st","extracted_subject":"MATH","extracted_chapter":"DIFFRENTIATION AND APPLICATIONS OF DERIVATIVES","extracted_topic":"DIFFERENTIATION BY SUBSTITUTION, HIGHER ORDER DERIVATIVES","questionPreviewUrl":"https://cdn.pureessence.tech/question-previews/packed_canvases/canvas_1638.png?top_left_x=532&top_left_y=630&width=532&height=215&app=solveeit"},"seo_metadata":{"keywords":[]},"doubt_solver_data":{"concept_summary":[]},"embeddings":[],"is_hidden":false,"_id":"66d23e6f425d76b7ab31c495","slug":"let-fx-2sup2x1sup-and-fx-2supxsup-2x-log-2-if-fx-f","content":"Let f(x) = 2^2x–1 and f(x) = –2^x + 2x log 2. If f´(x) \\>f¢(x), then","options":[{"_id":"68a37fe57f15af4716b6b5e3","text":"0 \\< x \\< 1","sequence":0,"isCorrect":false},{"_id":"68a37fe57f15af4716b6b5e4","text":"0 £ x \\< 1","sequence":1,"isCorrect":false},{"_id":"68a37fe57f15af4716b6b5e5","text":"x \\> 0","sequence":2,"isCorrect":true},{"_id":"68a37fe57f15af4716b6b5e6","text":"x ³ 0","sequence":3,"isCorrect":false}],"correct_answer":"x \\> 0","explanation":"f´(x) \\> f´(x)\n\n2^2x-1.2 log 2 \\> 2^x log 2 + 2 log 2\n\n2^2x \\> – 2^x + 2\n\n2^2x + 2^x – 2 \\> 0\n\n(2^x – 1)\\(\\frac{(2^{x} + 2)}{+ ve}\\) \\> 0\n\n2^x – 1 \\> 0 Ž 2^x \\> 1 Ž 2^x \\> 2⁰ Ž x \\> 0","question_type":"single_choice","appeared_in":[],"created_at":"2024-08-30T21:49:35.516Z","__v":0,"vector_store_id":"83ef7bd5-9f1b-434f-8145-ee157523ee08","updated_at":"2024-08-30T21:49:35.516Z","isStarred":false},"params":{"questionSlug":"let-fx-2sup2x1sup-and-fx-2supxsup-2x-log-2-if-fx-f"}}]}]]

Question: Let f(x) = 2<sup>2x–1</sup> and f(x) = –2<sup>x</sup> + 2x log 2. If f´(x) \>f¢(x), then...

Solution