If x + | y | = 2y, th e n y as a function of x is
- defined for all real x
- continuous at x = 0
- differentiable for all x
- such that $ \frac{dy}{dx} = \frac{1}{3} $ for x < 0
The Correct Option is D
Solution and Explanation
\ x + y = 2y, \\
x - y = 2y, \\
\end{array} \begin{array}
when y > 0 \\
when y < 0 \\
\end{array}$
$\Rightarrow \bigg \{ \begin{array}
\ y = x, \ when \ y > 0 \Rightarrow x > 0 \\
y = x/3, \ when \ y < 0 \Rightarrow x < 0 \\
\end{array}$
which could be plotted as,
Clearly, y is continuous for all x but not differentiable at x = 0,
Also, $ \frac{dy}{dx} = \bigg \{ \begin{array}
\ 1, \ x > 0 \\
1/3, \ x < 0 \\
\end{array} $
Thus, f (x) is defined for all x, continuous at x = 0,
differentiable for all x $ \in \ R - \{ 0 \}, \frac{dy}{dx}, = \frac{1}{3} $ for x < 0.
Top Questions on Differentiability
Let the function, \(f(x)\) = \(\begin{cases} -3ax^2 - 2, & x < 1 \\a^2 + bx, & x \geq 1 \end{cases}\) Be differentiable for all \( x \in \mathbb{R} \), where \( a > 1 \), \( b \in \mathbb{R} \). If the area of the region enclosed by \( y = f(x) \) and the line \( y = -20 \) is \( \alpha + \beta\sqrt{3} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha + \beta \) is:
- JEE Main - 2025
- Mathematics
- Differentiability
- Let the function \(f(x) = (x^2 - 1)|x^2 - ax + 2| + \cos|x| \) be not differentiable at the two points \( x = \alpha = 2 \) and \( x = \beta \). Then the distance of the point \((\alpha, \beta)\) from the line \(12x + 5y + 10 = 0\) is equal to:
- JEE Main - 2025
- Mathematics
- Differentiability
- The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is.
- JEE Main - 2025
- Mathematics
- Differentiability
- If \( f(x) = |x| + |x - 1| \), then which of the following is correct?
- CBSE CLASS XII - 2025
- Mathematics
- Differentiability
- Check the differentiability of the function $f(x) = |x|$ at $x = 0$.
- CBSE CLASS XII - 2025
- Mathematics
- Differentiability
Questions Asked in JEE Advanced exam
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
- JEE Advanced - 2025
- Waves and Oscillations
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____.
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80- JEE Advanced - 2025
- Organic Chemistry
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
- JEE Advanced - 2025
- Coordinate Geometry
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.