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Assertion (A): $y=2 x+3$ is a one to one real valued function. Reason $(\mathbf{R}): x_{1} \neq x_{2}$
$\Rightarrow y_{1} \neq y_{2}, y_{1}=2 x_{1}+3, y_{2}=2 x_{2}+3$, for any two real
$x_{1}$ and $x_{2}$
Options:
$\Rightarrow y_{1} \neq y_{2}, y_{1}=2 x_{1}+3, y_{2}=2 x_{2}+3$, for any two real
$x_{1}$ and $x_{2}$
Solution:
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Verified Answer
The correct answer is:
Both A and $\mathrm{R}$ are individually true and $\mathrm{R}$ is the correct explanation of A
(A) Given function is $y=2 x+3$
Let $y_{1}=y_{2}\left(\right.$ To show $\left.x_{1}=x_{2}\right)$
$\Rightarrow 2 x_{1}+3=2 x_{2}+3$
$\Rightarrow 2 x_{1}=2 x_{2}$
$\Rightarrow x_{1}=x_{2}$
Hence, $y=2 x+3$ is one-one real valued function.
(R) Since $y_{1}=y_{2} \Rightarrow x_{1}=x_{2}$
$\therefore x_{1} \neq x_{2} \Rightarrow y_{1} \neq y_{2}$
Thus, Both $(A)$ and $(R)$ are true and $R$ is the correct explanation of $A$.
Let $y_{1}=y_{2}\left(\right.$ To show $\left.x_{1}=x_{2}\right)$
$\Rightarrow 2 x_{1}+3=2 x_{2}+3$
$\Rightarrow 2 x_{1}=2 x_{2}$
$\Rightarrow x_{1}=x_{2}$
Hence, $y=2 x+3$ is one-one real valued function.
(R) Since $y_{1}=y_{2} \Rightarrow x_{1}=x_{2}$
$\therefore x_{1} \neq x_{2} \Rightarrow y_{1} \neq y_{2}$
Thus, Both $(A)$ and $(R)$ are true and $R$ is the correct explanation of $A$.
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