Q.OR Find the domain of cos−1(3𝑥 − 2) 2 22 If 𝑦 = log tan ( 𝜋 4 + 𝑥 2), then prove that 𝒅𝒚 𝒅𝒙 − 𝐬𝐞𝐜 𝒙 = 𝟎 2 23A 23B Find: ∫ (𝑥−3) (𝑥−1)3 𝑒𝑥 𝑑𝑥 OR Find out the area of shaded region in the enclosed figure. 2 23 B For Visually Impaired: Find out the area of the region enclosed by the curve 𝑦2 = 𝑥 , 𝑥 = 3 and 𝑥-axis in the first quadrant.

Showing the 12 most recent of 77 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.The area bounded by the curve $y = x|x|$, $x$-axis and the ordinates $x = -1$ and $x = 1$ is given by (A) $0$ (B) $\frac{1}{3}$ (C) $\frac{2}{3}$ (D) $3$

   ›Reveal solutionSolution

   The curve $y = x|x|$ is an odd function, so the signed area cancels to zero, but the bounded area (absolute area) is the sum of two equal positive lobes, giving $\frac{2}{3}$.

   The key here is to understand what $y = x|x|$ actually looks like. The absolute value on $x$ splits the definition into two cases:

   • When $x \ge 0$, $|x| = x$, so $y = x \cdot x = x^2$.
   • When $x < 0$, $|x| = -x$, so $y = x \cdot (-x) = -x^2$.

   So the curve is a parabola opening upward on the right side, and a parabola opening downward on the left side. It is an odd function: $f(-x) = -f(x)$. This symmetry is the heart of the problem.

   The question asks for the area bounded by the curve, the $x$-axis, and the vertical lines $x = -1$ and $x = 1$. "Area bounded" means geometric area — always positive — not signed area (the integral). This is a classic trap.

   Let’s work through it.

   1. Set up the absolute area integral. The geometric area between a curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$ is $\int_a^b |f(x)| \, dx$. Here:

   $\text{Area} = \int_{-1}^{1} |x|x|| \, dx.$

   2. Simplify $|x|x||$. Since $|x|x|| = |x| \cdot |x| = |x|^2 = x^2$ (because squaring removes the sign), we have:

   $|x|x|| = x^2 \quad \text{for all real } x.$

   That’s a neat simplification: the absolute value of the function is just $x^2$, a simple upward parabola.

   3. Compute the integral.

   $\text{Area} = \int_{-1}^{1} x^2 \, dx.$

   The antiderivative of $x^2$ is $\frac{x^3}{3}$. So:

   $\left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}.$

   Watch out

   A common mistake is to compute the signed integral $\int_{-1}^{1} x|x| \, dx$ directly. Since the function is odd, that integral is $0$ — which is option (A). But the question asks for area, not signed area. The area is always positive.

   Tip

   Whenever you see $|x|$ inside a function, split the domain at $x=0$ and handle each piece separately. For area problems, taking the absolute value of the whole function first often simplifies things — here it turned $x|x|$ into plain $x^2$.

   ✓Final answer

   The area bounded is $\boxed{\frac{2}{3}}$, which corresponds to option (C).

2. CBSE 2026Set 65/1/11 markMCQ
   Q.If $2\cos^{-1}x = y$, then (A) $0 \le y \le \pi$ (B) $-\pi \le y \le \pi$ (C) $0 \le y \le 2\pi$ (D) $-\pi \le y \le 0$

   ›Reveal solutionSolution

   The range of $\cos^{-1}x$ is $[0,\pi]$, so multiplying by 2 gives $y = 2\cos^{-1}x$ a range of $[0,2\pi]$. The correct option is (C).

   Concept and Intuition

   The key to this problem lies entirely in understanding the range of the inverse cosine function. $\cos^{-1}x$ (also written as $\arccos x$) is defined as the angle whose cosine is $x$, and by convention, that angle is always taken from the interval $[0, \pi]$. This is not arbitrary — it's the standard principal value branch that makes the function one-to-one and therefore invertible.

   Once you know that $\cos^{-1}x$ lives between $0$ and $\pi$ (inclusive), finding the range of $y = 2\cos^{-1}x$ is simply a matter of scaling that interval by a factor of 2. No tricky domain restrictions, no sign flips — just multiplication.

