and,

$g(1)=f(1)-1<0$                  [∵ f(1) < 1]

Thus, g(x) is continuous on [0, 1] such that g(0) g(1) < 0. Therefore, there exists a point $c \in(0,1)$ such that

Therefore, there exists a point $c \in(0,1)$ such that

$g(c)=0 \Rightarrow f(c)=c$

Hence, f(x) = x for at least one $x \in(0,1)$.