Available Questions 746 found Page 22 of 38
Standalone Questions
#925
Mathematics
Vector Algebra
VSA
APPLY
2023
KNOWLEDGE
2 Marks
24. If the projection of the vector $\hat{i}+\hat{j}+\hat{k}$ on the vector $p\hat{i}+\hat{j}-2\hat{k}$ is $\frac{1}{3}$, then find the value(s) of $p$.
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#924
Mathematics
Matrices and Determinants
LA
APPLY
2023
KNOWLEDGE
5 Marks
If $A=\begin{bmatrix}1&2&-2\\ -1&3&0\\ 0&-2&1\end{bmatrix}$ and $B^{-1}=\begin{bmatrix}3&-1&1\\ -15&6&-5\\ 5&-2&2\end{bmatrix},$ find $(AB)^{-1}$.
OR Solve the following system of equations by matrix method :$ x+2y+3z=6, 2x-y+z=2, 3x+2y-2z=3.$
OR Solve the following system of equations by matrix method :$ x+2y+3z=6, 2x-y+z=2, 3x+2y-2z=3.$
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#923
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
Show that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ defined as $f(x)=\frac{5x-3}{4}$ is both one-one and onto.
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Sol:
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#922
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
Let $f : \mathbb{R} - \left\{ \frac{4}{3} \right\} \to \mathbb{R}$ be a function defined as:$$f(x) = \frac{4x}{3x+4}$$Show that $f$ is a one-one function. Also, check whether $f$ is an onto function or not.
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#921
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
34. (a) If N denotes the set of all natural numbers and R is the relation on $N \times N$ defined by $(a, b) R (c, d)$, if $ad(b+c)=bc(a+d)$. Show that R is an equivalence relation.
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#920
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
A relation $R$ is defined on a set of real numbers $\mathbb{R}$ as:$$R = \{(x, y) : x \cdot y \text{ is an irrational number}\}$$Check whether $R$ is reflexive, symmetric, and transitive or not.
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#919
Mathematics
Differential Equations
SA
APPLY
2023
KNOWLEDGE
3 Marks
Solve the following differential equation : $xe^{\frac{y}{x}}-y+x\frac{dy}{dx}=0$
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#918
Mathematics
Differential Equations
SA
APPLY
2023
KNOWLEDGE
3 Marks
Find the general solution of the differential equation : $\frac{d}{dx}(xy^{2})=2y(1+x^{2})$
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#917
Mathematics
Differential Equations
SA
APPLY
2023
KNOWLEDGE
3 Marks
Find the general solution of the differential equation \(e^{x}\tan y~dx+(1-e^{x})\sec^{2}y~dy=0\).
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#916
Mathematics
Differential Equations
SA
APPLY
2023
KNOWLEDGE
3 Marks
29. (a) Find the particular solution of the differential equation $\frac{dy}{dx}=\frac{x+y}{x}, y(1)=0$.
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#915
Mathematics
Differential Equations
SA
APPLY
2023
KNOWLEDGE
3 Marks
Find the particular solution of the differential equation:$$\frac{dy}{dx} + \sec^{2}x \cdot y = \tan x \cdot \sec^{2}x$$given that $y(0) = 0$.
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#914
Mathematics
Three Dimensional Geometry
VSA
APPLY
2023
KNOWLEDGE
2 Marks
Position vectors of the points A, B and C as shown in the figure below are a, $\vec{b}$ and $\vec{c}$ respectively. If $\vec{AC}=\frac{5}{4}\vec{AB}$ , express $\vec{c}$ in terms of $\vec{a}$ and $\vec{b}$ .
OR Check whether the lines given by equations $x=2\lambda+2$, $y=7\lambda+1$, $z=-3\lambda-3$ and $x=-\mu-2,$ $y=2\mu+8,$ $z=4\mu+5$ are perpendicular to each other or not.
OR Check whether the lines given by equations $x=2\lambda+2$, $y=7\lambda+1$, $z=-3\lambda-3$ and $x=-\mu-2,$ $y=2\mu+8,$ $z=4\mu+5$ are perpendicular to each other or not.
