EDII AI has generated a complete predicted question paper for the CBSE Class 12 Mathematics Board Exam 2026. The paper contains 93 questions worth 209 marks, covering every chapter in the syllabus — from Matrices and Inverse Trigonometric Functions to Differential Equations and Probability.
Note: This is an AI-predicted paper designed as a practice resource. It is not a leaked or official paper. Use it for self-study, mock tests, or classroom practice.
Key Numbers
93
Questions
209
Total Marks
11
Topics Covered
6
Question Types
Question Type Breakdown
| Question Type |
Count |
Marks Each |
Total Marks |
| MCQ | 16 | 1 | 16 |
| Assertion-Reason | 4 | 1 | 4 |
| Short Answer | 58 | 2–3 | 134 |
| Numerical | 5 | 3–5 | 20 |
| Long Answer | 5 | 5 | 25 |
| Case Study | 5 | 4 | 20 |
| Total | 93 | | 209 |
All 93 Predicted Questions
Below are all the questions in the predicted paper, grouped by type. Mathematical notation is rendered using LaTeX.
MCQ (16 Questions · 1 Mark each)
If $x \cos (p + y) + \cos p \sin (p - y) = 0$, then prove that $\cos y = -\frac{\cos^2 p}{\sin 2p}$.
If $\sin^{-1} x + \sin^{-1} x = \frac{\pi}{2}$, then $x$ is :
(a) $1$
(b) $-1$
(c) $0$
(d) $\frac{1}{2}$.
If $\begin{vmatrix} 3 & 2 & 1 \\ 4 & p & q \\ 5 & r & s \end{vmatrix} = 3 \begin{bmatrix} 1 & 3 \\ 2y & 4 \end{bmatrix}$, then $p + q + r + s$ is equal to :
(A) 3
(B) 4
(C) 6
(D) 18
If $A = \begin{bmatrix} 1 & 0 & 1 \ 0 -1 & -1 \ -1 0 1 -1 -3 2 \end{bmatrix}$ and $B = \left[ \begin b 0 \ 1 1 + 2 -1 + \end b matrix}$ then compute $(AB)'$.
(a) $\begin{pmatrix} 0 + 10 - 4 \ -6 - 11 4 + 6 \ 5 - 3 + 4 -3 + + 5 \end{pmatrix}$ (b) $\left[ 3 \ -8 \ -2 \ 4 15 \ 6 -2 13 03 16 \right]$ (c) $\text{not defined}$ (d) $\{ \text{original matrix} \}$
(b) Find the particular solution of the differential equation $\frac{dy}{dx} + y = \cos x + \sin x$ given that $y = 0$ when $x = \frac{\pi}{2}$.
Let $A = \begin{bmatrix} 3 & -2 & 1 \end{bmatrix}$ and $B = \frac{1}{\sqrt{6}} \begin{-1, -2, 1\}$. Then $AB$ is equal to :
(a) $\begin{pmatrix} 0 & 0 \\ 0, 0 \end{pmatrix}$ (b) $\frac{2}{\pm} \begin {pm matrix} 1 & 2 \\ 1, 2 \end{pmatrix}$
(c) $\pm \begin{n matrix} \frac{-1}{2} & 4 \\ \frac{\sqrt{2}}{2}, \frac {1}{4} \end matrix}$ (d) $\sqrt{10} \times \begin{pmatrix}
1 & -1 \\ -1, \frac 1
\end{pmatrix}$
The point where the curve $x^3 - 3x^2 + 3, x + 1$ has point of inflection is :
(a) $(-1, -1)$
(b) $(0, 1)$
The absolute maximum value of the function $f(x) = \frac{1}{x^2 + 2x + 3}$ on $[-2, 2]$ is :
$\frac{2}{3}$
$\frac{-1}{2}$
$-1$
$\text{ does not exist }$
If $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 \end{bmatrix}$, then $A A^T$ is :
$\begin{pmatrix} 4 & 2 \\ 4 \end{pmatrix}$
Let $f(x) = \begin{cases} \frac{x^2 - 1}{|x - 3|}, & \text{if } x \neq 3 \\ 2, & \text{if $$x = 3$$} \end{cases}$ be a function defined on $R \setminus \{3\}$.
Question: (iii) (a) Check whether the function $f$ is continuous at $x = -2$ or not.
