Q.Find the area under the given curves and given lines:

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The area of the region bounded by $y = x^3$, x-axis, $x = -2$ and $x = 4$ is (A) 64 (B) $\dfrac{81}{4}$ (C) $\dfrac{66}{5}$ (D) 68

   ›Reveal solutionSolution

   The area is the sum of the absolute values of the definite integrals from $x=-2$ to $x=0$ and from $x=0$ to $x=4$, because the curve lies below the x-axis on the left interval and above it on the right. The total area is $\boxed{68}$, so the correct option is (D).

   The key concept here is that area between a curve and the x-axis is always taken as a positive quantity. When a function dips below the x-axis, the definite integral gives a negative number, but area is the absolute value of that signed area. So we must split the region at the x-intercept(s) where the curve crosses the axis, integrate each piece separately, take the absolute value, and then add them.

   For $y = x^3$, the only x-intercept is at $x = 0$. Between $x = -2$ and $x = 0$, the curve is below the axis (since $x^3$ is negative for negative $x$). Between $x = 0$ and $x = 4$, the curve is above the axis. Therefore, the total area is:

   $\text{Area} = \left| \int_{-2}^{0} x^3 \, dx \right| + \int_{0}^{4} x^3 \, dx$

   Now we compute step by step.

   1. Integrate $x^3$

      The antiderivative is $\frac{x^4}{4}$.

   2. Evaluate from $x = -2$ to $x = 0$

   $\int_{-2}^{0} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-2}^{0} = \frac{0^4}{4} - \frac{(-2)^4}{4} = 0 - \frac{16}{4} = -4$

   The signed area is $-4$, so the actual (positive) area is $| -4 | = 4$.

   3. Evaluate from $x = 0$ to $x = 4$

   $\int_{0}^{4} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{4} = \frac{4^4}{4} - \frac{0^4}{4} = \frac{256}{4} - 0 = 64$

   This is already positive, so the area here is $64$.

   4. Add the two areas

   $\text{Total area} = 4 + 64 = 68$

   Watch out

   A common mistake is to compute $\int_{-2}^{4} x^3 \, dx$ directly, which gives $\frac{4^4}{4} - \frac{(-2)^4}{4} = 64 - 4 = 60$. This is not the area — it’s the net signed area, which cancels the negative part. Always split at x-intercepts when finding area.

   Tip

   For odd functions like $x^3$, the area from $-a$ to $a$ is twice the area from $0$ to $a$, but here the limits are asymmetric ($-2$ to $4$), so you must compute each piece separately.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Area of the region bounded by the curves $y=2^x$, $y^2=4x$ and the lines $x=\dfrac{1}{2}$, $x=1$ is (A) $\dfrac{2-\sqrt{2}}{\log 2} - \dfrac{4-\sqrt{2}}{3}$ (B) $\dfrac{4-\sqrt{2}}{3} - \dfrac{2-\sqrt{2}}{\log 2}$ (C) $\dfrac{4+\sqrt{2}}{3} - \dfrac{2-\sqrt{2}}{\log 2}$ (D) $\dfrac{2-\sqrt{2}}{\log 2} - \dfrac{4+\sqrt{2}}{3}$

   ›Reveal solutionSolution

   The area is the difference between the integral of $y=2^x$ and the integral of $y=2\sqrt{x}$ (the upper branch of $y^2=4x$) from $x=1/2$ to $x=1$, giving option (B).

   The region is bounded by two curves and two vertical lines. The key is to see which curve lies above the other in the given interval. The curve $y^2 = 4x$ is a right-opening parabola; solving for $y$ gives $y = \pm 2\sqrt{x}$. Since the region is bounded also by $y = 2^x$ (which is always positive), we only consider the upper branch $y = 2\sqrt{x}$.

   Between $x = 1/2$ and $x = 1$, we need to compare $2^x$ and $2\sqrt{x}$. At $x = 1$, $2^1 = 2$ and $2\sqrt{1} = 2$, so they meet. At $x = 1/2$, $2^{1/2} = \sqrt{2} \approx 1.414$ and $2\sqrt{1/2} = 2/\sqrt{2} = \sqrt{2} \approx 1.414$ — they also meet! So the two curves actually intersect at both endpoints. But between them, which is larger? Test a point, say $x = 3/4$: $2^{3/4} = \sqrt[4]{8} \approx 1.68$, while $2\sqrt{3/4} = \sqrt{3} \approx 1.73$. So $2\sqrt{x} > 2^x$ in the interior. That means the parabola lies above the exponential curve in this interval.

