Verified Solution • 2025–26 Session moderate

QJEEM21Mab763: JEE Main 2021 Mathematics - Class 11 JEEM21Mab763 Solution: Let f : R → R be defined as f (x) = 2x – 1 and g : R - {1} → R…

Let f : R → R be defined as f (x) = 2x – 1 and g : R - {1} → R be defined as g(x) = x−12x−1{{x - {1 \over 2}} \over {x - 1}} . Then the composition function f(g(x)) is :

Multiple Choice Options

  • A one-one but not onto
  • B onto but not one-one
  • C both one-one and onto
  • D neither one-one nor onto

Step-by-Step Verified Solution

Given, f(x) = 2x −- 1; f : R →\to R

g(x)=x−1/2x−1;g:R−{1)→Rg(x) = {{x - 1/2} \over {x - 1}};g:R - \{ 1) \to R

f[g(x)]=2g(x)−1f[g(x)] = 2g(x) - 1

=2×(x−12x−1)−1=2×(2x−12(x−1))−1= 2 \times \left( {{{x - {1 \over 2}} \over {x - 1}}} \right) - 1 = 2 \times \left( {{{2x - 1} \over {2(x - 1)}}} \right) - 1

=2x−1x−1−1=2x−1−x+1x−1=xx−1= {{2x - 1} \over {x - 1}} - 1 = {{2x - 1 - x + 1} \over {x - 1}} = {x \over {x - 1}}

∴\therefore f[g(x)]=1+1x−1f[g(x)] = 1 + {1 \over {x - 1}}

Now, draw the graph of 1+1x−1′1 + {1 \over {x - 1'}}

JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Functions Question 100 English Explanation
∵\because Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.

Hence, the required function is one-one into.

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring