Question Description
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 40.(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 40.(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 40.(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?.

Solutions 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 40.(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? in English & in Hindi are available as part of our courses for UPSC. Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free.

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 40.(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 40.(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 40.(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 40.(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 40.(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.