JEE Main 2022 (24 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$ . Then a value of $2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})$ is

Select Answer:

Progressive Hints

Hint 1

Don't let the notation intimidate you. A binary operation is just a rule—a recipe. Treat the star symbol as a function

f (x, y) = x^{2} + y^{3}

. Before you jump into the algebra, visualize the structure: you are nesting these operations. Work from the inside out, just like you would with nested parentheses in arithmetic.

Hint 2

Once you expand both sides, you will land on a bi-quadratic equation. Remember, a bi-quadratic is just a quadratic in disguise! Use the substitution

t = x^{2}

to simplify your life. Don't try to solve for

x

directly; solve for

x^{2}

first.

Hint 3

When you solve

t^{2} + t - 6 = 0

, you will get two roots for

t

. Ask yourself: does

x^{2}

have any domain restrictions? Once you have the valid

x^{2}

, the final trigonometric expression will collapse beautifully. The starting point is

x^{4} + 2 x^{2} + 2 = x^{2} + 8

Visualized Solution

Understanding the Binary Operation

Given operation: $x * y = x^{2} + y^{3}$
Given equation: $(x * 1) * 1 = x * (1 * 1)$

Computing the Inner LHS: $(x * 1)$

Applying $x * y = x^{2} + y^{3}$ to $(x * 1)$ :
$(x * 1) = x^{2} + 1^{3}$
$(x * 1) = x^{2} + 1$

Computing the Full LHS: $((x * 1) * 1)$

Now compute $(x^{2} + 1) * 1$ :
$LHS = (x^{2} + 1)^{2} + 1^{3}$
$LHS = (x^{2} + 1)^{2} + 1$

Expanding the LHS Expression

Expand using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ :
$LHS = (x^{4} + 2 x^{2} + 1) + 1$
$LHS = x^{4} + 2 x^{2} + 2$

Computing the Inner RHS: $(1 * 1)$

Applying $x * y = x^{2} + y^{3}$ to $(1 * 1)$ :
$(1 * 1) = 1^{2} + 1^{3} = 1 + 1 = 2$

Computing the Full RHS: $x * (1 * 1)$

Now compute $x * 2$ :
$RHS = x^{2} + 2^{3}$
$RHS = x^{2} + 8$

Setting up the Polynomial Equation

Equating LHS and RHS:
$x^{4} + 2 x^{2} + 2 = x^{2} + 8$
Rearranging terms:
$x^{4} + x^{2} - 6 = 0$

Factorizing the Equation

Let $t = x^{2}$ , then $t^{2} + t - 6 = 0$
$(t + 3) (t - 2) = 0$
Substituting back: $(x^{2} + 3) (x^{2} - 2) = 0$

Finding the Value of $x^{2}$

Case 1: $x^{2} + 3 = 0 \Rightarrow x^{2} = - 3$ (Rejected as $x^{2} \geq 0$ )
Case 2: $x^{2} - 2 = 0 \Rightarrow x^{2} = 2$
Therefore, $x^{2} = 2$ and $x^{4} = 4$

Introducing the Target Expression

Target expression: $2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})$

Substitution into the Expression

Substitute $x^{2} = 2$ and $x^{4} = 4$ :
$= 2 sin^{- 1} (\frac{4 + 2 - 2}{4 + 2 + 2})$

Simplifying the Fraction

Simplify the fraction:
$= 2 sin^{- 1} (\frac{4}{8})$
$= 2 sin^{- 1} (\frac{1}{2})$

Evaluating the Inverse Sine

Since $sin (\frac{π}{6}) = \frac{1}{2}$ ,
$sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$

Final Calculation

Final computation:
$= 2 \cdot \frac{π}{6}$
$= \frac{π}{3}$

Conclusion and Key Takeaway

Key Takeaways:
1. Carefully apply custom binary operations step-by-step.
2. For real $x$ , $x^{2}$ must be non-negative; discard extraneous roots.
3. Simplify algebraic expressions before substituting into trigonometric functions.
Final Answer: $\frac{π}{3}$

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The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

We are dealing with a custom binary operation defined as

x * y = x^{2} + y^{3}

. Think of this as a mathematical machine: you feed it two numbers,

x

and

y

, and it outputs the square of the first plus the cube of the second.

