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CBSE Class 12-science Answered

  Consider the equation . Choose the CORRECT option(s). The sum of all solutions = 0< The number of solutions = 2< The number of solution = 1< The sum of squares of all solutions = 1<
Asked by harshsharma2531 | 09 Aug, 2020, 09:28: AM
answered-by-expert Expert Answer
To find solutions of the equation tan-1(x-1/x-2) + tan-1(x+1/x+2) = pi/4
tan to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 1 over denominator straight x minus 2 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction close parentheses equals straight pi over 4 rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator straight x minus 1 over denominator straight x minus 2 end fraction plus fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction end style over denominator 1 minus begin display style fraction numerator straight x minus 1 over denominator straight x minus 2 end fraction end style cross times begin display style fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction end style end fraction close parentheses equals straight pi over 4 rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style open parentheses straight x minus 1 close parentheses open parentheses straight x plus 2 close parentheses plus open parentheses straight x plus 1 close parentheses open parentheses straight x minus 2 close parentheses end style over denominator straight x squared minus 4 minus open parentheses straight x squared minus 1 close parentheses end fraction close parentheses equals straight pi over 4 rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style straight x squared plus straight x minus 2 plus straight x squared minus straight x minus 2 end style over denominator straight x squared minus 4 minus open parentheses straight x squared minus 1 close parentheses end fraction close parentheses equals straight pi over 4 rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 straight x squared minus 4 end style over denominator negative 3 end fraction close parentheses equals straight pi over 4 rightwards double arrow fraction numerator begin display style 2 straight x squared minus 4 end style over denominator negative 3 end fraction equals tan straight pi over 4 equals 1 rightwards double arrow fraction numerator 2 straight x squared minus 4 over denominator negative 3 end fraction equals 1 rightwards double arrow 2 straight x squared minus 4 equals negative 3 rightwards double arrow 2 straight x squared equals 1 rightwards double arrow straight x equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction Sum space of space all space solutions equals fraction numerator 1 over denominator square root of 2 end fraction minus fraction numerator 1 over denominator square root of 2 end fraction equals 0 Number space of space solutions space equals space 2 Sum space of space squares space of space all space solutions equals 1 half plus 1 half equals 1
Answered by Renu Varma | 09 Aug, 2020, 16:32: PM

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