   Thus the area is the integral of (upper curve minus lower curve) from $x = 1/2$ to $x = 1$:

   $\text{Area} = \int_{1/2}^{1} \left( 2\sqrt{x} - 2^x \right) dx$

   Now we compute step by step.

   1. Integrate $2\sqrt{x}$.

      Write $2\sqrt{x} = 2x^{1/2}$. Its antiderivative is $2 \cdot \frac{x^{3/2}}{3/2} = \frac{4}{3} x^{3/2}$.

   2. Integrate $2^x$.

      The antiderivative of $a^x$ is $\frac{a^x}{\log a}$. So $\int 2^x dx = \frac{2^x}{\log 2}$.

   3. Evaluate the definite integral.

   $\left[ \frac{4}{3} x^{3/2} - \frac{2^x}{\log 2} \right]_{1/2}^{1}$

   At $x = 1$: $\frac{4}{3}(1)^{3/2} - \frac{2^1}{\log 2} = \frac{4}{3} - \frac{2}{\log 2}$.

   At $x = 1/2$: $\frac{4}{3}\left(\frac{1}{2}\right)^{3/2} - \frac{2^{1/2}}{\log 2} = \frac{4}{3} \cdot \frac{1}{2\sqrt{2}} - \frac{\sqrt{2}}{\log 2} = \frac{4}{6\sqrt{2}} - \frac{\sqrt{2}}{\log 2} = \frac{2}{3\sqrt{2}} - \frac{\sqrt{2}}{\log 2}$.

   Simplify $\frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$ (multiply numerator and denominator by $\sqrt{2}$). So the lower limit value is $\frac{\sqrt{2}}{3} - \frac{\sqrt{2}}{\log 2}$.

   4. Subtract: upper limit minus lower limit.

   $\left( \frac{4}{3} - \frac{2}{\log 2} \right) - \left( \frac{\sqrt{2}}{3} - \frac{\sqrt{2}}{\log 2} \right)$

   Combine the constant terms: $\frac{4}{3} - \frac{\sqrt{2}}{3} = \frac{4 - \sqrt{2}}{3}$.

   Combine the $\frac{1}{\log 2}$ terms: $-\frac{2}{\log 2} + \frac{\sqrt{2}}{\log 2} = -\frac{2 - \sqrt{2}}{\log 2}$.

   So the area is:

   $\frac{4 - \sqrt{2}}{3} - \frac{2 - \sqrt{2}}{\log 2}$

   Watch out

   A common mistake is to assume $2^x$ is always above $2\sqrt{x}$ because exponentials grow fast. But here the interval is near $x=1$, where the parabola actually dominates. Always check a test point.

   ✓Final answer

   The correct option is (B).

3. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The area of the region bounded by the curves $y = x^3$, $y = x^2$ and the lines $x = 0$ and $x = 2$ is (A) $\dfrac{4}{3}$ (B) $\dfrac{3}{2}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{5}{3}$

   ›Reveal solutionSolution

   Area $=\dfrac{3}{2}$ sq. units — option (B).

   Compare the curves: on $[0,1]$, $x^{2}\ge x^{3}$; on $[1,2]$, $x^{3}\ge x^{2}$. So

   $\text{Area}=\int_{0}^{1}\!\left(x^{2}-x^{3}\right)dx+\int_{1}^{2}\!\left(x^{3}-x^{2}\right)dx.$

   $\int_{0}^{1}\!\left(x^{2}-x^{3}\right)dx=\frac{1}{3}-\frac{1}{4}=\frac{1}{12},$

   $\int_{1}^{2}\!\left(x^{3}-x^{2}\right)dx=\left[\frac{x^{4}}{4}-\frac{x^{3}}{3}\right]_{1}^{2}=\left(4-\frac{8}{3}\right)-\left(\frac{1}{4}-\frac{1}{3}\right)=\frac{4}{3}+\frac{1}{12}=\frac{17}{12}.$

   $\text{Area}=\frac{1}{12}+\frac{17}{12}=\frac{18}{12}=\frac{3}{2}.$

   ✓Final answer

   (B) $\dfrac{3}{2}$.

4. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.The area (in sq. units) bounded by the curve $y = 2x - x^2$ and the line $y = -x$ is (A) $\dfrac{9}{2}$ (B) $\dfrac{11}{2}$ (C) $\dfrac{16}{3}$ (D) $\dfrac{22}{5}$

   ›Reveal solutionSolution

   The area between the parabola $y = 2x - x^2$ and the line $y = -x$ is found by integrating the difference of the functions between their intersection points. The result is $\frac{9}{2}$, so the correct option is (A).

