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.
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Differentiability Questions
- \(\frac{dy}{dx}\) = \(\frac{(1-x-y^2)}{y}\) and \(x(1)=1\) , then \(5x(2)\) is equal to____.
- Let \( g : \mathbb{R} \rightarrow \mathbb{R} \) be a non-constant twice differentiable function such that \( g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right) \). If a real-valued function \( f \) is defined as \[ f(x) = \frac{1}{2} \left[ g(x) + g(2 - x) \right], \] then
- Let \( f(x) = (x + 3)^2 (x - 2)^3 \), \( x \in [-4, 4] \). If \( M \) and \( m \) are the maximum and minimum values of \( f \), respectively in \([-4, 4]\), then the value of \( M - m \) is:
- Let \( a \) and \( b \) be real constants such that the function \( f \) defined by \[f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 \\ bx + 2, & x>1 \end{cases}\]be differentiable on \( \mathbb{R} \). Then, the value of \( \int_{-2}^{2} f(x) \, dx \) equals
- Let \( g(x) \) be a linear function and \[ f(x) = \begin{cases} g(x), & x \leq 0 \\ \left( \frac{1 + x}{2 + x} \right)^{\frac{1}{x}}, & x > 0 \end{cases} \] is continuous at \( x = 0 \). If \( f'(1) = f(-1) \), then the value of \( g(3) \) is
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
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.