For every pair of continuous functions f, g: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$. Since $f$ and $g$ are continuous on $[0, 1]$,there exist $a, b \in [0

Vedclass · Accepted Answer

$(A, D)$

Vedclass · Answer

(D) Let $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$. Since $f$ and $g$ are continuous on $[0, 1]$,there exist $a, b \in [0, 1]$ such that $f(a) = \lambda$ and $g(b) = \lambda$. Define $h(x) = f(x) - g(x)$. Then $h(a) = f(a) - g(a) = \lambda - g(a) \ge 0$ and $h(b) = f(b) - g(b) = f(b) - \lambda \le 0$. By the Intermediate Value Theorem,there exists $c \in [0, 1]$ such that $h(c) = 0$,which implies $f(c) = g(c)$. For option $(A)$: $(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)$. Since $f(c) = g(c)$,this holds true. For option $(D)$: $(f(c))^2 = (g(c))^2$. Since $f(c) = g(c)$,this holds true. For options $(B)$ and $(C)$,consider $f(x) = g(x) = \lambda$ where $\lambda 
eq 0$. Then $(B)$ becomes $\lambda^2 + \lambda = \lambda^2 + 3\lambda$,which implies $\lambda = 3\lambda$,or $\lambda = 0$,which contradicts $\lambda 
eq 0$. Similarly for $(C)$. Thus,$(A)$ and $(D)$ are correct.

For every pair of continuous functions $f, g: [0, 1] \rightarrow \mathbb{R}$ such that $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$,the correct statement$(s)$ is (are):

Similar Questions

If $f(x) = \begin{cases} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}, & x \neq -1 \\ \frac{1}{\sqrt{\lambda \pi}}, & x = -1 \end{cases}$ is right continuous at $x = -1$,then $\lambda = $

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & x>0 \\ 2, & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & x < 0 \end{cases}$. If $f$ is continuous at $x=0$,find the value of $\alpha + \beta$.

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} \frac{x^{3}}{(1-\cos 2x)^{2}} \log_{e}\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^{2}}\right), & x \neq 0 \\ \alpha, & x=0 \end{cases}$. If $f$ is continuous at $x=0$,then $\alpha$ is equal to:

If $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & 0 \leq x \leq 1 \end{cases}$ is continuous in $[-1, 1]$, then $p = $

Given that $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $a = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module