   ---

   We want the area bounded by a downward-opening parabola and a straight line. The key idea: area between two curves is the integral of (top function minus bottom function) over the interval where they cross. So first, find where they intersect; then decide which curve is above the other; then integrate.

   1. Find intersection points

      Set $2x - x^2 = -x$.

      Bring all terms to one side:

      $2x - x^2 + x = 0 \quad\Rightarrow\quad 3x - x^2 = 0$

      Factor: $x(3 - x) = 0$

      So $x = 0$ and $x = 3$.

      These are the boundaries of the region.

   2. Determine which curve is on top

      Pick a test point between $0$ and $3$, say $x = 1$:

      • Parabola: $y = 2(1) - 1^2 = 1$
      • Line: $y = -1$ Since $1 > -1$, the parabola is above the line on $(0, 3)$. So the vertical height between them is $(2x - x^2) - (-x) = 2x - x^2 + x = 3x - x^2$.

   3. Set up the definite integral

      Area $= \int_{0}^{3} (3x - x^2) \, dx$

   4. Evaluate

      $\int (3x - x^2) \, dx = \frac{3x^2}{2} - \frac{x^3}{3}$

      From $0$ to $3$:

      $\left[ \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right] - \left[ 0 \right] = \frac{27}{2} - \frac{27}{3} = \frac{27}{2} - 9$

      $= \frac{27}{2} - \frac{18}{2} = \frac{9}{2}$

   Tip

   A common mistake is to forget that $y = -x$ is below the parabola in this interval. Always test a point between the intersections to confirm which function is greater.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The area of the region bounded by the curves $y = x^2 + 1$, $x = y^2 + 1$, X-axis, Y-axis and $x = 2$ (in Sq. units) is (A) $\dfrac{13}{3}$ (B) $4$ (C) $6$ (D) $\dfrac{15}{2}$

   ›Reveal solutionSolution

   Area $=\displaystyle\int_0^2 (x^2+1)\,dx-\int_1^2\sqrt{x-1}\,dx=\frac{14}{3}-\frac{2}{3}=4$ sq. units.

   The two curves are:

   • $y=x^2+1$ — an upward parabola, vertex $(0,1)$; at $x=2,\ y=5$.
   • $x=y^2+1$, i.e. $y=\sqrt{x-1}$ — a rightward parabola through $(1,0)$ and $(2,1)$.

   The region is bounded above by $y=x^2+1$, on the left by the $Y$-axis, on the right by $x=2$, below by the $X$-axis for $0\le x\le 1$ and by $y=\sqrt{x-1}$ for $1\le x\le 2$.

   Area under the top curve from $x=0$ to $2$:

   $\int_0^2 (x^2+1)\,dx=\left[\frac{x^3}{3}+x\right]_0^2=\frac{8}{3}+2=\frac{14}{3}.$

   Area under the lower boundary $y=\sqrt{x-1}$ from $x=1$ to $2$ (below the $X$-axis it contributes nothing since $y=0$ there):

   $\int_1^2\sqrt{x-1}\,dx=\left[\frac{2}{3}(x-1)^{3/2}\right]_1^2=\frac{2}{3}.$

   Required area:

   $\frac{14}{3}-\frac{2}{3}=\frac{12}{3}=4.$

   ✓Final answer

   Area $=4$ sq. units. Option (B) — $4$.

6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The area of the region enclosed by the curves $y^2 = 4(x+1)$ and $y^2 = 5(x-4)$ is (A) $\dfrac{280}{3}$ (B) $150$ (C) $140$ (D) $\dfrac{200}{3}$

   ›Reveal solutionSolution

   The parabolas meet at $y=\pm10$; integrating the horizontal width $5-\dfrac{y^2}{20}$ gives area $\dfrac{200}{3}$.

   Set up horizontal strips. Solving each curve for $x$:

   $x_1=\frac{y^2}{4}-1\quad(\text{from }y^2=4(x+1)),\qquad x_2=\frac{y^2}{5}+4\quad(\text{from }y^2=5(x-4)).$

   Intersection points. Setting $x_1=x_2$:

   $\frac{y^2}{4}-\frac{y^2}{5}=5\;\Longrightarrow\;\frac{y^2}{20}=5\;\Longrightarrow\;y^2=100\;\Longrightarrow\;y=\pm10,$

   with common $x=24$.

   Width of a strip. For $|y|<10$ the curve $x_2$ lies to the right of $x_1$:

   $x_2-x_1=\left(\frac{y^2}{5}+4\right)-\left(\frac{y^2}{4}-1\right)=5-\frac{y^2}{20}.$

   Integrate over $y\in[-10,10]$ (using symmetry):

   $A=2\int_0^{10}\left(5-\frac{y^2}{20}\right)dy=2\left[5y-\frac{y^3}{60}\right]_0^{10}=2\left(50-\frac{1000}{60}\right)=2\left(50-\frac{50}{3}\right)=\frac{200}{3}.$

   ✓Final answer

   The enclosed area is $\dfrac{200}{3}$ square units — option (D).

7. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The area of the region bounded by the curves $y = x^2 - 3x + 3$ and $y = 2x^2 - 1$ is (A) $\dfrac{403}{6}$ (B) $27$ (C) $19$ (D) $\dfrac{125}{6}$

   ›Reveal solutionSolution

   The region is bounded by two parabolas; find their intersection points, determine which curve is on top, then integrate the difference of the functions. The area is $\frac{125}{6}$, which corresponds to option (D).

   The problem asks for the area between two curves. The core idea is always the same: area between curves is the integral of (top function minus bottom function) over the interval where they cross. But the real work is in finding where they intersect and which one lies above the other.

   We have $y = x^2 - 3x + 3$ (call it $f(x)$) and $y = 2x^2 - 1$ (call it $g(x)$). Both are upward-opening parabolas. The first has vertex at $x = 1.5$, the second at $x = 0$. They will cross at two points, enclosing a finite region.

   1. Find the intersection points. Set $f(x) = g(x)$:

   $x^2 - 3x + 3 = 2x^2 - 1$

   Bring all terms to one side:

   $0 = 2x^2 - 1 - x^2 + 3x - 3 = x^2 + 3x - 4$

   So $x^2 + 3x - 4 = 0$. Factor:

   $(x + 4)(x - 1) = 0$

   Hence $x = -4$ and $x = 1$. These are the limits of integration.

   2. Determine which curve is on top in $[-4, 1]$. Pick a test point between the intersections, say $x = 0$:

   $f(0) = 0^2 - 3(0) + 3 = 3$

   $g(0) = 2(0)^2 - 1 = -1$

   Since $3 > -1$, $f(x)$ is above $g(x)$ throughout the interval. (Both are continuous parabolas crossing only at the endpoints, so the ordering cannot flip inside.)

   3. Set up the area integral. Area $A = \int_{x=-4}^{1} [f(x) - g(x)] \, dx$:

   $f(x) - g(x) = (x^2 - 3x + 3) - (2x^2 - 1) = -x^2 - 3x + 4$

   So

   $A = \int_{-4}^{1} (-x^2 - 3x + 4) \, dx$

   4. Integrate term by term.

   $\int (-x^2) \, dx = -\frac{x^3}{3}, \quad \int (-3x) \, dx = -\frac{3x^2}{2}, \quad \int 4 \, dx = 4x$

   Thus

   $A = \left[ -\frac{x^3}{3} - \frac{3x^2}{2} + 4x \right]_{-4}^{1}$

   5. Evaluate at the upper limit $x = 1$:

   $-\frac{1^3}{3} - \frac{3(1)^2}{2} + 4(1) = -\frac{1}{3} - \frac{3}{2} + 4$

   Convert to sixths: $-\frac{2}{6} - \frac{9}{6} + \frac{24}{6} = \frac{13}{6}$

   6. Evaluate at the lower limit $x = -4$:

   $-\frac{(-4)^3}{3} - \frac{3(-4)^2}{2} + 4(-4) = -\frac{-64}{3} - \frac{3(16)}{2} - 16$

   Simplify: $+\frac{64}{3} - \frac{48}{2} - 16 = \frac{64}{3} - 24 - 16 = \frac{64}{3} - 40$

   Convert to thirds: $\frac{64}{3} - \frac{120}{3} = -\frac{56}{3}$

   7. Subtract: upper minus lower.

   $A = \frac{13}{6} - \left( -\frac{56}{3} \right) = \frac{13}{6} + \frac{56}{3}$

   Write $\frac{56}{3}$ as $\frac{112}{6}$:

   $A = \frac{13}{6} + \frac{112}{6} = \frac{125}{6}$

   Watch out

   A common mistake is to forget that subtracting the lower-limit evaluation means subtracting a negative, which becomes addition. Also, be careful with signs when plugging $x = -4$ into $-x^3$ — it becomes $+64/3$, not $-64/3$.

   Tip

   If you ever doubt which function is on top, just test any $x$ between the intersections. The sign of $f(x) - g(x)$ tells you everything.

   ✓Final answer

   The area is $\boxed{\frac{125}{6}}$, which matches option (D).

8. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.The area of the triangle formed by the tangent and the normal drawn to the curve $y^2 = 4x$ at $(1, 2)$ with Y-axis is (in square units) (A) 4 (B) 3 (C) 2 (D) 1

   ›Reveal solutionSolution

   The area is found by computing the intercepts of the tangent and normal lines with the Y‑axis, then using the triangle’s base (distance between intercepts) and height (x‑coordinate of the point). The result is 3 square units.

   We are given the parabola $y^2 = 4x$ and the point $(1,2)$. The tangent and normal at this point, together with the Y‑axis, form a triangle. The Y‑axis is the line $x = 0$, so the triangle’s vertices are: the intersection of the tangent with the Y‑axis, the intersection of the normal with the Y‑axis, and the point of tangency $(1,2)$. The base of the triangle lies along the Y‑axis, and the height is the horizontal distance from the point to the Y‑axis (which is simply the x‑coordinate of the point).

   Why this approach works: Instead of finding all three vertices and using a determinant, we can note that the Y‑axis is vertical, so the triangle’s base is the segment between the two Y‑intercepts. The height is the perpendicular distance from the point to the Y‑axis, which is just $|x| = 1$. Then area = $\frac12 \times \text{base} \times \text{height}$.

   1. Find the equation of the tangent. For the parabola $y^2 = 4x$, the standard tangent at $(x_1, y_1)$ is $y y_1 = 2(x + x_1)$. Here $(x_1, y_1) = (1, 2)$, so:

   $y \cdot 2 = 2(x + 1) \quad\Rightarrow\quad 2y = 2x + 2 \quad\Rightarrow\quad y = x + 1.$

   The Y‑intercept (set $x = 0$) is $y = 1$. So the tangent meets the Y‑axis at $A(0, 1)$.

   2. Find the equation of the normal. The slope of the tangent is $1$ (from $y = x + 1$). The slope of the normal is the negative reciprocal: $-1$. Using point-slope form at $(1,2)$:

   $y - 2 = -1(x - 1) \quad\Rightarrow\quad y - 2 = -x + 1 \quad\Rightarrow\quad y = -x + 3.$

   The Y‑intercept (set $x = 0$) is $y = 3$. So the normal meets the Y‑axis at $B(0, 3)$.

   3. Determine the base of the triangle. The base is the segment $AB$ on the Y‑axis from $y = 1$ to $y = 3$. Its length is:

   $\text{base} = |3 - 1| = 2.$

   4. Determine the height.

      The height is the perpendicular distance from the point $(1,2)$ to the Y‑axis. Since the Y‑axis is the line $x = 0$, this distance is simply $|1| = 1$.

   5. Compute the area.

   $\text{Area} = \frac12 \times \text{base} \times \text{height} = \frac12 \times 2 \times 1 = 1.$

   Watch out

   A common mistake is to take the base as the distance between the two intercepts along the Y‑axis but forget that the triangle’s third vertex is $(1,2)$, not the origin. The height is not the y‑coordinate of the point; it’s the horizontal distance to the Y‑axis.

   Tip

   Whenever a triangle is formed by two lines through a given point and a coordinate axis, the area often simplifies to $\frac12 \times |\text{difference of intercepts}| \times |x\text{-coordinate of point}|$ if the axis is vertical, or $\frac12 \times |\text{difference of intercepts}| \times |y\text{-coordinate}|$ if the axis is horizontal.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If the points of intersection of the parabolas $y^2 = 5x$ and $x^2 = 5y$ lie on the line L, then the area of the triangle formed by the directrix of one parabola, latus rectum of another parabola and the line L is (A) $\dfrac{15}{32}$ (B) $\dfrac{12}{25}$ (C) $\dfrac{25}{8}$ (D) $\dfrac{25}{32}$

   ›Reveal solutionSolution

   The two parabolas meet on the line $L:y=x$. The directrix of one, the latus rectum of the other, and $L$ form a right triangle with both legs $\tfrac{5}{2}$, so its area is $\tfrac{25}{8}$.

   Step 1 — line $L$. Solving $y^2=5x$ and $x^2=5y$: substituting $x=\tfrac{y^2}{5}$ gives $y^4=125y$, so $y=0$ or $y=5$, i.e. intersections $(0,0)$ and $(5,5)$. Thus $L:y=x$.

   Step 2 — directrix and latus rectum. Each parabola has $4a=5$, so $a=\tfrac{5}{4}$.

   • For $y^2=5x$: directrix $x=-\tfrac{5}{4}$; latus rectum $x=\tfrac{5}{4}$.
   • For $x^2=5y$: directrix $y=-\tfrac{5}{4}$; latus rectum $y=\tfrac{5}{4}$.

