Let fx=limn→∞(x^2+2x+4+sin⁡πx)/(n-1x^2+2x+4+sin⁡πxn+1), then

Correct answer: (A) fx is continuous and differentiable for all x∈R.. ∵ x^2+2x+4+sinπx=x+1^2+3+sin⁡πx≥2,∀x∈R. ∴ fx=limn→∞(1-x+1^2+3+sin⁡πx-n)/(1+x+1^2+3+sin⁡πx-n) =(1-0)/(1+0)=1 Clearly, fx=1, ∀x∈R Hence, f(x) is continuous and differentiable ∀x∈R

Q49 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

Let $f (x) = \underset{n \to \infty}{l i m} \frac{{(x^{2} + 2 x + 4 + \sin π x)}^{n} - 1}{{(x^{2} + 2 x + 4 + \sin π x)}^{n} + 1},$ then

Correct answer
A
$f (x)$ is continuous and differentiable for all $x \in R .$
B
$f (x)$ is continuous but not differentiable for all $x \in R .$
C
$f (x)$ is discontinuous at infinite number of points.
D
$f (x)$ is discontinuous at two points.

Solution

∵ x^{2} + 2 x + 4 + s i n π x = {(x + 1)}^{2} + 3 + \sin π x \geq 2, \forall x \in R .

∴ f (x) = \underset{n \to \infty}{l i m} \frac{1 - {\{{(x + 1)}^{2} + 3 + \sin π x\}}^{- n}}{1 + {\{{(x + 1)}^{2} + 3 + \sin π x\}}^{- n}}

= \frac{1 - 0}{1 + 0} = 1

Clearly,

f (x) = 1, \forall x \in R

Hence,

f (x)

is continuous and differentiable

\forall x \in R

Hence, the correct answer is option: 'A'.

Let fx=limn→∞x2+2x+4+sin⁡πxn-1x2+2x+4+sin⁡πxn+1, then

Solution

More Continuity and Differentiability Questions

Let $f (x) = \underset{n \to \infty}{l i m} \frac{{(x^{2} + 2 x + 4 + \sin π x)}^{n} - 1}{{(x^{2} + 2 x + 4 + \sin π x)}^{n} + 1},$ then