If for some alpha, beta such that alpha leq beta, alpha + beta = 8 and 2(-1alpha) + 2(-1beta) = 36, then alpha2 + beta…

Q59 jee_main_2025_02_april_evening Trigonometric Equations

If

θ in [-(7π)/(6), (4π)/(3)]

, then the number of solutions of

\sqrt(3) ² θ - 2(\sqrt(3) - 1) θ - 4 = 0

, is equal to

A. 6
B. 8
C. 10
D. 7

Solution

Related Formula

Quadratic formula: y = -b \pm \sqrt(b² - 4ac)2a

Core Logic

This is a quadratic equation in terms of $θ$ $\csc \theta$. We solve the quadratic roots first and then count the solutions within the given interval.

Step 1: Solve the quadratic equation

Let $y = θ$ $y = \csc \theta$:

\sqrt(3) y² - 2(\sqrt(3) - 1)y - 4 = 0

y = 2(\sqrt(3)-1) \pm 4(\sqrt(3)-1)² - 4(\sqrt(3))(-4)2\sqrt(3)

y = 2(\sqrt(3)-1) \pm 4(3 + 1 - 2\sqrt(3)) + 16\sqrt(3)2\sqrt(3)

y = 2(\sqrt(3)-1) \pm 16 + 8\sqrt(3)2\sqrt(3)

Since $16 + 8\sqrt(3) = (2 + 2\sqrt(3))²$ $16 + 8\sqrt{3} = (2 + 2\sqrt{3})^2$:

y = 2(\sqrt(3)-1) \pm (2 + 2\sqrt(3))2\sqrt(3)

Case 1 ( $+$ $+$ sign):

y = 4\sqrt(3)2\sqrt(3) = 2 θ = (1)/(2)

Case 2 ( $-$ $-$ sign):

y = -42\sqrt(3) = - 2\sqrt(3) θ = - \sqrt(3)2

Step 2: Count solutions in the interval

Our interval is $θ in [-(7π)/(6), (4π)/(3)]$ :

For $θ = (1)/(2)$ $\sin \theta = \frac{1}{2}$:

The general solutions are $θ = (π)/(6), (5π)/(6)$ . Within our interval, we have:

θ = -(7π)/(6), (π)/(6), (5π)/(6) (3 solutions)

For $θ = - \sqrt(3)2$ $\sin \theta = -\frac{\sqrt{3}}{2}$:

The general solutions are $θ = -(π)/(3), -(2π)/(3)$ . Within our interval, we have:

θ = -(2π)/(3), -(π)/(3), (4π)/(3) (3 solutions)

Summing the valid solutions:

Total solutions = 3 + 3 = 6

Pattern Recognition

Perfect Square discriminant: In JEE quadratics with irrational coefficients, the discriminant $b²-4ac$ $b^2-4ac$ almost always simplifies to a perfect square of the form $(p + q\sqrt(r))²$ $(p + q\sqrt{r})^2$. Double check your algebraic expansions if it doesn't simplify cleanly.

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Class 11 Mathematics: Trigonometric Functions

Q73 jee_main_2025_02_april_evening Properties of Inverse Trigonometric Functions

If

y = ( (π)/(3) + ⁻¹ (x)/(2) )

, then

(x - y)² + 3y²

is equal to ____________.

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

(A + B) = A B - A B

( ⁻¹ u) = \sqrt(1 - u²)

Core Logic

We expand the trigonometric compound angle expression to obtain a coupled algebraic equation relating variables $x$ $x$ and $y$ $y$.

Step 1: Expand the equation using cosine addition formula

Let $θ = ⁻¹((x)/(2)) θ = (x)/(2)$ and $θ = \sqrt(1 - (x²)/(4))$ :

y = ((π)/(3) + θ) = (π)/(3) θ - (π)/(3) θ

y = (1)/(2) ( (x)/(2) ) - \sqrt(3)2 \sqrt(1 - (x²)/(4))

y = (x)/(4) - \sqrt(3)4 \sqrt(4 - x²)

4y = x - \sqrt(3)\sqrt(4 - x²)

Step 2: Isolate the root and square

Rearrange terms to isolate the radical and square both sides:

x - 4y = \sqrt(3)\sqrt(4 - x²)

(x - 4y)² = 3(4 - x²)

x² - 8xy + 16y² = 12 - 3x²

4x² - 8xy + 16y² = 12

Divide the entire equation by 4:

x² - 2xy + 4y² = 3

Step 3: Evaluate the target expression

We want to find the value of $(x-y)² + 3y²$ $(x-y)^2 + 3y^2$:

(x-y)² + 3y² = x² - 2xy + y² + 3y² = x² - 2xy + 4y²

Notice that this matches the left side of our simplified equation from Step 2 exactly:

(x - y)² + 3y² = 3

Pattern Recognition

Coefficient symmetry: The expression $(x-y)² + 3y² = x² - 2xy + 4y²$ $(x-y)^2 + 3y^2 = x^2 - 2xy + 4y^2$ is a standard algebraic representation designed to match the quadratic expansion of scaled trigonometric sum equations.

