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\(\lim\limits_{x\to3^-} f(x) = \lim\limits_{x\to3^-} (x^2+5x+1)\)

= (3)² + 5(3) + 1 

= 9 + 15 + 1 

= 25

\(\lim\limits_{x\to3^+} f(x) = \lim\limits_{x\to3^+} (x^3+x+5)\)

= (3)³ + 3 + 5 

= 27 + 3 + 5 

= 35

∴\(\lim\limits_{x\to3^-} f(x) \neq \lim\limits_{x\to3^+} f(x)\)

∴\(\lim\limits_{x\to3} f(x) \) does not exist.

∴ f(x) is discontinuous at x = 3. 

∴ f(x) has a jump discontinuity at x = 3.

Correct answer is

(C) 0

LHL: = \(\lim\limits_{x \to2^-} \)f(x) =  \(\lim\limits_{x \to2^-} \) 2 + x 

= 4

RHL: = \(\lim\limits_{x \to2^+} \)f(x) =  \(\lim\limits_{x \to2^+} \) 2 - x 

= 0

 = \(\lim\limits_{x \to2^-} \)f(x) ≠  \(\lim\limits_{x \to2^+} \) f(x)

f(x) is discontinuous at x=2

LHL: = \(\lim\limits_{x \to0^-} \)f(x) =  \(\lim\limits_{x \to0^-} \)  3 - x 

= 3

RHL: = \(\lim\limits_{x \to3^+} \)f(x) =  \(\lim\limits_{x \to3^+} \) x² 

= 0 

  \(\lim\limits_{x \to3^-} \)f(x) ≠  \(\lim\limits_{x \to3^+} \) f(x)

f(x) is discontinuous at x = 0

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

\(\lim\limits_{h \to 0}f(c-h)\) = \(\lim\limits_{h \to 0}f(c+h)\) = f(c)

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise,

We can summarise it as, 

A function is continuous at x = c if  :

\(\lim\limits_{x \to c}f(x)\) = f(c)

Here we have,

\(f(x) = \begin{cases} x^{10}&, \quad \text{if } x ≤{1} \\x^2&,\quad \text{if } x >{1} \end{cases} \) ……equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain 

(domain = set of numbers for which f is defined)

Let c is any random number such that c < 1 

[thus c being random number, it is able to include all numbers less than 1]

f(c) = c¹⁰ [ using eqn 1]

\(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}x^{10}\)= c¹⁰

Clearly,

\(\lim\limits_{x \to c}f(x)\) = f(c)

∴ We can say that f(x) is continuous for all x < 1

As x = 1 is a point at which function is changing its nature so we need to check the continuity here.

f(1) = 1¹⁰ = 1 [using eqn 1]

LHL = \(\lim\limits_{h \to 0}f(1-h)\) 

= \(\lim\limits_{h \to 0}(1-h)^{10}\) = 1

RHL = \(\lim\limits_{h \to 0}f(1+h)\) 

= \(\lim\limits_{h \to 0}(1+h)^{2}\) = 1

Thus, 

LHL = RHL = f(1) 

∴ f(x) is continuous at x = 1 

Let m is any random number such that m > 1 

[thus m being random number, it is able to include all numbers greater than 1]

f(m) = m² [ using eqn 1]

And,

 \(\lim\limits_{x \to m}f(x)\) = \(\lim\limits_{x \to m}x^{2}\)=m²

Clearly,

\(\lim\limits_{x \to m}f(x)\) = f(m)

∴ We can say that f(x) is continuous for all m > 1 

Hence, 

f(x) is continuous for all real x 

There no point of discontinuity. It is everywhere continuous

Option : (D)

Formula :-

(i) A function f(x) is said to be continuous at a point x = a of its domain, if
\(\lim\limits_{x \to a}f(x)\) = f(a)
\(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a)

Given :-

\(f(x) = \begin{cases} ax^2+b &,0≤x<{1}\\ 4 &, x=1\\x+3&,1<x≤2\end{cases} \) 

\(\lim\limits_{x \to 1^-}f(x)\) = \(\lim\limits_{h\to0}f(1-h)\)²+ b

= a + b

Function f(x) is discontinuous at x = 1

\(\lim\limits_{x \to 1^-}f(x)\) ≠ f(1)

⇒ a + b ≠ 4

Checking option,

(5,2) is answer.

