JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let f and g be two functions defined by f(x)={x+1,∣x−1∣,amp;xamp;x≥0lt;0andg(x)={x+1,1,amp;xamp;x≥0lt;0 Then, (g∘f)(x) is:
Choose the correct answer:
- A.
continuous everywhere but not differentiable at x=1
- B.
continuous everywhere but not differentiable exactly at one point
(Correct Answer) - C.
differentiable everywhere
- D.
not continuous at x=−1
continuous everywhere but not differentiable exactly at one point
Explanation
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Definition of f(x): Expanding the modulus, f(x)=x+1 for x < 0, 1−x for 0 \leq x < 1, and x−1 for x≥1.
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Definition of g(f(x)): By rule, g(f(x))=f(x)+1 if f(x) < 0, and 1 if f(x)≥0.
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Determining Intervals:
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For x < -1, f(x)=x+1, which is negative. Thus, g(f(x))=(x+1)+1=x+2.
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For x≥−1, f(x) is always ≥0. Thus, g(f(x))=1.
-
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Final Function:
(g∘f)(x)={x+2,1,amp;xamp;x≥−1lt;−1 -
Conclusion: The function is continuous at x=−1 (since LHL=RHL=1), but it has a sharp corner, making it non-differentiable at exactly one point (x=−1).
Correct Option: (2)
Explanation
-
Definition of f(x): Expanding the modulus, f(x)=x+1 for x < 0, 1−x for 0 \leq x < 1, and x−1 for x≥1.
-
Definition of g(f(x)): By rule, g(f(x))=f(x)+1 if f(x) < 0, and 1 if f(x)≥0.
-
Determining Intervals:
-
For x < -1, f(x)=x+1, which is negative. Thus, g(f(x))=(x+1)+1=x+2.
-
For x≥−1, f(x) is always ≥0. Thus, g(f(x))=1.
-
-
Final Function:
(g∘f)(x)={x+2,1,amp;xamp;x≥−1lt;−1 -
Conclusion: The function is continuous at x=−1 (since LHL=RHL=1), but it has a sharp corner, making it non-differentiable at exactly one point (x=−1).
Correct Option: (2)