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f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x) has local maxima at x = –1 and f '(x) has local minima at x = 0, then
  • a)
    the distance between (–1, 2) and (a f(a)), where x = a is the point of local minima is 2√5
  • b)
    f(x) is increasing for x ∈ [1, 2 5]
  • c)
    f(x) has local minima at x = 1
  • d)
    the value of f(0) = 15
Correct answer is option 'B,C'. Can you explain this answer?
Verified Answer
f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x)...
Let f (x) = ax3 + bx2 + cx + d
Then, f (2) = 18  ⇒ 8a + 4b + 2c + d = 18 … (1)
f (1) = – 1 ⇒ a + b + c + d = – 1 … (2)
f (x) has local max. at x = – 1
⇒ 3a – 2b + c = 0 … (3)
f '(x) has local min. at x = 0 ⇒ b = 0 … (4)
Solving (1), (2), (3) and (4), we get
f ''(-1) <0, f ''1) > 0 ⇒ x =-1 is a point of local max. and x = 1 is a point of local min. Distance between (– 1, 2) and (1, f (1)), i.e. (1, – 1) is 
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Most Upvoted Answer
f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x)...
We can find the equation of the cubic polynomial by substituting the given values into the general form of a cubic polynomial: f(x) = ax^3 + bx^2 + cx + d.

Given f(2) = 18, we have:
18 = a(2^3) + b(2^2) + c(2) + d
18 = 8a + 4b + 2c + d

And given f(1) = ?, we have:
? = a(1^3) + b(1^2) + c(1) + d
? = a + b + c + d

We have two equations with four unknowns (a, b, c, d), so we cannot determine the exact value of f(1) unless we have more information.
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f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x) has local maxima at x = –1 and f '(x) has local minima at x = 0, thena)the distance between (–1, 2) and (a f(a)), where x = a is the point of local minima is 2√5b)f(x) is increasing for x ∈ [1, 2 5]c)f(x) has local minima at x = 1d)the value of f(0) = 15Correct answer is option 'B,C'. Can you explain this answer?
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f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x) has local maxima at x = –1 and f '(x) has local minima at x = 0, thena)the distance between (–1, 2) and (a f(a)), where x = a is the point of local minima is 2√5b)f(x) is increasing for x ∈ [1, 2 5]c)f(x) has local minima at x = 1d)the value of f(0) = 15Correct answer is option 'B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x) has local maxima at x = –1 and f '(x) has local minima at x = 0, thena)the distance between (–1, 2) and (a f(a)), where x = a is the point of local minima is 2√5b)f(x) is increasing for x ∈ [1, 2 5]c)f(x) has local minima at x = 1d)the value of f(0) = 15Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x) has local maxima at x = –1 and f '(x) has local minima at x = 0, thena)the distance between (–1, 2) and (a f(a)), where x = a is the point of local minima is 2√5b)f(x) is increasing for x ∈ [1, 2 5]c)f(x) has local minima at x = 1d)the value of f(0) = 15Correct answer is option 'B,C'. Can you explain this answer?.
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