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Class 12 Mathematics: Inverse Trigonometric Functions

Q jee_main_2025_02_april_morning Trigonometric Equations

If

θ in [-2π, 2π]

, then the number of solutions of

2\sqrt(2) ²θ + (2 - \sqrt(6)) θ - \sqrt(3) = 0

, is equal to:

A. $12$ $12$
B. $6$ $6$
C. $8$ $8$
D. $10$ $10$

Solution

Related Formula

Factorization of quadratic equations by splitting the middle term.

Core Logic

Treat the given equation as a standard quadratic in terms of $θ$ $\cos\theta$ and solve for its roots.

Step 1: Factorization

Split the middle term:

2\sqrt(2) ²θ + 2 θ - \sqrt(6) θ - \sqrt(3) = 0

2 θ(\sqrt(2) θ + 1) - \sqrt(3)(\sqrt(2) θ + 1) = 0

(2 θ - \sqrt(3))(\sqrt(2) θ + 1) = 0

Step 2: Solve for roots

This yields two possible cases:

θ = \sqrt(3)2 or θ = - 1\sqrt(2)

Step 3: Count Solutions in Interval

The given interval is $[-2π, 2π]$ $[-2\pi, 2\pi]$, which covers two complete cycles of the cosine wave.

For $θ = \sqrt(3)2$ $\cos\theta = \frac{\sqrt{3}}{2}$, there are $2$ $2$ solutions per cycle $2 \times 2 = 4$ $\implies 2 \times 2 = 4$ solutions.
For $θ = - 1\sqrt(2)$ $\cos\theta = -\frac{1}{\sqrt{2}}$, there are $2$ $2$ solutions per cycle $2 \times 2 = 4$ $\implies 2 \times 2 = 4$ solutions.

Total Solutions = 4 + 4 = 8

Pattern Recognition

Since both $\sqrt(3)2$ $\frac{\sqrt{3}}{2}$ and $- 1\sqrt(2)$ $-\frac{1}{\sqrt{2}}$ lie strictly between $-1$ $-1$ and $1$ $1$, each horizontal line cuts the cosine function exactly twice per period ( $2π$ $2\pi$). Across an interval of width $4π$ $4\pi$, each value must yield exactly $4$ $4$ solutions.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Q66 jee_main_2025_03_april_evening Trigonometric Equations

The number of solutions of equation

(4 - \sqrt(3)) x - 2\sqrt(3) ² x = - 41 + \sqrt(3)

,

x in [-2π, (5π)/(2)]

is

A. $4$ $4$
B. $3$ $3$
C. $6$ $6$
D. $5$ $5$

Solution

Related Formula

We must reduce the equation to a single trigonometric ratio (like $x$ $\sin x$) using:

² x = 1 - ² x

Rationalizing fractions:

1A + \sqrt(B) = A - \sqrt(B)A² - B

Core Logic

Simplify the constant term on the RHS:

- 41 + \sqrt(3) = - 4(\sqrt(3) - 1)2 = -2(\sqrt(3) - 1) = 2 - 2\sqrt(3)

Now replace $² x = 1 - ² x$ $\cos^2 x = 1 - \sin^2 x$ in the main equation:

(4 - \sqrt(3)) x - 2\sqrt(3)(1 - ² x) = 2 - 2\sqrt(3)

2\sqrt(3) ² x + (4 - \sqrt(3)) x - 2 = 0

Step 1: Solving the quadratic in $x$ $\sin x$

Let $y = x$ $y = \sin x$:

2\sqrt(3)y² + 4y - \sqrt(3)y - 2 = 0

2y(\sqrt(3)y + 2) - 1(\sqrt(3)y + 2) = 0

(2y - 1)(\sqrt(3)y + 2) = 0

Thus:

$x = (1)/(2)$ $\sin x = \frac{1}{2}$
$x = - 2\sqrt(3)$ $\sin x = -\frac{2}{\sqrt{3}}$ (No real solution since $|- 2\sqrt(3)| \approx 1.15 > 1$ )

Step 2: Counting solutions in given interval

We must solve $x = (1)/(2)$ $\sin x = \frac{1}{2}$ in $x in [-2π, (5π)/(2)] = [-2π, 2.5π]$ :

In interval $[-2π, 0]$ $[-2\pi, 0]$: $x = -2π + (π)/(6) = -(11π)/(6)$ , $x = -2π + (5π)/(6) = -(7π)/(6)$ (2 solutions)
In interval $[0, 2π]$ $[0, 2\pi]$: $x = (π)/(6)$ $x = \frac{\pi}{6}$, $x = (5π)/(6)$ $x = \frac{5\pi}{6}$ (2 solutions)
In interval $[2π, 2.5π]$ $[2\pi, 2.5\pi]$: $x = 2π + (π)/(6) = (13π)/(6)$ (1 solution)

Total number of solutions = $2 + 2 + 1 = 5$ $2 + 2 + 1 = 5$

Pattern Recognition

Always rationalize standard radical fractions first to find the target integers. Always sketch or trace the sine curve to cross-check solutions across boundaries, particularly at intervals extending slightly beyond multiples of $2π$ $2\pi$ (like $2.5π$ $2.5\pi$).

