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Question: The function f (x) = \(\left( {{x^2} - 1} \right)\left| {{x^2} - 3x + 2} \right| + \cos \left| x \ri...

The function f (x) = (x2−1)∣x2−3x+2∣+cos⁡∣x∣\left( {{x^2} - 1} \right)\left| {{x^2} - 3x + 2} \right| + \cos \left| x \right| is non-differentiable at
(a)\left( a \right) -1
(b)\left( b \right) 0
(c)\left( c \right) 1
(d)\left( d \right) 2

Explanation

Solution

Hint : In this particular type of question use the concept if there is any modulus part in the equation equate the function which is inside the modulus to zero and evaluate the condition in which function opens as positive and in which function opens as negative, then factorize the equation so use these concepts to reach the solution of the question.

Complete step by step solution :
Given function –
f(x)=(x2−1)∣x2−3x+2∣+cos⁡∣x∣f\left( x \right) = \left( {{x^2} - 1} \right)\left| {{x^2} - 3x + 2} \right| + \cos \left| x \right|
Now we have to check this function is not differentiable at,
According to differentiability at a particular point (say, x = a) if left hand derivative = right hand derivation then the function is differentiable.
I.e. f′(a)L.H.D=f′(a)R.H.Df'{\left( a \right)_{L.H.D}} = f'{\left( a \right)_{R.H.D}}
To differentiable
So as we see that the above function is having modulus.
So first equate the modulus part to zero we have,
⇒x2−3x+2=0\Rightarrow {x^2} - 3x + 2 = 0
Now factorize this equation we have
⇒x2−x−2x+2=0\Rightarrow {x^2} - x - 2x + 2 = 0
⇒x(x−1)−2(x−1)=0\Rightarrow x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0
⇒(x−1)(x−2)=0\Rightarrow \left( {x - 1} \right)\left( {x - 2} \right) = 0
⇒x=1,2\Rightarrow x = 1,2
Now when, 1 < x < 2 function is negative so the modulus function open as negative, so this is the condition of L.H.D
And when, x > 2 function is positive so the modulus function open as positive, so this is the condition of R.H.D
And we all know that cos (-x) = cos x, so cos |x| = cos x, for L.H.D as well as for R.H.D respectively.
L.H.D
f′(x)=ddx((x2−1)(−(x2−3x+2))+cos⁡x)f'\left( x \right) = \dfrac{d}{{dx}}\left( {\left( {{x^2} - 1} \right)\left( { - \left( {{x^2} - 3x + 2} \right)} \right) + \cos x} \right)
And R.H.D
f′(x)=ddx((x2−1)(x2−3x+2)+cos⁡x)f'\left( x \right) = \dfrac{d}{{dx}}\left( {\left( {{x^2} - 1} \right)\left( {{x^2} - 3x + 2} \right) + \cos x} \right)
Now when x < 1, the modulus function opens negative and for 1 < x < 2, the modulus function again opens as negative.
So when, x = 1 function is always differentiable.
Now check for, x = 2
So, L.H.D
f′(x)=ddx((x2−1)(−(x2−3x+2))+cos⁡x)f'\left( x \right) = \dfrac{d}{{dx}}\left( {\left( {{x^2} - 1} \right)\left( { - \left( {{x^2} - 3x + 2} \right)} \right) + \cos x} \right)
Now differentiate it w.r.t x , according to property that [ddxab=addxb+bddxa]\left[ {\dfrac{d}{{dx}}ab = a\dfrac{d}{{dx}}b + b\dfrac{d}{{dx}}a} \right], so we have,
f′(x)=−[(x2−1)ddx(x2−3x+2)+(x2−3x+2)ddx(x2−1)]+ddxcos⁡xf'\left( x \right) = - \left[ {\left( {{x^2} - 1} \right)\dfrac{d}{{dx}}\left( {{x^2} - 3x + 2} \right) + \left( {{x^2} - 3x + 2} \right)\dfrac{d}{{dx}}\left( {{x^2} - 1} \right)} \right] + \dfrac{d}{{dx}}\cos x
Now as we know that, [ddxxn=nxn−1,ddxcos⁡x=−sin⁡x]\left[ {\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}\cos x = - \sin x} \right], so we have,
f′(x)=−[(x2−1)(2x−3)+(x2−3x+2)(2x)]−sin⁡xf'\left( x \right) = - \left[ {\left( {{x^2} - 1} \right)\left( {2x - 3} \right) + \left( {{x^2} - 3x + 2} \right)\left( {2x} \right)} \right] - \sin x
Now for x = 2 we have,
f′(2)=−[(22−1)(2(2)−3)+(22−3(2)+2)(2×2)]−sin⁡2f'\left( 2 \right) = - \left[ {\left( {{2^2} - 1} \right)\left( {2\left( 2 \right) - 3} \right) + \left( {{2^2} - 3\left( 2 \right) + 2} \right)\left( {2 \times 2} \right)} \right] - \sin 2
f′(2)=−[3+0]−sin⁡2f'\left( 2 \right) = - \left[ {3 + 0} \right] - \sin 2
f′(2)=−3−sin⁡2f'\left( 2 \right) = - 3 - \sin 2
So L.H.D = -3 – sin 2
Now, R.H.D
f′(x)=ddx((x2−1)((x2−3x+2))+cos⁡x)f'\left( x \right) = \dfrac{d}{{dx}}\left( {\left( {{x^2} - 1} \right)\left( {\left( {{x^2} - 3x + 2} \right)} \right) + \cos x} \right)
Now differentiate it w.r.t x so we have,
f′(x)=[(x2−1)ddx(x2−3x+2)+(x2−3x+2)ddx(x2−1)]+ddxcos⁡xf'\left( x \right) = \left[ {\left( {{x^2} - 1} \right)\dfrac{d}{{dx}}\left( {{x^2} - 3x + 2} \right) + \left( {{x^2} - 3x + 2} \right)\dfrac{d}{{dx}}\left( {{x^2} - 1} \right)} \right] + \dfrac{d}{{dx}}\cos x
Now as we know that, [ddxxn=nxn−1,ddxcos⁡x=−sin⁡x]\left[ {\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}\cos x = - \sin x} \right], so we have,
f′(x)=[(x2−1)(2x−3)+(x2−3x+2)(2x)]−sin⁡xf'\left( x \right) = \left[ {\left( {{x^2} - 1} \right)\left( {2x - 3} \right) + \left( {{x^2} - 3x + 2} \right)\left( {2x} \right)} \right] - \sin x
Now for x = 2 we have,
f′(2)=[(22−1)(2(2)−3)+(22−3(2)+2)(2×2)]−sin⁡2f'\left( 2 \right) = \left[ {\left( {{2^2} - 1} \right)\left( {2\left( 2 \right) - 3} \right) + \left( {{2^2} - 3\left( 2 \right) + 2} \right)\left( {2 \times 2} \right)} \right] - \sin 2
f′(2)=[3+0]−sin⁡2f'\left( 2 \right) = \left[ {3 + 0} \right] - \sin 2
f′(2)=3−sin⁡2f'\left( 2 \right) = 3 - \sin 2
So R.H.D = 3 – sin 2
So as we see that, L.H.D ≠\neR.H.D, at x = 2.
So the given function is not differentiable at x = 2.
So this is the required answer.
Hence option (d) is the correct answer.

Note : Whenever we face such types of questions the key concept we have to remember is that always recall that, function to be differentiable L.H.D should be equal to R.H.D, so first simplify the modulus part and find out L.H.S and R.H.S function as above then differentiate it as above and check that at, x = 2 whether L.H.D is equal to the R.H.D or not if yes then it is differentiable otherwise not.