   Watch out

   A common mistake is to confuse the range of $\cos^{-1}x$ with that of $\sin^{-1}x$ (which is $[-\pi/2, \pi/2]$). Always recall: $\cos^{-1}x \in [0,\pi]$, not $[-\pi/2,\pi/2]$.

   Step-by-step solution

   1. Recall the range of $\cos^{-1}x$ The inverse cosine function $\cos^{-1}: [-1,1] \to [0,\pi]$ gives an output angle in radians. This means:

   $0 \le \cos^{-1}x \le \pi \quad \text{for all } x \in [-1,1].$

   2. Multiply the inequality by 2 Since 2 is positive, multiplying through preserves the direction of the inequalities:

   $2 \cdot 0 \le 2\cos^{-1}x \le 2 \cdot \pi$

   which simplifies to:

   $0 \le y \le 2\pi.$

   3. Check if every value in $[0,2\pi]$ is actually attained

      As $x$ varies continuously from $-1$ to $1$, $\cos^{-1}x$ varies continuously from $\pi$ down to $0$. So $y = 2\cos^{-1}x$ varies continuously from $2\pi$ down to $0$, covering every number in between. The range is exactly the closed interval $[0,2\pi]$.

   4. Match with the given options

      • (A) $0 \le y \le \pi$ — too narrow, misses values between $\pi$ and $2\pi$.
      • (B) $-\pi \le y \le \pi$ — includes negative values, which are impossible since $\cos^{-1}x \ge 0$.
      • (C) $0 \le y \le 2\pi$ — exactly matches our result.
      • (D) $-\pi \le y \le 0$ — entirely negative, completely wrong.

   Tip

   You can also think geometrically: $\cos^{-1}x$ is the angle in the upper half of the unit circle (from $0$ to $\pi$ radians). Doubling that angle sweeps out the full circle's worth of angles — from $0$ all the way around to $2\pi$.

   ✓Final answer

   The correct option is (C), since $y = 2\cos^{-1}x$ lies in $[0, 2\pi]$.

3. CBSE 2026Set 65/2/11 markMCQ
   Q.Which of the following expressions will give the area of region bounded by the curve $y = x^2$ and line $y = 16$? (A) $\int_{0}^{4} x^2\, dx$ (B) $2 \int_{0}^{4} x^2\, dx$ (C) $\int_{0}^{16} \sqrt{y}\, dy$ (D) $2 \int_{0}^{16} \sqrt{y}\, dy$

   ›Reveal solutionSolution

   The region between $y = x^2$ and $y = 16$ is symmetric about the $y$-axis; integrating horizontally from $y = 0$ to $y = 16$ with $x = \sqrt{y}$ and doubling for both sides gives $2 \int_{0}^{16} \sqrt{y}\, dy$.

   The parabola $y = x^2$ opens upward with vertex at the origin, and the horizontal line $y = 16$ cuts it at two points. Finding those intersection points: $x^2 = 16$ gives $x = \pm 4$. So the bounded region sits between $x = -4$ and $x = 4$, below the line and above the parabola.

   The key decision is whether to integrate with respect to $x$ (vertical slices) or $y$ (horizontal slices). Both are valid, but the setup differs.

   Vertical slices (integrating with respect to $x$):

   At any $x$ between $-4$ and $4$, a vertical strip runs from the parabola $y = x^2$ up to the line $y = 16$. The height of that strip is $16 - x^2$. The area is

   $A = \int_{-4}^{4} (16 - x^2)\, dx.$

   Because the integrand $16 - x^2$ is even (symmetric about $x = 0$), this equals

   $A = 2 \int_{0}^{4} (16 - x^2)\, dx = 2 \int_{0}^{4} 16\, dx - 2 \int_{0}^{4} x^2\, dx.$

   Notice that $\int_{0}^{4} x^2\, dx$ alone is not the area; it gives the area under the parabola from $0$ to $4$, not the region between the parabola and the line. So option (A) is incorrect, and option (B) is also incorrect (it's twice the area under the parabola, not the region we want).

