MCQ
A function $f,$ defined for all positive real numbers, satisfies the equation $f(x^2) = x^3$ for every $x > 0$ . Then the value of $f ‘ (4) =$
  • A
    $12$
  • $3$
  • C
    $3/2$
  • D
    cannot be determined

Answer

Correct option: B.
$3$
b
$2x f ‘ (x^2) = 3 x^2 ; 4 f ‘ (x) = 12 ==> f ‘ (4) = 3$ 

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

$\int(1+2\text{x}+3\text{x}^2+4\text{x}^3+ ... )\text{dx }(\mid\text{x}\mid<1)$
  1. $-(1+\text{x})^{-1}+\text{c}$
  2. $(1-\text{x})^{-1}+\text{c}$
  3. $-(1-\text{x})^{-2}+\text{c}$
  4. None of these
If $l(m,\,n) = \int_0^1 {{t^m}{{(1 + t)}^n}dt,} $ then the expression for $l(m,\,n)$ in terms of $l(m + 1,\,\,n - 1)$ is
The direction ratios of the line perprndicular to the lines $\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$ and, $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$ are proportional to:
Let $\phi (x) = (f(x))^3 -3(f(x))^2 + 4f(x) + 5x + 3 \sin x + 4 \cos x\, \forall \, x \in R$, then -
The solution of $x \frac{d y}{d x}+y=e^x$ is
Integrate the following functions with respect to x: $\int\frac{\text{dx}}{4\text{x}+5}$
  1. $\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}$
  2. $\frac{1}{4}\text{ In }(4\text{x}+5)-\text{c}$
  3. $\frac{-1}{4}\text{ In }(4\text{x}+5)-\text{c}$
  4. $4\text{ In }(4\text{x}-5)-\text{c}$
$f\left( x \right) = \left[ {\frac{{{x^2} + 1}}{{{x^2}\left[ {\left| x \right|} \right] + 1}}} \right]$ is discontinuous at (where $[.]$ denotes greatest integer function)
A unit vector perpendicular to the plane determined by the points $P\,(1,\,\, - 1,\,\,2),\,\,Q\,(2,\,\,0,\, - 1)$ and $R\,(0,\,\,2,\,\,1)$ is
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors inclined at an angle $\theta,$ then the value of $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ is:
  1.  $2\sin\frac{\theta}{2}$
  2. $2\sin\theta$
  3. $2\cos\frac{\theta}{2}$
  4. $2\cos\theta$
If $\text{f(x)}=|\log_{10}\text{x}|\text{fx}=\log_{10}\text{x},$ then at x = 1:
  1. f(x) is continuous and $\text{f}'(1^+)=\log_{10}\text{e}$
  2. f(x) is continuous and $\text{f}'(1^+)=\log_{10}\text{e}$
  3. f(x) is continuous and $\text{f}'(1^-)=-\log_{10}\text{e}$
  4. f(x) is continuous and $\text{f}'(1^-)=-\log_{10}\text{e}$