Quantitative Aptitude PYQs

Calculation:

f(x) = x2 - 1                ------(1)

gof(x) = x - √x + 1      -----(2)

Now, checking each options, for g [f(x)] = x - √x + 1

Option 1: \(\sqrt{x+1}-\sqrt[4]{x+1} \)

Let,  \(g(x) = \sqrt{x+1}-\sqrt[4]{x+1} \)

 \(g(f(x)) = \sqrt{(x^2-1)+1}-\sqrt[4]{(x^2-1)+1}\)

\(g(f(x)) = \sqrt{x^2}-\sqrt[4]{x^2}\)

\(g(f(x)) = x-\sqrt{x}\)

Hence, option 1 is not correct.

Option 2: \(\sqrt{x+1}-\sqrt[4]{x+1} +1\)

Let,  \(g(x) = \sqrt{x+1}-\sqrt[4]{x+1} +1\)

 \(g(f(x)) = \sqrt{(x^2-1)+1}-\sqrt[4]{(x^2-1)+1} +1\)

\(g(f(x)) = \sqrt{x^2}-\sqrt[4]{x^2}+1\)

\(g(f(x)) = x-\sqrt{x}+1\)

Hence, option 2 is correct.

Option 3: \(\sqrt{x+1}+\sqrt[4]{x+1} \)

Let,  \(g(x) = \sqrt{x+1}+\sqrt[4]{x+1} \)

 \(g(f(x)) = \sqrt{(x^2-1)+1}+\sqrt[4]{(x^2-1)+1}\)

\(g(f(x)) = \sqrt{x^2}+\sqrt[4]{x^2}\)

\(g(f(x)) = x+\sqrt{x}\)

Hence, option 3 is not correct.

Option 4: \( x+1-\sqrt{x+1}+1\)

Let,  \(g(x) = x+1-\sqrt{x+1}+1\)

 \(g(f(x)) = x^2-1+1-\sqrt{x^2-1+1}+1\)

\(g(f(x)) = x^2 -\sqrt{x^2}+1\)

\(g(f(x)) = x^2 -x+1\)

Hence, option 4 is not correct.

From the above question, \(g(x) = \sqrt{x+1}-\sqrt[4]{x+1}+1 \)

To find: g(15)

Explanation:

Now, putting the value of x as 15 in the function g(x).

\(g(x) = \sqrt{15+1}-\sqrt[4]{15+1}+1 \)

\(g(x) = \sqrt{16}-\sqrt[4]{16}+1 \)

\(g(x) = 4-2+1 \)

g(x) = 3

Hence, g(15) is 3.

Calculation:

Given:

\(\rm 2 f(x)+f\left(\frac{1}{x}\right)=x+3 \)        -----(1)

Put x = 1/x

Now, we get

\(2f\left(\frac{1}{x}\right) + \rm f(x)= {1 \over x}+3 \)        -----(2)

Now,

Equation (1) x 2 - Equation (2)

\(\rm 4 f(x)+2f\left(\frac{1}{x}\right) - 2f\left(\frac{1}{x}\right) - \rm f(x)=2x+6 - ({1 \over x}+3 )\)

\(\rm 3 f(x)= 2x - {1 \over x}+3 \)

Put x = 0.5

\(\rm 3 f(0.5)= 2(0.5) - {1 \over 0.5}+3 \)

\(\rm 3 f(0.5)= 1 - 2+3 \)

∴ \(\rm f(0.5)= {2 \over 3}\)

Formula Used:

\(\frac{d}{dx}x^n = nx^{n-1}\)

Explanation:

Given that 

\(\rm 2 f(x)+f\left(\frac{1}{x}\right)=x+3 \)        ------(1)

Put x = 1/x

Now, we get

\(2f\left(\frac{1}{x}\right) + \rm f(x)= {1 \over x}+3 \)       ------(2)

Now, eq (1) x 2 - eq (2)

\(\rm 4 f(x)+2f\left(\frac{1}{x}\right) - 2f\left(\frac{1}{x}\right) - \rm f(x)=2x+6 - ({1 \over x}+3 )\)

\(\rm 3 f(x)= 2x - {1 \over x}+3 \)

\(\rm f(x)={2 \over 3}x - {1 \over 3x}+1 \)

Differentiating with respect to x,

\(\rm f'(x)={2 \over 3}- {1 \over 3}{-1\over x^2}\)

