Question
Download Solution PDF\(\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}}\) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFगणना:
हमें \(\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}}\) का मूल्य खोजना होगा
\( \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}} = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}2\sin x - \sin x - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}2\sin x - \sin x + {\rm{\;}}1}}\)
\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{2\sin x(\sin x + 1) - 1\;(\sin x + 1)}}{{2\sin x(\sin x - 1) - 1\;(\sin x - 1)}}\)
\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{\left( {2\sin x - 1} \right)\;(\sin x + 1)}}{{\left( {2\sin x - 1} \right)\;(\sin x - 1)}}\)
\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{(\sin x + 1)}}{{(\sin x - 1)}} = \;\frac{{\left( {\frac{1}{2} + 1} \right)}}{{\left( {\frac{1}{2} - 1} \right)}} = \;\frac{{\left( {\frac{3}{2}} \right)}}{{\left( {\frac{{ - 1}}{2}} \right)}} = \; - 3\)
∴ विकल्प 4 सही है।Last updated on Jun 18, 2025
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