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CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

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Section - B

Q. Nos. 21 to 26 carry 2 marks each.
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Question : 21
Total: 36
If f(x)=‌
4x+3
6x−4
,x≠‌
2
3
, then show that (fof) (x)=x, for all x≠‌
2
3
. Also, write inverse of f.
OR
Check if the relation R in the set ℝ of real numbers defined as R={(a,b):a<b} is (i) symmetric, (ii) transitive
Solution:     👈: Video Solution
f(x)=‌
4x+3
6x−4

f∘f(x)=‌
4(
4x+3
6x−4
)+3
6(‌
4x+3
6x−4
)−4

=‌
16x+12+18x−12
24x+18−24+16

=‌
34x
34
=x
y=‌
4x+3
6x−4

6xy−4y=4x+3
6xy−4x=4y+3
x=‌
4y+3
6y−4

∴f(x)=‌
4x+3
6x−4
OR
Given : R={(a,b):a<b}
Let A be the set of real numbers.
(i) For symmetric : Let a,b∈A
Then, (a,b)∈R such that a<b
⇒b<a which is false
⇒‌‌(b,a)∉R
∴R is not symmetric.
(ii) For transitive : Let a,b,c∈A
Then, (a,b)∈R and (b,c)∈R
⇒a<b‌ and ‌b<c
⇒a<c
⇒(a,c)∈R
∴R is transitive.
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