Put x = 0.5

\(\rm f'(0.5)={2 \over 3}- {1 \over 3}{-1\over (0.5)^2}\)

\(\rm f'(0.5)={2 \over 3}- {(-4) \over 3}\)

∴ \(\rm f'(0.5)={2+4 \over 3} = {6 \over 3} = 2\)

Concept:

A function y = f(x) is called decreasing (non-increasing) on the interval (a, b) if 

∀ x1, x2 ∈ (a, b): x1 < x2 ⇒ f(x1) ≥ f(x2)

\(\frac{{dy}}{{dx}} \le 0;or;f'\left( x \right) \le 0\)

Formula used:

\(\frac{d}{dx}x^n\ =\ nx^{n-1}\)

Calculation:

We have

\(f(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \ 2 & x+4 & 6 \ 3 & 6 & x+9 \end{array}\right|\)

Function f(x) is the Determinant of the given matrix,

⇒ f(x) = (x+1) [(x+4)(x+9) - 36] - 2 [2(x+9) - 18] + 3[12 - 3(x+4)]

⇒ f(x) = (x+1) [x² + 9x + 4x + 36 - 36] - 2 [2x + 18 - 18] + 3[12 - 3x-12]

⇒ f(x) = (x+1) [x2 + 13x] - 2 [2x] + 3[- 3x]

⇒ f(x) = x³ + 13x2 + x2 + 13x - 4x - 9x

⇒ f(x) = x3 + 14x2 

To find the function is decreasing, \(f'(x)\leq0\)

 f(x) = x3 + 14x2 

\(⇒ f'(x) = 3x^2 + 28x \leq0\)

\(⇒ f'(x) = x(x + {28 \over3}) \leq0\)

\(⇒ x \leq0, x \geq-{28 \over3}\)

∴ The function decreases in \([-{28 \over3},0]\).

Concept:

A function y = f(x) is called decreasing (non-increasing) on the interval (a, b) if 

∀ x1, x2 ∈ (a, b): x1 < x2 ⇒ f(x1) ≥ f(x2)

\(\frac{{dy}}{{dx}} \le 0;or;f'\left( x \right) \le 0\)

 

For a function y = f(x):

• Relative (local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
• Relative (local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
• At the points of relative (local) maxima or minima, f'(x) = 0.
• At the points of relative (local) maxima, f''(x) < 0.
• At the points of relative (local) minima, f''(x) > 0.

 

Formula used:

\(\frac{d}{dx}x^n\ =\ nx^{n-1}\)

Calculation:

\(f(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \ 2 & x+4 & 6 \ 3 & 6 & x+9 \end{array}\right|\) 

Function f(x) is the Determinant of the given matrix,

⇒ f(x) = (x+1) [(x+4)(x+9) - 36] - 2 [2(x+9) - 18] + 3[12 - 3(x+4)]

⇒ f(x) = (x+1) [x2 + 9x + 4x + 36 - 36] - 2 [2x + 18 - 18] + 3[12 - 3x-12]

⇒ f(x) = (x+1) [x2 + 13x] - 2 [2x] + 3[- 3x]

⇒ f(x) = x3 + 13x2 + x2 + 13x - 4x - 9x

⇒ f(x) = x3 + 14x2 

To find the function is decreasing, \(f'(x)\leq0\)

 f(x) = x3 + 14x2 

\(⇒ f'(x) = 3x^2 + 28x \leq0\)

\(⇒ f'(x) = x(x + {28 \over3}) \leq0\)

\(⇒ x \leq0, x \geq-{28 \over3}\)

As the sign changes from positive to negative at \({x}=-\frac{28}{3}\) . Hence, the function attains a local maximum value at \({x}=-\frac{28}{3}\)

As the sign changes from negative to positive at x = 0. Hence, the function attains a local minimum value at x = 0.

Concept:

Minima and maxima of a function: These are the values of functions where its first derivative (slope) becomes zero.

We can calculate the minima or maxima of function f(x) by the following steps.

Step:1 Calculate the first derivative f'(x) of a function.

Step:2 Equate it with zero and find the critical values (let x1 and x2).

Step:3 Find out the second derivative f''(x) and check its sign-on critical values.

Step:4 If f''(x) < 0 then there will be maxima and if f''(x) > 0 then minima at the corresponding value.