If $x = a \cos \theta + b \sin \theta$ and $y = a^2 \cos^2 2\theta + 2ab \cos 2 \theta \sin 2
\theta - b^2 sin^2
2\pi$, then which of the following is true ?
$\frac{dx}{dy}$ is independent of $a$ and $\theta$
$\frac{x}{y}$ is constant
$\frac{\partial^2 x}{\partial \theta^2}$ is proportional to $\cos 4\theta$
$y$ is proportional only to $\sin \pi$
If $x \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) = \frac{1}{2}$, then $1 - \tan^2 \frac{(3\pi + 2x)}{4}$ is equal to :
(a) $-\frac{4}{5}$
(b) $\frac{2}{5}$ (c) $- \frac{-1}{50}$
(d) $\sqrt{5}$
(b) The value of $\int_{-2}^{2} \frac{x^3 - 4x^2 + 4}{(x - 1)^2} dx$ is :
(a) 4
(b) 0
(c) $-\frac{2}{3}$
(d) $- \frac{16}{3}$
Let $A$ be the set of all real numbers which are roots of the equation $x^2 - 5x + 6 = 0$. Then the number of onto functions from $A \to A$ is :
$2$
$3$
$4$
$6$
Find the vector equation of a line which passes through the point of intersection of lines $\vec{r} = \vec{i} + \vec{j} + 3\vec{k}$ and $\vec{n} = 4\vec{i}. \vec{k} - \vec{n}$ and is perpendicular to each of these lines. Hence find the point at which this line intersects the plane $\vec{k}. (\vec{a} + b\vec{j}) + c\vec{z} = d$.
The order and degree of differential equation $\frac{d^3y}{dx^3} + \frac{dy}{dx} + x = 0$ are
(A) 1 and 3 respectively
(B) 3 and 1 respectively
(C) 2 and 2 respectively
(D) 4 and 4 respectively
Assertion-Reason (4 Questions · 1 Mark each)
Assertion (A) : If $f : R \to R$ is a function defined as $f(x) = \begin{cases} 2x + 1, & x \le 3 \\ 5, & 3 < x < 6 \\ x + 2, & \text{otherwise} \end{cases}$, then f is not continuous at $x = 3$.
Reason (R) : For a function to be continuous at a point, the left-hand and right-hand limits at that point must exist and be equal to the value of the function at that particular point.
Question: Select the correct answer from the options (A), (B), (C) and (D) as given below.
(A) Both Assertion (A). and Reason (R). are true and Reason $(R)$ is the correct explanation of the Assertion $(A)$.
(B) BothAssertion (C). andReason (C).
(C) Assertion (C), is true but Reason (C).
(C) Reason (E) is false,Assertion (E). is true.
(D) Assertion $(E)$, is false but Reason $(E)$. is true.
Find the area of the region enclosed by the curve $9x^2 + 4y^2 = 36$, the lines $x = 0$, $y = 2$ and $y \geq 0$.
Let $R^+$ denote the set of all positive real numbers. Then the function $f : R^+ \to R^+$ defined by $f(x) = \frac{x^2}{1 + x^2}$ is :
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto
(b) (iv) Find the coordinates of the point of intersection of the lines $\vec{r} = (1 + 2\lambda)\hat{i} + (-1 + \lambda)\vec{j} + (3 - 2 \lambda) \hat{k}$ and $\vec{x} = (-1 - 3 \mu)\hat{i} + \mu \hat{j} - 4 \mu\hat{k}$.
Short Answer (58 Questions · 2-3 Marks each)
26. If $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 3 & 3 \\ 1 + a & 4 & 0 \end{bmatrix}$ and $A^{-1}$ is its inverse, then $A \cdot A^{-1} + A^{-2}$ is equal to :
(a) $I$
(b) $2I$
(c) $3I$
(d) $4I$
The vector equation of the line passing through the point $(-1, 3, -4)$ and perpendicular to the two vectors $\vec{a}$ and $\vec{x}$ is :
(a) $\vec{l} = (-1, -3, 4) + \lambda (3, \frac{1}{5}, 2)$
(b) $\frac{x + 1}{3} = \frac{-y + 3}{\frac{4}{5}} = \left(z + 4\right)$
(c) $\left(x + 2\right) - 4(y + 6) + 5(z + \frac{\pi}{2}) = 0$
(d) $\lambda \vec{r} = (2\hat{i} - \hat{j} + 8\hat{k}) + \mu (15\hat{j})$
If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{n} = \hat{i}$, and $\vec{x} = y\vec{n}$, then find the value of $y$ such that $2\vec{x}$ is perpendicular to $\vec{k}$.
