Let X = {x ∈ N : 1 ≤ x ≤ 17} and
Question: Let $X=\{x \in N: 1 \leq x \leq 17\}$ and $\mathrm{Y}=\{\mathrm{ax}+\mathrm{b}: \mathrm{x} \in \mathrm{X}$ and $\mathrm{a}, \mathrm{b} \in \mathrm{R}, \mathrm{a}0\}$. If mean and variance of elements of $\mathrm{Y}$ are 17 and 216 respectively then $a+b$ is equal to :-779-27Correct Option: 1 Solution: $\sigma^{2}=$ variance $\mu=$ mean $\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}$ $\mu=17$ $\Rightarrow \frac{\sum_{x=1}^{17}(a x+b)}{17}=17$ $\Rightarrow 9 \mathrm{a}+\m...
Read More →If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :
Question: If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :$\frac{1}{4}$$\frac{1}{5}$$\frac{1}{7}$$\frac{1}{6}$Correct Option: , 4 Solution: Sum of 1 st 25 terms $=$ sum of its next 15 terms $\Rightarrow\left(\mathrm{T}_{1}+\ldots . .+\mathrm{T}_{25}\right)=\left(\mathrm{T}_{26}+\ldots . .+\mathrm{T}_{40}\right)$ $\Rightarrow\left(\mathrm{T}_{1}+\ldots . .+\mathrm{T}_{40}\right)=2\left(\mathr...
Read More →The area (in sq. units) of the region
Question: The area (in sq. units) of the region $\left\{(x, y): 0 \leq y \leq x^{2}+1,0 \leq y \leq x+1\right.$ $\left.\frac{1}{2} \leq x \leq 2\right\}$ is :$\frac{79}{16}$$\frac{23}{6}$$\frac{79}{24}$$\frac{23}{16}$Correct Option: , 3 Solution: $0 \leq y \leq x^{2}+1,0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2$ Required area $=\int_{1 / 2}^{1}\left(x^{2}+1\right) d x+\frac{1}{2}(2+3) \times 1$ $=\frac{19}{24}+\frac{5}{2}=\frac{79}{24}$...
Read More →The area (in sq. units) of the region
Question: The area (in sq. units) of the region $\left\{(x, y): 0 \leq y \leq x^{2}+1,0 \leq y \leq x+1\right.$ $\left.\frac{1}{2} \leq x \leq 2\right\}$ is :$\frac{79}{16}$$\frac{23}{6}$$\frac{79}{24}$$\frac{23}{16}$Correct Option: , 3 Solution: $0 \leq y \leq x^{2}+1,0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2$ Required area $=\int_{1 / 2}^{1}\left(x^{2}+1\right) d x+\frac{1}{2}(2+3) \times 1$ $=\frac{19}{24}+\frac{5}{2}=\frac{79}{24}$...
Read More →Let y=y(x) be the solution of the differential equation,
Question: Let $y=y(x)$ be the solution of the differential equation, $\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y0, y(0)=1 .$ If $y(\pi)=a$ and $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=\pi$ is $\mathrm{b}$, then the ordered pair (a, b) is equal to :$(2,1)$$\left(2, \frac{3}{2}\right)$$(1,-1)$$(1,1)$Correct Option: , 4 Solution: $\frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x, y0$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}+1}=\frac{-\cos \mathrm{x}}{2+\sin \mathrm{x}} \mathrm{dx}$...
Read More →A hyperbola having the transverse axis of length
Question: A hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3 x^{2}+4 y^{2}=12$, then this hyperbola does not pass through which of the following points ?$\left(1,-\frac{1}{\sqrt{2}}\right)$$\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$$\left(\frac{1}{\sqrt{2}}, 0\right)$$\left(-\sqrt{\frac{3}{2}}, 1\right)$Correct Option: , 2 Solution: Ellipse : $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ eccentricity $=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$ $\there...
Read More →Given the following two statements:
Question: Given the following two statements: $\left(S_{1}\right):(q \vee p) \rightarrow(p \leftrightarrow \sim q)$ is a tautology. $\left(S_{2}\right): \sim q \wedge(\sim p \leftrightarrow q)$ is a fallacy. Then :only $\left(\mathrm{S}_{1}\right)$ is correct.both $\left(S_{1}\right)$ and $\left(S_{2}\right)$ are correct.both $\left(S_{1}\right)$ and $\left(S_{2}\right)$ are not correct.only $\left(S_{2}\right)$ is correct.Correct Option: , 3 Solution: Let TV(r) denotes truth value of a statemen...
Read More →The contrapositive of the statement
Question: The contrapositive of the statement "If I reach the station in time, then I will catch the train" is :If I will catch the train, then I reach the station in time.If I do not reach the station in time, then I will not catch the train.If I will not catch the train, then I do not reach the station in time.If I do not reach the station in time, then I will catch the train.Correct Option: , 3 Solution: Let $\mathrm{p}$ denotes statement $\mathrm{p}$ : I reach the station in time. q: I will ...
