Relations and Functions MCQ Quiz - Objective Question with Answer for Relations and Functions - Download Free PDF

Concept:

Convert decimal to binary

Conversion steps:

• Divide the number by 2.
• Get the integer quotient for the next iteration.
• Get the remainder for the binary digit.
• Repeat the steps until the quotient is equal to 0.

Calculation:

∴ (45)₁₀ = 101101

Concept:

One-To-One Function: A function f: A → B is one-to-one if every element of the range B corresponds to exactly one element of the domain A.

Calculation:

Let's say that f(x₁) = f(x₂).

⇒ 2x₁ = 2x₂

⇒ x₁ = x₂

Hence, the given function is a One to One function.

Additional Information

• One-One/Injective function: A function that maps distinct elements of its domain to distinct elements of its range.
• Onto/Surjective function: A function in which every element of the domain has at least one image in the domain.
• One-One Onto/Bijective function: is a function which is both one-one and onto. i.e. each element of the domain is paired with exactly one element of the range, and each element of the range is paired with exactly one element of the domain.
• If a set S has n elements, then the total number of functions from S to S is nⁿ.
• If a set S has n elements, then the total number of one-one onto functions from S to S is n!
• The total number of injective functions from a set containing m elements into the set containing n elements is: n × (n -1) × (n - 2) × … × (n - m + 1) = ⁿP_m.

Calculation:

Given, f(x) = |x – 1| = \(\left\{\begin{array}{cl} x-1& x \geq 1 \\ 1-x & x \leq 1 \end{array}\right.\)

Also, \(\rm g(x)=\left\{\begin{matrix}e^x,& x\ge0\\\ x+1, &x\le0\end{matrix}\right.\)

∴ f(g(x)) = |g(x) – 1| 

⇒ f(g(x)) = \(\left\{\begin{array}{cl} \left|e^x-1\right| & x \geq 0 \\ |x+1-1| & x \leq 0 \end{array}\right.\)

⇒ f(g(x)) = \(\left\{\begin{array}{cc} e^x-1 & x \geq 0 \\ -x & x \leq 0 \end{array}\right.\)

∴ f(g(x)) is neither one-one nor onto.

The correct answer is Option 1.

Calculation

\(f(g(x))=\left\{\begin{array}{lll} 2+2 g(x) & , & -1 \leq g(x)<0 \\ 1-\frac{g(x)}{3} & , & 0 \leq g(x) \leq 3 \end{array}\right.\) 

By (1) x ∈ ϕ 

And by (2) x∈ [ 3, 0] and x∈ [0, 1] 

Range of f(g(x)) is [0, 1] 

Hence option 3 is correct

Concept:

• log A + log B = log AB
• log A - log B = log(A/B)
• log a^k = k log a
• \(\sum_{k=1}^n k = {n(n+1)\over 2}\)
• \(\sum_{k=1}^n k^2 = {n(n+1)(2n+1)\over 6}\)

Calculation:

Given,

\(\displaystyle \frac{logx+logx^4+logx^9+....+logx^{n^2}}{logx+logx^2+logx^3+....+logx^n}\)

⇒ \(\displaystyle \frac{log[x(x^4)(x^9)....(x^{n^2})]}{log[x(x^2)(x^3)...(x^n)]}\)

⇒ \(\displaystyle \frac{log[x^{1 +4 + 9+ ...n^2}]}{log[x^{1+2+3+...+n}]}\)

⇒ \(\displaystyle \frac{log[x^{\sum_{k=1}^n k^2} ]}{log[x^{\sum_{k=1}^n k}]}\)

⇒ \(\displaystyle \frac{{\sum_{k=1}^n k^2} \ log[x]}{{\sum_{k=1}^n k}\ log[x]}\)

⇒ \(\displaystyle \frac{{\sum_{k=1}^n k^2} }{{\sum_{k=1}^n k}\ }\)

⇒ \(\displaystyle \frac{{n(n+1)(2n+1)\over 6}}{{n(n+1)\over 2}}\)

⇒ \({2n+1\over 3}\)

∴ The correct answer is option (1).

Concept:

• If a function repeats over at a constant period we say that is a periodic function.
• It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.
• The period of sin x and cos x is 2π

Calculation:

To Find: Period of 4cos3 x - 3cos x

As we know 4cos3 x - 3cos x = cos 3x

Period of cos x is 2π

Therefore, the Period of cos 3x is \(\rm \frac {2\pi}{3}\)

Concept:

If f(x)  is odd function then f(-x) = -f(x)

Calculation:

Given: f(x) = x2 + 4x + 4

Replace x by -x,

⇒ f(-x) = (-x)² + 4(-x) + 4

= x2 - 4x + 4                       (∵ (-x)² = x²)

⇒ f(-x) ≠ ± f(x)

Hence function is neither odd nor even.

Concept:

• Inverse Function:

Let f: A → B be one-one and onto (bijective) function. Then f^-1 exists which is a function f^-1: B → A, which maps each element b ∈ B with an element

a ∈ A such that f(a) = b is called the inverse function of f: A → B.

• Methods to find inverse:

Let f : A → B be a bijective function.

Step – I Put f (x) = y

Step – II Solve the equation y = f (x) to obtain x in terms of y.

Interchange x and y to obtain the inverse of the given function f.

