Solve : log - x 2 - 6 x 10 sin x + sin 3 x = log - x 2 - 6 x 10 sin 2 x .

log - x 2 - 6 x 10 sin x + sin 3 x is defined only when sin x + sin 3 x > 0 . and - x 2 - 6 x 10 > 0 , - x 2 - 6 x 10 ≠ 1 So, log - x 2 - 6 x 10 sin x + sin 3 x = log - x 2 - 6 x 10 sin 2 x if, i - x 2 - 6 x 10 > 0 , - x 2 - 6 x 10 ≠ 1 ii sin x + sin 3 x > 0 iii sin 2 x > 0 and iv sin x + sin 3 x = sin 2 x From iv we have, 2 sin 2 x cos x = sin 2 x or, cos x = 1 2 [ ∵ by iii sin 2 x > 0 ] or, x = 2 n π ± π 3 ; n = 0 , ± 1 , ± 2 , … Those values of x which satisfy the above four conditions are the roots of the given equation. From i we have - 6 < x < 0 . Now, sin 2 n π - π 3 < 0 ; which shows that the values of x = 2 n π - π 3 are not the roots of given equation, since condition ii is not satisfied. Putting n = 0 , ± 1 , ± 2 , … in sin 2 n π + π 3 we can verify that x = - 5 π 3 is the only value of x which satisfies all the four conditions. Thus the solution of the given equation is x = - 5 π 3 .

Find the domain of definition the following function. (Read the symbol · as fractional part function respectively) f x = 1 - sin x log 5 1 - 4 x 2 + cos - 1 1 - x

We have, f x = 1 - sin x log 5 1 - 4 x 2 + cos - 1 1 - x Now, log a x is defined for x , a > 0 and a ≠ 1 . log 5 1 - 4 x 2 ≠ 0 ⇒ 1 - 4 x 2 ≠ 1 ⇒ - 4 x 2 ≠ 0 ⇒ x ≠ 0 Also, 1 - 4 x 2 > 0 ⇒ x 2 < 1 4 ⇒ x < 1 2 ⇒ x ∈ - 1 2 , 1 2 And, domain of cos - 1 x is - 1 , 1 and 0 ≤ x < 1 ⇒ - 1 < - x ≤ 0 ⇒ 0 < 1 - x ≤ 1 Hence, cos - 1 1 - x is already defined for all x . Also, 1 - sin x ≥ 0 ⇒ sin x ≤ 1 And, sin x ∈ - 1 , 1 ∀ x ∈ R , hence x ∈ 2 n π - π 2 , 2 nπ + π 2 , n ∈ Z . Hence, domain of given function is x ∈ - 1 2 , 1 2 - 0 .

The graph of a function f is given. State the value of f - 1 .

Graph of y = f x We can clearly see that, the vertical line x = - 1 intersects the curve at the point - 1 , - 2 , So, f - 1 = - 2 .

The graph of a function f is given. For what values of x is f x = 2

If we draw the line y = 2 , it intersects the curve at two different point - 3 , 2 and 1 , 2 . So, for x = - 3 and x = 1 , the value of y = f x = 2 .

The graph of a function f is given. State the domain and range of f .

Domain: The set of all points over which the function is defined. We can see here, this function is defined from x = - 3 to x = 3 . Hence, the domain is - 3 , 3 . Range: The set of values the function takes on as output. We can see here, the output, the value of y , varies from y = - 2 to y = 3 . Hence, the range is - 2 , 3

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 25 - Functions>EXERCISE-3>Q 31

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Let $\{x\}$ & $\{x\}$ denote the fractional and integral part of a real number $x$ , respectively, then number of solutions of the equations $4 \{x\} = x + [x]$ , is

44.44% studentsanswered this correctly

Important Questions on Functions

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 25 - Functions>EXERCISE-4>Q 1

If $f (x) = - 1 + | x - 1 |, - 1 \leq x \leq 3; g (x) = 2 - | x + 1 |, - 2 \leq x \leq 2$ , then calculate $(f o g) (x)$ & $(g o f) (x)$ . Draw their graph. Discuss the continuity of $(f o g) (x)$ at $x = - 1$ & the differentiability of $(g o f) (x)$ at $x = 1$ .