   Take the directrix of the first, $x=-\tfrac{5}{4}$, and the latus rectum of the second, $y=\tfrac{5}{4}$ (any one-from-each choice gives the same area by symmetry).

   Step 3 — vertices.

   • $x=-\tfrac{5}{4}\cap y=x:\;\left(-\tfrac{5}{4},-\tfrac{5}{4}\right)$
   • $y=\tfrac{5}{4}\cap y=x:\;\left(\tfrac{5}{4},\tfrac{5}{4}\right)$
   • $x=-\tfrac{5}{4}\cap y=\tfrac{5}{4}:\;\left(-\tfrac{5}{4},\tfrac{5}{4}\right)$

   Step 4 — area. The right angle is at $\left(-\tfrac{5}{4},\tfrac{5}{4}\right)$; both legs have length $\tfrac{5}{4}-(-\tfrac{5}{4})=\tfrac{5}{2}$.

   $\text{Area}=\frac{1}{2}\cdot\frac{5}{2}\cdot\frac{5}{2}=\frac{25}{8}.$

   ✓Final answer

   The area is $\dfrac{25}{8}$ — option (C).

10. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
    Q.The area of the region bounded by the curve $y = |x-2| + |x-8|$, X-axis and the lines $x = 0$, $x = 10$ is (A) 50 (B) 68 (C) 100 (D) 98

    ›Reveal solutionSolution

    The area is found by splitting the absolute-value expression into linear pieces over intervals where the signs of the terms are constant, then integrating the resulting piecewise linear function from 0 to 10. The total area is 68, so the correct option is (B).

    Concept & Intuition

    The function $y = |x-2| + |x-8|$ is a sum of two absolute values. Each absolute value changes slope at its “corner” (where the inside is zero). So the whole function is piecewise linear, with possible corners at $x=2$ and $x=8$. Instead of integrating an absolute value directly, we break the domain $[0,10]$ into subintervals where the expressions inside the absolute values have constant sign. On each subinterval, the absolute values become simple linear expressions, and the area under the curve (above the X‑axis) is just the integral of that linear function. Since the curve is always non‑negative, the area is exactly the definite integral from 0 to 10.

    ---

    Step‑by‑step solution

    1. Identify the critical points The expressions inside the absolute values are zero at $x=2$ and $x=8$. These split the real line into three intervals:

    $(-\infty, 2),\quad [2,8),\quad [8,\infty).$

    Our integration limits are $x=0$ to $x=10$, so we consider the subintervals $[0,2]$, $[2,8]$, and $[8,10]$.

    2. Write the function without absolute values on each subinterval

       • On $[0,2]$: $x-2 \le 0$ so $|x-2| = -(x-2) = 2-x$. $x-8 < 0$ so $|x-8| = -(x-8) = 8-x$. Hence

    $y = (2-x) + (8-x) = 10 - 2x.$

    • On $[2,8]$: $x-2 \ge 0$ so $|x-2| = x-2$. $x-8 \le 0$ so $|x-8| = 8-x$. Hence

    $y = (x-2) + (8-x) = 6.$

    • On $[8,10]$: $x-2 > 0$ so $|x-2| = x-2$. $x-8 \ge 0$ so $|x-8| = x-8$. Hence

    $y = (x-2) + (x-8) = 2x - 10.$

    3. Integrate piecewise from 0 to 10

    $\text{Area} = \int_{0}^{2} (10 - 2x)\,dx \;+\; \int_{2}^{8} 6\,dx \;+\; \int_{8}^{10} (2x - 10)\,dx.$

    • First integral:

    $\int_{0}^{2} (10 - 2x)\,dx = \left[10x - x^2\right]_{0}^{2} = (20 - 4) - 0 = 16.$

    • Second integral:

    $\int_{2}^{8} 6\,dx = 6 \times (8-2) = 6 \times 6 = 36.$

    • Third integral:

    $\int_{8}^{10} (2x - 10)\,dx = \left[x^2 - 10x\right]_{8}^{10} = (100 - 100) - (64 - 80) = 0 - (-16) = 16.$

    4. Add the three areas

    $\text{Total area} = 16 + 36 + 16 = 68.$

    Tip

    Notice the symmetry: the function is V‑shaped on the ends and flat in the middle. The two end pieces each contribute 16, and the middle rectangle contributes 36. This symmetry can save time if you compute one end and double it.

