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If f x = 1 + s i n x p sin ⁡ x : - π 6 < x < 0 q : x = 0 e tan ⁡ 3 x . cot ⁡ 5 x : 0 < x < π 6 is continuous at x = 0 , then the value of 2 p + 10 ln ⁡ q is equal to

Correct answer

7.20

Step-by-step solution

l i m x &#8594; 0 + f x = q &#8658; l i m x &#8594; 0 + e tan &#8289; 3 x tan &#8289; 5 x = q = l i m x &#8594; 0 + e tan &#8289; 3 x 3 x . 5 x tan &#8289; 5 x . 3 x 5 x &#8658; e 3 5 = q Now, l i m x &#8594; 0 - f x = q &#8658; e lim x &#8594; 0 - 1 - sin &#8289; x - p sin x = q = e l i m x &#8594; 0 - - p sin &#8289; x 1 - sin &#8289; x - 1 = e p &#8756; e p = e 3 5 &#8658; p = 3 5 , &#160; q = e 3 5 2 p + 10 . ln &#8289; q = 6 5 + 10 3 5 = 36 5

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