The domain of the function f(x) =...

MathematicsDomain and Range of Inverse Trigonometric FunctionsJEE Advanced 1984Moderate

The domain of the function $f (x) = sin^{- 1} (lo g_{2} \frac{x ^{2}}{2})$ is given by .........

Answer:

[- 2, - 1] \cup [1, 2]

Visualized Solution (Hindi)

Analyze the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x)

Function: $f (x) = sin^{- 1} (lo g_{2} \frac{x ^{2}}{2})$
Goal: Find the set of all real values of $x$ for which $f (x)$ is defined.
Constraint 1: The argument of $sin^{- 1} (u)$ must satisfy $- 1 \leq u \leq 1$ .
Constraint 2: The argument of $lo g_{b} (v)$ must satisfy $v > 0$ .

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin^{- 1} Constraint

Applying the range of $sin^{- 1}$ :
$- 1 \leq lo g_{2} (\frac{x ^{2}}{2}) \leq 1$

Remove the Logarithm

Since the base $2 > 1$ , the inequality direction remains the same when taking the exponent:
$2^{- 1} \leq \frac{x ^{2}}{2} \leq 2^{1}$
$\frac{1}{2} \leq \frac{x ^{2}}{2} \leq 2$

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x^{2}

Multiply the entire inequality by $2$ :
$1 \leq x^{2} \leq 4$

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x : Part 1

Solving $x^{2} \geq 1$ :
$∣ x ∣ \geq 1 \Rightarrow x \in (- \infty, - 1] \cup [1, \infty)$

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x : Part 2

Solving $x^{2} \leq 4$ :
$∣ x ∣ \leq 2 \Rightarrow x \in [- 2, 2]$

Final Intersection and Domain

Intersection of $x \in (- \infty, - 1] \cup [1, \infty)$ and $x \in [- 2, 2]$ :
Domain: $x \in [- 2, - 1] \cup [1, 2]$

00:00 / 00:00

Conceptually Similar Problems

MathematicsDomain and Range of a FunctionJEE Advanced 1985Easy

View in:English Hindi

If $f (x) = sin ln (\frac{4 - x ^{2}}{1 - x})$ , then domain of $f (x)$ is .... and its range is .........

Answer:(-2, 1), [-1, 1]

MathematicsDomain and Range of a FunctionJEE Main 2004Easy

View in:English Hindi

The domain of the function $f (x) = \frac{s i n ^{- 1} ( x - 3 )}{9 - x ^{2}}$ is

(A)

[1, 2]

(B)

[2, 3)

(C)

[1, 2]

(D)

[2, 3]

Answer:B

MathematicsDomain and Range of Inverse Trigonometric FunctionsJEE Advanced 2003Easy

View in:English Hindi

Domain of definition of the function $f (x) = sin^{- 1} (2 x) + π /6$ for real valued $x$ , is

(A)

[- 1/4, 1/2]

(B)

[- 1/2, 1/2]

(C)

[- 1/2, 1/9]

(D)

[- 1/4, 1/4]

Answer:A

MathematicsDomain and Range of Inverse Trigonometric FunctionsJEE Main 2002Easy

View in:English Hindi

The domain of $sin^{- 1} [lo g_{3} (x /3)]$ is

(A)

[1, 9]

(B)

[- 1, 9]

(C)

[- 9, 1]

(D)

[- 9, - 1]

Answer:A

MathematicsDomain and Range of a FunctionJEE Advanced 1983Easy

View in:English Hindi

The domain of definition of the function $y = \frac{1}{l o g _{10} ( 1 - x )} + x + 2$ is

(A)

(- 3, - 2)

excluding

- 2.5

(B)

[0, 1]

excluding

0.5

(C)

[- 2, 1)

excluding

0

(D)

none of these

Answer:C

MathematicsDomain and Range of a FunctionJEE Main 2003Easy

View in:English Hindi

Domain of definition of the function $f (x) = \frac{3}{4 - x ^{2}} + lo g_{10} (x^{3} - x)$ , is

(A)

