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11th Standard English Medium Maths Subject Sets, Relations and Functions Creative 2 Mark Questions with Solution Part - II updated Creative Questions

Published on: 13/05/2022

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Std: 11th Standard - Maths

chapter: - Creative Questions

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Questions:

Generic 2

5 x 2 = 10

Draw the curves of
(i) y = x² + 1
(ii) Y = (x + 1)² by using the graph of curve y = x.

If A = { 0, 1, 2, 3, 4, 5, 6, 7 } is a set. Then,

Find the domain and range of the function f(x) = \(\frac { { x }^{ 2 }-9 }{ x-3 } \).

Find the domain of each of the following functions given by:
f(x) = \(\frac { { x }^{ 3 }-x+3 }{ { x }^{ 2 }-1 } \).

For n, m\(\in \)N, n/m means that n is a factor of n & m. Then find whether the given relation is an equivalence relation.

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Answers Key:

Generic 2

5 x 2 = 10

f (x) = x^{2 + 1} causes the graph of the function
f(x) = x² shifts to the upward for one unit.
f (x) = (x + 1)² causes the graph of the
function f(x) = x²shifts to the left for one unit.

A = { x | x is a whole number less than or equal to 7 } is the set-bilder form of A.

We have f(x) = \(\frac { { x }^{ 2 }-9 }{ x-3 } \)
Domain of f : Clearly f(x) is not defined for x - 3 = 0 i.e. x = 3. Therefore, Domain (f) = R- {3}
Range of f: Let f(x) = y. Then,
f(x) = y ⇒ \(\frac { { x }^{ 2 }-9 }{ x-3 } \)=y ⇒ x+3 = y
it follows from the above relation that y takes all real values except 6 when x takes values in the ser R - {3}. Therefore, Range (f) = R {6}.

Given that f(x) = \(\frac { { x }^{ 3 }-x+3 }{ { x }^{ 2 }-1 } \)
Here, f(x) is not defined if x² - 1 ≠ 0
(x - 1)(x + 1) ≠ 0
x ≠1, x ≠ -1
Hence, the domain of f = R - {-1, 1}

Since n is a factor of n. So the relation is reflexive.
When n is a factor of m (where m \(\neq\) n) then m cannot be a factor of n.
So the relation is not symmetric when n is a factor of m and m is a factor of p then n will be a factor of p. So the given relation is transitive. So it is not an equivalence relation.