   Horizontal slices (integrating with respect to $y$):

   At any height $y$ between $0$ and $16$, a horizontal strip extends from the left branch of the parabola to the right branch. Solving $y = x^2$ for $x$ gives $x = \pm \sqrt{y}$. The width of the strip is

   $\sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}.$

   The area is then

   $A = \int_{0}^{16} 2\sqrt{y}\, dy = 2 \int_{0}^{16} \sqrt{y}\, dy.$

   This matches option (D).

   Tip

   When a region is symmetric about an axis, integrating along that axis (here, the $y$-axis) often simplifies the setup: you capture both halves at once by doubling the contribution from one side.

   Let's verify the options:

   1. (A) $\int_{0}^{4} x^2\, dx$ computes the area under the parabola from $x = 0$ to $x = 4$, not the region between the parabola and the line.

   2. (B) $2 \int_{0}^{4} x^2\, dx$ doubles that, giving the area under the parabola from $x = -4$ to $x = 4$. Still not what we want.

   3. (C) $\int_{0}^{16} \sqrt{y}\, dy$ gives the area under the right branch of the parabola (from $x = 0$ to $x = 4$) when viewed as a function $x = \sqrt{y}$. This is only half the region.

   4. (D) $2 \int_{0}^{16} \sqrt{y}\, dy$ accounts for both branches, giving the full area between the parabola and the line.

   ✓Final answer

   The correct option is (D) $2 \int_{0}^{16} \sqrt{y}\, dy$.

4. CBSE 2026Set V11 markMCQ
   Q.The domain of $\tan^{-1} x$ is

   (a) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$

   (b) $(0, \pi)$

   (c) $[-1, 1]$

   (d) $(-\infty, \infty)$

   ›Reveal solutionSolution

   The tangent function maps $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ onto all of $\mathbb{R}$, so $\tan^{-1}x$ accepts every real $x$; answer (d).

   The principal-branch tangent $\tan:\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\to\mathbb{R}$ is a bijection onto $\mathbb{R}$. Its inverse $\tan^{-1}$ therefore has domain equal to the range of $\tan$, namely all real numbers.

   $\text{Domain}(\tan^{-1}x)=(-\infty,\infty),\qquad \text{Range}=\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right).$

   ✓Final answer

   (d) $(-\infty,\infty)$

5. CBSE 2026Set CX1 mark
   Q.Find the value of $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$.

   ›Reveal solutionSolution

   $\tan^{-1}\sqrt3=\tfrac{\pi}{3}$, $\sec^{-1}(-2)=\tfrac{2\pi}{3}$, giving $-\tfrac{\pi}{3}$.

   Concept: Use the principal-value ranges: $\tan^{-1}\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and $\sec^{-1}\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$.

   $\tan^{-1}\sqrt3=\dfrac{\pi}{3}\qquad(\tan\tfrac{\pi}{3}=\sqrt3).$

   For $\sec^{-1}(-2)$ we need $\theta\in[0,\pi]$ with $\sec\theta=-2$, i.e. $\cos\theta=-\tfrac12$, giving $\theta=\dfrac{2\pi}{3}$.

   $\tan^{-1}\sqrt3-\sec^{-1}(-2)=\dfrac{\pi}{3}-\dfrac{2\pi}{3}=-\dfrac{\pi}{3}.$

   ✓Final answer

   $\tan^{-1}\sqrt3-\sec^{-1}(-2)=-\dfrac{\pi}{3}$.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is

   (a) $2$

   (b) $4/9$

   (c) $9/4$

   (d) $9/2$

   ›Reveal solutionSolution

   Integrate $x$ as a function of $y$ (since the boundary is the $y$-axis and the line $y=3$) using $x=y^2/4$ from the parabola.

   From $y^2=4x$, $x=\dfrac{y^2}{4}$.

   Area $=\displaystyle\int_0^3 x\,dy = \int_0^3 \dfrac{y^2}{4}dy = \left[\dfrac{y^3}{12}\right]_0^3 = \dfrac{27}{12}=\dfrac{9}{4}$.

   ✓Final answer

   The correct option is (c) $9/4$.