Read More →Let A be a 2 x 2 real matrix with entries from
Question: Let $A$ be a $2 \times 2$ real matrix with entries from $\{0,1\}$ and $|\mathrm{A}| \neq 0$. Consider the following two statements : (P) If $A \neq I_{2}$, then $|A|=-1$ (Q) If $|\mathrm{A}|=1$, then $\operatorname{tr}(\mathrm{A})=2$, where $\mathrm{I}_{2}$ denotes $2 \times 2$ identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then:(P) is true and (Q) is falseBoth (P) and (Q) are falseBoth (P) and (Q) are true(P) is false and (Q) is trueCorrect Option: , 4 Solutio...
Read More →The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points
Question: The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane :$x+2 y-z=1$$x-2 y+z=1$$x-y-2 z=1$$2 x+y-z=1$Correct Option: , 4 Solution: Equation of $\mathrm{AB}=\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ Let coordinates of M $=(1,(1+3 \lambda),-3 \lambda)$. $\overrightarrow{\mathrm{PM}}=-3 \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}-3(\lambd...
Read More →The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points
Question: The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane :$x+2 y-z=1$$x-2 y+z=1$$x-y-2 z=1$$2 x+y-z=1$Correct Option: , 4 Solution: Equation of $\mathrm{AB}=\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ Let coordinates of M $=(1,(1+3 \lambda),-3 \lambda)$. $\overrightarrow{\mathrm{PM}}=-3 \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}-3(\lambd...
Read More →Solve this following
Question: Let $f(x)=|x-2|$ and $g(x)=f(f(x)), x \in[0,4]$. Then $\int_{0}^{3}(g(x)-f(x)) d x$ is equal to : $\frac{3}{2}$0$\frac{1}{2}$ 1Correct Option: , 4 Solution: $\int_{0}^{3} g(x)-f(x)=\int_{0}^{3}|| x-2|-2| d x-\int_{0}^{3}|x-2| d x$ $=\left(\frac{1}{2} \times 2 \times 2+1+\frac{1}{2} \times 1 \times 1\right)-\left(\frac{1}{2} \times 2 \times 2+\frac{1}{2} \times 1 \times 1\right)$ $=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$...
Read More →Let S be the set of all
Question: Let $S$ be the set of all $\lambda \in R$ for which the system of linear equations $2 x-y+2 z=2$ $x-2 y+\lambda z=-4$ $x+\lambda y+z=4$ has no solution. Then the set $\mathrm{S}$contains more than two elements.is a singleton.contains exactly two elements.is an empty set.Correct Option: , 3 Solution: $2 x-y+2 z=2$ $x-2 y+\lambda z=-4$ $x+\lambda y+z=4$ For no solution : $D=\left|\begin{array}{ccc}2 -1 2 \\ 1 -2 \lambda \\ 1 \lambda 1\end{array}\right|=0$ $\Rightarrow 2\left(-2-\lambda^{...
Read More →Solve this
Question: A triangle $A B C$ lying in the first quadrant has two vertices as $\mathrm{A}(1,2)$ and $\mathrm{B}(3,1)$. If $\angle \mathrm{BAC}=90^{\circ}$, and $\operatorname{ar}(\triangle \mathrm{ABC})=5 \sqrt{5}$ sq. units, then the abscissa of the vertex $C$ is: $2+\sqrt{5}$$1+\sqrt{5}$$1+2 \sqrt{5}$$2 \sqrt{5}-1$Correct Option: , 3 Solution: $\left(\frac{\mathrm{K}-2}{\mathrm{~h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1 \Rightarrow \mathrm{K}=2 \mathrm{~h}$ ..........(1) $\sqrt{5}|h-1|=10$ $\...
Read More →Area (in sq. units) of the region outside
Question: Area (in sq. units) of the region outside $\frac{|x|}{2}+\frac{|y|}{3}=1$ and inside the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :$3(4-\pi)$$6(\pi-2)$$3(\pi-2)$$6(4-\pi)$Correct Option: , 2 Solution: $\frac{|x|}{2}+\frac{|y|}{3}=1$ $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ Area of Ellipse $=\pi a b=6 \pi$ Required area, $=\pi \times 2 \times 3-($ Area of quadrilateral $)$ $=6 \pi-\frac{1}{2} 6 \times 4$ $=6 \pi-12$ $=6(\pi-2)$...
Read More →Solve The Following Questions
Question: The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane :$x+2 y-z=1$$x-2 y+z=1$$x-y-2 z=1$$2 x+y-z=1$Correct Option: , 4 Solution: Equation of $\mathrm{AB}=\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ Let coordinates of M $=(1,(1+3 \lambda),-3 \lambda)$. $\overrightarrow{\mathrm{PM}}=-3 \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}-3(\lambd...
Read More →Solve this following
Question: Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(ab)$ be a given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, $\phi(t)=\frac{5}{12}+t-t^{2}$, then $a^{2}+b^{2}$ is equal to : 126135145116Correct Option: 1 Solution: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(ab) ; \frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a \ldots(i)$ Now, $\phi(\mathrm{t})=\frac{5}{12}+\mathrm{t}-\mathrm{t}^{2}=\frac{8}{12}-\left(\mathrm{t}-\frac{1}{2}\right)^{2}$...