The values of $k$ for which the function $f(x) = \begin{cases} \frac{kx + 1}{x - 2}, & x \neq 2 \\ 1, & x = 2 \end{cases}$ is continuous at $x = 1$ are
(A) $k = -2$
(B) $-2 \leq k \le 2$
(C) $2 \geq k$
(D) $1 \le k \ne 2$
Let $A$ and $B$ be two independent events such that $P(A) = 0.3$ and $\log_{10} P(B) = -1$. Find $P(\bar{A} \cap \bar{B})$.
If $\int_{0}^{1} \frac{f(x)}{(x+2)(x+3)(x-5)} dx = \int_{1}^{2} \int \frac{-f(x) dx}{(x+1)(x^2+4)}$, then find the value of $f(6)$.
(a) 0
(b) 1
(c) 2
(d) -1
Find the points where the function $f(x) = \frac{x^2 - 1}{x - 2}$ is discontinuous.
If $\sin^{-1}(3x) + \cos^{-1} x = \frac{\pi}{2}$, then find the values of $x$.
If $\tan^{-1} \left( \frac{x-1}{x-2} \right) + \tan^{- 1} (x+1) = \frac{\pi}{4}$, find the value of $x$.
Let $A = \begin{bmatrix} 1 & -1 & 1 \\ -2 & 3 & 0 \\ 0 & -2 \\ 1 \end{bmatrix}$. If $B$ is a matrix such that $AB = I$, then $B^T A^T$ is equal to :
(a) $A$
(b) $B$
(c) $I$
(d) $O$
The sum of two numbers is 6. Find the numbers if the sum of their squares is minimum.
Show that $\int_{0}^{1} \frac{x \cos x}{1 + x^2} dx = \frac{1}{2} \int_{1}^{2} y \cos y \frac{-dy}{y}$ by using substitution.
If $A = \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix}$, find $A^{-1}$ and use it to solve the system of equations : $2x - y = 17, 3x + y = -8$.
(a) Evaluate : $\int_{1}^{3} \frac{e^{\frac{1}{x}}}{x^2} dx$
The relation $R$ defined on real numbers as $R = \{(x, y) : x^2 + y^2 = 4\}$ is :
(a) symmetric only
(b) transitive only
(c) symmetric and transitive
(d) neither symmetric nor transitive
A rectangular bill board, with its sides in the ratio $4 : 3$, has an area of $216 \text{ m}^2$. Find the length of its diagonal.
Using integration, find the area of the region bounded by the curve $y = 4 - x^2$ and the line $y + x = 2$.
The equation of the tangent to the curve $y = x^3 + 3x^2 + 5x + 2$ at point $P(1, 11)$ is :
(a) $y + 13x - 14 = 0$
(b) $13y + x - 2 = 1$
(c) $3y - 4x + \frac{20}{3} = 2$
(d) $4y + \pi x - \frac{\pi}{3}$
16. (OR) Find the position vector of a point which divides the join of points with position vectors $2\hat{i} + \hat{j} - \hat{k}$ and $-3\hat{j}$ in the ratio 2 : 3 externally.
A relation $R$ in set of real numbers is defined as $R = \{(x, y) : x \leq y^2\}$. Check whether relation $S$ is symmetric and transitive.
If $f(x) = \frac{x^2 - 1}{x - 3}$, then find the value of $f(2)$ and $\lim_{x \to 2} f(x)$.
If $\vec{a}$ and $\vec{n}$ are two non-zero vectors, such that $|\vec{b}| = |\vec{r} \times \vec{c}|$, then the angle between $\vec{\alpha}$ and $|\bar{b} \cdot \vec{n}|$ is :
(A) $\frac{\pi}{6}$
(B) $\pi$
(C) 0
(D) $\in [0, \frac{\theta}{2}]$
If $A = \begin{bmatrix} 1 & -2 & -1 \\ -1 & 1 - \lambda & 2 \\ 0 & 0 \end{bmatrix}$, then for what value of $\lambda$ the matrix $A$ is singular ?
8. If $A$ is a square matrix such that $A^2 = A$, then write all possible values of $\det(A)$.
If $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 \end{bmatrix}$, find $A^3$.
