First slide

Functions (XII)

Question

Let f (x) = sin^–1(log[x]) + log(sin^–1[x]), where [ ] denotes the greatest integer function. Then,

Moderate

A

domain of f is [1, 2)
B

domain of f is [1, 3)
C

range of f is $\{\log \frac{π}{2}\}$
D

range of f is {0}

Solution

We have,
f (x) = sin^–1(log [x]) + log(sin^–1[x]) (1)
Let g(x) = sin^–1(log [x]) (2)
and, h(x) = log(sin^–1[x]) (3)
Now for g(x);
– 1 ≤ log [x] ≤ 1 {as sin^–1 q exists when – 1 ≤ q ≤ 1}
and, [x] > 0 {as log [x] exists when [x] > 0}
⇒ 1/e ≤ [x] ≤ e and [x] > 0
⇒ [x] = 1, 2
⇒ x ∈ [1, 3) (4)
Again, from (3), we have
h(x) = log (sin^–1[x]) exists when;
sin^–1 [x] > 0 and – 1 ≤ [x] ≤ 1
⇒ [x] > 0 and –1 ≤ [x] ≤ 1
⇒ 0 < [x] ≤ 1 ⇒ [x] = 1
⇒ x ∈ [1, 2) (5)
⇒ Domain of f (x) is [1, 2)
Now, for range,
we know, f (x) = sin^–1(log[x]) + log(sin^–1[x])
where x ∈ [1, 2) ⇒ [x] = 1
$∴$ Range of f (x) = sin^–1(log 1) + log(sin^–1 1)
$\begin{array}{l} = \sin^{- 1} (0) + \log (\frac{π}{2}) \\ = \log (π / 2) \end{array}$
⇒ Range of $f (x) = \{\log \frac{π}{2}\}$
The correct option is (1) and (3)

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