The median of the numbers 84, 78, 54, 56, 68, 22, 34, 45, 39, 54 is
Question: The median of the numbers 84, 78, 54, 56, 68, 22, 34, 45, 39, 54 is(a) 45(b) 49.5(c) 54(d) 56 Solution: (c) 54We will arrange the data in ascending order as:22, 34, 39, 45, 54, 54, 56, 68, 78, 84Here, n is 10, which is an even number.Thus, we have: Median $=$ Mean of $\left(\frac{n}{2}\right)$ th $\left(\frac{n}{2}+1\right)$ th observations $=\frac{1}{2}$ (5 th observation $+6$ th observation) $=\frac{1}{2}(54+54)=54$...
Read More →Solve the following
Question: The molecular geometry of $\mathrm{SF}_{6}$ is octahedral. What is the geometry of $\mathrm{SF}_{4}$ (including lone pair(s) of electrons, if any)?TetrahedralTrigonal bipyramidalPyramidalSquare planarCorrect Option: , 2 Solution: Geometry is trigonal bipyramidal but shape is "See Saw"....
Read More →The number of points,
Question: The number of points, at which the function $f(x)=|2 x+1|-3|x+2|+\left|x^{2}+x-2\right|, x \in R$ is not differentiable, is Solution: $f(x)=|2 x+1|-3|x+2|+\left|x^{2}+x-2\right|$ $f(x)= \begin{cases}x^{2}-7 ; x1 \\ -x^{2}-2 x-3 ; -\frac{1}{2}x1 \\ -x^{2}-6 x-5 ; -2x\frac{-1}{2} \\ x^{2}+2 x+3 ; x-2\end{cases}$ $\therefore f^{\prime}(x)=\left\{\begin{array}{lc}2 x ; x1 \\ -2 x-3 ; -\frac{1}{2}x1 \\ -2 x-6 ; -2x\frac{-1}{2} \\ 2 x+2 ; x-2\end{array}\right.$ Check at $1,-2$ and $\frac{-1}...
Read More →The median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is
Question: The median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is(a) 4(b) 5(c) 6(d) 7 Solution: (c) 6We will arrange the given data in ascending order as:3, 4, 4, 5, 6, 7, 7, 7, 12Here, nis 9, which is an odd number.Thus, we have Median $=$ Value of $\frac{1}{2}(n+1)$ th term Median score $=\frac{1}{2}(9+1)$ th term $=5$ th term $=6$...
Read More →The weight of 10 students (in kgs) are
Question: The weight of 10 students (in kgs) are55, 40, 35, 52, 60, 38, 36, 45, 31, 44The median weight is(a) 40 kg(b) 41 kg(c) 42 kg(d) 44 kg Solution: (c) 42 kgArranging the numbers in ascending order, we have:31, 35, 36, 38, 40, 44, 45, 52, 55, 60Here, nis 10, which is an even number.Thus, we have: Median $=$ Mean of $\left(\frac{n}{2}\right)$ th observation \ $\left(\frac{n}{2}+1\right)$ th observation Median weight $=$ Mean of the weights of $\left(\frac{10}{2}\right)$ th student \ $\left(\...
Read More →If $f: R --- R is a function defined by
Question: If $f: R \rightarrow R$ is a function defined by $f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi$, where [.] denotes the greatest integer function, then $\boldsymbol{f}$ is :(1) discontinuous only at $x=1$(2) discontinuous at all integral values of $x$ except at $x=1$(3) continuous only at $x=1$(4) continuous for every real $x$Correct Option: , 4 Solution: Doubtful points are $\mathrm{x}=\mathrm{n}, \mathrm{n} \in \mathrm{I}$ L.H.L $=\lim _{x \rightarrow n^{-}}[x-1] \cos \left(\frac{...
