k के किन मानों के लिए निम्नलिखित प्रणाली स्थिर है: y = (k² − 3k − 4)log(x) + sin(x)?
उत्तर और चरणबद्ध हल
D — k = 4, −1। x → 0⁺ पर log(x) अपरिबद्ध हो जाता है, इसलिए स्थिरता के लिए उसका गुणांक शून्य होना चाहिए: k² − 3k − 4 = 0 → (k − 4)(k + 1) = 0 → k = 4 या −1।
विस्तृत हल (मूल भाषा में)
Concept & Reasoning
Factor the parameter polynomial: k²−3k−4=(k−4)(k+1). If its coefficient is nonzero, log(x) can grow without bound for bounded positive inputs approaching zero, violating a BIBO-style bounded-output requirement. Setting the coefficient to zero leaves sin(x), which is bounded between −1 and 1.
How to Solve It in the Exam
Unbounded function term? Force its coefficient to zero.
Important Exam Point
After cancelling log(x), confirm the remaining sin(x) term is bounded.
Quick Trick
Unbounded function term? Force its coefficient to zero.
Common Mistake
Solving the quadratic incorrectly or checking only positive k.
For what values of k is the following system stable, y = (k² − 3k − 4)log(x) + sin(x)?
Answer & step-by-step solution
The term multiplying log(x) can become unbounded as x approaches 0⁺, so for the system to avoid that unbounded contribution its coefficient must be zero. Solve k²−3k−4=0, giving k=4 or k=−1.
Concept & Reasoning
Factor the parameter polynomial: k²−3k−4=(k−4)(k+1). If its coefficient is nonzero, log(x) can grow without bound for bounded positive inputs approaching zero, violating a BIBO-style bounded-output requirement. Setting the coefficient to zero leaves sin(x), which is bounded between −1 and 1.
How to Solve It in the Exam
Unbounded function term? Force its coefficient to zero.
Important Exam Point
After cancelling log(x), confirm the remaining sin(x) term is bounded.
Quick Trick
Unbounded function term? Force its coefficient to zero.
Common Mistake
Solving the quadratic incorrectly or checking only positive k.
k ના કયા મૂલ્યો માટે નીચેની પ્રણાલી સ્થિર છે: y = (k² − 3k − 4)log(x) + sin(x)?
જવાબ અને પગલાવાર ઉકેલ
D — k = 4, −1. x → 0⁺ પર log(x) અમર્યાદિત બને છે, તેથી સ્થિરતા માટે તેનો ગુણાંક શૂન્ય હોવો જોઈએ: k² − 3k − 4 = 0 → (k − 4)(k + 1) = 0 → k = 4 કે −1.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept & Reasoning
Factor the parameter polynomial: k²−3k−4=(k−4)(k+1). If its coefficient is nonzero, log(x) can grow without bound for bounded positive inputs approaching zero, violating a BIBO-style bounded-output requirement. Setting the coefficient to zero leaves sin(x), which is bounded between −1 and 1.
How to Solve It in the Exam
Unbounded function term? Force its coefficient to zero.
Important Exam Point
After cancelling log(x), confirm the remaining sin(x) term is bounded.
Quick Trick
Unbounded function term? Force its coefficient to zero.
Common Mistake
Solving the quadratic incorrectly or checking only positive k.
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