Calculation:

Given: \(f(x) = \frac{1 + 2x}{x + 7}\)

Let y = f(x) = 2x + 1 / x + 7

⇒ xy + 7y = 2x + 1

⇒ 7y – 1 = 2x – xy

⇒ x(2 - y) = 7y – 1

⇒ x = (7y - 1) / (2 - y)

⇒ f^-1 (x) = \(\frac{7x - 1}{2 - x}\)

Concept:

• If a function repeats over at a constant period we say that is a periodic function.
• It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.
• The period of sin x and cos x is 2π

Calculation:

To Find: Period of 3sin x - 4sin3 x

As we know 3sin x - 4sin3 x = sin 3x

Period of sin x is 2π

Therefore, Period of sin 3x is \(\rm \frac {2\pi}{3}\)

Concept:

Logarithm properties:  

Product rule: The log of a product equals the sum of two logs.

\(\rm {\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

Quotient rule: The log of a quotient equals the difference of two logs.

\(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

Power rule: In the log of power the exponent becomes a coefficient.

\(\rm {\log _a}{m^n} = n{\log _a}m\)

Formula of Logarithms:

If \(\rm lo{g_a}x = b \) then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: \(\rm \log_{3}{(x^{4} - x^3)} - \log_{3} (x - 1) = 3\)

\(\rm \Rightarrow \log_{3} \left[{\frac{(x^{4} - x^3)}{(x - 1)}} \right ] = 3\)        (∵ \(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\))

\(\rm \Rightarrow \log_{3} \left[{\frac{x^3(x-1)}{(x - 1)}} \right ] = 3\)

\(\rm \Rightarrow \log_{3} x^3 = 3\)

\(\Rightarrow \rm 3\log_3 x = 3\)               (∵ \(\rm {\log _a}{m^n} = n{\log _a}m\)) 

\(\Rightarrow \rm \log_3 x = 1 \\\therefore x=3\)

Concept:

Range: The range of a function is the set of all possible values it can produce, i.e., all values of y for which x is defined.

Note:

The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.

Calculation:

Let, y = f(x) = \(\rm \frac{x+1}{x-3}\)

⇒y(x - 3) = x + 1

⇒yx - 3y - x = 1

⇒ x(y - 1) - 3y = 1

⇒ x(y - 1) = 1 + 3y

⇒\(\rm x = \frac{1+3y}{y-1}\)

It is clear that x is not defined when y - 1 = 0, i.e, when  y = 1

∴ Range (f) = R - {1}

Hence, option (2) is correct.

Mistake PointsIt is given in the Question that, f(x) is real function. So,

f(x) has real values for value of x other than x = 3

∴ Domain of given function = R - {3}, where R is set of all real numbers

Concept:

Domin of sin-1 x is [-1, 1]

Adding or subtracting the same quantity from both sides of an inequality leaves the inequality symbol unchanged.

Calculation:

Given:  f(x) = sin-1 (x + 1) 

As we know, domin of sin1 x is [-1, 1]

Therefore, -1 ≤ (x + 1) ≤ 1

subtracting 1 in above inequality, 

⇒ -1 - 1 ≤ x + 1 - 1 ≤ 1 - 1

⇒ -2 ≤ x ≤ 0

∴ Domin of sin-1 (x + 1) is [-2, 0] 

Mistake Points[-2, 0] is different from [-2, 0). '[' and ']' indicates that the end number (2 and 0) is also included. '(' and ')' indicates that 2 and 0 are not taken into consideration.

Concept:

• (x + y)(x - y) = x² - y²
• \(\ln { 1\over a} = \ln 1 - \ln a = 0 - \ln a = -\ln a\)

Calculation:

Given: f(x) = ln (x + \(\sqrt{1+\text{x}^2}\)),__(i)

Replace x by (-x) in (i),

⇒  f(-x) = ln (-x + \(\sqrt{1+(\text{-x})^2}\)),

⇒  f(-x) = ln (-x + \(\sqrt{1+\text{x}^2}\)),

Multiply and divide by (x + \(\sqrt{1+\text{x}^2}\)) inside the ln function,

⇒  f(-x) =  \(\ln (-x + \sqrt{1+\text{x}^2}.{ x + \sqrt{1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({- x^2 + {1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({1 \over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(-\ln ({ x + \sqrt{1+\text{x}^2}})\),

From (i),

⇒  f(-x) =  - f(x)\

⇒  f(-x) + f(x) = 0

∴ The correct answer is option (1).

Given:

f(x) = x²

Calculation:

f(x) = x²

⇒ y = x² ⇒ x = \(\pm\sqrt y\)

⇒ y < 0 there is no real value of x 

⇒ y ≥ 0 

⇒ Range of f = [0, ∞] 

∴ The range of f will be non negative numbers (Since Zero is also in the range)

Concept:

1. Domain of a  functions:

• The domain of a function is the set of all possible values of the independent variable. That is all the possible inputs for a function.

Calculation:

Observe that the given function is in the form of numerator and denominator. The function will be well defined for all non zero values of the denominator.

Therefore, \({\rm{x}} - 2{\rm{\;}} \ne 0\) that implies that \({\rm{x\;}} \ne 2\).

Similarly square root function is well defined for all non-negative values.

Therefore, \({\rm{x}} - 2 > 0\) that implies \({\rm{x}} > 2.\)

Thus, domain of the given function is \(\left( {2,{\rm{\;}}\infty } \right).\)