If p and q are positive real numbers such tha | vedclass

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(D) Given $p^{2} + q^{2} = 1$ where $p, q > 0$.We know the identity $(p+q)^{2} = p^{2} + q^{2} + 2pq$.Substituting the given value: $(p+q)^{2} = 1 + 2

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$\sqrt{2}$

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(D) Given $p^{2} + q^{2} = 1$ where $p, q > 0$.We know the identity $(p+q)^{2} = p^{2} + q^{2} + 2pq$.Substituting the given value: $(p+q)^{2} = 1 + 2pq$.By the $AM \geq GM$ inequality,for positive real numbers $p^{2}$ and $q^{2}$,we have $\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2}q^{2}} = pq$.Since $p^{2} + q^{2} = 1$,we get $\frac{1}{2} \geq pq$,which implies $2pq \leq 1$.Substituting this into the identity: $(p+q)^{2} = 1 + 2pq \leq 1 + 1 = 2$.Taking the square root on both sides,we get $p+q \leq \sqrt{2}$.Thus,the maximum value of $(p+q)$ is $\sqrt{2}$.

If $p$ and $q$ are positive real numbers such that $p^{2} + q^{2} = 1$,then the maximum value of $(p+q)$ is:

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