Read More →Box I contains 30 cards numbered 1 to 30 and Box II
Question: Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50 . A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is :$\frac{8}{17}$$\frac{2}{3}$$\frac{4}{17}$$\frac{2}{5}$Correct Option: 1 Solution: Let $\mathrm{B}_{1}$ be the event where Box-I is selected. $\ \mathrm{~B}_{2} \rightarrow$ where box-II selected $\mathrm{P}\left(\mathrm{B}_{1}\r...
Read More →If a function f(x) defined by
Question: If a function $f(x)$ defined by $f(x)= \begin{cases}a e^{x}+b e^{-x}, -1 \leq x1 \\ c x^{2} , \quad 1 \leq x \leq 3 \\ a x^{2}+2 c x, 3x \leq 4\end{cases}$ be continuous for some $a, b, c \in R$ and $\mathrm{f}^{\prime}(0)+\mathrm{f}^{\prime}(2)=\mathrm{e}$, then the value of of $\mathrm{a}$ is :$\frac{e}{e^{2}-3 e-13}$$\frac{e}{e^{2}+3 e+13}$$\frac{1}{\mathrm{e}^{2}-3 \mathrm{e}+13}$$\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}$Correct Option: , 4 Solution: $f(x)= \begin{cases}a ...
Read More →Solve The Following Questions
Question: The lines\ $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\ell(2 \hat{\mathrm{i}}+\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}})+\mathrm{m}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$Intersect when $\ell=1$ and $\mathrm{m}=2$Intersect when $\ell=2$ and $m=\frac{1}{2}$Do not intersect for any values of $\ell$ and $\mathrm{m}$Intersect for all values of $\ell$ and $\mathrm{m}$Correct Option: 3, Solution: $\overrightarr...
Read More →Solve this following
Question: Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $\mathrm{x}^{2}-6 \mathrm{x}+\mathrm{q}=0$. If $\alpha$, $\beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 \mathrm{q}+\mathrm{p}):(2 \mathrm{q}-\mathrm{p})$ is : $3: 1$$33: 31$ $9: 7$$5: 3$Correct Option: , 3 Solution: $x^{2}-3 x+p=0\begin{aligned}\alpha \\\beta\end{aligned}$ $\alpha, \beta, \gamma, \delta$ in G.P. $\alpha+\alpha r=3 \quad \ldots(1)$ $x^{2}-6 x+q=0\begi...
Read More →A die is thrown two times and the sum of the scores appearing
Question: A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4 . Then the conditional probability that the score 4 has appeared atleast once is :$\frac{1}{8}$$\frac{1}{9}$$\frac{1}{3}$$\frac{1}{4}$Correct Option: , 2 Solution: A: Sum obtained is a multiple of $4 .$ $A=\{(1,3),(2,2),(3,1),(2,6),(3,5),(4,4),$, $(5,3),(6,2),(6,6)\}$ B : Score of 4 has appeared at least once. $B=\{(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)$ $(4,1),(4,2),(4,3),(4,5),(4,6...
Read More →Prove the following
Question: Let $\alpha0, \beta0$ be such that $\alpha^{3}+\beta^{2}=4$. If the maximum value of the term independent of $x$ in the binomial expansion of $\left(\alpha x^{\frac{1}{4}}+\beta x^{-\frac{1}{6}}\right)^{10}$ is $10 k$, then $\mathrm{k}$ is equal to :(1) 17633635284Correct Option: , 2 Solution: Let $t_{r}+1$ denotes $r+1^{\text {th }}$ term of $\left(\alpha x^{\frac{1}{9}}+\beta x^{-\frac{1}{6}}\right)^{10}$ $t_{r+1}={ }^{10} C_{r} \alpha^{10-r}(x)^{\frac{10-r}{9}} \cdot \beta^{r} x^{-\...
Read More →If |x|<1, |y|<1 and
Question: If $|x|1$, $|y|1$ and $x \neq y$, then the sum to infinity of the following series $(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots \ldots$ $\frac{x+y-x y}{(1-x)(1-y)}$$\frac{x+y-x y}{(1+x)(1+y)}$$\frac{x+y+x y}{(1+x)(1+y)}$$\frac{x+y+x y}{(1-x)(1-y)}$Correct Option: 1, Solution: $|x|1,|y|1, x \neq y$ $(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)$........ By multiplying and dividing $x-y$ : $\frac{\left(x^{2}-y^{2}\right...
Read More →Solve this
Question: Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x,[x]^{2}+2[x+2]-7=0$ has :no integral solution exactly four integral solutions exactly two solutions infinitely many solutionsCorrect Option: , 4 Solution: $[x]^{2}+2[x+2]-7=0$ $\Rightarrow[x]^{2}+2[x]+4-7=0$ $\Rightarrow[x]=1,-3$ $\Rightarrow x \in[1,2) \cup[-3,-2)$...
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