(- 1, 0) \cup (1, 2) \cup (2, \infty)

(B)

(a, 2)

(C)

(- 1, 0) \cup (a, 2)

(D)

(1, 2) \cup (2, \infty)

Answer:A

MathematicsDomain and Range of a FunctionJEE Advanced 2001Easy

View in:English Hindi

The domain of definition of $f (x) = \frac{l o g _{2} ( x + 3 )}{x ^{2} + 3 x + 2}$ is

(A)

R ∖ {- 1, - 2}

(B)

(- 2, \infty)

(C)

R ∖ {- 1, - 2, - 3}

(D)

(- 3, \infty) ∖ {- 1, - 2}

Answer:D

MathematicsInverse of a FunctionJEE Advanced 1999Moderate

View in:English Hindi

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

(A)

(1/2)^{x (x - 1)}

(B)

\frac{1}{2} (1 + 1 + 4 lo g_{2} x)

(C)

\frac{1}{2} (1 - 1 + 4 lo g_{2} x)

(D)

not defined

Answer:B

MathematicsDomain and Range of a FunctionJEE Main 2007Moderate

View in:English Hindi

The largest interval lying in $(- π /2, π /2)$ for which the function, $f (x) = 4^{- x^{2}} + cos^{- 1} (x /2 - 1) + lo g (cos x)$ , is defined, is

(A)

[- π /4, π /2]

(B)

[0, π /2]

(C)

[0, π]

(D)

(- π /2, π /2)

Answer:B

MathematicsInverse of a FunctionJEE Advanced 2004Easy

View in:English Hindi

If $f (x) = sin x + cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain

(A)

[0, π /2]

(B)

[- π /4, π /4]

(C)

[- π /2, π /2]

(D)

[0, π]

Answer:B

The domain of the function $f (x) = sin^{- 1} (lo g_{2} \frac{x ^{2}}{2})$ is given by .........

Visualized Solution (Hindi)

Analyze the Function $f (x)$

Apply $sin^{- 1}$ Constraint

Remove the Logarithm

Isolate $x^{2}$

Solve for $x$ : Part 1

Solve for $x$ : Part 2

Final Intersection and Domain

Conceptually Similar Problems

If $f (x) = sin ln (\frac{4 - x ^{2}}{1 - x})$ , then domain of $f (x)$ is .... and its range is .........

The domain of the function $f (x) = \frac{s i n ^{- 1} ( x - 3 )}{9 - x ^{2}}$ is

Domain of definition of the function $f (x) = sin^{- 1} (2 x) + π /6$ for real valued $x$ , is

The domain of $sin^{- 1} [lo g_{3} (x /3)]$ is

The domain of definition of the function $y = \frac{1}{l o g _{10} ( 1 - x )} + x + 2$ is

Domain of definition of the function $f (x) = \frac{3}{4 - x ^{2}} + lo g_{10} (x^{3} - x)$ , is

The domain of definition of $f (x) = \frac{l o g _{2} ( x + 3 )}{x ^{2} + 3 x + 2}$ is

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The largest interval lying in $(- π /2, π /2)$ for which the function, $f (x) = 4^{- x^{2}} + cos^{- 1} (x /2 - 1) + lo g (cos x)$ , is defined, is

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The domain of the function $f (x) = sin^{- 1} (lo g_{2} \frac{x ^{2}}{2})$ is given by .........

If $f (x) = sin ln (\frac{4 - x ^{2}}{1 - x})$ , then domain of $f (x)$ is .... and its range is .........

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If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The domain of the function f(x)=sin−1(log2​2x2​) is given by .........

Visualized Solution (Hindi)

Analyze the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin−1 Constraint

Remove the Logarithm

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x: Part 1

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x: Part 2

Final Intersection and Domain

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Analyze the Function $f (x)$

Apply $sin^{- 1}$ Constraint

Isolate $x^{2}$

Solve for $x$ : Part 1

Solve for $x$ : Part 2

If $f (x) = sin ln (\frac{4 - x ^{2}}{1 - x})$ , then domain of $f (x)$ is .... and its range is .........

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