UPSC Exam > UPSC Questions > Let f(x) = cos(|pi - x|) + (x - pi) * sin |x|...
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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 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) 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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Let f(x) = cos(|pi - x|) + (x - pi) * sin |x| and g(x) = x ^ 2 for x \...
Introduction
The functions f(x) and g(x) are defined as follows:
- f(x) = cos(|pi - x|) + (x - pi) * sin |x|
- g(x) = x^2
We analyze the function h(x) = f(g(x)).
Analysis of h(x)
- h is not differentiable at x = 0:
At x = 0, g(0) = 0. The function f(x) involves absolute values, which can lead to non-differentiability at points where the argument of the absolute value is zero. Since f(x) has terms involving |pi - x| and |x|, f(g(x)) inherits this property at x = 0, making h not differentiable.
h'(sqrt(pi)) = 0?
- Evaluating h'(sqrt(pi)):
To check if h' at sqrt(pi) equals zero, differentiate h(x) using the chain rule. g'(x) = 2x, and evaluate f'(g(x)) at x = sqrt(pi). If f'(g(sqrt(pi))) = 0 or if the derivative of g(x) leads to a cancellation, then h' will be zero. However, detailed calculations are needed to confirm this.
h''(x) = 0 in (-pi, pi)
- Existence of solution:
The second derivative of h(x) can be analyzed to find where it equals zero. If h''(x) changes sign in the interval (-pi, pi), the Intermediate Value Theorem guarantees a solution.
Existence of x0 such that h(x0) = x0
- Finding x0 in (-pi, pi):
By the continuity of h(x), apply the Intermediate Value Theorem. Since h(0) = f(0) and f(0) is likely positive, while h(pi) = f(pi^2) is also positive, there exists some x0 where h(x0) crosses x0, confirming a solution in (-pi, pi).
The functions f(x) and g(x) are defined as follows:
- f(x) = cos(|pi - x|) + (x - pi) * sin |x|
- g(x) = x^2
We analyze the function h(x) = f(g(x)).
Analysis of h(x)
- h is not differentiable at x = 0:
At x = 0, g(0) = 0. The function f(x) involves absolute values, which can lead to non-differentiability at points where the argument of the absolute value is zero. Since f(x) has terms involving |pi - x| and |x|, f(g(x)) inherits this property at x = 0, making h not differentiable.
h'(sqrt(pi)) = 0?
- Evaluating h'(sqrt(pi)):
To check if h' at sqrt(pi) equals zero, differentiate h(x) using the chain rule. g'(x) = 2x, and evaluate f'(g(x)) at x = sqrt(pi). If f'(g(sqrt(pi))) = 0 or if the derivative of g(x) leads to a cancellation, then h' will be zero. However, detailed calculations are needed to confirm this.
h''(x) = 0 in (-pi, pi)
- Existence of solution:
The second derivative of h(x) can be analyzed to find where it equals zero. If h''(x) changes sign in the interval (-pi, pi), the Intermediate Value Theorem guarantees a solution.
Existence of x0 such that h(x0) = x0
- Finding x0 in (-pi, pi):
By the continuity of h(x), apply the Intermediate Value Theorem. Since h(0) = f(0) and f(0) is likely positive, while h(pi) = f(pi^2) is also positive, there exists some x0 where h(x0) crosses x0, confirming a solution in (-pi, pi).
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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 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?
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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 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?.
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?.
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