Let $y = f (x)$ and $y = g (x)$ be two differentiable function in $[0,2]$ such that $f(0) = 3,$ $f(2) = 5$ , $g (0) = 1$ and $g(2) = 2$. If there exist atlellst one $c \in \left( {0,2} \right)$ such that $f'(c)=kg'(c)$,then $k$ must be
Let $y = f (x)$ and $y = g (x)$ be two differentiable function in $[0,2]$ such that $f(0) = 3,$ $f(2) = 5$ , $g (0) = 1$ and $g(2) = 2$. If there exist atlellst one $c \in \left( {0,2} \right)$ such that $f'(c)=kg'(c)$,then $k$ must be
- A
$2$
- B
$3$
- C
$\frac{1}{2}$
- D
$1$
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- [JEE MAIN 2023]
Examine the applicability of Mean Value Theorem:
$(i)$ $f(x)=[x]$ for $x \in[5,9]$
$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$
$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$
Examine the applicability of Mean Value Theorem:
$(i)$ $f(x)=[x]$ for $x \in[5,9]$
$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$
$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$
Consider $f (x) = | 1 - x | \,;\,1 \le x \le 2 $ and $g (x) = f (x) + b sin\,\frac{\pi }{2}\,x$, $1 \le x \le 2$ then which of the following is correct ?
Consider $f (x) = | 1 - x | \,;\,1 \le x \le 2 $ and $g (x) = f (x) + b sin\,\frac{\pi }{2}\,x$, $1 \le x \le 2$ then which of the following is correct ?