\(f(x)=\) \(tan^{-1}(\frac{1-x}{1+x})-\)\(tan^{-1}(\frac{x+2}{1-2x})\)

  \(=tan^{-1}(\frac{1-x}{1+x.1})-\) \(tan^{-1}(\frac{x+2}{1-2.x})\)

= (tan^-11 - tan^-1x) - (tan^-1x +tan^-12)

( ∵ \(tan^{-1}\frac{a-b}{1+ab}=\)tan^-1a - tan^-1b)

Differentiating with respect to x, we get

f'(x) = \(-\frac{2}{1+x^2}\)

Given that y = 3e^2x + 2e^3x 

Differentiating both sides w.r.t. x, we get

dy/dx = 6e^2x + 6e^3x = 6(e^2x + e^3x)

Again, Differentiating both sides w.r.t. x, we get

d²y/dx² = 6(2e^2x + 3e^3x)

Now, d²y/dx² - 5dy/dx + 6y = 6(2e^2x + 3e^3x) - 5(6(e^2x + e^3x)) + 6(3e^2x + 2e^3x)

= 12e^2x + 18e^3x - 30e^2x - 30e^3x + 18e^2x + 12e^3x = 0

Given that e^y(x + 1) = 1

⇒ e^y = 1/(x + 1)

Differentiating both sides w.r.t. x, we get

e^y(dy/dx) = - 1/(x + 1)² ⇒ 1/(x + 1)dy/dx = - 1/(x + 1)²

⇒ dy/dx = - 1/(x + 1)

Again, Differentiating both sides w.r.t. x, we get

d²y/dx² = 1/(x + 1)² = (- 1/(x + 1))² = (dy/dx)²

A real function f is said to be continuous at x = c, 

Where c is any point in the domain of f 

If :

\(\lim\limits_{h \to 0}f(c-h)\) = \(\lim\limits_{h \to 0}f(c+h)\) = f(c)

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as,

A function is continuous at x = c if :

\(\lim\limits_{x \to c}f(x)\) = f(c)

Here we have,

\(f(x) = \begin{cases}x^3-x^2+2x-2&,\quad if\, x ≠1\\4&,\quad if\,x=1\end{cases} \) ….equation 1

Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain 

(domain = set of numbers for which f is defined )

Function is changing its nature (or expression) at x = 1, 

So we need to check its continuity at x = 1 first.

Clearly,

f(1) = 4 [using eqn 1]

\(\lim\limits_{x \to 1}f(x)\) = \(\lim\limits_{x \to 1}(x^3-x^2+2x-2)\) 

= 1³ - 1² + 2 x 1 - 2 = 0

Clearly,

 \(\lim\limits_{x \to c}f(x)\) ≠ f(c)

∴ f(x) is discontinuous at x = 1. 

Let c be any real number such that c ≠ 0 

f(c) = c³ – c² + 2c – 2 [using eqn 1]

 \(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}(x^3-x^2+2x-2)\) 

= c³ – c² + 2c – 2

Clearly,

 \(\lim\limits_{x \to c}f(x)\) = f(c)

∴ f(x) is continuous for all real x except x = 1

A real function f is said to be continuous at x = c, 

Where c is any point in the domain of f 

If :

\(\lim\limits_{h \to 0}f(c-h)\) = \(\lim\limits_{h \to 0}f(c+h)\) = f(c)

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

A function is continuous at x = c if :

\(\lim\limits_{x \to c}f(x)\) = f(c)

Here we have,

\(f(x) = \begin{cases}\frac{x}{|x|}&,\quad x ≠0\\0&,\quad x=0\end{cases} \) …….equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain 

(domain = set of numbers for which f is defined)

Function is changing its nature (or expression) at x = 0, 

So we need to check its continuity at x = 0 first.

NOTE : 

Definition of mod function : 

|x| = \(\begin{cases}-x,x <0\\x,x≥0\end{cases} \)

 LHL = \(\lim\limits_{h \to 0}f(0-h)\) 

= \(\lim\limits_{h \to 0}f(-h)\) 

= \(\frac{-h}{-(-h)}\)

= \(\frac{-h}{h}\) 

= -1

[using eqn1 and idea of mod fn]

f(0) = 0 

[using eqn 1] 

Clearly,

LHL ≠ RHL ≠ f(0) 

∴ function is discontinuous at x = 0 

Let c be any real number such that c > 0

∴ f(c) = \(\frac{c}{|c|}\)

= \(\frac{c}{c}\) = 1

[using eqn 1]

And,

\(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}\frac{c}{|c|}\) 

= \(\lim\limits_{x \to c}\frac{c}{c}\) = 1

Thus,

\(\lim\limits_{x \to c}f(x)\) = f(c)

∴ f(x) is continuous everywhere for x > 0. 

Let c be any real number such that c < 0

∴ f(c) = \(\frac{c}{|c|}\) = \(\frac{c}{-c}\) = -1

[using eqn 1 and idea of mod fn]

And,

 \(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}\frac{c}{|c|}\) 

= \(\lim\limits_{x \to c}\frac{c}{-c}\) = -1

Thus,

 \(\lim\limits_{x \to c}f(x)\) = f(c)

∴ f(x) is continuous everywhere for x < 0.

Hence, 

We can conclude by stating that f(x) is continuous for all Real numbers except zero.

