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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\mathop {Lim}\limits_{n\, \to \infty } \,\,\frac{\pi }{{6n}}\left[ {{{\sec }^2}\left( {\frac{\pi }{{6n}}} \right) + {{\sec }^2}\left( {2\,\cdot\,\frac{\pi }{{6n}}} \right) + ..... + {{\sec }^2}(n - 1)\frac{\pi }{{6n}} + \frac{4}{3}} \right]$ has the value equal to
  • ✓
    $\frac{{\sqrt 3 }}{3}$
  • B
    $\sqrt 3 $
  • C
    $2$
  • D
    $\frac{2}{{\sqrt 3 }}$
✓

Answer

Correct option: A.
$\frac{{\sqrt 3 }}{3}$
a
$T_r =$ $\frac{\pi }{{6n}}{\sec ^2}\frac{{r\pi }}{{6n}}$
$S =$ $\sum {{T_r}}  = \frac{\pi }{{6n}}\sum\limits_{r = 1}^n {{{\sec }^2}\frac{{r\pi }}{{6n}}} $ $ =$ $\frac{\pi }{6}\int\limits_0^1 {{{\sec }^2}\frac{{\pi x}}{6}dx} $ $  =$ $\left. {\tan \frac{{\pi x}}{6}} \right|_{\,0}^{\,1}$ $ = $$\frac{1}{{\sqrt 3 }}$ $ = $ $\frac{{\sqrt 3 }}{3}$

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