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Asked by sarveshvibrantacademy | 12 May, 2019, 14:14: PM

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Finding space the space range space of space space cos space 2 straight x space minus space 2 sinx space plus 3 rightwards double arrow 1 minus 2 sin squared straight x space minus space 2 sinx space plus 3 rightwards double arrow negative 2 open parentheses sin squared straight x plus sinx minus 2 close parentheses rightwards double arrow negative 2 open parentheses open parentheses sinx plus 1 half close parentheses squared minus 2 minus 1 fourth close parentheses rightwards double arrow negative 2 open parentheses open parentheses sinx plus 1 half close parentheses squared minus 9 over 4 close parentheses rightwards double arrow negative 2 open parentheses sinx plus 1 half close parentheses squared plus 9 over 2 1 greater or equal than sin space straight x space greater or equal than negative 1 3 over 2 greater or equal than sin space straight x plus 1 half space greater or equal than negative 1 half 9 over 4 greater or equal than open parentheses sin space straight x plus 1 half close parentheses squared space greater or equal than 1 fourth fraction numerator negative 9 over denominator 2 end fraction less or equal than negative 2 open parentheses sin space straight x plus 1 half close parentheses squared space less or equal than negative 1 half 0 less or equal than negative 2 open parentheses sin space straight x plus 1 half close parentheses squared space plus 9 over 2 less or equal than 4 for space signum space function sgn left parenthesis straight x right parenthesis equals 1 space for space straight x greater than 0 sgn left parenthesis straight x right parenthesis equals 0 space for space straight x equals 0 sgn left parenthesis straight x right parenthesis equals negative 1 space for space straight x less than 0 so space sgn left parenthesis cos space 2 straight x space minus space 2 sinx space plus 3 right parenthesis equals 1 space for space straight x greater than 0 sgn left parenthesis cos space 2 straight x space minus space 2 sinx space plus 3 right parenthesis equals 0 space for space straight x equals 0 so space isolated space point space discontinuity since space limit space exists space but space not space equal space to space the space value space at space straight x equals 0

Answered by Renu Varma | 13 May, 2019, 10:50: AM

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