Mathematics Test - 16

Concept:

• A function is said to continuous at a given value if the function does not have a sudden change at that value.
• If two functions are individually continuous then it’s summation difference, the product will be also continuous.
• A function is continuous at x = a
  If L.H.L = R.H.L = f(a)
  i.e. \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = f\left( a \right)\) 
• If f(x) and g(x) are individually continuous then it’s f(x) ± g(x), f(x) ⋅ g(x) and \(\frac{{f\left( x \right)}}{{g\left( x \right)}}\;;g\left( x \right) \ne 0\) will be continuous.

Calculation:

Option (A):

Let f(x) = sin x

From graph, at origin i.e. x = 0, sin x is continuous.

Option (B):

Let g(x) = x

From graph at origin i.e. x = 0

g(x) = x is continuous.

Option (C):

Let h(x) = x sin x

h(x) = g(x) ⋅ f(x)

Since f(x) and g(x) is individually continuous at origin, so h(x) will be also continuous.

Option (D):

Let \(h\left( x \right) = \frac{{\sin x}}{x} = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) 

Since sin x is continuous at origin and x is also continuous at origin and values are zero.

i.e. \(h\left( x \right) = \frac{{\sin x}}{x}\) 

At origin:

\(h\left( {x = 0} \right) = \frac{{\sin \left( 0 \right)}}{0} = \frac{0}{0}\)  is a indeterminate form)

Therefore, we can say that.

h(x) will be discontinuous at origin as at x = 0, h(x) become undefined.

So, option (D) is correct as \(\frac{{\sin x}}{x}\) is discontinuous at origin.

Concept:

The function is continuous at any point x when

Left hand limit  = Right hand limit = f(x)

Where LHL = \(\rm\lim_{α \rightarrow 0}\) f(x - α) and RHL = \(\rm\lim_{α \rightarrow 0}\) f(x + α)

Calculation:

Given f(x) = \(\begin{Bmatrix} \rm 3x + 2, x \geq1\\ \rm 5, \;\;\;\;\;\;\;\;x<1 \end{Bmatrix}\)

LHL = \(\rm\lim_{α \rightarrow 0}\)f(1 - α)

LHL = 5

f(1) = 3(1) + 2

f(1) = 5

RHL = \(\rm\lim_{α \rightarrow 0}\)f(1 + α)

RHL = \(\rm\lim_{α \rightarrow 0}\)[3(1 + α) + 2]

RLH = \(\rm\lim_{α \rightarrow 0}\)[5 + 3α]

RHL = 5

∵ LHL = RHL = f(x)

∴ f(x) is continuous and f(1) = 5

Concept:

\(\rm \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

\(\rm \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

Steps for derivatives of functions expressed in the parametric form:

1. First of all, we write the given functions u and v in terms of the parameter x.
2. Using differentiation find out du/dx and dv/dx.
3. Then by using the formula used for solving functions in parametric form i.e.
4.  Lastly substituting the values of du/dx and dv/dx and simplify to obtain the result.

Calculation:

Let u = tan-1 x and v = cot-1 x

Differentiating with respect to x, we get

\(\rm \frac{du}{dx} = \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

\(\rm \frac{dv}{dx} = \frac{d(\cot^{-1} x)}{dx} = \frac{-1}{1+x^2}\)

Now,

\(\rm \frac{d\tan^{-1} x}{d\cot^{-1} x} = \frac{du}{dv}\\=\frac{\frac{du}{dx}}{\frac{dv}{dx}}\\=\frac{\frac{1}{1+x^2}}{\frac{-1}{1+x^2}}=-1\)

CONCEPT:

• By Lagrange mean value theorem, since f is continuous in the interval [a,b] and diff in (a, b) so there exists at least one point ε in between (a,b) for which,

\(f'(ε ) = \frac{f(b) - f(a)}{(b - a)}\;\;\;\;\; \ldots \left( 1 \right)\)

Where a < ε  < b

EXPLANATION:

Given:  f(b) - f(a) = (b - a) f'(ε) for f(x) = Ax² + Bx + C in (a, b)

f(x) = Ax2 + Bx + C

⇒ f'(x) = 2Ax + B

⇒ f'(ε) = 2Aε + B

• But given that f(b) - f(a) = (b - a) f'(ε)
• Using equation 1,

 \(⇒ 2Aε + B =\frac{(Ab^2 + Bb + C) - (Aa^2 + Ba + C)}{b - a}\)

 \(\Rightarrow ε = \frac{b + a}{2} \in (a, b)\)

∴ Option (3) is correct

Concept:

L’Hospital’s Rule

Suppose that we have one of the following cases,

I.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule as:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Calculation:

Since, the function is continuous at x = 1.

Then, f(1) = \(\displaystyle\lim_{x\rightarrow1}{f(x)}\) 

k = \(\displaystyle\lim_{x\rightarrow1}\frac{log x}{x-1}\) = \(\frac{0}{0}\) ( is in indeterminate form )

Apply L´Hospital rule,

k = \(\displaystyle\lim_{x\rightarrow1}\frac{1/ x}{1}\) = 1 as x tends to 1

∴ k = 1

Hence, the correct answer is option 3)

CONCEPT:

• By Lagrange mean value theorem, since f is continuous in the interval [1,5] and diff in (1, 5) so there exists at least one point in between (1,5) for which,

\(f'(c) = \frac{f(5) - f(1)}{(5 - 1)}\;\;\;\;\; \ldots \left( 1 \right)\)

Where,1 < c < 5

CALCULATION:

Given: For x ∈ (1, 5) ⇒ f'(x) ≥ 9

Let x = c ∈ (1, 5)

⇒ f'(c) ≥ 9

\(⇒ f'(c) = \frac{f(5) - f(1)}{(5 - 1)}≥ 9\)

\(⇒\frac{f(5) - (- 3)}{4} ≥ 9\)

\(⇒\frac{f(5) + 3}{4} ≥ 9\)

⇒ f(5) + 3 ≥ 36

⇒ f(5) ≥ 33

So, the correct answer is option 1.

