Question:
For $ x \epsilon R , f (x) = | \log 2 - \sin x|$ and $g(x) = f(f(x))$, then :
For $ x \epsilon R , f (x) = | \log 2 - \sin x|$ and $g(x) = f(f(x))$, then :
Updated On: July 22, 2025
- $g$ is not differentiable at $x = 0$
- $g'(0)$ = $cos(log2)$
- $g'(0)$ = $-cos(log2)$
- $g$ is differentiable at $x = 0$ and $g'(0) = -sin(log2)$
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
$g(x)=\left|\log _{e} 2-\sin \left(\left|\log _{e} 2-\sin x\right|\right)\right|$
At $x=0, g(x)=\log _{e}(2)-\sin \left(\log _{e} 2-\sin x\right)$
$\therefore g'(x)=\cos \left(\log _{e}(2)-\sin x\right) \times \cos (x)$
$\Rightarrow g'(0)=\cos \left(\log _{e}(2)\right)$
At $x=0, g(x)=\log _{e}(2)-\sin \left(\log _{e} 2-\sin x\right)$
$\therefore g'(x)=\cos \left(\log _{e}(2)-\sin x\right) \times \cos (x)$
$\Rightarrow g'(0)=\cos \left(\log _{e}(2)\right)$
Was this answer helpful?
0
0
Top Questions on Differentiability
- If \( f(x) = |x| + |x - 1| \), then which of the following is correct?
- CBSE CLASS XII - 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
- Check the differentiability of the function $f(x) = |x|$ at $x = 0$.
- CBSE CLASS XII - 2025
- Mathematics
- Differentiability
- Consider the quadratic equation \( ax^2 + bx + c = 0 \), where \( 2a + 3b + 6c = 0 \) and let \[ g(x) = \frac{a x^3}{3} + \frac{b x^2}{2} + c x \] Statement-I: The given quadratic equation \( ax^2 + bx + c = 0 \) has at least one root in \( (0, 1) \). Statement-II: Rolle's theorem is applicable to \( g(x) \) on \( [0, 1] \). Then:
- AP EAPCET - 2025
- Mathematics
- Differentiability
- If \(f(x) = \sec^{-1} \left(\frac{1}{2x^2 -1}\right)\) and \(g(x) = \tan^{-1} \left(\frac{\sqrt{1 + x^2} - 1}{x}\right)\), then the derivative of \(f(x)\) with respect to \(g(x)\) is?
- AP EAPCET - 2025
- Mathematics
- Differentiability
View More Questions
Questions Asked in JEE Main exam
- The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be ________________________ A.
- JEE Main - 2025
- Electrical Circuits
- Let \( M \) denote the set of all real matrices of order 3 x 3 and let \( S = \{-3, -2, -1, 1, 2\} \). Let
\( S_1 = \{A = [a_{ij}] \in M : A = A^T \text{ and } a_{ij} \in S, \forall i, j\} \),
\( S_2 = \{A = [a_{ij}] \in M : A = -A^T \text{ and } a_{ij} \in S, \forall i, j\} \),
\( S_3 = \{A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j\} \).
If \(n(S_1 \cup S_2 \cup S_3) = 125\), then \( \alpha \) equals:- JEE Main - 2025
- Linear Algebra
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
- JEE Main - 2025
- 3D Geometry
- Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of \(16((sec^{-1}x)^{2}+(cosec^{-1}x)^{2})\) is:
- JEE Main - 2025
- Inverse Trigonometric Functions
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
View More Questions
JEE Main Notification
GMC Azamgarh Admission 2025: Dates, Courses, Fees, Eligibility, SelectJuly 22, 2025GMC Azamgarh Admission 2025
Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.