Download App

STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 5. continuity and differentiation→STD 12 - 5. continuity and differentiation — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If a curve passes through the origin and the slope of the tangent to it at any point $( x , y )$ is $\frac{x^{2}-4 x+y+8}{x-2},$ then this curve also passes through the point
→
If $P = \left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&4\\3&4&5\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 2}&0\\0&{ - 4}\end{array}} \right]$$\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 5}&{ - 6}\\0&0&1\end{array}} \right]$ then ${P_{22}} = $
→
The probability of selecting integers $a \in[-5,30]$ such that $x^{2}+2(a+4) x-5 a+64>0$, for all $x \in R$, is:
→
Let $y=f(x)$ be the solution of the differential equation $y(x+1) d x-x^2 d y=0, y(1)=e$. Then $\lim _{x \rightarrow 0^{+}} f(x)$ is equal to
→
On the interval $\left( {0,{\pi \over 2}} \right)$, the function $ log \,sin \,x $ is
→
For the function $f(x) = {e^x},a = 0,b = 1$, the value of $ c$ in mean value theorem will be
→
Let $A = \{x : -1 \leq x \leq 1\}$ and $f : A \rightarrow A$ is a function defined by $f(x) = x |x|$ then $f$ is:
→
$\tan \left[ {{{\sec }^{ - 1}}\sqrt {1 + {x^2}} } \right] = $
→
If the lines $\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$ and $\frac{{x - 3}}{1} = \frac{{y - k}}{1} = \frac{z}{1}$ intersect, then $k =$
→
The line $y = mx$ bisects the area enclosed by the curve $y = 1 + 4x - x^2\,\, \&$ the lines $x = 0, x = \frac{3}{2} \& \,\,y = 0.$ Then the value of $m$ is:
→