MCQ
Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be functions defined by$f(x)=\left\{\begin{array}{cl}x|x| \sin \left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x=0,\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}1-2 x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text { otherwise }\end{array}\right.$
Let $a, b, c, d \in R$. Define the function $h: R \rightarrow R$ by
$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in R$
Match each entry in List $-I$ to the correct entry in List $-II$.
List $-I$ List $-II$
$(P)$ If $a=0, b=1, c=0$ and $d=0,$ then  $(1) \ h$ is one $-$ one
$(Q)$ If $a=1, b=0, c=0$ and $d=0,$ then  $(2) \ h$ is onto.
$(R)$ If $a=0, b=0, c=1$ and $d=0,$ then  $(3) \ h$ is differentiable on $R$.
$(S)$ If $a=0, b=0, c=0$ and $d=1,$ then $(4) $ the range of $h$ is $[0,1]$
  $(5)$ the range of $h$ is $\{0,1\}$
The correct option is
  • A
    $(P) \rightarrow(4)( Q ) \rightarrow(3)( R ) \rightarrow(1)( S ) \rightarrow(2)$
  • B
    $(P) \rightarrow (5) (Q) \rightarrow (2) (R) \rightarrow (4) (S) \rightarrow (3)$
  • C
    $(P) \rightarrow (5) (Q) \rightarrow (3) (R) \rightarrow (2) (S) \rightarrow (4)$
  • D
    $(P) \rightarrow (4) (Q) \rightarrow (2) (R) \rightarrow (1) (S) \rightarrow (3)$

Answer

$f(x)=\left\{\begin{array}{cl}x|x| \sin \frac{1}{x} ; & x \neq 0 \\ 0 ; & x=0\end{array} \quad g(x)=\left\{\begin{array}{cc}1-2 x ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{array}\right.\right. $
$ g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{cc}2 x ; & 0 \leq \frac{1}{2}-x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{array}=\left\{\begin{array}{cc}2 x ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{array}\right\}\right. $
$ g(x)+g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{lll}1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{array}\right\}$
$(P)$ Now a $=0, b=1, c=0, d=0$
$\because h ( x )= g ( x )+ g \left(\frac{1}{2}- x \right)=\left\{\begin{array}{ll}1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{array}\right.$
$($image$)$
Hence Range of $h ( x )$ is $\{0,1\}$
$(Q)a=1, b=0, c=0, d=0$
$h(x)=f(x)=\left\{\begin{array}{ccc}x|x| \sin \frac{1}{x} & ; & x \neq 0 \\ 0 & ; \quad x=0\end{array}\right.$
$\text{RHD}=\lim _{x \rightarrow 0} \frac{x^2 \sin \frac{1}{x}-0}{x}=0$
$\text{LHD}=\lim _{x \rightarrow 0} \frac{-x^2 \sin \frac{1}{x}-0}{x}=0$
Hence $h(x)$ is differentiable on $R$
$(R)$ $a =0, b =0, c =1, d =0$
$h(x)=x-g(x)=\left\{\begin{array}{cc}3 x-1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{array}\right.$
$($image$)$
$\begin{array}{l}\therefore h ( x ) \text { is ONTO } \\ \text { (S) } \quad a =0, b =0, c =0, d =1 \\ h ( x )= g ( x )=\left\{\begin{array}{cl}1-2 x ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{array}\right.\end{array}$
$($image$)$

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