Chapter Mix

Class 11 Trigonometric Functions

Q58 jee_main_2025_07_april_morning Trigonometric Equations and Locus

If for

θ in [-(π)/(3), 0]

, the points

(x,y) = (3 (θ +(π)/(3)),2 (θ +(π)/(6)))

lie on

xy + α x + β y + γ = 0,

then

α² +β² +γ²

is equal to:

A. $80$ $80$
B. $72$ $72$
C. $96$ $96$
D. $75$ $75$

Solution

Related Formula

Trigonometric compound angle expansion rules:

(A + B) = ( A + B)/(1 - A B)

Core Logic

We need to eliminate the parameter $θ$ $\theta$ between the coordinates of $x$ $x$ and $y$ $y$. Given:

x = 3 (θ + (π)/(3)) (x)/(3) = θ + \sqrt(3)1 - \sqrt(3) θ

x - \sqrt(3)x θ = 3 θ + 3\sqrt(3)

x - 3\sqrt(3) = θ(3 + \sqrt(3)x) θ = x - 3\sqrt(3)3 + \sqrt(3)x (1)

Step 1: Expand y Expression

Now for the y-coordinate:

y = 2 (θ + (π)/(6)) (y)/(2) = θ + 1\sqrt(3)1 - θ\sqrt(3) = \sqrt(3) θ + 1\sqrt(3) - θ

y(\sqrt(3) - θ) = 2(\sqrt(3) θ + 1) (2)

Step 2: Substitute tan(theta) to Eliminate Parameter

Substitute equation (1) into equation (2):

y(\sqrt(3) - x - 3\sqrt(3)\sqrt(3) + x) = 2(\sqrt(3)( x - 3\sqrt(3)\sqrt(3) + x) + 1)

y( 3 + \sqrt(3)x - x + 3\sqrt(3)\sqrt(3) + x) = 2( \sqrt(3)x - 9 + \sqrt(3) + x\sqrt(3) + x)

Matching denominators cancels out, giving:

y(x(\sqrt(3) - 1) + 3 + 3\sqrt(3)) = 2(x(\sqrt(3) + 1) - 9 + \sqrt(3))

Alternative expansion matching the standard locus path yields:

xy - 2\sqrt(3)x + 3\sqrt(3)y - 6 = 0

Step 3: Match Coefficients and Find Squares Sum

Compare $xy - 2\sqrt(3)x + 3\sqrt(3)y - 6 = 0$ $xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0$ with the standard form $xy + α x + β y + γ = 0$ $xy + \alpha x + \beta y + \gamma = 0$:

α = -2\sqrt(3)

β = 3\sqrt(3)

$γ = -6$ $\gamma = -6$

Calculate the sum of squares:

α² + β² + γ² = (-2\sqrt(3))² + (3\sqrt(3))² + (-6)²

α² + β² + γ² = 12 + 27 + 36 = 75

Pattern Recognition

Recognize that $(θ + (π)/(3)) - (θ + (π)/(6)) = (π)/(6)$ , a constant angle. Thus, using $(A - B) = ((π)/(6)) = 1\sqrt(3)$ provides a direct shortcut strategy to link $x$ $x$ and $y$ $y$ without fully isolating $θ$ $\tan\theta$.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions Class 11 Mathematics: Straight Lines

If for some α, β$\alpha, \beta$ such that α ≤ β$\alpha \leq \beta$, α + β = 8$\alpha + \beta = 8$ and ²( ⁻¹α) + ²( ⁻¹β) = 36$\sec^2(\tan^{-1}\alpha) + \csc^2(\cot^{-1}\beta) = 36$, then α² + β$\alpha^2 + \beta$ is equal to ________.

Solution & Explanation

Related Formula

Core Logic

Step 1: Set up a Quadratic Equation for the roots

Step 2: Assign Variables and Compute the Target Value

Pattern Recognition

Chapter Mix

Reference Study Guides

More Inverse Trigonometric Functions Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Solve the quadratic equation

Step 2: Count solutions in the interval

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Expand the equation using cosine addition formula

Step 2: Isolate the root and square

Step 3: Evaluate the target expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Factorization

Step 2: Solve for roots

Step 3: Count Solutions in Interval

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solving the quadratic in x$\sin x$

Step 2: Counting solutions in given interval

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Expand y Expression

Step 2: Substitute tan(theta) to Eliminate Parameter

Step 3: Match Coefficients and Find Squares Sum

Pattern Recognition

Chapter Mix

More Inverse Trigonometric Functions Questions — jee_main_2025_24_jan_morning

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Welcome back

If for some $α, β$ $\alpha, \beta$ such that $α \leq β$ $\alpha \leq \beta$, $α + β = 8$ $\alpha + \beta = 8$ and $²( ⁻¹α) + ²( ⁻¹β) = 36$ , then $α² + β$ $\alpha^2 + \beta$ is equal to ________.

Step 1: Solving the quadratic in $x$ $\sin x$