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f(x) is a polynomial of degree 4, vanishes at x = -1 and has local minimum/maximum at x = 1, x = 2, and x = 3. If, -2∫^2 f(x) dx = -1348/15. Then what is the value of f(x)?(a) x^4 – 8x^3 + 22x^2 – 24x – 55(b) x^4 – 8x^3 + 22x^2 – 24x + 55(c) x^4 – 8x^3 – 22x^2 – 24x – 55(d) Data not sufficientThis question was addressed to me during an online exam.Asked question is from First Order Derivative in chapter Limits and Derivatives of Mathematics – Class 11

Answer»

The CORRECT answer is (a) x^4 – 8x^3 + 22x^2 – 24x – 55

Explanation: Here it is given that f(x) is a polynomial of degree 4

So, it has critical POINTS at x=1, x=2 and x=3

So, f’(x) = A(x – 1)(x – 2)(x – 3)

 = A(x^3 – 6x^2 + 11x -6)

On INTEGRATING this, gives,

f(x) = A/4(x^4 – 8x^3 + 22x^2 – 24x) + B

Also, f(-1) = 0

So, we get,

B = -554A/4

Also, it is given that -2∫^2 f(x) dx = -1348/15

So, A/4 -2∫^2(x^4 – 8x^3 + 22x^2 – 24x – 55)dx = -1348/15

By changing the RANGE of the integrals, we get,

=> A/2 0∫^2(x^4 + 22x^2 – 55) dx = -1348/15

=> -337A/15 = -1348/15

=> A = 4

Thus, f(x) = x^4 – 8x^3 + 22x^2 – 24x – 55.


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