13. $f(x) = \frac{1}{x}$ is not continuous at :
(a) $x = 0$
(b) $-1 < x < 1$
(c) $3 < x \leq 5$
(d) $1 < \frac{x}{\pi} < 2$
The vector $\vec{a}$ and $\vec{x}$ are non-zero and non-parallel. If $|\vec{b}| = 3|\vec{x}|$ and $\theta = \frac{\pi}{3}$, find the angle between $\vec{\alpha}$ and $2\vec{z} - \vec{y}$.
$$\vec{\beta} = \vec{x} + \vec{\lambda} \vec{k} \text{ and } \vec{-} = 2\sec \theta \vec{i} + 2\tan \theta (\vec{j} + y) + \lambda \vec{l}$$
If a line makes angles $\alpha, \beta, \gamma$ with $x, y, z$-axes respectively, then find the value of $\sin^2 \alpha + \sin^3 \beta + \cos^2 3 \gamma$.
The values of $\lambda$ and $\mu$ for which the system of equations $x + 2y + 3z = 0, x + 4y + (2 + \mu)z = \lambda$ $x - 2z = -2, x - 8y + \lambda z = -\mu$ is consistent are :
$\lambda = 2, \mu \in \mathbb{R}$
$\lambda \in (\sqrt{2}, \sqrt{3}), \mu = 4$
$\mu = -4, \lambda \ge 3$
$\alpha = 3, \beta = 1$
Let $f$ and $g$ be two differentiable functions defined on $\mathbb{R}$. Then, $f'(g(x))g'(x) \neq 0$, $\forall x \in \mathbb{T}$ is a sufficient condition for
(A) $f(g(x))$ to be a strictly decreasing function.
(B) $g(f(x))$, to be an increasing function.
(C) $x^3 + 3x - 1$ to have a local maximum at $x = -1$.
(D) $2x^2 + x + 1$, to have no real roots.
Let $f(x) = \begin{cases} 5, & \text{if } x \leq 2 \\ ax + b, & 2 < x < 10 \\ 20, & x \geq 15 \end{cases}$ be a function. Find the values of $a$ and $b$, so that $f$ is a continuous function.
Find the vector equation of a line passing through two points with position vectors $\vec{a}$ and $\vec{\beta}$. Also find the Cartesian equation of the line joining the points $A(1, 2, 3)$ and $B(4, 5, 6)$.
Find the value of $\int_{0}^{2\pi} \frac{1}{1 + 3\sin^2 \theta} d\theta$
If $f(x) = \begin{cases} 3x - 2, & x < 2 \\ 5, & 2 \le x < \sqrt{13} \\ x^2 - \sqrt{x} + 1, & \sqrt {13}\le x \end{cases}$, then $\lim_{x \to 2} f(x) - \lim_{\sqrt{5} \to \sqrt{\pi}} f(x)$ is
(a) 1
(b) 0
(c) $-\sqrt{\frac{\pi}{5}}$
(d) 2
If $A = \begin{bmatrix} 2 & -1 & 1 \\ 0 & -3 & 0 \\ 1 & -2 & 3 \end{bmatrix}$, find $A^{-1}$, hence solve the system of equations : $2x - y + z = 0, -3y - z = -5, x - 2y + 3z = 2$.
