Let f: R rightarrow (0, infty) and g: R right | vedclass

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

Question

(A,D) Given $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$. Since $f^{\prime}(2) = 0$ and $g(2) = 0$,this is a $\frac{0}{0

Vedclass · Accepted Answer

$f$ has a local minimum at $x=2$

Vedclass · Answer

(A,D) Given $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$. Since $f^{\prime}(2) = 0$ and $g(2) = 0$,this is a $\frac{0}{0}$ form. Applying $L$'Hopital's rule:$\lim_{x \rightarrow 2} \frac{f^{\prime}(x) g(x) + f(x) g^{\prime}(x)}{f^{\prime \prime}(x) g^{\prime}(x) + f^{\prime}(x) g^{\prime \prime}(x)} = 1$.Substituting $x=2$ and using $f^{\prime}(2) = g(2) = 0$:$\frac{0 + f(2) g^{\prime}(2)}{f^{\prime \prime}(2) g^{\prime}(2) + 0} = 1 \implies \frac{f(2)}{f^{\prime \prime}(2)} = 1 \implies f(2) = f^{\prime \prime}(2)$.Since $f: R \rightarrow (0, \infty)$,$f(2) > 0$,so $f^{\prime \prime}(2) > 0$. By the second derivative test,$f$ has a local minimum at $x=2$ (Option $A$).Consider $h(x) = f(x) e^{-x^2/2}$. This is not directly helpful,but consider $h(x) = f(x) e^{-x/2}$ or similar. Actually,consider $h(x) = f(x) - f^{\prime \prime}(x)$. Since $f(2) = f^{\prime \prime}(2)$,$h(2) = 0$. Thus,$f(x) - f^{\prime \prime}(x) = 0$ for at least $x=2$ (Option $D$).Thus,both $A$ and $D$ are correct.

Similar Questions

If a function $f(x)$ defined by $f(x)=\begin{cases} a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4 \end{cases}$ is continuous for some $a, b, c \in R$ and $f'(0)+f'(2)=e$,then the value of $a$ is:

Let $f(x)$ be defined in $[-2, 2]$ by
$f(x) = \begin{cases} \max(4 - x^2, 1 + x^2), & -2 < x < 0 \\ \min(4 - x^2, 1 + x^2), & 0 < x < 2 \end{cases}$
Then $f(x)$:

Which of the following is not true?

$A$ function $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ is

$A$ curve is represented by the equations $x = \sec^2 t$ and $y = \cot t$,where $t$ is a parameter. If the tangent at the point $P$ on the curve where $t = \pi / 4$ meets the curve again at the point $Q$,then $|PQ|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module