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#913
Mathematics
Relations and Functions
VSA
APPLY
2023
KNOWLEDGE
2 Marks
A function $f:A\rightarrow B$ defined as $f(x)=2x$ is both one-one and onto. If $A=\{1,2,3,4\}$, then find the set $B$.
OR
Evaluate : $\sin^{-1}(\sin\frac{3\pi}{4})+\cos^{-1}(\cos\frac{3\pi}{4})+\tan^{-1}(1)$
OR
Evaluate : $\sin^{-1}(\sin\frac{3\pi}{4})+\cos^{-1}(\cos\frac{3\pi}{4})+\tan^{-1}(1)$
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To find the set $B$, we need to determine the range of the function $f$.Reasoning:Since the function $f: A \rightarrow B$ is given to be onto (surjective), every element in the codomain ($B$) must have a corresponding element in the domain ($A$). This implies that the set $B$ is equal to the range of $f$.
Calculation:
Given the domain $A = \{1, 2, 3, 4\}$ and the rule $f(x) = 2x$, we calculate the image for each element in $A$: For $x = 1$: $f(1) = 2(1) = 2$ For $x = 2$: $f(2) = 2(2) = 4$ For $x = 3$: $f(3) = 2(3) = 6$ For $x = 4$: $f(4) = 2(4) = 8$ Hence Collecting these values gives us the set $B$.$$B = \{2, 4, 6, 8\}$$
#912
Mathematics
Probability
SA
APPLY
2023
Competency
3 Marks
There are two coins. One of them is a biased coin such that P (head): P (tail) is 1:3 and the other coin is a fair coin. A coin is selected at random and tossed once. If the coin showed head, then find the probability that it is a biased coin.
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#911
Mathematics
Probability
SA
APPLY
2023
Competency
3 Marks
A pair of dice is thrown simultaneously. If $X$ denotes the absolute difference of numbers obtained on the pair of dice, then find the probability distribution of $X$.
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#910
Mathematics
Probability
SA
APPLY
2023
KNOWLEDGE
3 Marks
31. From a lot of 30 bulbs which include 6 defective bulbs, a sample of 2 bulbs is drawn at random one by one with replacement. Find the probability distribution of the number of defective bulbs and hence find the mean number of defective bulbs.
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#909
Mathematics
Probability
SA
APPLY
2023
Competency
3 Marks
The probability distribution of a random variable X is given below :
$$\begin{array}{|c|c|c|c|}
\hline
X & 1 & 2 & 3 \\
\hline
P(X) & \frac{k}{2} & \frac{k}{3} & \frac{k}{6} \\
\hline
\end{array}$$
(i) Find the value of $k$.
(ii) Find $P(1\le X<3)$.
(iii) Find $E(X)$, the mean of $X$.
OR
$A$ and $B$ are independent events such that $P(A\cap\overline{B})=\frac{1}{4}$ and $P(\overline{A}\cap B)=\frac{1}{6}$ Find $P(A)$ and $P(B)$.
$$\begin{array}{|c|c|c|c|}
\hline
X & 1 & 2 & 3 \\
\hline
P(X) & \frac{k}{2} & \frac{k}{3} & \frac{k}{6} \\
\hline
\end{array}$$
(i) Find the value of $k$.
(ii) Find $P(1\le X<3)$.
(iii) Find $E(X)$, the mean of $X$.
OR
$A$ and $B$ are independent events such that $P(A\cap\overline{B})=\frac{1}{4}$ and $P(\overline{A}\cap B)=\frac{1}{6}$ Find $P(A)$ and $P(B)$.