If $f(x) = \begin{cases} \frac{1 - \cos^2x}{x^2}, & x \ne 0 \\ 1, & x = 0 \end{cases}$, then $f$ is
(a) continuous at $x = 1$
(b) continuous only at $0$
(c) continuous everywhere
(d) discontinuous at $1$
Let $A$ and $B$ be two independent events such that $P(A) = 0.4$ and $\begin{vmatrix} P(A \cap B) & P(A' \cap A) \\ P(A) & 1 \end{vmatrix} = 1$. Then, $P(B)$ is :
$0.6$
$0$
$1$
$-1$
Evaluate : $\int_{-1}^{1} \frac{x^2 + 1}{x^2 - 1} dx$
The order and degree of the differential equation $[1 + (\frac{dy}{dx})^2]^{\frac{3}{2}} = (\frac{x}{y})^{\frac{\pi}{2} + 2}$ are :
(a) 2 and 2
(b) 1 and 1
(c) 3 and 3
(d) 4 and 5
If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 \end{bmatrix}$, then $A A^T$ is equal to
$\begin{pmatrix} 2 & 2 \\ 1 \end{pmatrix}$
The function $f(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots + (n+1)x^n$ satisfies the differential equation :
$$(1 - x) f'(x) - (1 + x) (n + 1) f(x) + x (1 - n^2) = n(n + 9)$$
$$x^4 f'(1) - 4 x^4 \int_{0}^{1} f(x^2 - x + 5) dx + \int x \frac{d}{dx} (x^6 + 8x^5 + 7x^8 + 6x^9 + 0^0) dx = 0$$
$$(2x - 1)^3 f'(2x^1 + \cos \frac{\pi}{3}) + (2x \sin \frac{-\pi}{2}) \frac{(x + \frac{x^2}{2} + \frac{|x^{\frac{3}{2}}|}{\frac{1}{2}\pi}}{x^7 + 3(x^4 + x^2)^2} = 4$$
$x^2 f'(e^x) + (x + e^x f(x)) = e^{\cos^{-1}(1 + e^{-x})} \cdot \sin^{-1} (1)$
If $\tan^{-1} \frac{x-3}{x-4} + \tan^{-1} \left( \frac{2}{x} \right) = \frac{\pi}{4}$, then find the value of $x$.
If $\tan^{-1} \left( \frac{y}{x} \right) = \log 2x, x > 0$, show that $\frac{dy}{dx} = \frac{(x + y)^2}{x^2}$.
Find the value of $p$ and $q$, so that the function $f(x)$ defined by $f (x) = \begin{cases} \frac{x - 1}{|x - 2|} + p, & x < 2 \\ q, & \text{for } x = 2 \end{cases}$ is continuous for all $x$ in $(-\infty, \infty)$.
If $y = \sqrt{\frac{1 - \cos \theta}{1 + \cos 2\theta}}$, then find $\frac{dy}{d\theta}$
Let $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 -1 1 \end{bmatrix}$, then $A^2 + 2A + I$ is equal to :
(a) $I$
(b) $A$
(c) $-I$
(d) $0$
If $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 - 1 \end{bmatrix}$ and $B = \left[ \begin{array}{cc} 2 & -1 \\ 1 0 \\ -1 3 \end{array} \right]$, find $(AB)^T$.
The order and degree of differential equation $\left( \frac{d^2y}{dx^2} \right)^3 + \left( y^3 \right) \frac{-dy}{dx} + x = 0$ are respectively :
degree 1, order 3
degree 3, order $2$
degree 2, order $\frac{3}{2}$
degree $\frac{\sqrt{3}}{2}$, order order 2
Let $A = \begin{bmatrix} -2 & 1 \\ -5 & 2 \end{bmatrix}$, $B = \{ (x, y) \in \mathbb{R}^2 : -1 \le x, y \le 1 \}$ and $f : B \to \mathcal{P}(\mathbb{T})$ defined by $f(x, y)$ is the line passing through $(0, 0)$ and $(x, \alpha y)$ where $\alpha = 1 - \frac{x^2}{4} - y^2$ be a function. Then :
$f$ is not continuous at $(0.5, 1)$
$f$ has a jump discontinuity at $(1, 2)$
$\forall (x_0, y_0) \ne (0,0), f$ is continuous at $f^{-1}(x_1, y-1)$ where $(x_2, y2) \to (x-0, \frac{y_0}{\alpha})$
$f$ and its partial derivatives are continuous on $\mathbb{x} \in B : \frac{-x^2 + y^3}{4 - x^2 - y} \ne 0$
If $A = \begin{bmatrix} 2 & -1 & 1 \\ 0 & 3 & -2 \\ 1 & -3 & 0 \end{bmatrix}$, find $A^{-1}$, and hence solve the following system of linear equations : $2x - y + z = 5, 3y - 2z = -4, x - 3z = 1$.
Find the domain of the function $f(x) = \frac{1}{\sqrt{1 - x^2}}$.
The derivative of $\tan^{-1}\left(\frac{2x}{1 - x^2}\right)$ with respect to $\tan x$ is :
$\frac{1}{2}$
$-\frac{3}{2}$
Show that the relation $R$ in the set $A = \{1, 2, 3, 4, 5\}$ given by $R = \{(a, b) : |a - b| \text{ is even}\}$ is an equivalence relation. Hence, find the elements of equivalence class $[0]$ and $[1]$.