CONCEPT:

• The function is to be differentiable at a given point if at that point the left-hand derivative and right-hand derivatives are equal.

CALCULATION:

Given: f(x) = x|x| and g(x) = sin x

⇒ g(f(x)) = gof (x) = sin x|x| =  F(x) (Let)

\(F(x) = \left\{ \begin{matrix} \sin x^2 & x \ge 0 \\\ sin(-x^2) = -sinx^2 & x < 0 \end{matrix} \right.\)

At x = 0

 \(RHD =\lim_{h \rightarrow 0} \frac{ \sin (0 + h)^2-0}{h }\)

\(RHD =\lim_{h \rightarrow 0} \frac{ \sin ( h)^2}{h \times h } = 0\)

At x = 0

\(LHD = \lim_{h \rightarrow 0} \frac{-sin ( 0 - h)^2 - 0}{ - h}\)

\(LHD = \lim_{h \rightarrow 0} \frac{-hsin ( - h)^2 }{ - h \times h} = 0\)

• So f(x) is diff at x = 0 because LHD = RHD

So, the correct answer is option 1.

Concept:

f(x) is continuous at x = a, if LHL = RHL = f(a)

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

 f(x) is differentiable if LHD = RHD

\(\begin{array}{l} \rm L H D=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

Discontinuity of the First Kind: A function f(x) is said to have a discontinuity of the first kind from the right at x = a if the right hand of the function exists but not equal to f(a).

Discontinuity of the Second Kind: A function f(x) is said to have discontinuity of the second kind at x = a, if neither left-hand limit of f(x) at x = a nor right-hand limit of f(x) at x = a exists.

Removable Discontinuity: A function f(x) is said to have a removable discontinuity at x = a if the left-hand limit at x tends to point ‘a’ is equal to the right-hand limit at x tends to point ‘a’ but their common value is not equal to f(a). 

Calculation:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\left| x \right|}}{x}}\\ {0,} \end{array}\begin{array}{*{20}{c}} {,x ≠ 0}\\ {x = 0} \end{array}} \right.\)

For x ≠ 0,

f(x) = -x/x = -1,   if x < 0

f(x) = x/x = 1. if x > 0

Now,

LHL = \(\mathop {\lim }\limits_{x \to 0^- }f(x)=\mathop {\lim }\limits_{x \to 0^- }-x/x=-1\)

RHL = \(\mathop {\lim }\limits_{x \to 0^+ }f(x)=\mathop {\lim }\limits_{x \to 0^+}x/x=1\)

\(\mathop {\lim }\limits_{x \to 0 }f(x) = 0\)

Since,

LHL ≠ RHL we can say that the function is not continuous at x = 0

Only x = 0, is the point of discontinuity. 

Based on the definition, the function has discontinuity of the first kind. 

Concept:

Since the above function f(x) satisfy the assumption of Rolle’s theorem in interval [-1, 1]

∴ f(x) is continuous in [-1, 1]

F(x) is differentiable in (-1, 1)

F(-1) = f(1)

∴ for continuity

LHL = RHL = Functional value at 0 i.e f(0)

\(LHL = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {{\rm{lim}}}\limits_{x \to 0} \left( {1 + x} \right)\)

= 1

\(\begin{array}{l} RHL = \mathop {{\rm{lim}}}\limits_{x \to 0} f\left( x \right) = \mathop {{\rm{lim}}}\limits_{x \to 0} \left( {1 - x} \right)\left( {px + q} \right)\\ = \mathop {{\rm{lim}}}\limits_{x \to {0^ + }} \left( {px + q - p{x^2} - qx} \right) \end{array}\)  

= q

∴ LHL = RHL

⇒ q = 1

Also for differentiability

LHD = RHD

\(\Rightarrow \mathop {{\rm{lim}}}\limits_{x \to {0^ - }} f\left( x \right) = \mathop {{\rm{lim}}}\limits_{x \to {0^ + }} f\left( x \right)\)

\(\frac{d}{{dx}}\left( {1 + x} \right) = \frac{d}{{dx}}\left[ {\left( {1 - x} \right)\left( {px + q} \right)} \right]\;\)

at x = 0

\(\Rightarrow 1 = \frac{d}{{dx}}\left[ {px + q - p{x^2} - qx} \right]\)

\(⇒[p+0-2px-q]=1 \)

at x = 0

p – q = 1

⇒ p = 1 + q

⇒ p = 1 + 1

⇒ p = 2

Explanation:

To find the point of discontinuity of the function f(x) = [x], draw the graph of the function f(x) = [x].

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

The domain of [x] is R and range is I, where R is the set of real numbers and I is the set of integers.

From the graph, we can say that the function is discontinuous at every integer.

Hence, the function f(x) = [x] discontinuous is at infinite points.