Let f be a real-valued function of a real variable defined as f(x) = x^{2 }for x ≥ 0, and f(x) = -x^{2} for x < 0.

Which one of the following statements is true?

Option 4 : f(x) is differentiable but its first derivative is not differentiable at x = 0.

Given that,

A function f(x) is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

i.e.,

\(\rm\lim _{x \rightarrow a^{-}} f'(x)=\lim _{x \rightarrow a^{+}} f'(x)\)

**Analysis:**

f(x) = x^{2}, x ≥ 0

= -x^{2}, x ≤ 0

f^{'}(x) = 2x, x ≥ 0

= -2x, x < 0

f^{'}(x) = 2|x|

f^{’}(x) is continuous but not differentiable at x = 0.

Hence f(x) is differentiable but its first derivative is not differentiable at x = 0.

Option 3 : Continuous and non – differentiable

__Concept:__

A **function f(x) is continuous** at x = a if,

**Left limit = Right limit = Function value = Real and finite**

A **function is said to be differentiable** at x =a if,

**Left derivative = Right derivative = Well defined**

__Calculation:__

__Given:__

f(x) = |x|

|x| = x for x ≥ 0

|x|= -x for x < 0

At x = 0

Left limit = 0, Right limit = 0, f(0) = 0

As

**Left limit = Right limit = Function value = 0**

∴ **|X| is continuous at x = 0.**

Now

Left derivative (at x = 0) = -1

Right derivative (at x = 0) = 1

**Left derivative ≠ Right derivative**

**∴ |x| is not differentiable at x = 0**

Option 3 : Discontinuous at exactly three point

__Concept:__

A function is written in the form of the **ratio of two polynomial functions** is called a rational function.

Rational functions are **continuous at all the points except for the points where denominator becomes zero**.

\(f(x)=\dfrac{P(x)}{Q(x)}\)

where P(x) and Q(x) are polynomials and Q(x) ≠ 0.

f(x) will be **discontinuous at points where Q(x) = 0**.

__Calculation:__

__Given:__

\(f(x)=\dfrac{4-x^2}{4x-x^3}\)

This is a rational function, so it will be discontinuous at points where denominator becomes zero.

4x - x^{3} = 0

x(4 - x^{2}) = 0

x(2^{2} - x2) = 0

x(2 + x)(2 - x) = 0

x = 0, x = - 2 and x = 2

**Hence the function \(f(x)=\dfrac{4-x^2}{4x-x^3}\) will be discontinuous at exactly three points 0, - 2 and 2.**

Option 3 : \(f\left( a \right).f\left( b \right) > 0\)

We know that, (Intermediate value theorem)

If \(f\left( a \right)\;f\left( b \right) < 0\) then f(x) has at least one root in (a, b)

Since there is no root in [a, b] this implies f(a) and f(b) are of same sign

i.e. either they both are positive or they both are negative

In both cases \(f\left( a \right)f\left( b \right) > 0\)

Option 3 : Continuous in the interval (0, 1) and (1, ∞)

__Concept:__

__Given:__

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

A function will be **discontinuous** at the point where the **denominator becomes zero**.

X^{2} – 1 will become zero at x = +1 and x = -1

∴ Function f(x) will be continuous in the interval (0, 1),(1, ∞)

Option 4 : The limit must exist at the point and the value of limit should be same as the value of the function at that point

__Explanation:__

- For a function say f(x), \(\mathop {\lim }\limits_{x \to c} f(x)\) will exist when
**left and limit and right-hand limit of the function at point 'x = c' will be the same and it will be equal, to the limit of the function at the point 'x = c'**where l is a finite value. - Any function say f(x) is said to be continuous at a point say 'x = c' if and only if
**the limit exists at point 'x = c' and it will be equal to the value of that function at 'x = c'**.

\(⇒ \mathop {\lim }\limits_{x \to c^{-}} f(x) =\mathop {\lim }\limits_{x \to c^{+}} f(x) = f(c) =l\)

Where,

\(LHL = \mathop {\lim }\limits_{x \to c^{-}} f(x) \) and \(RHL = \mathop {\lim }\limits_{x \to c^{+}} f(x) \)

Referring to the graph below,

If LHL = RHL ≠ f(c) ⇒ Limit will exist at 'x = c' but f(x) at 'x= c' will not be continuous.

