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Manipal Entrance Test 2014 Math Paper

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Question : 64 of 80

Marks: +1, -0

The function

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

+\cos|x|

is non-differentiable at

-1
0
1
2

Solution:

we know that funtion

|x|

is not differentiable at

x=0

∴

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

Hence, it is not differentiable at

x=1

and 2

Now,

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

is not differentiable at

x=2

For

1 < x<2, f(x)=-(x^{2}-1)

(x^{2}-3x+2)+\cos x

For

2 < x<3, f(x)=(x^{2}-1)(x^{2}-3x+2)+\cos x

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

(x^{2}-3x+2)-\sin x

L'(2)=-3\sin 2

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

(x^{2}-3x+2)-\sin x

R f'(2)=(4-1)(4-3)+0-\sin 2

=3-\sin 2

Hence,

L'(2) \neq R f'(2)

So,

f(x)

is not differentable at

x=2

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