Our mission is to solve the equation

(x * 1) * 1 = x * (1 * 1)

The Algebraic Battlefield

Let us tackle the Left Hand Side (LHS) first:

(x * 1) * 1

Following the rule, we first evaluate the inner bracket:

(x * 1) = x^{2} + 1^{3} = x^{2} + 1

Now, we take this result and apply the operation with

1

again:

(x^{2} + 1) * 1 = (x^{2} + 1)^{2} + 1^{3}

Expanding this using the identity

(a + b)^{2} = a^{2} + 2 ab + b^{2}

, we get:

(x^{4} + 2 x^{2} + 1) + 1 = x^{4} + 2 x^{2} + 2

Now, let us look at the Right Hand Side (RHS):

x * (1 * 1)

Again, start with the inner bracket:

(1 * 1) = 1^{2} + 1^{3} = 2

Now, we evaluate

x * 2

x * 2 = x^{2} + 2^{3} = x^{2} + 8

The Master Equation

Equating the two sides, we get:

x^{4} + 2 x^{2} + 2 = x^{2} + 8

Rearranging this, we arrive at the bi-quadratic equation:

x^{4} + x^{2} - 6 = 0

The Elegant Simplification

Let

t = x^{2}

. Our equation becomes:

t^{2} + t - 6 = 0

Factoring this quadratic, we find:

(t + 3) (t - 2) = 0

This gives us

t = - 3

t = 2

. Since

t = x^{2}

and

x

must be a real number,

x^{2}

cannot be negative. We reject

t = - 3

and accept

t = 2

. Thus,

x^{2} = 2

and

x^{4} = 4

Final Calculation

Now, we turn to the final expression:

2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})

Substituting our values

x^{2} = 2

and

x^{4} = 4

2 sin^{- 1} (\frac{4 + 2 - 2}{4 + 2 + 2}) = 2 sin^{- 1} (\frac{4}{8}) = 2 sin^{- 1} (\frac{1}{2})

We know that

sin (\frac{π}{6}) = \frac{1}{2}

, so

sin^{- 1} (\frac{1}{2}) = \frac{π}{6}

Finally, the result is:

2 \cdot \frac{π}{6} = \frac{π}{3}

Common Traps & FAQs

Why can't I just distribute the star operation? Like, is

(x * 1) * 1

the same as

x * (1 * 1)

That is a classic trap! Binary operations are not necessarily associative. You cannot assume properties like associativity unless they are proven. Always evaluate step-by-step, working from the innermost brackets outwards.

I got

x^{2} = - 3

and

x^{2} = 2

. Why do we reject

- 3

In the context of real numbers, the square of any real number

x

must be non-negative (

x^{2} \geq 0

). Since

- 3

is negative, it cannot be the square of a real number. Always check your constraints!

Is there a trick to the inverse sine part? It looks so complicated.

It looks intimidating, but it is a test of your patience. Once you have

x^{2} = 2

, the expression simplifies to a constant. Trust the algebra and perform the substitution carefully.

Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$ . Then a value of $2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})$ is

Select Answer:

Visualized Solution

Understanding the Binary Operation

Computing the Inner LHS: $(x * 1)$

Computing the Full LHS: $((x * 1) * 1)$

Expanding the LHS Expression

Computing the Inner RHS: $(1 * 1)$

Computing the Full RHS: $x * (1 * 1)$

Setting up the Polynomial Equation

Factorizing the Equation

Finding the Value of $x^{2}$

Introducing the Target Expression

Substitution into the Expression

Simplifying the Fraction

Evaluating the Inverse Sine

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Algebraic Battlefield

The Master Equation

The Elegant Simplification

Final Calculation

Similar Questions

If $0 < a, b < 1$ , and $tan^{- 1} a + tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - \frac{( a ^{2} + b ^{2} )}{2} + \frac{( a ^{3} + b ^{3} )}{3} - \frac{( a ^{4} + b ^{4} )}{4} + \dots$ is :

If $S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}$ then $\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))$ is equal to _________.