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The sum of probabilities in a probability distribution must equal $1$.$$\sum P(X) = 1$$$$\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1$$Find the common denominator (which is $6$):$$\frac{3k}{6} + \frac{2k}{6} + \frac{1k}{6} = 1$$$$\frac{6k}{6} = 1$$$$k = 1$$(ii) Find $P(1 \le X < 3)$The range $1 \le X < 3$ includes $X=1$ and $X=2$, but not $X=3$.$$P(1 \le X < 3) = P(X=1) + P(X=2)$$Substituting $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}$$$$P(1 \le X < 3) = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$(iii) Find $E(X)$, the mean of $X$The expectation (mean) is defined as $E(X) = \sum X \cdot P(X)$.Using $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}, \quad P(X=3) = \frac{1}{6}$$$$E(X) = \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{3}\right) + \left(3 \times \frac{1}{6}\right)$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{3}{6}$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{1}{2}$$$$E(X) = 1 + \frac{2}{3} = \frac{5}{3}$$OR Part: Independent EventsProblem:$A$ and $B$ are independent events such that:$P(A \cap \overline{B}) = \frac{1}{4}$$P(\overline{A} \cap B) = \frac{1}{6}$Solution:Let $P(A) = x$ and $P(B) = y$.Since events are independent:$$P(A \cap \overline{B}) = P(A) \cdot P(\overline{B}) = x(1-y)$$$$P(\overline{A} \cap B) = P(\overline{A}) \cdot P(B) = (1-x)y$$We have the system of equations:$x(1-y) = \frac{1}{4} \implies x - xy = \frac{1}{4}$$(1-x)y = \frac{1}{6} \implies y - xy = \frac{1}{6}$Subtracting equation (2) from equation (1):$$(x - xy) - (y - xy) = \frac{1}{4} - \frac{1}{6}$$$$x - y = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$$$$x = y + \frac{1}{12}$$Substitute $x$ back into equation (2):$$\left(1 - \left(y + \frac{1}{12}\right)\right)y = \frac{1}{6}$$$$\left(\frac{11}{12} - y\right)y = \frac{1}{6}$$$$\frac{11y}{12} - y^2 = \frac{1}{6}$$Multiply by 12 to clear the denominator:$$11y - 12y^2 = 2$$$$12y^2 - 11y + 2 = 0$$Solving the quadratic equation for $y$:$$y = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(12)(2)}}{2(12)}$$$$y = \frac{11 \pm \sqrt{121 - 96}}{24} = \frac{11 \pm \sqrt{25}}{24} = \frac{11 \pm 5}{24}$$Two possible values for $y$:$y_1 = \frac{16}{24} = \frac{2}{3}$$y_2 = \frac{6}{24} = \frac{1}{4}$Finding corresponding $x$ values using $x = y + \frac{1}{12}$:If $y = \frac{2}{3}$: $x = \frac{2}{3} + \frac{1}{12} = \frac{8+1}{12} = \frac{9}{12} = \frac{3}{4}$If $y = \frac{1}{4}$: $x = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$Answer:There are two possible sets of values:$P(A) = \frac{3}{4}$ and $P(B) = \frac{2}{3}$$P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$
#908
Mathematics
Probability
LA
APPLY
2023
Competency
5 Marks
(a) In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/5 be the probability that he knows the answer and 2/5 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/3. What is the probability that the student knows the answer, given that he answered it correctly? OR (b) A box contains 10 tickets, 2 of which carry a prize of ₹8 each, 5 of which carry a prize of ₹4 each, and remaining 3 carry a prize of ₹2 each. If one ticket is drawn at random, find the mean value of the prize.
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#907
Mathematics
Applications of Derivatives
VSA
APPLY
2023
KNOWLEDGE
2 Marks
Show that the function $f(x)=\frac{16\sin x}{4+\cos x}-x$, is strictly decreasing in $(\frac{\pi}{2},\pi)$
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Sol:
#906
Mathematics
Applications of Derivatives
LA
UNDERSTAND
2023
Competency
5 Marks
(a) The median of an equilateral triangle is increasing at the rate of 2√3 cm/s. Find the rate at which its side is increasing. OR (b) Sum of two numbers is 5. If the sum of the cubes of these numbers is least, then find the sum of the squares of these numbers.
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