Give that f(x) =\(\frac {1-cos4x}{x^2}\) if x < 0 ,f(x) = a if x = 0 , f(x) =\(\frac {\sqrt {x}}{\sqrt {16 + \sqrt {x} }- 4}\) if x > 0, is continuous at x = 0, then a will be
Give that f(x) =\(\frac {1-cos4x}{x^2}\) if x < 0 ,f(x) = a if x = 0 , f(x) =\(\frac {\sqrt {x}}{\sqrt {16 + \sqrt {x} }- 4}\) if x > 0, is continuous at x = 0, then a will be
- 16
- 2
- 4
- 8
The Correct Option is D
Solution and Explanation
As we know that function is continuous at x = 0, then:
lim (x→0-) f(x)= lim (x→0+) f(x) = f(a)
lim (x→0-) f(x) = lim (x→0-) \(\frac {1-cos4x}{x^2}\)
lim (x→0-) f(x) = lim (x→0-) \(\frac {(2 sin^2 2x)}{(2x)^2}\ .4\) = 8
lim (x→0+) f(x) = lim (x→0+)\(\frac {\sqrt {x}}{\sqrt {16 + \sqrt {x} }- 4}\)
lim (x→0+) = \({\sqrt {16 + \sqrt {x} } + 4}\)= 8
and f(0)= 8
so, a = 8
Therefore the correct option is (D) 8.
Top Questions on Continuity and differentiability
Sports car racing is a form of motorsport which uses sports car prototypes. The competition is held on special tracks designed in various shapes. The equation of one such track is given as
(i) Find \(f'(x)\) for \(0<x>3\).
(ii) Find \(f'(4)\).
(iii)(a) Test for continuity of \(f(x)\) at \(x=3\).
OR
(iii)(b) Test for differentiability of \(f(x)\) at \(x=3\).
- CBSE CLASS XII - 2026
- Mathematics
- Continuity and differentiability
- Let \[ f(x)=\lim_{\theta\to 0}\frac{\cos(\pi x)-x^{2/\theta}\sin(x-1)}{1-x^{2/\theta}(x-1)}. \] Statement 1: \(f(x)\) is discontinuous at \(x=1\).
Statement 2: \(f(x)\) is continuous at \(x=-1\).- JEE Main - 2026
- Mathematics
- Continuity and differentiability
Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}
- JEE Main - 2026
- Mathematics
- Continuity and differentiability
- If \[ f(x) = \begin{cases} \dfrac{a|x| + x^2 - 2(\sin|x|)(\cos|x|)}{x}, & x \ne 0, \\ b, & x = 0, \end{cases} \] is continuous at \( x = 0 \), then \( a + b \) is equal to.
- JEE Main - 2026
- Mathematics
- Continuity and differentiability
- (a) Differentiate $\sqrt{e^{\sqrt{2x}}}$ with respect to $e^{\sqrt{2x}}$ for $x>0$.
- CBSE CLASS XII - 2025
- Mathematics
- Continuity and differentiability
Questions Asked in MHT CET exam
- A particle is moving with a constant velocity of 5 m/s in a circular path of radius 2 m. What is the centripetal acceleration of the particle?
- MHT CET - 2025
- centripetal acceleration
- A ball is thrown vertically upwards with an initial velocity of 20 m/s. Calculate the time it takes for the ball to reach the highest point. (Assume \( g = 9.8 \, \text{m/s}^2 \))
- MHT CET - 2025
- laws of motion
- A charge of 2 $\mu C$ is placed in an electric field of intensity \( 4 \times 10^3 \, \text{N/C} \). What is the force experienced by the charge?
- MHT CET - 2025
- Electric charges and fields
- A particle is moving with a constant velocity of \( 5 \, \text{m/s} \) in a circular path of radius \( 2 \, \text{m} \). What is the centripetal acceleration of the particle?
- MHT CET - 2025
- Uniform Circular Motion
- Which of the following has the lowest boiling point?
- MHT CET - 2025
- intermolecular forces
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.