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Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0?.

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Here you can find the meaning of Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0? defined & explained in the simplest way possible. Besides giving the explanation of Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0?, a detailed solution for Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0? has been provided alongside types of Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0? theory, EduRev gives you an ample number of questions to practice Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \in R If h(x) = f(g(x)) then (a) h is not differentiable at x = 0(c) h^ prime prime (x) = 0 has a solution in (- pi, pi)(d) There exists x_{0} \in (- pi, pi) such that h(x_{0}) = x_{0}(b) h' * (sqrt(pi)) = 0? tests, examples and also practice UPSC tests.