If the domain of the function $f(x) = \log_e (4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1} \left(\frac{10x+6}{3}\right)$ is $(\alpha, \beta]$, then $36|\alpha + \beta|$ is equal to Options: (A) 54 (B) 72 (C) 63 (D) 45

### Analyzing the Setup To find the domain of the function $f(x) = \log_e (4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1} \left( \frac{10x+6}{3} \right)$, we must ensure that every component of the function is defined simultaneously. The function exists only in the intersection of the domains of its individual parts. If any term is undefined, the entire function $f(x)$ fails to exist. ### The Logarithmic Gatekeeper For the term $\log_e (4x^2 + 11x + 6)$, the argument must be strictly positive. We set up the inequality: $$4x^2 + 11x + 6 > 0$$ Factoring the quadratic expression, we obtain: $$(4x + 3)(x + 2) > 0$$ Using the wavy curve method with critical points at $x = -2$ and $x = -\frac{3}{4}$, we determine the valid region for the logarithm: $$D_{\log}: x \in (-\infty, -2) \cup \left( -\frac{3}{4}, \infty \right)$$ ### The Inverse Trigonometric Boundaries For the inverse trigonometric functions $\sin^{-1}(u)$ and $\cos^{-1}(u)$, the input $u$ must satisfy the condition $-1 \le u \le 1$. For $\sin^{-1}(4x + 3)$, we solve: $$-1 \le 4x + 3 \le 1$$ $$-4 \le 4x \le -2$$ $$D_{\sin^{-1}}: x \in [-1, -0.5]$$ For $\cos^{-1} \left( \frac{10x+6}{3} \right)$, we solve: $$-1 \le \frac{10x+6}{3} \le 1$$ $$-3 \le 10x + 6 \le 3$$ $$-9 \le 10x \le -3$$ $$D_{\cos^{-1}}: x \in \left[ -\frac{9}{10}, -\frac{3}{10} \right]$$ ### The Intersection - Finding the Sweet Spot We must now find the intersection of the three domains: 1. $x \in (-\infty, -2) \cup (-0.75, \infty)$ 2. $x \in [-1, -0.5]$ 3. $x \in [-0.9, -0.3]$ First, we intersect the trigonometric domains: $$[-1, -0.5] \cap [-0.9, -0.3] = [-0.9, -0.5]$$ Next, we intersect this result with the logarithmic domain $(-\infty, -2) \cup (-0.75, \infty)$. The values in $[-0.9, -0.5]$ that fall within the logarithmic domain are restricted to the interval $(-0.75, -0.5]$. Thus, the final domain is $(-0.75, -0.5]$. ### Final Calculation Given the domain is $(\alpha, \beta]$, we identify: $$\alpha = -\frac{3}{4}, \quad \beta = -\frac{1}{2}$$ We are asked to calculate $36|\alpha + \beta|$: $$\alpha + \beta = -\frac{3}{4} - \frac{1}{2} = -\frac{5}{4}$$ Finally, we compute the result: $$36 \times \left| -\frac{5}{4} \right| = 36 \times \frac{5}{4} = 9 \times 5 = 45$$ The final answer is **45**.

JEE Main 2023 (15 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Inverse Trigonometric Functions: If the domain of the function $f (x) = lo g_{e} (4 x^{2} + 11 x + 6) + sin^{- 1} (4 x + 3) + cos^{- 1} (\frac{10 x + 6}{3})$ is $(α, β]$ , then $36∣ α + β ∣$ is equal to

Select Answer:

Progressive Hints

Hint 1

Think of the domain as a 'permission slip'. For the function

f (x)

to exist, every single component—the logarithm, the sine inverse, and the cosine inverse—must grant permission simultaneously. This means you are looking for the intersection of three distinct sets of real numbers on the number line.

Hint 2

Recall the 'Gatekeepers' of calculus: For

lo g_{e} (u)

, we require

u > 0

because the logarithm of a non-positive number is undefined in the real domain. For

sin^{- 1} (u)

and

cos^{- 1} (u)

, the input

u

is strictly bound by the interval

[- 1, 1]

. Write down these three inequalities separately before you even think about combining them.

Hint 3

Start by solving the quadratic inequality

4 x^{2} + 11 x + 6 > 0

using the wavy curve method to find the intervals where the expression is positive. Then, solve the two linear inequalities for the inverse trigonometric functions. Once you have three intervals on your number line, find the region where all three overlap. That region is your

(α, β]

Visualized Solution

The Master Strategy

$f (x) = lo g_{e} (4 x^{2} + 11 x + 6) + sin^{- 1} (4 x + 3) + cos^{- 1} (\frac{10 x + 6}{3})$
To find the domain of $f (x)$ , we must find the domain of each individual term.
The final domain is the intersection: $D_{f} = D_{l o g} \cap D_{s i n^{- 1}} \cap D_{c o s^{- 1}}$

Domain of the Logarithmic Part

For $lo g_{e} (4 x^{2} + 11 x + 6)$ to be defined, the argument must be strictly positive.
$4 x^{2} + 11 x + 6 > 0$

