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STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
Let $b$ be a nonzero real number. Suppose $f: R \rightarrow R$ is a differentiable function such that $(0)=1$.

If the derivative $f^{\prime}$ of $f$ satisfies the equation $f ^{\prime}( x )=\frac{ f ( x )}{ b ^2+ x ^2}$ for all $x \in R$, then which of the following statements is/are TRUE?

$(A)$ If $b>0$, then $f$ is an increasing function

$(B)$ If $b<0$, then $f$ is a decreasing function

$(C)$ $(x)(-x)=1$ for all $x \in R$

$(D)$ $(x)-f(-x)=0$ for all $x \in R$

  • A
    $A,B$
  • B
    $A,D$
  • C
    $B,C$
  • ✓
    $A,C$
✓

Answer

Correct option: D.
$A,C$
d
$f^{\prime}( x )=\frac{f( x )}{ b ^2+ x ^2}$

$\int \frac{f^{\prime}( x )}{f( x )} dx =\int \frac{ dx }{ x ^2+ b ^2}$

$\Rightarrow \ell n |f( x )|=\frac{1}{ b } \tan ^{-1}\left(\frac{ x }{ b }\right)+ c$

Now $f(0)=1$

$\therefore c=0$

$\therefore|f( x )|= e ^{\frac{1}{b} \cdot \tan ^{-1}\left(\frac{x}{b}\right)}$

$\Rightarrow f( x )= \pm e ^{\frac{1}{ b } \tan ^{-1}\left(\frac{x}{ b }\right)}$

since $f(0)=1 \therefore \quad f( x )= e ^{\frac{1}{b} \tan ^{-1}\left(\frac{x}{b}\right)}$

$x \rightarrow- x$

$f(- x )= e ^{-\frac{1}{ b } \tan ^{-1}\left(\frac{ x }{ b }\right)}$

$\therefore f( x ) \cdot f(- x )= e ^0=1 \text { (option } C \text { ) }$

and for $b>0$

$f( x )= e ^{\frac{1}{6} \operatorname{tax}-1\left(\frac{x}{6}\right)}$

$\Rightarrow f( x )$ is increasing for all $x \in R$ (option $A$ )

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