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Published on: 23/05/2021
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Questions + Answers key
Take MCQ Maths Test1.
Let the function f:\(R\rightarrow R\) to be defined by:
\(f(x)=cosx\) for all \(x\in R\).
Show that 'f' is neither one-one nor onto.
2.
Let f : \(R\rightarrow R\) be defined by \(f(X)=X^{ 2 }+1\) Find the pre-image of
(i) 17
(ii) -3.
3.
If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write fog.
4.
If A = {1,2,3} and f,g are relations corresponding to the subset \(A\times A\) indicated against them, which of f,g is a function? why?
f = {(1, 3) (2, 3), (3, 2)}; g = {(1, 2), (1, 3), (3, 1)}.
1.
Let \(x_{ 1 },x_{ 2 }\in R\).
Now \(f(x_{ 1 })=f(x_{ 2 })\Rightarrow cosx_{ 1 }=cosx_{ 2 }\)
\(\Rightarrow \) \(x_{ 1 }=(2n\pi +x_{ 2 })\)
\(\Rightarrow \) '\(f\)' is not one-one.
(ii) Since cos \(x\) lies in [-1,1],
\(\therefore \) R is not fully covered.
Hence, '\(f\)' is not onto.
2.
(i) Here \(f(x)=17\)
\(\Rightarrow \) \(x^{ 2 }+1=17\Rightarrow x^{ 2 }=16\Rightarrow x=\pm 4\).
\(\therefore \) Pre-image of 17 = {-4, 4}.
(ii) \(f(x)=-3\)
\(\Rightarrow \) \(x^{ 2 }+1=-3\Rightarrow x^{ 2 }=-4\Rightarrow x\) is not real.
\(\therefore \) Pre-image of -3 = \(\phi \)
3.
In g : \(2\rightarrow 5\) and in f, \(5\rightarrow 2\); etc.
fog = {(2, 2), (3, 3)}.
4.
(i) 'f' is a function.
[∵ each element of A in the first place in the ordered pair is related to only one element of A in the second place]
(ii) 'g' is a not function.
[∵ 1 is related to two elements of A namely 2 and 3
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