Calculation:

Calculation:

\(y = \sqrt {9-4x^2}\)

Differentiating the function with respect to x

\({dy \over dx} = {1\over2{\sqrt {9-4x^2}}} (-8x) \)

To find the maximum value of y, \({dy \over dx} = 0\)

\({1\over2{\sqrt {9-4x^2}}} (-8x) = 0\)

⇒ \(x = 0\)

y is maximum at x = 0 if \({d^2y \over dx^2} < 0\).

\({dy \over dx} ={-4x\over{\sqrt {9-4x^2}}}\)

\(⇒ {d^2y \over dx^2} = {-4 {\sqrt {9-4x^2}-(-4x)({-8x \over {2 \sqrt {9-4x^2} }})}\over {9-4x^2}}\)

\(⇒ {d^2y \over dx^2} = {-4 {\sqrt {9-4x^2}-({32x^2 \over {2 \sqrt {9-4x^2} }})}\over {9-4x^2}}\)

\(⇒ {d^2y \over dx^2} = {-4 {({9-4x^2)}-({16x^2})}\over {(9-4x^2)}^{3\over 2}}\)

\(⇒ {d^2y \over dx^2} = {{{-36+16x^2}-{16x^2}}\over {(9-4x^2)}^{3\over 2}}\)

At x = 0, 

\(⇒ {d^2y \over dx^2} = -{{4}\over {3}} \leq0\)

Hence, y is maximum at x = 0

Maximum value of \(y = \sqrt {9-4(0)^2}\)

∴ \(y = \sqrt {9} = 3\)

Concept:

Minima and maxima of a function: These are the values of functions where its first derivative (slope) becomes zero.

We can calculate the minima or maxima of function f(x) by the following steps.

 

Step:1 Calculate the first derivative f'(x) of a function.

Step:2 Equate it with zero and find the critical values (let x1 and x2).

Step:3 Find out the second derivative f''(x) and check its sign-on critical values.

Step:4 If f''(x) < 0 then there will be maxima and if f''(x) > 0 then minima at the corresponding value.

Calculation:

\(x = {1\over4}\sqrt {9-y^2}\)

Differentiating the function with respect to y

\({dx \over dy} = {1\over8{\sqrt {9-y^2}}} (-2y) \)

To find the maximum value of x, \({dx \over dy} = 0\)

\( {1\over8{\sqrt {9-y^2}}} (-2y) = 0\)

⇒ y = 0

x is maximum at y = 0 if \({d^2x \over dy^2} < 0\).

\({dx \over dy} ={-y\over{4\sqrt {9-y^2}}}\)

\(⇒ {d^2x \over dy^2} ={1\over4} {-1 {\sqrt {9-y^2}-(-y)({-2y \over {2 \sqrt {9-y^2} }})}\over {9-y^2}}\)

\(⇒ -{d^2x \over dy^2} = {1\over4}{- {\sqrt {9-y^2}-({y^2 \over { \sqrt {9-y^2} }})}\over {9-y^2}}\)

\(⇒ {d^2x \over dy^2} =\frac{1}{4}\times { {-({9-y^2)}-({y^2})}\over {(9-y^2)}^{3\over 2}}\)

\(⇒ {d^2x \over dy^2} =\frac{1}{4} {{{-9+y^2}-{y^2}}\over {(9-y^2)}^{3\over 2}}\)

At y = 0, 

\(⇒ {d^2x \over dy^2} = -{{1}\over {12}} \leq0\)

Hence, x is maximum at y = 0

Maximum value of \(x = {1\over4}\sqrt {9-(0)^2}\)

\(x = {1\over 4} \sqrt {9} = {3\over 4}\)

∴ The maximum value of xy = xmax × ymax = \( 3×{3 \over 4} = {9 \over 4}\)

Concept:

Range of sin θ is [-1, 1]

Calculation:

Given that

f(x) = π + sin2 x.

The given function is a combination of a constant term (π) and the

square of the sine function sin2 x.

We know that 

-1 ≤ sin x ≤ 1 

⇒ 0 ≤ sin²x ≤ 1 

⇒ π + 0  ≤ π + sin2x ≤ π + 1 

Therefore, the range of π + sin2 x is [π, π + 1].

Concept: 

The period of the function f(x) = π + sin2 x is determined by the

periodicity of the sine function, as the sin2 x.

