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UPSC NDA I 2016 Mathematics Solved Paper

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For the next two (2) items that follow :

Consider the function

f(x)=(x-1)^2(x+1)(x-2)^3

[NDA I 2016]

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Question : 29 of 120

Marks: +1, -0

What is the number of points of local minima of the function f(x) ?

None
One
Two
Three

Solution:

f(x)=(x-1)^{2}(x+1)(x-2)^{3}

f'(x)=2(x-1)(x+1)(x-2)^{3}+(x-1)^{2}(x-2)^{3}+(x-1)^{2}(x+1)3(x-2)^{2}

=(x-1)(x-2)^{2}[2(x+1)(x-2)+(x-1)(x-2)+3(x-1)(x+1)]

f'(x)=(x-1)(x-2)^{2}[2x^{2}-2x-4+x^{2}-3x+2+3x^{2}-3]

=(x-1)(x-2)^{2}[6x^{2}-5x-5]

For maxima and minima

f'(x)=0

(x-1)(x-2)^{2}[6x^{2}-5x-5]=0

x=1,2,2,\; \;\frac{5 \pm \sqrt{145}}{12}

The change in signs of f(x) for different values of x is shown:

∴ Local Minima are

x=\; \;\frac{5 - \sqrt{145}}{12}

&

x=\; \;\frac{5 + \sqrt{145}}{12}

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