The number of points, where the function $f: \mathbb{R} \to \mathbb{R}$, $$ f(x) = |x - 1|\cos|x - 2|\sin|x - 1| + (x - 3)|x^2 - 5x + 4|, $$ is NOT differentiable, is
- 1
- 2
- 3
- 4
The Correct Option is B
Approach Solution - 1
To determine the number of points where the function \( f(x) = |x - 1|\cos|x - 2|\sin|x - 1| + (x - 3)|x^2 - 5x + 4| \) is not differentiable, we need to analyze the points of non-differentiability in each term of the function.
The full function can be split into two components:
- \( g(x) = |x - 1|\cos|x - 2|\sin|x - 1| \)
- \( h(x) = (x - 3)|x^2 - 5x + 4| \)
We shall determine points of non-differentiability in each part and then combine them to find the overall points of non-differentiability for \( f(x) \).
Step 1: Analyzing \( g(x) = |x - 1|\cos|x - 2|\sin|x - 1| \)
- The expression involves absolute values of \( |x - 1| \) and \( |x - 2| \). Points \( x = 1 \) and \( x = 2 \) are potential points of non-differentiability (because the absolute value function has corners at arguments where the argument becomes zero).
Step 2: Analyzing \( h(x) = (x - 3)|x^2 - 5x + 4| \)
- The expression \( x^2 - 5x + 4 = (x - 1)(x - 4) \), and it involves the absolute value, impacting points where argument is zero, i.e., \( x = 1 \) and \( x = 4 \).
- Thus, points of interest for non-differentiability due to absolute values are \( x = 1, 3, \) and \( 4 \), because \( x - 3 \) itself does not make this term non-differentiable when \( x \neq 3 \).
Step 3: Conclusion
Combining results from both components, we look at the union of all potential non-differentiable points:
- The points \( x = 1, 2, 4 \) are where absolute values cause non-differentiability.
- Point \( x = 3 \) in \( h(x) \) does not actually introduce non-differentiability due to the linear factor cancelling out the \( x - 3 \) in the expression \( (x - 3)|x^2 - 5x + 4| \).
Thus, the number of points where \( f(x) \) is not differentiable is 3.
Approach Solution -2
f(x) = |x – 1| cos |x – 2| sin |x – 1| + (x – 3) |x2 – 5x + 4|
= |x – 1| cos |x – 2| sin |x – 1| + (x – 3) |x – 1||x – 4|
= |x – 1| [cos |x – 2| sin |x – 1| + (x – 3) |x – 4|]
Sharp edges at x = 1 and x = 4
∴ Non-differentiable at x = 1 and x = 4.
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Concepts Used:
Differentiability
Differentiability of a function A function f(x) is said to be differentiable at a point of its domain if it has a finite derivative at that point. Thus f(x) is differentiable at x = a
\(\frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
⇒ f'(a – 0) = f'(a + 0)
⇒ left-hand derivative = right-hand derivative.
Thus function f is said to be differentiable if left hand derivative & right hand derivative both exist finitely and are equal.
If f(x) is differentiable then its graph must be smooth i.e. there should be no break or corner.
Note:
(i) Every differentiable function is necessarily continuous but every continuous function is not necessarily differentiable i.e. Differentiability ⇒ continuity but continuity ⇏ differentiability
(ii) For any curve y = f(x), if at any point \(\frac{d y}{d x}\) = 0 or does not exist then, the point is called “critical point”.
3. Differentiability in an interval
(a) A function fx) is said to be differentiable in an open interval (a, b), if it is differentiable at every point of the interval.
(b) A function f(x) is differentiable in a closed interval [a, b] if it is
- Differentiable at every point of interval (a, b)
- Right derivative exists at x = a
- Left derivative exists at x = b.