Here, LHL = RHL = p but f(c) = q

Referring to the graph below,

If LHL = RHL = f(c) ⇒ Limit will exist at 'x = c' and the function f(x) at 'x = c' will be continuous.

Here, LHL = RHL = f(c) = l ⇒ f(x) is continuous at 'x = c'

The values of x for which the function

\(f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} + 3x - 4}}\)

is **NOT** continuous are

Option 3 : -4 and 1

__Concept:__

A function f(x) is continuous at x = a, if the function is defined at x = a and,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

__Analysis:__

\(f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} + 3x - 4}}\)

This function is not defined for:

x^{2} + 3x – 4 = 0

⇒ (x + 4)(x - 1) = 0 i.e.

At x = 1 and x = - 4

∴ This function f(x) is not continuous at x = 1, -4Option 3 : is continuous ∀x ϵ R and differentiable ∀x ϵ R except at x=2/3

__Concept:__

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

__Calculation:__

__Given:__

f(x) = |2 − 3x|

|2 − 3x| = 2 - 3x for x < 2/3

|2 − 3x| = -2 + 3x for x ≥ 2/3

At x = 2/3,

Left limit = 0, Right limit = 0, f(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 2/3.

Now

Left derivative (at x = 2/3) = -3

Right derivative (at x = 2/3) = 3

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 2/3

Let \(g\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - x,}&{x \le 1}\\ {x + 1,}&{x \ge 1} \end{array}} \right.\) and \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {1 - x,}&{x \le 0}\\ {{x^2},}&{x > 0} \end{array}} \right.\).

Consider the composition of f and g, i.e. (fog)(x) = f(g(x)). The number of discontinuities in (fog)(x) present in the interval (-∞, 0) is:

Option 1 : 0

**Concept:**

A function f(x) is said to be continuous at a point x = a, in its domain if,

\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\) exists or its graph is a single unbroken curve.

In other words,

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

**Analysis:**

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {1 - x,}&{x \le 0}\\ {{x^2},}&{x > 0} \end{array}} \right.\)

\(g\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - x,}&{x \le 1}\\ {x + 1,}&{x \ge 1} \end{array}} \right.\)

For x ≤ 0, f(x) = 1 – x

g (x) = -x

(fog)(x) = f(g(x)) = f(-x) = 1 + x

y = 1 + x is a continuous function.

Hence, the number of discontinuities = 0

Option 2 : \(\left[ {\frac{{ - 1}}{2},1} \right]\)

We know that

(a + b + c)^{2} ≥ 0

a^{2} + b^{2} + c^{2} + 2(ab + bc + ac) ≥ 0

1 + 2(ab + bc + ac) ≥ 0

**(ab + bc + ac) ≥ -1/2**

(a - b)^{2} + (b - c)^{2} + (c - a)^{2} ≥ 0

2(a2 + b2 + c2) - 2(ab + bc + ac) ≥ 0

(a2 + b2 + c2) - (ab + bc + ac) ≥ 0

(1) - (ab + bc + ac) ≥ 0

**(ab + bc + ac) ≤ 1**

**⇒ -1/2 ≤ **(ab + bc + ac) ≤ 1

(ab + bc + ac) ∈ [-1/2, 1]

Option 3 : e

**Concept:**

A function is said to be continuous if

\(\rm \displaystyle \lim_{x\rightarrow a} f(x) = f(a)\)

**Calculation:**

**Given:**

f(x) = (x + 1)cotx

Taking log on both sides

log (f(x)) = cot x log (x + 1)

For checking continuity at x = 0

\(\rm \displaystyle \lim_{x\rightarrow 0} log(f(x)) = \rm \displaystyle \lim_{x\rightarrow 0} \cot x \ log(x\ +\ 1)\)