A real function f is said to be continuous at x = c, 

Where c is any point in the domain of f 

If :

\(\lim\limits_{h \to 0}f(c-h)\) = \(\lim\limits_{h \to 0}f(c+h)\) = f(c)

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as

A function is continuous at x = c if :

\(\lim\limits_{x \to c}f(x)\) = f(c)

Here we have,

\(f(x) = \begin{cases}\frac{sin\,x}{x}&,\quad x <0\\x+1&,\quad x≥0\end{cases} \) …….equation 1

To prove it everywhere continuous we need to show that at every point in the domain of f(x) 

[domain is nothing but a set of real numbers for which function is defined ]

 \(\lim\limits_{x \to c}f(x)\) = f(c),

Where c is any random point from domain of f.

Clearly from definition of f(x) {see from equation 1}, 

f(x) is defined for all real numbers.

∴ we need to check continuity for all real numbers.

Let c is any random number such that c < 0 

[thus c being a random number, it can include all negative numbers ]

f(c) = \(\frac{sin\,c}{c}\) [ using eqn 1]

\(\lim\limits_{x \to c}f(x)\) =\(\lim\limits_{x \to c}\frac{sin\,x}{x}\) = \(\frac{sin\,c}{c}\)

∴ We can say that f(x) is continuous for all x < 0 

Now, 

Let m be any random number from the domain of f such that m > 0 thus m being a random number, it can include all positive numbers.

f(m) = m+1 [using eqn 1]

\(\lim\limits_{x \to m}f(x)\) = \(\lim\limits_{x \to m}x+1\) = m + 1

Clearly,

\(\lim\limits_{x \to c}f(x)\) = f(c) = m + 1

∴ We can say that f(x) is continuous for all x > 0

As zero is a point at which function is changing its nature so we need to check LHL, RHL separately 

f(0) = 0+1 = 1 [using eqn 1]

LHL = \(\lim\limits_{h \to 0}f(0-h)\) 

= \(\lim\limits_{h \to 0}\frac{sin\,(-h)}{-h}\) 

= \(\lim\limits_{h \to 0}\frac{sin\,h}{h}\) = 1

[∵ sin – θ = – sin θ and \(\lim\limits_{h \to 0}\frac{sin\,h}{h}\) = 1 ]

RHL = \(\lim\limits_{h \to 0}f(0+h)\) 

= \(\lim\limits_{h \to 0}h+1\) = 1

Thus, 

LHL = RHL = f(0). 

∴ f(x) is continuous at x = 0 

Hence, 

We proved that f is continuous for x < 0 ; 

x > 0 and x = 0 

Thus, 

f(x) is continuous everywhere. 

Hence, proved

y = (tan^-1 x)²
⇒ y₁ = 2(tan^-1 x) \(\frac{1}{1+x^2}\)
⇒ (1 + x²)y₁ = 2(tan^-1 x)
⇒ (1 + x²)y₂ + y₁2x = 2 \(\frac{1}{1+x^2}\)
⇒ (1 + x²)² y₂ + x(1 + x²)y₁ = 2.

(i) (b) f(x) = \(\begin{cases} \frac{|x-2|}{x-2}, & \quad x≠2\\ 0, & \quad x=2 \end{cases} \)

(ii) From the figure we can see that
f(2^–) = 1, f(2⁺) = -1 and f(2) = 0
Therefore, f(2^–) = 1 ≠ f(2⁺) = -1 ≠ f(2) = 0. Discontinuous.

1. To find lim_x→2f(x)
we have to find f(2^–) and f(2⁺)
f(2^–) = lim_x→2x−[x] = 2 -1 = 1,
f(2⁺) = lim_x→2 3x – 5 = 6 -5 = 1
f(2^–) = f(2⁺) = 1.
Therefore lim_x→2 f(x) = 1

2. Here, f(2) = 0 ≠ f(2^–) = f(2⁺) = 1.
Therefore discontinuity at x = 2.

∵ x = 2 at²

⇒ \(\frac{dx}{dy}=4at\)

Again,

y = at⁴

\(\frac{dx}{dy}=4at^3\)

Now,

\(\frac{dx}{dy}=\frac{dy/dt}{dx/dt}\)

= \(\frac{4at^3}{4at}\) = t²

Given, f(x) = 32^x - 15^x

It is the difference of two exponential functions

∴ It is continuous on the set of real numbers i.e., x ∈ R.

Given, f(x) = 7x⁴ - 5x³ - 3x + 1

It is a polynomial function

∴ f(x) is continuous on the set of real numbers i.e., x ∈ R.

Correct option is:(C) x = 0, 1 

If f(x) =[x] for x ∈ (-1, 2)

This function is discontinuous at all integer values of x between -1 and 2. 

∴ f(x) is discontinuous at x = 0 and x = 1.