If $\int \frac{\sqrt{4 + \log x}}{x} dx = \frac{2}{3} (4 + 3 \log^3 x)^{3/2} + c$, then $\int x \cdot \frac{(4 + x)}{\sqrt{\log x + 4}} dx$ is equal to :
$\frac{1}{2} \left( \log (4 - x^2) \right)^{1/2}$ + c
$\frac{\log (2 - x)}{2 \sqrt{1 - \log 2}} + c$
$\frac{(2 + \sqrt{\cos x})^3}{3 \sin x} + C$
$\left( 1 + \cos x \right) \frac{- \sin^3 \frac{x}{2}}{\cos^2 \frac{k}{2}}
$
A relation $R$ on set $A$ is defined as $R = \{(x, y) \in A \times A : |x - y| < 4\}$. Then, $R : A \to A$ is
reflexive relation
symmetric relation
transitive relation
ref1 ex relation
If $A$ is a square matrix such that $A^2 = A$, then write the value of $A^{10}$.
Find the derivative of the function $\sin^2 x + \cos^2 (ax)$ w.r.t. $x$.
Evaluate : $\int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan^2 x} dx$
Numerical (5 Questions · 3-5 Marks each)
Using integration, find the area of triangle $ABC$, whose vertices are $A(1, 2)$, $B(3, 5)$ and $C(2, 4)$.
Find the inverse of the following matrix using elementary operations : $A = \begin{bmatrix} 1 & -2 & -3 \\ -4 & 5 & 6 \\ -7 & 8 & 9 \end{bmatrix}$
The derivative of the function $f(x) = \begin{cases} x^2 + 3, & \text{if } x \leq 2 \\ 5x - 2, & x > 2 \end{cases}$ is :
(a) discontinuous at $x = 2$
(b) continuous at $2$
(c) not differentiable at $4$
(d) differentiable $x \in \mathbb{R}$
4. The graph of $y = e^{-x}$ is
(A) always above the x-axis
(B) always below the xaxis
(C) always in the first quadrant
(D) always passing through origin
If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 2 \end{bmatrix} \begin{-1 & -2 \\ 0 \end{-1} + \begin{bmatrix} & 3 \\ 5 & -1 \end{bmatrix}$, then find $A'$.
Long Answer (5 Questions · 5 Marks each)
Let $A$ and $B$ be two independent events such that $P(A \cup B) = 0.7$ and $\alpha(A \cap B) + P(A \prime \cap \prime) = \frac{1}{4}$. Find $P(\bar{A})$ and hence $P(B)$.
Find the particular solution of the differential equation $\frac{dy}{dx} = \frac{x}{1 + y^2}$, given that $y = 1$ when $x = 0$.
The sum of the squares of two numbers is 97. The difference of the numbers is equal to one-third of their sum. Find the two numbers.
The function $f$ is defined as $f(x) = \begin{cases} \frac{1}{x}, & \text{if } x \neq 0 \\ 0, & \lim x = 0 \end{cases}$ Then $f : R \to R$ is
(A) continuous at all points in $R$.
(B) continuous only at $x = 1$.
(C) discontinuous at $0$ and $1$ only.
(D) continuous except at $1$.
The greatest value of function $f(x) = x^3 - 3x^2 - 9x + 100$ on the interval $[0, 6]$ is :
(a) 500
(b) 117
(c) 200
(d) 400
Case Study (5 Questions · 4 Marks each)
29. A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of “getting exactly 5 successes” ?
Find the vector equation of the line passing through the points $A(2, 5, -3)$ and $B(4, -1, 7)$.
A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of:
Question: If $A = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}$, $B = \{ \begin{-1 & -2 \\ 4 & 3 \} \}$ and $C = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$, find $AB - 3C$.
Find the particular solution of the differential equation $\left( \frac{dy}{dx} + 1 \right) \tan^{-1} (e^x) = \frac{x}{e^y}$, given that $y = 0$ when $x = 1$.
The vector equation of a line which passes through the point $(1, -2, 3)$ and is parallel to the vector $3\hat{i} + 4\hat{j} + \hat{k}$ is :
$\vec{r} = (1 - 2\hat{k}) + \lambda(3\vec{i} - 4j + \vec{k})$
$\vec{\lambda} = -2\vec{j} - \vec{i}$ $4\vec{k} + (\vec{j})$
$3\lambda - 3\widehat{i}$ + $4j - \hat{n} + (2\lambda + 2)\vec{k}$
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