For $α, β, γ \neq = 0$ . If $sin^{- 1} α + sin^{- 1} β + sin^{- 1} γ = π$ and $(α + β + γ) (α - γ + β) = 3 α β$ , then $γ$ equal to

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$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

Match List I with List II:

List-I

List-II

Let $S$ be the set of all solutions of the equation $cos^{- 1} (2 x) - 2 cos^{- 1} (1 - x^{2}) = π, x \in [- \frac{1}{2}, \frac{1}{2}]$ . Then $\sum_{x \in S} 2 sin^{- 1} (x^{2} - 1)$ is equal to

Let $(x, y)$ be such that $sin^{- 1} (a x) + cos^{- 1} (y) + cos^{- 1} (b x y) = \frac{π}{2}$ . Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

List-II

Let $S = {x : cos^{- 1} x = π + sin^{- 1} x + sin^{- 1} (2 x + 1)}$ . Then $\sum_{x \in S} (2 x - 1)^{2}$ is equal to

The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ for $x \in [- 1, 1]$ , and $[x]$ denotes the greatest integer less than or equal to $x$ , is :

Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let x∗y=x2+y3 and (x∗1)∗1=x∗(1∗1). Then a value of 2sin−1(x4+x2+2x4+x2−2​) is

Select Answer:

Visualized Solution

Understanding the Binary Operation

Computing the Inner LHS: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x∗1)

Computing the Full LHS: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ((x∗1)∗1)

Expanding the LHS Expression

Computing the Inner RHS: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1∗1)

Computing the Full RHS: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∗(1∗1)

Setting up the Polynomial Equation

Factorizing the Equation

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

Introducing the Target Expression

Substitution into the Expression

Simplifying the Fraction

Evaluating the Inverse Sine

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Algebraic Battlefield

The Master Equation

The Elegant Simplification

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin−1(5x​)+cosec−1(45​)=2π​, then the values of x is

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Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$ . Then a value of $2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})$ is

Computing the Inner LHS: $(x * 1)$

Computing the Full LHS: $((x * 1) * 1)$

Computing the Inner RHS: $(1 * 1)$

Computing the Full RHS: $x * (1 * 1)$

Finding the Value of $x^{2}$

If $0 < a, b < 1$ , and $tan^{- 1} a + tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - \frac{( a ^{2} + b ^{2} )}{2} + \frac{( a ^{3} + b ^{3} )}{3} - \frac{( a ^{4} + b ^{4} )}{4} + \dots$ is :

If $S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}$ then $\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))$ is equal to _________.

For $α, β, γ \neq = 0$ . If $sin^{- 1} α + sin^{- 1} β + sin^{- 1} γ = π$ and $(α + β + γ) (α - γ + β) = 3 α β$ , then $γ$ equal to

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

Let $S$ be the set of all solutions of the equation $cos^{- 1} (2 x) - 2 cos^{- 1} (1 - x^{2}) = π, x \in [- \frac{1}{2}, \frac{1}{2}]$ . Then $\sum_{x \in S} 2 sin^{- 1} (x^{2} - 1)$ is equal to

Let $S = {x : cos^{- 1} x = π + sin^{- 1} x + sin^{- 1} (2 x + 1)}$ . Then $\sum_{x \in S} (2 x - 1)^{2}$ is equal to

The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ for $x \in [- 1, 1]$ , and $[x]$ denotes the greatest integer less than or equal to $x$ , is :