A) 1
B) 3
C) 4
D) 5
Correct Answer: B
Solution :
\[{{\sin }^{-1}}\frac{x}{5}+\cos e{{c}^{-1}}\frac{5}{4}=\frac{\pi }{4}\] \[\Rightarrow \] \[{{\sin }^{-1}}\frac{x}{5}+{{\sin }^{-1}}\left( \frac{4}{5} \right)=\frac{\pi }{2}\] \[\Rightarrow \]\[{{\sin }^{-1}}\left\{ \frac{x}{5}\sqrt{1-{{\left( \frac{4}{5} \right)}^{2}}}+\frac{4}{5}\sqrt{1-{{\left( \frac{x}{5} \right)}^{2}}} \right\}=\frac{\pi }{2}\] \[\Rightarrow \]\[\frac{x}{5}\sqrt{\frac{25-16}{25}}+\frac{4}{5}\sqrt{\frac{25-{{x}^{2}}}{25}}=\sin \frac{\pi }{2}\] \[\Rightarrow \] \[\frac{x}{5}.\frac{3}{5}+\frac{4}{5}.\frac{1}{5}\sqrt{25-{{x}^{2}}}=1\] \[\Rightarrow \] \[\frac{3x}{25}+\frac{4}{25}\sqrt{25-{{x}^{2}}}=1\] \[\Rightarrow \] \[3x+4\sqrt{25-{{x}^{2}}}=25\] \[\Rightarrow \] \[4\sqrt{(25-{{x}^{2}})}=25-3x\] On squaring both sides, we get \[16(25-{{x}^{2}})={{(25)}^{2}}+9{{x}^{2}}-150x\] \[\Rightarrow \] \[400-16{{x}^{2}}=625+9{{x}^{2}}-150x\] \[\Rightarrow \] \[25{{x}^{2}}-150x+225=0\] \[\Rightarrow \] \[{{x}^{2}}-6x+9=0\] \[\Rightarrow \] \[{{(x-3)}^{2}}=0\] \[\Rightarrow \] \[x=3\]
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