    Watch out

    A common mistake is to forget that the absolute value changes sign at both corners, not just one. If you only split at $x=2$ and treat the rest as a single piece, you’ll get the wrong linear expression for $x>8$.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
    Q.Let $P\left(\dfrac{\pi}{4}\right), Q\left(\dfrac{5\pi}{4}\right), R\left(\dfrac{3\pi}{4}\right), T\left(\dfrac{7\pi}{4}\right)$ be the points on the hyperbola $x^2 - 4y^2 - 4 = 0$ in the parametric form. Then the area of the quadrilateral $PQRT$ is (in square units) (A) $4\sqrt{2}$ (B) $16\sqrt{2}$ (C) $32\sqrt{2}$ (D) $8\sqrt{2}$

    ›Reveal solutionSolution

    The quadrilateral is a rectangle whose sides are aligned with the asymptotes of the hyperbola. Its area is $8\sqrt{2}$ square units, option (D).

    The hyperbola is $x^2 - 4y^2 = 4$, which we rewrite in standard form:

    $\frac{x^2}{4} - \frac{y^2}{1} = 1.$

    Its standard parametric form is $x = 2\sec\theta$, $y = \tan\theta$. The given points use parameters $\frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}$ — these are spaced by $\frac{\pi}{2}$ in the parameter. That spacing is the key: on a rectangular hyperbola, equal parametric spacing often gives a rectangle, but here the hyperbola is not rectangular ($a \neq b$). Still, the geometry of the secant/tangent parametrization creates a special shape.

    Let’s find the coordinates.

    1. Compute the four points.

       For $\theta = \frac{\pi}{4}$:

       $x = 2\sec\frac{\pi}{4} = 2\cdot\sqrt{2} = 2\sqrt{2}$,

       $y = \tan\frac{\pi}{4} = 1$.

       So $P = (2\sqrt{2}, 1)$.

       For $\theta = \frac{5\pi}{4}$:

       $\sec\frac{5\pi}{4} = \sec(\pi + \frac{\pi}{4}) = -\sec\frac{\pi}{4} = -\sqrt{2}$,

       $\tan\frac{5\pi}{4} = \tan(\pi + \frac{\pi}{4}) = \tan\frac{\pi}{4} = 1$.

       So $Q = (-2\sqrt{2}, 1)$.

       For $\theta = \frac{3\pi}{4}$:

       $\sec\frac{3\pi}{4} = \sec(\pi - \frac{\pi}{4}) = -\sec\frac{\pi}{4} = -\sqrt{2}$,

       $\tan\frac{3\pi}{4} = \tan(\pi - \frac{\pi}{4}) = -\tan\frac{\pi}{4} = -1$.

       So $R = (-2\sqrt{2}, -1)$.

       For $\theta = \frac{7\pi}{4}$:

       $\sec\frac{7\pi}{4} = \sec(2\pi - \frac{\pi}{4}) = \sec\frac{\pi}{4} = \sqrt{2}$,

       $\tan\frac{7\pi}{4} = \tan(2\pi - \frac{\pi}{4}) = -\tan\frac{\pi}{4} = -1$.

       So $T = (2\sqrt{2}, -1)$.

    2. Identify the shape.

       The four points are:

    $P(2\sqrt{2}, 1),\quad Q(-2\sqrt{2}, 1),\quad R(-2\sqrt{2}, -1),\quad T(2\sqrt{2}, -1).$

    These are the vertices of a rectangle centered at the origin, with sides parallel to the axes.

    Horizontal side length: from $x = -2\sqrt{2}$ to $x = 2\sqrt{2}$ gives $4\sqrt{2}$.

    Vertical side length: from $y = -1$ to $y = 1$ gives $2$.

    3. Compute the area. Area of rectangle = length $\times$ width = $(4\sqrt{2}) \times 2 = 8\sqrt{2}$.

    Watch out

    A common mistake is to assume the quadrilateral is a parallelogram and use the determinant formula with the parametric form directly, but the points here are not in cyclic order as given — the order $P, Q, R, T$ goes around the rectangle correctly, so the simple rectangle area works.

    Tip

    Notice that the parameters differ by $\frac{\pi}{2}$, which on the standard parametrization $x = a\sec\theta$, $y = b\tan\theta$ always gives points symmetric about both axes — forming a rectangle centered at the origin.

    ✓Final answer

    The area of quadrilateral $PQRT$ is $\boxed{8\sqrt{2}}$ square units, which corresponds to option (D).

12. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.The area of a triangle (in sq. units) formed by the latus rectum of the parabola $x^2 = 16y$ and the lines joining the vertex of the parabola to the ends of the latus rectum is (A) 24 (B) 28 (C) 32 (D) 64

    ›Reveal solutionSolution

    The problem asks for the area of a triangle formed by the vertex of the parabola $x^2 = 16y$ and the endpoints of its latus rectum. By identifying the parabola's properties, finding the coordinates of the vertex and the latus rectum's endpoints, and then using the formula for the area of a triangle, we find the area to be $\boxed{32}$ square units.

