Sketch the solution set of the...

MathematicsStandard and General Equation of a CircleJEE Advanced 1978Moderate

Sketch the solution set of the following system of inequalities: $x^{2} + y^{2} - 2 x \leq 0; 3 x - y - 12 \leq 0; y - x \leq 0; y \geq 0$ .

Visualized Solution (Hindi)

Introduction to the System

We need to find the intersection of four inequalities:
1. $x^{2} + y^{2} - 2 x \leq 0$
2. $3 x - y - 12 \leq 0$
3. $y - x \leq 0$
4. $y \geq 0$

Analyzing the Circle: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x^{2} + y^{2} - 2 x \leq 0

Rearrange the first inequality: $x^{2} - 2 x + y^{2} \leq 0$
Complete the square for $x$ : $(x^{2} - 2 x + 1) + y^{2} \leq 1$
Standard form: $(x - 1)^{2} + y^{2} \leq 1$
This represents the interior and boundary of a circle with:
Center: $(1, 0)$
Radius: $1$

The Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y - x \leq 0

Consider $y - x \leq 0 ⟹ y \leq x$
The boundary is the line $y = x$ .
This line passes through $(0, 0)$ and $(1, 1)$ .
The inequality $y \leq x$ represents the half-plane below the line $y = x$ .

Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y \geq 0

The condition $y \geq 0$ restricts the solution to the upper half-plane (on or above the x-axis).

The Redundant Line: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3 x - y - 12 \leq 0

Rearrange: $y \geq 3 x - 12$
The line $y = 3 x - 12$ has an x-intercept at $(4, 0)$ .
Since our circle is bounded within $0 \leq x \leq 2$ , the entire circle region already satisfies $y \geq 3 x - 12$ .
This constraint is redundant for the bounded region.

Final Solution Set

The final solution set is the intersection of:
1. Interior of $(x - 1)^{2} + y^{2} \leq 1$
2. Region below $y = x$
3. Region above $y = 0$
The vertices of this region are $(0, 0)$ , $(2, 0)$ , and the intersection point $(1, 1)$ .

00:00 / 00:00

Conceptually Similar Problems

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Advanced 1980Easy

View in:English Hindi

Find the solution set of the system $x + 2 y + z = 1; 2 x - 3 y - w = 2; x \geq 0; y \geq 0; z \geq 0; w \geq 0$ .

Answer:

x = 1, y = 0, z = 0, w = 0

MathematicsSolution of Quadratic EquationsJEE Advanced 1983Easy

View in:English Hindi

Find all real values of $x$ which satisfy $x^{2} - 3 x + 2 > 0$ and $x^{2} - 2 x - 4 \leq 0$

Answer:

x \in [1 - 5, 1) \cup (2, 1 + 5]

MathematicsStandard Equations of Parabola, Ellipse, and HyperbolaJEE Advanced 1981Moderate

View in:English Hindi

Each of the four inequalities given below defines a region in the xy plane. One of these four regions does not have the following property: For any two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in the region, the point $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ is also in the region. The inequality defining this region is

(A)

x^{2} + 2 y^{2} \leq 1

(B)

Max

{∣ x ∣, ∣ y ∣} \leq 1

(C)

x^{2} - y^{2} \leq 1

(D)

y^{2} - x \leq 0

Answer:C

MathematicsLinear InequalitiesJEE Advanced 1987Moderate

View in:English Hindi

Find the set of all $x$ for which $\frac{2 x}{2 x ^{2} + 5 x + 2} > \frac{1}{x + 1}$

Answer:

(- 2, - 1) \cup (- 2/3, - 1/2)

MathematicsArea Bounded by CurvesJEE Advanced 1992Moderate

View in:English Hindi

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

Answer:

(π - \frac{2}{3})

sq. units

MathematicsArea Bounded by CurvesJEE Advanced 1991Moderate

View in:English Hindi

Sketch the curves and identify the region bounded by $x = \frac{1}{2}, x = 2, y = ln x$ and $y = 2^{x}$ . Find the area of this region.

Answer:

\frac{4 - 2}{l o g 2} - \frac{5}{2} lo g 2 + \frac{3}{2}

MathematicsStandard and General Equation of a CircleJEE Advanced 1983Easy

View in:English Hindi

The points of intersection of the line $4 x - 3 y - 10 = 0$ and the circle $x^{2} + y^{2} - 2 x + 4 y - 20 = 0$ are ......... and .........

Answer:(4, 2), (-2, -6)

MathematicsArea Bounded by CurvesJEE Advanced 2005Moderate

View in:English Hindi

Find the area bounded by the curves $x^{2} = y, x^{2} = - y$ and $y^{2} = 4 x - 3$ .

Answer:

\frac{1}{3}

sq. units

MathematicsArea Bounded by CurvesJEE Main 2014Moderate

View in:English Hindi

The area of the region described by $A = {(x, y) : x^{2} + y^{2} \leq 1 and y^{2} \leq 1 - x}$ is:

(A)

\frac{π}{2} - \frac{2}{3}

(B)

\frac{π}{2} + \frac{2}{3}

(C)

\frac{π}{2} + \frac{4}{3}

(D)

\frac{π}{4} - \frac{2}{3}

Answer:C

MathematicsProperties of DeterminantsJEE Advanced 1981Easy

View in:English Hindi

The solution set of the equation $111 4 - 2 2 x 205 5 x^{2} = 0$ is .........

Answer:

{- 1, 2}