Calculation:

Given that 

f(x) = π + sin2 x    ----(1)

The cos x function has a period of 2π.

The cos 2x function has a period of π.

We know that 

sin²x = (1 = cos 2x)/2

As, adding or subtracting a constant number does not affect the period

of the function.

Hence, The period of the function π + sin2 x is π.

Given:

A parabola passes through (1, 2) and satisfies the differential equation \(\frac{d y}{d x}=\frac{2 y}{x}, x>0, y>0 .\)

Concept: 

The general form of the equation of a parabola is \(x^2 = 4 ay\) 

Directrix of the parabola is y + a = 0

Calculation:

\(\frac{d y}{2y}=\frac{dx}{x}\)

Integrating both sides,

\(\int\frac{d y}{y}=\int\frac{2dx}{x}\)

\({ln:y}=2{ln:x}+{ln:c}\)

\({ln:y}={ln:x^2}+{ln:c}\)

\({ln:y}={ln:x^2c}\)

\(y=x^2c\)

Parabola passes through (1, 2)

So, x = 1, y = 2

2 = 1²c

c = 2

The equation of parabola is \(x^2 = {y \over2}\)

Now comparing the given equation of parabola with the general form,

\( {y \over2} = 4ay\)

\( {1 \over8} =a\)

Therefore, the directrix of the parabola is given by:

⇒ \(y + {1 \over8} = 0\)

∴  \(y = - {1 \over8} \)

Given: 

A parabola passes through (1, 2) and satisfies the differential equation \(\frac{d y}{d x}=\frac{2 y}{x}, x>0, y>0 .\)

Concept: 

The general form of the equation of a parabola is \(x^2 = 4 ay\) 

Directrix of the parabola is y + a = 0

 

Latus Rectum is 4a.

 

Calculation:

\(\frac{d y}{2y}=\frac{dx}{x}\)

Integrating both sides,

\(\int\frac{d y}{y}=\int\frac{2dx}{x}\)

\({ln:y}=2{ln:x}+{ln:c}\)

\({ln:y}={ln:x^2}+{ln:c}\)

\({ln:y}={ln:x^2c}\)

\(y=x^2c\)

Parabola passes through (1, 2)

So, x = 1, y = 2

2 = 12c

c = 2

The equation of parabola is \(x^2 = {y \over2}\)

Now comparing the given equation of parabola with the general form,

\( {y \over2} = 4ay\)

\( {1 \over8} =a\)

Therefore, the Latus Rectum of the parabola is given by:

∴ \(4a = {1 \over2}\)

Concept:

L’Hopital’s Rule: If we have an indeterminate form 0/0 or ∞/∞ we can

apply L'Hopital's rule. We need to differentiate the numerator

and denominator and then take the limit.

Formula Used:

(d/dx)xⁿ = nx^n-1

(d/dx)a^x = x ln a

Calculation:

Given that 

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

g(x) = x - 1                  -----(2)

\(⇒L= \displaystyle \lim _{x \rightarrow 1} \frac{f(x)-1}{g(x)}\)

Putting the value from equation (1) & (2)

\(⇒ L= \displaystyle \lim _{x \rightarrow 1} \frac{\frac{a^{x-1}+b^{x-1}}{2}-1}{x-1}\)

Let us assume, x - 1 = y so limits changes as y → 0.

\(⇒ L=\displaystyle \lim _{y \rightarrow 0} \frac{\frac{a^{y}+b^{y}}{2}-1}{y}\)

\(⇒L= \displaystyle \lim _{y \rightarrow 0} \frac{a^{y}+b^{y}-2}{2y}\)

\(⇒ L={1\over 2}\displaystyle \lim _{y \rightarrow 0} \frac{(a^{y}-1)+(b^{y}-1)}{y}\)

\(⇒L= {1\over 2}\displaystyle \lim _{y \rightarrow 0} \frac{a^{y}-1}{y}+\frac{b^{y}-1}{y}\)

By L-Hospital's Rule, 

differentiating numerator and denominator separately,

\(⇒L= {1\over 2}\displaystyle \lim _{y \rightarrow 0} \frac{a^{y}-1}{y}+\frac{b^{y}-1}{y}\)

\(\displaystyle \lim _{y \rightarrow 0} \frac{a^{y}-1}{y} = ln:a,: : \displaystyle \lim _{y \rightarrow 0} \frac{b^{y}-1}{y} = ln:b\)

\(⇒ {1\over 2} (ln:a + ln:b)\)

Using the property of logarithm,

\(⇒ {1\over 2} (ln:a + ln:b) = {1\over 2} (ln:ab )\)

Concept:

L’Hopital’s Rule: If we have an indeterminate form 0/0 or ∞/∞ we can

apply L'Hopital's rule. We need to differentiate the numerator

and denominator and then take the limit.

