Question

Solve Questions 41 and 43

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Answer 1

I got answer of 43) b

Answer 2

$\underline{\mathrm{41.}}$

$\star \:\mathrm{A=\frac{1}{\sqrt{19-\sqrt{360}}}}$

$\mathrm{=\frac{\sqrt{19+\sqrt{360}}}{\sqrt{(19-\sqrt{360})(19+\sqrt{360}})}}$

$\mathrm{=\frac{\sqrt{19+\sqrt{360}}}{\sqrt{361-360}}}$

$\mathrm{=\sqrt{19+\sqrt{360}}}$

$\mathrm{=\sqrt{19+6\sqrt{10}}}$

$\mathrm{=\sqrt{9+6\sqrt{10}+10}}$

$\mathrm{=\sqrt{(3+\sqrt{10})^{2}}}$

$\mathrm{=3+\sqrt{10}}$

$\star \:\mathrm{B=\frac{1}{\sqrt{21-\sqrt{440}}}}$

$\mathrm{=\frac{\sqrt{21+\sqrt{440}}}{\sqrt{(21-\sqrt{440})(21-\sqrt{440}})}}$

$\mathrm{=\frac{\sqrt{21+\sqrt{440}}}{\sqrt{441-440}}}$

$\mathrm{=\sqrt{21+\sqrt{440}}}$

$\mathrm{=\sqrt{21+2\sqrt{110}}}$

$\mathrm{=\sqrt{11+2\sqrt{110}+10}}$

$\mathrm{=\sqrt{(\sqrt{11}+\sqrt{10})^{2}}}$

$\mathrm{=\sqrt{11}+\sqrt{10}}$

$\star \:\mathrm{C=\frac{2}{\sqrt{20+\sqrt{396}}}}$

$\mathrm{=2\frac{\sqrt{20-\sqrt{396}}}{\sqrt{(20+\sqrt{396})(20-\sqrt{396}})}}$

$\mathrm{=2\frac{\sqrt{20-\sqrt{396}}}{\sqrt{440-396}}}$

$\mathrm{=2\frac{\sqrt{20-\sqrt{396}}}{2}}$

$\mathrm{=\sqrt{20-\sqrt{396}}}$

$\mathrm{=\sqrt{20-6\sqrt{11}}}$

$\mathrm{=\sqrt{11-6\sqrt{11}+9}}$

$\mathrm{=\sqrt{(\sqrt{11}-3)^{2}}}$

$\mathrm{=\sqrt{11}-3}$

$\text{Now, A - B + C}$

$\mathrm{=(3+\sqrt{10})-(\sqrt{11}+\sqrt{10})+(\sqrt{11}-3)}$

$\mathrm{=3+\sqrt{10}-\sqrt{11}-\sqrt{10}+\sqrt{11}-3}$

$\mathrm{=0}$

$\to \boxed{\mathrm{\frac{1}{\sqrt{19-\sqrt{360}}}-\frac{1}{\sqrt{21-\sqrt{440}}}+\frac{2}{\sqrt{20+\sqrt{396}}}=0}}$

$\text{Option (C) is correct.}$

$\underline{\mathrm{43.}}$

$\mathrm{Now,\:(\sqrt[6]{15-2\sqrt{56}})\times (\sqrt[3]{\sqrt{7}+2\sqrt{2}})}$

$\mathrm{=\sqrt[6]{(\sqrt{8}-\sqrt{7})^{2}}\times \sqrt[3]{\sqrt{8}+\sqrt{7}}}$

$\mathrm{=\{(\sqrt{8}-\sqrt{7})^{2}\}^{1/6}\times \{\sqrt{8}+\sqrt{7}\}^{1/3}}$

$\mathrm{=(\sqrt{8}-\sqrt{7})^{1/3}\times (\sqrt{8}+\sqrt{7})^{1/3}}$

$\mathrm{=\{(\sqrt{8}-\sqrt{7})(\sqrt{8}+\sqrt{7})\}^{1/3}}$

$\mathrm{=(8-7)^{1/3}}$

$\mathrm{=1}$

$\to \boxed{\mathrm{(\sqrt[6]{15-2\sqrt{56}})\times (\sqrt[3]{\sqrt{7}+2\sqrt{2}})=1}}$

$\text{Option (B) is correct.}$