Solving the Log Inequality

Factorize the quadratic: $4 x^{2} + 8 x + 3 x + 6 > 0$
$4 x (x + 2) + 3 (x + 2) > 0$
$(4 x + 3) (x + 2) > 0$

Visualizing the Log Domain

Critical points: $x = - 2$ and $x = - \frac{3}{4}$
Using the wavy curve method, the expression is positive outside the roots.
$D_{l o g} : x \in (- \infty, - 2) \cup (- \frac{3}{4}, \infty)$

Domain of the Sine Inverse Part

For $sin^{- 1} (4 x + 3)$ to be defined, the argument must lie in $[- 1, 1]$ .
$- 1 \leq 4 x + 3 \leq 1$

Solving the Sine Inverse Inequality

Subtract $3$ from all sides: $- 4 \leq 4 x \leq - 2$
Divide by $4$ : $- 1 \leq x \leq - \frac{1}{2}$
$D_{s i n^{- 1}} : x \in [- 1, - \frac{1}{2}]$

Domain of the Cosine Inverse Part

For $cos^{- 1} (\frac{10 x + 6}{3})$ to be defined, the argument must also lie in $[- 1, 1]$ .
$- 1 \leq \frac{10 x + 6}{3} \leq 1$

Solving the Cosine Inverse Inequality

Multiply by $3$ : $- 3 \leq 10 x + 6 \leq 3$
Subtract $6$ : $- 9 \leq 10 x \leq - 3$
Divide by $10$ : $- \frac{9}{10} \leq x \leq - \frac{3}{10}$
$D_{c o s^{- 1}} : x \in [- \frac{9}{10}, - \frac{3}{10}]$

Finding the Intersection

We need the common region for all three domains:
1. $D_{l o g} : (- \infty, - 2) \cup (- \frac{3}{4}, \infty)$
2. $D_{s i n^{- 1}} : [- 1, - \frac{1}{2}]$
3. $D_{c o s^{- 1}} : [- \frac{9}{10}, - \frac{3}{10}]$

The Overlapping Region

Observing the number line, all three domains overlap between $- \frac{3}{4}$ and $- \frac{1}{2}$ .
At $x = - \frac{3}{4}$ , the log domain is open (exclusive).
At $x = - \frac{1}{2}$ , the sine inverse domain is closed (inclusive).
Intersection: $x \in (- \frac{3}{4}, - \frac{1}{2}]$

Identifying $α$ and $β$

The problem states the domain is $(α, β]$ .
Comparing this with our result $(- \frac{3}{4}, - \frac{1}{2}]$ :
$α = - \frac{3}{4}$
$β = - \frac{1}{2}$

Calculating $α + β$

We need to find the value of $36∣ α + β ∣$ .
First, calculate the sum: $α + β = - \frac{3}{4} - \frac{1}{2}$
$α + β = - \frac{3}{4} - \frac{2}{4} = - \frac{5}{4}$

The Final Answer

Substitute $α + β = - \frac{5}{4}$ into the expression.
$36∣ α + β ∣ = 36 - \frac{5}{4}$
$= 36 \times \frac{5}{4}$
$= 9 \times 5 = 45$

00:00 / 00:00

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

Solution Diagram

Analyzing the Setup

To find the domain of the function

f (x) = lo g_{e} (4 x^{2} + 11 x + 6) + sin^{- 1} (4 x + 3) + cos^{- 1} (\frac{10 x + 6}{3})

, we must ensure that every component of the function is defined simultaneously.

The function exists only in the intersection of the domains of its individual parts. If any term is undefined, the entire function

f (x)

fails to exist.

The Logarithmic Gatekeeper

For the term

lo g_{e} (4 x^{2} + 11 x + 6)

, the argument must be strictly positive. We set up the inequality:

4 x^{2} + 11 x + 6 > 0

Factoring the quadratic expression, we obtain:

(4 x + 3) (x + 2) > 0

Using the wavy curve method with critical points at

x = - 2

and

x = - \frac{3}{4}

, we determine the valid region for the logarithm:

D_{l o g} : x \in (- \infty, - 2) \cup (- \frac{3}{4}, \infty)

The Inverse Trigonometric Boundaries

For the inverse trigonometric functions

sin^{- 1} (u)

and

cos^{- 1} (u)

, the input

u

must satisfy the condition

- 1 \leq u \leq 1

For

sin^{- 1} (4 x + 3)

, we solve:

- 1 \leq 4 x + 3 \leq 1

- 4 \leq 4 x \leq - 2

D_{s i n^{- 1}} : x \in [- 1, - 0.5]

For

cos^{- 1} (\frac{10 x + 6}{3})

, we solve:

- 1 \leq \frac{10 x + 6}{3} \leq 1

- 3 \leq 10 x + 6 \leq 3

- 9 \leq 10 x \leq - 3

D_{c o s^{- 1}} : x \in [- \frac{9}{10}, - \frac{3}{10}]

The Intersection - Finding the Sweet Spot

We must now find the intersection of the three domains: 1.

