JEE Advanced (Single Correct MCQs): Functions - Airforce X Y / Indian Navy SSR MCQ

f (x) = x² is many one as f (1) = f (–1) = 1 Also f is into as – ve real number have no pre-image.
∴ F is neither injective nor surjective.

For entire graph to be above x-axis, we should have

∴ Put x = 2
LHS = f (2²) = |4 – 1| = 3 and RHS = ( f (2))2 = 1
∴ (a) is not correct
Consider f (x + y) = f (x) + f (y)
Put x = 2, y = 5 we get
f (7) = 6; f (2) + f (5) = 1 + 4 = 5
∴ (b) is not correct
Consider f (| x |) = | f (x) |
Put x = – 5 then f (| –5 |) = f (5) = 4
| f (– 5) | = | – 5 – 1| = 6
∴ (c) is not correct.
Hence (d) is the correct alternative.

Graph of f (x) shows f (x) > 6 for x < 0 or x > 4

Clearly it is a periodic function with period 1.
∴ (a) is the correct alternative.

From graph (i) infinite many points for min value of f (x)

From graph (ii) only pt. of min of f (x) at x = q/p

From graph (iii) only one pt. of min of f (x) at x = 0

f (x) is continuous and defined for all x > 0 and

⇒ Clearly f (x) = ln x which satisfies all these properties
∴ f (x) = ℓnx

Let h (x) = |x| then g (x) = |f (x)| = h (f (x))
Since composition of two continuous functions is continuous, therefore g is continuous if f is continuous.

For integral values of x; g (x) = 1
For x < 0; (but not integral value) x – [x] > 0 ⇒ g (x) > 1
For x > 0;(but not integral value) x – [x] > 0 ⇒ g (x) >1

From E to F we can define, in all, 2 × 2 × 2 × 2 = 16 functions (2 options for each element of E) out of which 2 are into, when all the elements  of E either map to 1 or to 2.
∴ No. of onto functions = 16  – 2 = 14

Given that f (x) = (x + 1)2, x > – 1
Now if g (x) is the reflection of f (x) in the line y = x then it can be obtained by interchanging x and y in f (x) i.e., y = (x + 1)² changes to x = (y + 1)²

∴ f is an increasing function ⇒ f is one-one.

 to be defined and real

    ...(1) But we know that – π/2 < sin^–1 2x < π/2 ...(2) Combining (1) and (2), we get

We have 

We can see here that as x → ∞ , f (x) → 1 which is the min value of f (x). i.e. f_min = 1. Also f (x) is max when
 min which is so when x  = – 1/2

Since f  –  g : R → R for any x there is only one value of (f (x) – g(x)) whether x is rational or irrational.
Moreover as x∈R, f (x) – g (x) also belongs to R.
Therefore, (f – g) is one-one onto.

Given that X and Y are two sets and f : X → Y.
{f (c) = y; c ⊂ X, y ⊂ Y} and
{f ^–1(d) = x : d ⊂ Y, x ⊂ X }
The pictorial representation of given information is as shown:

Since f ^–1 (d) =  x ⇒ f (x) = d Now if a ⊂ x
⇒ f (a) ⊂ f (x) = d ⇒ f ^–1 [f (a)]  = a
∴ f ^–1 (f (a)) = a, a ⊂ x is the correct option.