Formula Used:

(d/dx)xn = nxn-1

(d/dx)ax = x ln a

Calculation:

Given that 

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

g(x) = x - 1                  -----(2)

\(⇒L= \displaystyle \lim _{x \rightarrow 1} \frac{f(x)-1}{g(x)}\)

Putting the value from equation (1) & (2)

\(⇒ L= \displaystyle \lim _{x \rightarrow 1} \frac{\frac{a^{x-1}+b^{x-1}}{2}-1}{x-1}\)

Let us assume, x - 1 = y so limits changes as y → 0.

\(⇒ L=\displaystyle \lim _{y \rightarrow 0} \frac{\frac{a^{y}+b^{y}}{2}-1}{y}\)

\(⇒L= \displaystyle \lim _{y \rightarrow 0} \frac{a^{y}+b^{y}-2}{2y}\)

\(⇒ L={1\over 2}\displaystyle \lim _{y \rightarrow 0} \frac{(a^{y}-1)+(b^{y}-1)}{y}\)

\(⇒L= {1\over 2}\displaystyle \lim _{y \rightarrow 0} \frac{a^{y}-1}{y}+\frac{b^{y}-1}{y}\)

By L-Hospital's Rule, 

differentiating numerator and denominator separately,

\(⇒L= {1\over 2}\displaystyle \lim _{y \rightarrow 0} \frac{a^{y}-1}{y}+\frac{b^{y}-1}{y}\)

\(\displaystyle \lim _{y \rightarrow 0} \frac{a^{y}-1}{y} = ln:a,: : \displaystyle \lim _{y \rightarrow 0} \frac{b^{y}-1}{y} = ln:b\)

\(⇒ {1\over 2} (ln:a + ln:b)\)

\(⇒ \displaystyle \lim _{x \rightarrow 1} \frac{\frac{a^{x-1}+b^{x-1}}{2}-1}{x-1} = {1\over 2} (ln:ab )\)

\(⇒ \displaystyle \lim _{x \rightarrow 1} \frac{\frac{a^{x-1}+b^{x-1}}{2}-1}{x-1} = ln{\sqrt {ab}}\)

Putting this value in eq (1)

\(⇒ \displaystyle \lim _{x \rightarrow 1} f(x)^{\frac{1}{g(x)}} =e^ {ln{\sqrt {ab}}}\)

\(⇒ \displaystyle \lim _{x \rightarrow 1} f(x)^{\frac{1}{g(x)}} = {\sqrt {ab}}\)

Concept: 

The domain of a function is the set of all possible inputs for the function.

For example, the domain of f(x) = x² is all real numbers, and the domain

of g(x)=1/x is all real numbers except for x = 0.

Explanation:

Given that

\(f(x)=\sqrt{2-x}+\sqrt{2+x}\)

To find the domain of the function f(x),

\( {2-x} \geq 0\) & \( {2+x} \geq 0\)        

\(2 \geq x\) & \(2 \geq -x ⇒x \geq -2\)

∴ The domain of x is [-2, 2]

Calculation:

\(f(x)=\sqrt{2-x}+\sqrt{2+x}\)

Let f(x) = y

Squaring both sides,

\([f(x)]^2= y^2={2-x}+{2+x}+2\sqrt{(2-x)(2+x)}\)

\(y^2=4+2\sqrt{(4-x^2)}\)

Therefore, y² is maximum when x is minimum, so x is minimum at 0

Hence, the given function is greatest at x = 0,

The greatest value is given by,

\(y^2_{max}=4+2\sqrt{(4-0^2)}\)

\(y^2_{max}=8\)

\(y_{max}={\sqrt8}\)

The greatest value of the function \( \sqrt{2-x}+\sqrt{2+x} : \ is\ : {\sqrt8}\)

Concept:

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

• In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• This means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

 

Calculation:

Let

f(x) = |x| and g(x) = [x] - 1, where [.] is the greatest integer function.