x \in (- \infty, - 2) \cup (- 0.75, \infty)

x \in [- 1, - 0.5]

x \in [- 0.9, - 0.3]

First, we intersect the trigonometric domains:

[- 1, - 0.5] \cap [- 0.9, - 0.3] = [- 0.9, - 0.5]

Next, we intersect this result with the logarithmic domain

(- \infty, - 2) \cup (- 0.75, \infty)

. The values in

[- 0.9, - 0.5]

that fall within the logarithmic domain are restricted to the interval

(- 0.75, - 0.5]

Thus, the final domain is

(- 0.75, - 0.5]

Final Calculation

Given the domain is

(α, β]

, we identify:

α = - \frac{3}{4}, β = - \frac{1}{2}

We are asked to calculate

36∣ α + β ∣

α + β = - \frac{3}{4} - \frac{1}{2} = - \frac{5}{4}

Finally, we compute the result:

36 \times - \frac{5}{4} = 36 \times \frac{5}{4} = 9 \times 5 = 45

The final answer is 45.

Common Traps & FAQs

Why do we use intersection and not union? Shouldn't we combine all possible values?

Think of the function as a machine with three parts connected in series. If one part breaks (i.e., is undefined), the whole machine stops working. We need the values of

x

where *all* parts are defined simultaneously. Therefore, we must find the intersection, not the union.

Why is the log inequality strict (

> 0

) while the inverse trig inequalities are inclusive (

\leq

\geq

Great observation! Logarithms are undefined at zero, so the argument must be strictly positive. However, inverse trigonometric functions like

sin^{- 1} (x)

and

cos^{- 1} (x)

are perfectly well-defined at their boundaries,

- 1

and

1

. Always check the definition of the function to see if the boundary is included.

What if my intervals don't overlap at all? Does that mean the answer is zero?

If the intervals don't overlap, the domain is the empty set, meaning the function is undefined everywhere. However, in JEE Advanced problems, there is almost always a valid domain. If you find no overlap, don't panic—re-check your arithmetic, especially the signs when dividing or multiplying inequalities!

Animated Solution for Mathematics - Inverse Trigonometric Functions: If the domain of the function $f (x) = lo g_{e} (4 x^{2} + 11 x + 6) + sin^{- 1} (4 x + 3) + cos^{- 1} (\frac{10 x + 6}{3})$ is $(α, β]$ , then $36∣ α + β ∣$ is equal to

Select Answer:

Visualized Solution

The Master Strategy

Domain of the Logarithmic Part

Solving the Log Inequality

Visualizing the Log Domain

Domain of the Sine Inverse Part

Solving the Sine Inverse Inequality

Domain of the Cosine Inverse Part

Solving the Cosine Inverse Inequality

Finding the Intersection

The Overlapping Region

Identifying $α$ and $β$

Calculating $α + β$

The Final Answer

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

Analyzing the Setup

The Logarithmic Gatekeeper

The Inverse Trigonometric Boundaries

The Intersection - Finding the Sweet Spot

Final Calculation

Similar Questions

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If the domain of the function $f (x) = cos^{- 1} (\frac{2 - ∣ x ∣}{4}) + (lo g_{e} (3 - x))^{- 1}$ is $[- α, β) - {γ}$ , then $α + β + γ$ is equal to :

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Select Answer:

Visualized Solution

The Master Strategy

Domain of the Logarithmic Part

Solving the Log Inequality

Visualizing the Log Domain

Domain of the Sine Inverse Part

Solving the Sine Inverse Inequality

Domain of the Cosine Inverse Part

Solving the Cosine Inverse Inequality

Finding the Intersection

The Overlapping Region

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+β

The Final Answer

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

Analyzing the Setup

The Logarithmic Gatekeeper

The Inverse Trigonometric Boundaries

The Intersection - Finding the Sweet Spot

Final Calculation

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If the domain of the function $f (x) = cos^{- 1} (\frac{2 - ∣ x ∣}{4}) + (lo g_{e} (3 - x))^{- 1}$ is $[- α, β) - {γ}$ , then $α + β + γ$ is equal to :

If the domain of the function $f (x) = cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + sin^{- 1} (2 x^{2} - 3 x + 1)$ is the interval $[α, β]$ , then $α + 2 β$ is equal to :

If the domain of the function $f (x) = \frac{c o s ^{- 1} x ^{2} - x + 1}{s i n ^{- 1} ( \frac{2 x - 1}{2} )}$ is the interval $(α, β]$ , then $α + β$ is equal to :

If the domain of the function $f (x) = sin^{- 1} (\frac{1}{x ^{2} - 2 x - 2})$ , is $(- \infty, α] \cup [β, γ] \cup [δ, \infty)$ , then $α + β + γ + δ$ is equal to

The domain of the function $f (x) = sin^{- 1} (\frac{3 x ^{2} + x - 1}{( x - 1 ) ^{2}}) + cos^{- 1} (\frac{x - 1}{x + 1})$ is :

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The domain of the function $cos^{- 1} (\frac{2 s i n ^{- 1} ( \frac{1}{4 x ^{2} - 1} )}{π})$ is: