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Detailed solution

Asked by g_archanasharma | 04 Feb, 2019, 06:05: PM

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f(x) = y =

begin mathsize 16px style square root of 25 minus straight x squared end root end style

begin mathsize 16px style straight y equals negative square root of 25 minus straight x squared end root 25 minus straight x squared greater or equal than 0 left parenthesis 5 plus straight x right parenthesis left parenthesis 5 minus straight x right parenthesis greater or equal than 0 straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight x plus 5 right parenthesis left parenthesis 5 minus straight x right parenthesis If space straight x space is space positive space then space straight x squared less or equal than 25 straight x less or equal than 5 Hence comma space straight x element of left parenthesis negative infinity comma space 5 right square bracket If space straight x space is space negative space straight x squared less or equal than 25 straight x greater or equal than negative 5 Hence comma space straight x element of left square bracket negative 5 comma space infinity right parenthesis The space intersection space of space teh space two space intervals space colon left parenthesis negative infinity comma space 5 right square bracket intersection left square bracket negative 5 comma space infinity right parenthesis end style

Answered by Sneha shidid | 05 Feb, 2019, 10:37: AM

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Q) Find the range of f(x) = x + 1/x Ans) See the image which I have uploaded.. I have just broken the eqn into √x² + 1/√x² +2 -2 ..and have formed the identity (a+b)² and the range is coming (-∞, 2 ] U [ -2, ∞ ) which is wrong.. the correct range is coming from (a-b)² which is (-∞, -2 ] U [ 2, ∞ ). My Doubt- Why (a+b)² is not giving the correct range? Request- Please send the solution using (a+b)². I've attached an image and let me know if I'm wrong somewhere or why the correct solution isn't coming by using (a+b)². Thanks, Samit Future Top 50 ranker in JEE ADVANCE

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