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The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer?.

Solutions for The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this methodQ.The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 isa)b)c)d)None of theseCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.