MCQ
Let $f :(0, \infty) \rightarrow(0, \infty)$ be a differentiable function such that $f(1)= e$ and $\lim \limits_{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0$ If $f ( x )=1,$ then $x$ is equal to
  • A
    $2e$
  • B
    $\frac{1}{2 e }$
  • C
    $e$
  • $\frac{1}{ e }$

Answer

Correct option: D.
$\frac{1}{ e }$
d
$L=\lim _{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}$

using L.H. rule

$L =\lim _{t \rightarrow x} \frac{2 tf ^{2}( x )- x ^{2} \cdot 2 f ^{\prime}( t ) \cdot f ( t )}{1}$

$\Rightarrow L =2 xf ( x )\left( f ( x )- x f ^{\prime}( x )\right)=0$ (given)

$\Rightarrow f(x)=x f^{\prime}(x) \Rightarrow \int \frac{f^{\prime}(x) d x}{f(x)}=\int \frac{d x}{x}$

$\Rightarrow \ell n | f ( x )|=\ell n | x |+ C$

$\because f (1)= e , x >0, f ( x )>0$

$\Rightarrow f ( x )= ex , \quad$ if $f ( x )=1 \Rightarrow x =\frac{1}{ e }$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}\,} \right| = \lambda $, then the value of $\lambda $is
If $y = {\tan ^{ - 1}}(\sec x - \tan x)$ then ${{dy} \over {dx}} = $
Objective of LPP is:
The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y - z = -5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is:
The vectors $2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$ and $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ are perpendicular if:
  1. a = 2, b = 3, c = -4
  2. a = 4, b = 4, c = 5
  3. a = 4, b = 4, c = -5
  4. a = -4, b = 4, c = -5
A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is
  1. $\frac{44}{85\times49}$
  2. $\frac{11}{85\times49}$
  3. $\frac{13\times24}{17\times25\times49}$
  4. None of these.
Let $A$ and $B$ be two events such that $P\left( {A \cap B} \right) = \frac{1}{6},\,P(A \cup B) = \frac{{31}}{{45}},\,P(\bar B) = \frac{7}{{10}}$ , then
$\int_{\; - \pi }^\pi {\frac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}\;dx} = $
Let $\Gamma$ denote a curve $y = y ( x )$ which is in the first quadrant and let the point $(1,0)$ lie on it. Let the tangent to $\Gamma$ at a point $P$ intersect the $y$-axis at $Y_p$. If $P Y_p$ has length $1$ for each point $P$ on $\Gamma$, then which of the following options is/are correct?

$(1)$ $y=\log _0\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$

$(2)$ $x y^{\prime}-\sqrt{1-x^2}=0$

$(3)$ $y=-\log _0\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}$

$(4)$ $x y^{\prime}+\sqrt{1-x^2}=0$

For $0 < a < 1$, the value of the integral $\int_0^\pi \frac{\mathrm{d} x}{1-2 \mathrm{a} \cos x+\mathrm{a}^2}$