\( h(x)=\frac{f(g(x))}{g(f(x))}\)

\( h(x)=\frac{|[x] - 1|}{[|x|]-1}\)

\(⇒ \displaystyle \lim _{x \rightarrow 0+}\frac{|[x] - 1|}{[|x|]-1}\)

For a number greater than 0, Let us say x = 0.1, so the greatest integer

[x] = 0

\(⇒ \displaystyle \lim _{x \rightarrow 0+}\frac{|0 - 1|}{[x]-1}\)

|-1| = 1, For a number greater than 0, Let us say x = 0.1, so the greatest

integer [x] = 0

\(⇒\frac{1}{0-1}\)

\(\therefore \displaystyle \lim _{x \rightarrow 0+}\frac{|0 - 1|}{[x]-1}=-1\)

Concept:

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

• In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• This means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

 

Calculation:

Let

f(x) = |x| and g(x) = [x] - 1, where [.] is the greatest integer function.

\( h(x)=\frac{f(g(x))}{g(f(x))}\).

\( h(x)=\frac{|[x] - 1|}{[|x|]-1}\)

\(⇒ \displaystyle \lim _{x \rightarrow 0-}\frac{|[x] - 1|}{[|x|]-1}\)

For a number lesser than 0, Let us say x = - 0.1, so the greatest integer [x] = -1

\(⇒ \displaystyle \lim _{x \rightarrow 0-}\frac{|-1 - 1|}{[-x]-1}\)

\(⇒ \displaystyle \lim _{x \rightarrow 0-}\frac{|-2|}{[-x]-1}\)

|-2| = 2, For a number lesser than 0, Let us say x = - 0.1, so the greatest integer [-x] = 0

\(⇒\frac{2}{0-1}\)

\(\therefore \displaystyle \lim _{x \rightarrow 0-}\frac{|-2|}{[-x]-1} = -2\)

Concept:

If a function is continuous at a particular value x. 

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

Calculation:

\(f(x)= \begin{cases}\frac{x-3}{|x-3|}+a ; & x<3 \ a-b ; & x=3 \ \frac{x-3}{|x-3|}+b ; & x>3\end{cases}\)

And f(x) be continuous at x = 3.

For x < 3, |x-3| = -(x-3)

Left-hand Limit,

\(⇒ LHL = \frac{x-3}{|x-3|}+a\)

\(⇒ LHL = -\frac{x-3}{x-3}+a\)

⇒ LHL = -1 + a

For x > 3, |x-3| = (x-3)

Right-hand Limit,

\(⇒ RHL = \frac{x-3}{|x-3|}+b\)

\(⇒ RHL =\frac{x-3}{x-3}+b\)

⇒ RHL = 1 + b

For x = 3, f(x) = a - b,

As the function is continuous at x = 3.

So, LHL = RHL = f(x = 3)

⇒ -1 + a = 1 + b = a - b

⇒ -1 + a = a - b & 1 + b = a - b

⇒ -1 = - b & 1 + b = a - b

⇒ b = 1 & 1 + 1 = a - 1

⇒ b = 1 & a = 3

Concept:

If a function is continuous at a particular value x. 

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

Calculation:

\(f(x)= \begin{cases}\frac{x-3}{|x-3|}+a ; & x<3 \ a-b ; & x=3 \ \frac{x-3}{|x-3|}+b ; & x>3\end{cases}\)

And f(x) be continuous at x = 3.

For x < 3, |x-3| = -(x-3)

Left-hand Limit,

\(⇒ LHL = \frac{x-3}{|x-3|}+a\)

\(⇒ LHL = -\frac{x-3}{x-3}+a\)

⇒ LHL = -1 + a

For x > 3, |x-3| = (x-3)

Right-hand Limit,

\(⇒ RHL = \frac{x-3}{|x-3|}+b\)

\(⇒ RHL =\frac{x-3}{x-3}+b\)

⇒ RHL = 1 + b

For x = 3, f(x) = a - b,

As the function is continuous at x = 3.

So, LHL = RHL = f(x = 3)

⇒ -1 + a = 1 + b = a - b

⇒ -1 + a = a - b & 1 + b = a - b

⇒ -1 = - b & 1 + b = a - b

⇒ b = 1 & 1 + 1 = a - 1

⇒ b = 1 & a = 3