JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 11)
For three positive integers p, q, r, $${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$$ and r = pq + 1 such that 3, 3 log$$_yx$$, 3 log$$_zy$$, 7 log$$_xz$$ are in A.P. with common difference $$\frac{1}{2}$$. Then r-p-q is equal to
12
$$-$$6
6
2
Explanation
$x^{p q^{2}}=y^{q r}=z^{p^{2} r}$
$$ 3 \log _{y} x=\frac{7}{2}, 3 \log _{z} y=4,7 \log _{x} z=\frac{9}{2} $$
$$ \begin{aligned} & \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\\\ & y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\\\ & \Rightarrow \frac{7}{6} p q^{2}=q r=\frac{3}{4} p^{2} r \\\\ & \therefore 7 p q=6 r, 4 q=3 p^{2} \\\\ & r=p q+1 \\\\ & r=\frac{6 r}{7}+1 \Rightarrow r=7 \\\\ & p q=6 \\\\ & p\left(\frac{3 p^{2}}{4}\right)=6 \\\\ & p=2, q=3 \\\\ & r-p-q=7-5=2 \end{aligned} $$
$$ 3 \log _{y} x=\frac{7}{2}, 3 \log _{z} y=4,7 \log _{x} z=\frac{9}{2} $$
$$ \begin{aligned} & \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\\\ & y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\\\ & \Rightarrow \frac{7}{6} p q^{2}=q r=\frac{3}{4} p^{2} r \\\\ & \therefore 7 p q=6 r, 4 q=3 p^{2} \\\\ & r=p q+1 \\\\ & r=\frac{6 r}{7}+1 \Rightarrow r=7 \\\\ & p q=6 \\\\ & p\left(\frac{3 p^{2}}{4}\right)=6 \\\\ & p=2, q=3 \\\\ & r-p-q=7-5=2 \end{aligned} $$
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