\(\rm \displaystyle \lim_{x\rightarrow 0} log(f(x)) = \rm \displaystyle \lim_{x\rightarrow 0} \frac { \ log(x\ +\ 1)}{tan \ x}\) = 0/0

Using L-Hopital rule

\(\rm \displaystyle \lim_{x\rightarrow 0} log(f(x)) = \rm \displaystyle \lim_{x\rightarrow 0} \frac { 1}{sec^2 \ x\ (x+1)} \)

\(= \rm \displaystyle \lim_{x\rightarrow 0} \frac {cos^2x }{(x+1)} \)

⇒ log (f(0)) = 1

**⇒ f(0) = e ^{1} = e**

Option 4 : continuous but not differentiable

__Concept:__

\(ϕ \left( x \right) = \left| x \right| = \left\{ {\begin{array}{*{20}{c}} {x,x > 0}\\ { - x,x < 0} \end{array}} \right.\)

or given function \(y = 5 |x|\), **the slope of the graph will only change. without shifting along X-axis and Y-axis.**

\(ϕ \left( x \right) = 5\left| x \right| = \left\{ {\begin{array}{*{20}{c}} {5x,x > 0}\\ { - 5x,x < 0} \end{array}} \right.\)

Hence, the given function is continuous about x = 0.

Left-hand side value of derivative of the function,

\({f'}\left( {0^ - } \right) = \mathop {\lim }\limits_{h \to 0^-} \dfrac{{f\left( {0\; -\;h} \right) - f\left( h \right)}}{h}\)

\(= \mathop {\lim }\limits_{h \to 0} \frac{{ - {5h}}}{h} = -5\)

Right-hand side value of derivative of the function,

\({f'}\left( {0^ + } \right) = \mathop {\lim }\limits_{h \to 0^+} \dfrac{{f\left( {0 + h} \right) - f\left( h \right)}}{h}\)

\(= \mathop {\lim }\limits_{h \to 0} \dfrac{{ +{5h}}}{h} = + \;5\)

LHS ≠ RHS

Hence , f(x) is continuous and not differentiable at x = 0.If 0 < x < 1, which of the following terms increases as x increases?

(i) 1 – x^{3}

(ii) x – 1

(iii) 1/x^{2}

Option 2 : (ii) only

__Concept:__

1. The graph for the function **1 - x ^{3}** will be as follows:

We can see that the value of y = 1 - x^{3} **decreases **as **x increases **in the interval [0,1]

2. The graph for the function **x - 1** will be as follows:

The value of y = x - 1 is **increasing** in the interval [0,1]

3. The graph for the function **1/x ^{2}** is as follows:

The value of y = 1/x^{2} decreases as **x increases** in the interval [0,1].

**From the above options, only the graph of "x-1" is increasing in the interval [0,1]**

Option 1 : There exists a y in the interval (0, 1) such that f(y) = f(y + 1)

Let’s define a new function t : t(y) = f(y) – f(y+1)

We know that f is continuous in [ 0 , 2 ], t will also be continuous in [ 0 , 1 ].

In question it is given that f(0)= - 1 , f(2) = -1 and f(1) = 1, putting into above function we get t(0) and t(1)

t(0) = f(0) - f(1) = -1 - 1 = - 2 and t(1) = f(1)- f(2) = 1+1 = 2

so we get t(0)=-2 and t(1) = 2 since function is changing its value from -2 negative to +2 positive, it means function t is a increasing function and there exists a point where t will be 0 in (0,1)

t=0 then f(y)= f(y+1)

Hence option 1 is the correct answer.

Same logic can be applied to option d:

h(y) = f(y) + f(y+2)

We know that function f is continuous in [0,2], function h will also be continuous in [0,1].

In question it is given that f(0)=-1, f(2)=-1 and f(1)=1, putting into above function.

h(0) = f(0) - f(1) = -1 - 1 = - 2 and h(1) = f(1)- f(2) = 1+1 = 2

So we get h(0)=-2 and h(1) = 2, since function is changing its value from -2 to +2. It means function h is a increasing function and there exists a point where h would be 0 in (0,1)

h=0 then f(y) = -f(y+2)

Hence option 1 and 4 both are correct answer.