    The core idea here is to first understand the geometry of a parabola and its key features: the vertex and the latus rectum. Once these geometric elements are identified and their coordinates determined, calculating the area of the triangle becomes a straightforward application of the area formula.

    The given parabola is $x^2 = 16y$. This equation is in the standard form $x^2 = 4ay$, which describes a parabola opening upwards with its vertex at the origin $(0,0)$.

    The latus rectum of a parabola is a line segment that passes through the focus, is perpendicular to the axis of symmetry, and has its endpoints on the parabola. For a parabola of the form $x^2 = 4ay$, the focus is at $(0, a)$, and the latus rectum is the horizontal line segment $y=a$ with length $4a$.

    The triangle in question has three vertices:

    1. The vertex of the parabola.
    2. One end of the latus rectum.
    3. The other end of the latus rectum.

    We can find the coordinates of these three points and then use the standard formula for the area of a triangle given its vertices. Alternatively, recognizing that the base of this triangle (the latus rectum) is horizontal and the vertex of the parabola is on the y-axis, we can use the simple formula $\frac{1}{2} \times \text{base} \times \text{height}$.

    1. Identify the properties of the parabola:

       The given equation of the parabola is $x^2 = 16y$.

       Comparing this with the standard form $x^2 = 4ay$, we can determine the value of $a$.

       $4a = 16$

       $a = \frac{16}{4}$

       $a = 4$

       For a parabola of the form $x^2 = 4ay$:

       • The vertex is at the origin, $V = (0, 0)$.
       • The focus is at $(0, a)$, which is $F = (0, 4)$.
       • The axis of the parabola is the y-axis.

    2. Determine the endpoints of the latus rectum:

       The latus rectum is a line segment passing through the focus $(0, 4)$ and perpendicular to the axis of the parabola (the y-axis). Therefore, the equation of the line containing the latus rectum is $y = 4$.

       To find the endpoints of the latus rectum, substitute $y=4$ into the parabola's equation:

       $x^2 = 16(4)$

       $x^2 = 64$

       Taking the square root of both sides:

       $x = \pm\sqrt{64}$

       $x = \pm 8$

       So, the endpoints of the latus rectum are $L_1 = (-8, 4)$ and $L_2 = (8, 4)$.

       Tip

       The length of the latus rectum for $x^2=4ay$ is $4a$. In this case, $4a = 16$. The endpoints are $(\pm 2a, a)$. Here, $2a = 2(4) = 8$, so the endpoints are $(\pm 8, 4)$, which matches our calculation.

    3. Identify the vertices of the triangle:

       The triangle is formed by the vertex of the parabola and the ends of the latus rectum.

       The vertices are:

       • $V = (0, 0)$
       • $L_1 = (-8, 4)$
       • $L_2 = (8, 4)$

    4. Calculate the area of the triangle:

       We can use the formula for the area of a triangle given its vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

    $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

    Let $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (-8, 4)$, and $(x_3, y_3) = (8, 4)$.
    

    $\text{Area} = \frac{1}{2} |0(4 - 4) + (-8)(4 - 0) + 8(0 - 4)|$

    $\text{Area} = \frac{1}{2} |0(0) - 8(4) + 8(-4)|$

    $\text{Area} = \frac{1}{2} |0 - 32 - 32|$

    $\text{Area} = \frac{1}{2} |-64|$

    $\text{Area} = \frac{1}{2} \times 64$

    $\text{Area} = 32 \text{ square units}$

    Alternatively, we can use the base-height formula.
    The base of the triangle can be taken as the segment $L_1L_2$. Since $L_1 = (-8, 4)$ and $L_2 = (8, 4)$, the base is a horizontal segment.
    Length of base $L_1L_2 = |8 - (-8)| = |8 + 8| = 16$ units.
    The height of the triangle is the perpendicular distance from the vertex $V(0, 0)$ to the line containing the base $L_1L_2$. The line containing $L_1L_2$ is $y=4$.
    The perpendicular distance from $(0, 0)$ to $y=4$ is $|4 - 0| = 4$ units.
    

    $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

    $\text{Area} = \frac{1}{2} \times 16 \times 4$

    $\text{Area} = 8 \times 4$

    $\text{Area} = 32 \text{ square units}$

    Both methods yield the same result.
    

    ✓Final answer

    The area of the triangle is $\boxed{32}$ square units.