7. CBSE 2026Set ANNUAL1 mark
   Q.$\sin^{-1}x$ is a function whose domain is __________.

   ›Reveal solutionSolution

   $\sin^{-1}x$ is defined only where $\sin\theta=x$ has a solution, i.e. for $x\in[-1,1]$.

   The sine function takes values only in $[-1,1]$, so its inverse $\sin^{-1}x$ can only accept inputs in that range.

   ✓Final answer

   The domain of $\sin^{-1}x$ is $[-1,1]$.

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $y = \cos^{-1} x$ then

   (a) $0 \le y \le \pi$

   (b) $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$

   (c) $-\pi \le y \le \pi$

   (d) None of these

   ›Reveal solutionSolution

   $\cos^{-1}x$ is defined so that its principal value always lies in $[0,\pi]$.

   The function $\cos x$ is one-one and onto from $[0,\pi]$ to $[-1,1]$, so its inverse $\cos^{-1}x$ is defined on domain $[-1,1]$ with range (principal value branch) $[0,\pi]$. Thus if $y=\cos^{-1}x$, then $0 \le y \le \pi$.

   ✓Final answer

   (a) $0 \le y \le \pi$.

9. CBSE 2026Set ANNUAL1 markMCQ
   Q.Principal value of $\tan^{-1}(-1)$ is

   (a) $\dfrac{\pi}{4}$

   (b) $-\dfrac{\pi}{4}$

   (c) $\dfrac{3\pi}{4}$

   (d) None of these

   ›Reveal solutionSolution

   The principal value of $\tan^{-1}x$ always lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

   We need $y$ such that $\tan y = -1$ and $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.

   Since $\tan\left(\dfrac{\pi}{4}\right) = 1$, we get $\tan\left(-\dfrac{\pi}{4}\right) = -1$, and $-\dfrac{\pi}{4}$ lies within the principal branch.

   So $\tan^{-1}(-1) = -\dfrac{\pi}{4}$.

   ✓Final answer

   (b) $-\dfrac{\pi}{4}$.

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.The principal value of $\cos^{-1} x$ is:

    (a) $[0, \pi]$

    (b) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

    (c) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

    (d) None of these

    ›Reveal solutionSolution

    The principal value branch of $\cos^{-1}x$ is $[0,\pi]$ by definition.

    The function $\cos:[0,\pi]\to[-1,1]$ is a bijection, so its inverse $\cos^{-1}:[-1,1]\to[0,\pi]$ is defined with range (principal value branch) $[0,\pi]$.

    ✓Final answer

    Option (a): $[0,\pi]$.

11. CBSE 2026Set ANNUAL1 markMCQ
    Q.Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is:

    (a) 2

    (b) $\dfrac{9}{4}$

    (c) $\dfrac{9}{3}$

    (d) $\dfrac{9}{2}$

    ›Reveal solutionSolution

    Integrate $x=\dfrac{y^2}{4}$ with respect to $y$ from $0$ to $3$, since the region is bounded by the $y$-axis.

    For $y^2=4x$, we have $x=\dfrac{y^2}{4}$.

    The area bounded by the curve, the $y$-axis, and $y=3$ (from $y=0$ to $y=3$) is

    $A=\displaystyle\int_0^3 x\,dy=\int_0^3\dfrac{y^2}{4}\,dy=\left[\dfrac{y^3}{12}\right]_0^3=\dfrac{27}{12}=\dfrac{9}{4}$

    ✓Final answer

    Option (b): $\dfrac{9}{4}$ square units

12. CBSE 2026Set ANNUAL1 markMCQ
    Q.Principal value of cos⁻¹(1/2) is:

    (a) π/2

    (b) π/3

    (c) π/4

    (d) π/6

    ›Reveal solutionSolution

    The principal value of $\cos^{-1}x$ lies in $[0,\pi]$, and $\cos\left(\frac{\pi}{3}\right) = \frac12$.

    We need $\theta \in [0,\pi]$ such that $\cos\theta = \dfrac12$.

    Since $\cos\left(\dfrac{\pi}{3}\right) = \dfrac12$ and $\dfrac{\pi}{3} \in [0,\pi]$, this is the principal value.

    ✓Final answer

    $\cos^{-1}\left(\dfrac12\right) = \dfrac{\pi}{3}$ (option b).