Functions and Logarithms (JEE Advanced)

Question 1

Let f(x)=√(log_(1/3)(log₄(lfloor xrfloor²−5))), where lfloor trfloor denotes the greatest integer not exceeding t. The domain of f (the set of all real x for which f(x) is a real number) is ______. Ex

Accepted Answer

B. The square root needs log_(1/3)(u)≥ 0 with u=log₄(lfloor xrfloor²−5). Since the base 1/3 is less than 1, log_(1/3)u≥ 0 exactly when 0<u≤ 1.
0<log₄(lfloor xrfloor²−5)≤ 1 means 1<lfloor xrfloor²−5≤ 4, i.e. 6<lfloor xrfloor²≤ 9.
lfloor xrfloor is an integer, and the only perfect square in (6,9] is 9, so lfloor xrfloor=± 3, giving x∈[−3,−2)∪[3,4).
Trap: if you forget the upper bound u≤1 you would wron

Question 2

Let f:mathbbR→mathbbR be defined by f(x)=(x−1)lvert x−1rvert+2x. Here sgn(t)=1 for t>0, sgn(0)=0 and sgn(t)=−1 for t<0. Which of the following is correct? Exactly one option is correct.

Accepted Answer

D. Put u=x−1. Then f(x)=φ(u)+2, where φ(u)=ulvert urvert+2u.
φ is odd: φ(−u)=−ulvert urvert−2u=−φ(u). For u≥ 0, φ(u)=u²+2u=(u+1)²−1, which is strictly increasing on [0,∞) and takes every value in [0,∞). By oddness, φ is strictly increasing on (−∞,0] and takes every value in (−∞,0]. Hence φ, and therefore f, is a bijection of mathbbR onto mathbbR. So (A) is false.
Inverse of φ: for v≥0, (u+1)²−1=v wit

Question 3

Let f(x)=√(x−1) and g(x)=log₂(3−x), each on its natural domain, and let h=f° g+g° f, defined exactly where both f° g and g° f are defined. Which of the following is correct? Exactly one option is corr

Accepted Answer

C. Domain of f° g: need 3−x>0 and g(x)≥ 1, i.e. log₂(3−x)≥1, so 3−x≥2, i.e. x≤1. Domain (−∞,1].
Domain of g° f: need x≥1 and 3−√(x−1)>0, i.e. x−1<9. Domain [1,10).
The domain of h is the intersection (−∞,1]∩[1,10)={1}.
h(1)=f(g(1))+g(f(1))=f(log₂ 2)+g(0)=f(1)+log₂ 3=0+log₂ 3.
So h has domain {1} and range {log₂ 3}. Options (A) and (B) are the domains of the separate compositions; (D) overlooks that t

Functions and Logarithms: questions with solutions

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