Relations and Functions MCQ Quiz in मल्याळम - Objective Question with Answer for Relations and Functions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

To get the value of any function at any point with respect to some other function, we need to manipulate the given function/ point in such a way that we get the function that is needed.

Calculation:

Given, f(x + 1) = x2 - 3x + 2

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

f(x + 1) = x(x - 2) - 1(x - 2)

f(x + 1) = (x - 2)(x - 1)

f(x + 1) = (x + 1 - 3)(x + 1 - 2).......[-2 = 1 - 3 & -1 = 1 - 2]

Now replace (x + 1) terms with (x)

∴ f(x) = (x - 3)(x - 2)

f(x) = x² -3x - 2x + 6

f(x) = x² - 5x + 6

Additional Information

This problem can also be solved by putting (x - 1) in place of x.

Concept:

Logarithm properties

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

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

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

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

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

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

4. Change of base rule

\({\log _m}n = \frac{{{{\log }_a}n}}{{{{\log }_a}m}}\)

If m = n;
⇒ \({\log _m}m = \frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1\)

5. \({\log _m}n = \frac{1}{{{{\log }_n}m}}\)

Calculation:

Here, we have to find the value of \({\log _7}{\rm{\;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } } \)

\({\log _7}{\rm{\;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } } \)

= log₇ log₇ (7^1/2 × 7^1/4 × 7^1/8)

= log₇ log₇ (7(^{1/2 + 1/4 + 1/8)})

= log₇ log₇ (7^{(4 + 2 + 1)/8})

= log₇ log₇ (7^7/8)

From power rule;

= log₇ (7/8) log₇7

= log₇ (7/8) × 1 = log₇ (7/8) = log₇ 7 – log₇ 8

= 1 – log₇ 8 = 1 – log₇ 2³

Concept:

Reflexive Set:

A set, A is Reflexive when for any x ∈ R

⇒ (x , x) ∈ R.

Symmetric Set:

A set, A is Symmetric when for any x, y ∈ R 

(x, y) ∈ R ⇒ (y, x) ∈ R.

Transitive Set:

A set, A is Transitive when for any x, y, z ∈ R 

(x, y) ∈ R and (y, z) ∈ R  ⇒ (x, z) ∈ R.

Explanation:

Let us check all the conditions:

Reflexivity:

Let x be an arbitrary element of R. For Reflexive, let y + x

i.e. for any x ∈ R

⇒ 2x + x = 41

⇒ x = \(\frac{41}{3}\) ∉ N

⇒(x, x) ∉ R

Thus, it is NOT reflexive.

Symmetry:

Let (x, y) ∈ R. Then,

2x + y = 41 Which is not equal to 2y + x = 41

i.e. (y , x) ∉ R

So, R is NOT symmetric.

Transitivity:

Let (x, y) and (y, z) ∈ R

⇒ 2x + y = 41 and 2y + z = 41

which is not equal to 2x + z = 41

i.e. (x , z) ∉ R

Thus, R is NOT transitive.

Given:

⇒ 2log(x+3) = log81

Concept used:

a log(b) = log(b)^a

Calculation:

Here,

⇒ 2log(x+3) = log81

⇒ log(x + 3)² = log81

⇒ (x + 3)² = 81

⇒ (x + 3) = \({ \sqrt{81}}\)

x + 3 = 9 

x = 9 - 3 = 6

∴ x = 6

Concept:

For any two functions f and g, f o g is defined as f[g(x)].

Calculation:

Given that,

f(x) = 4x + 3   

Using the above concept

⇒ fof(x) = 4(4x + 3) + 3

⇒ fof(x) = 16x + 15 

Again using the same concept

⇒ fofof(x) = 16(4x + 3) + 15

⇒ fofof(x) = 64x + 63

Put x = -1 in the above function

⇒ fofof(-1) = 64(-1) + 63 = -1

∴  f o f o f(-1) equal to -1.

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: \(\rm \log_{4}{(x^{2} - 1)} - \log_{4} (x + 1) = 1\)

\(\rm ⇒ \log_{4} \left[{\frac{(x^{2} - 1)}{(x + 1)}} \right ] = 1\)        (∵ \(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\))

\(\rm ⇒ \log_{4} \left[{\frac{(x - 1)(x+1)}{(x + 1)}} \right ] = 1\)

\(\rm ⇒ \log_{4} (x -1) = 1\)

⇒ (x - 1) = 4

∴ x = 5

Calculation:

Given:

f (x + 1/x) = x² + 1/x²

⇒ f (x + 1/x) = x² + 1/x² + 2 - 2

⇒ f (x + 1/x) = (x + 1/x)² - 2

Substituting x in place of x + 1/x, we get

Given :

If f = {(1, 2)} and g = {(2, 3)} 

Concept used:

Mapping Diagram consists of two columns in which one denotes the domain of a function f whereas the other column denotes the Range.

Example:  f = {(2, -5), (3, 4)} means 2 maps to -5 and 3 maps to 4 

Calculations:

gof(1) 

⇒ g(2) because f(1) = 2

⇒ g(2) = 3

∴ gof(1) = 3 

∴ option 2 is correct.

Concept:

One-One Function / Injective Function:

A function f: A → B is said to be a one-one function if different elements in A have different images or are associated with different elements in B.

i.e., if f(x1) = f(x2) then x1 = x2, ∀ x1, x2 ∈ R

Onto Function / Surjective Function:

A function f: A → B is said to be an onto function if each element in B has at least one pre-image in A.

i.e., If-Range of function f = Codomain of function f, then f is onto.

Bijective Function:

A function f: A → B is said to be a B a bijective function if it is both one-one and onto function.

Calculation:

Given: \(f(x) = x + \sqrt{x^2}\)

The given function can also be written as,

f(x) = x + |x|

This function can be converted into a piecewise-defined function.

For, x > 0

⇒ f(x) = x + (+ x) = 2x

For, x < 0

⇒ f(x) = x + (- x) = 0

Clearly, f(x) ≥ 0 for all x. So, it is not surjective or onto as,

Range of function f ≠  Codomain of function f.

Also, f(x) = x + |x|. Clearly, f(-1) = f(-2) = 0

This means for one tow values of x, we get the same value of y as 0  so, f is not one or we can say that it is many -one

Concept:

A function y = √f(x)  is define for f(x) > 0.

If (x - a)(x - b) ≤ 0 ⇒ x ∈ [a, b]

Calculation:

Given  that, \(f(x) =\sqrt{1 - (x-1)^2} \ \)

For a function to be defined,

1 - (x - 1)² ≥ 0

⇒ (x - 1)2 - 1 ≤ 0

Since, (a - b)² = a² - 2ab + b²

⇒ x² + 1  - 2x - 1 ≤ 0

⇒ x² - 2x ≤ 0

⇒ x(x - 2) ≤ 0

∴ x ∈ [0, 2]