Read More →Match the type of interaction in column A with the distance dependence of their interaction energy in column B :
Question: Match the type of interaction in column A with the distance dependence of their interaction energy in column B : (I)-(B), (II)-(D), (III)-(C)(I)-(A),(II)-(B), (III)-(D)(I)-(A), (II)-(B), (III)-(C)(I)-(A), (II)-(C), (III)-(D)Correct Option: , 4 Solution: (I) Ion-ion interaction energy $\propto\left(\frac{1}{r}\right)$. (II) Dipole-dipole interaction energy $\propto\left(\frac{1}{r^{3}}\right)$. (III) London dispersion $\propto\left(\frac{1}{r^{6}}\right)$....
Read More →The runs scored by 11 members of a cricket team are
Question: The runs scored by 11 members of a cricket team are15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0The median score is(a) 27(b) 29(c) 31(d) 20 Solution: (b) 29Arranging the weight of 10 students in ascending order, we have:0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56Here,nis 11, which is an odd number.Thus, we have: Median $=$ Value of $\left(\frac{n+1}{2}\right)$ th observation Median score $=$ Value of $\left(\frac{11+1}{2}\right)$ th term $=$ Value of 6 th term $=29$...
Read More →Let f
Question: Let $f: \mathrm{R} \rightarrow \mathrm{R}$ satisfy the equation $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$ and $f(x) \neq 0$ for any $x \in R$. If the function $f$ is differentiable at $\mathrm{x}=0$ and $f^{\prime}(0)=3$, then $\lim _{h \rightarrow 0} \frac{1}{h}(f(h)-1)$ is equal to_________. Solution: If $f(x+y)=f(x) \cdot f(y) \ f^{\prime}(0)=3$ then $f(x)=a^{x} \Rightarrow f^{\prime}(x)=a^{x} \cdot \ell n a$ $\Rightarrow f^{\prime}(0)=\ell \mathrm{na}=3 \Rightarrow \mathrm{a}=\...
Read More →The mean of the following data is 8.
Question: The mean of the following data is 8. The value ofpis(a) 23(b) 24(c) 25(d) 21 Solution: (c) 25 Now, Mean $=\frac{303+9 p}{41+p}$ Given : Mean $=8$ $\therefore \frac{303+9 p}{41+p}=8$ $\Rightarrow 303+9 p=328+8 p$ $\Rightarrow p=25$...
Read More →Solve the following
Question: The shape/ structure of $\left[\mathrm{XeF}_{5}\right]^{-}$and $\mathrm{XeO}_{3} \mathrm{~F}_{2}$, respectively, are:pentagonal planar and trigonal bipyramidaloctahedral and square pyramidaltrigonal bipyramidal and pentagonal planartrigonal bipyramidal and trigonal bipyramidalCorrect Option: 1 Solution:...
Read More →be a function defined as
Question: Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be a function defined as $f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x} , \text { if } x0 \\ b , \text { if } x=0 \\ \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}} , \text { if } x0\end{array}\right.$ If $f$ is continuous at $x=0$, then the value of $\mathrm{a}+\mathrm{b}$ is equal to:(1) $-\frac{5}{2}$(2) $-2$(3) $-3$(4) $-\frac{3}{2}$Correct Option: , 4 Solution: $f(x)$ is continuous at $x=0$ $\lim _{x \rightarrow 0^{+}} f(x)...
Read More →Solve this
Question: If $\bar{x}_{1}, \bar{x}_{2}, \ldots, \bar{x}_{n}$ are the means of $n$ groups with $n_{1}, n_{2}, \ldots, n_{n}$ number of observations respectively, then the mean $\bar{x}$ of all the groups taken together is (a) $\sum_{i=1}^{n} n_{i} \bar{x}_{i}$ (b) $\sum_{\frac{i=1}{n^{2}}}^{n} n_{i} \bar{x}_{i}$ (c) $\frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{\sum_{i=1}^{n} n_{i}}$ (d) $\frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{2 n}$ Solution: $(\mathrm{c}) \frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{\sum...
Read More →If $mathrm{AB}_{4}$ molecule is a polar molecule,
Question: If $\mathrm{AB}_{4}$ molecule is a polar molecule, a possible geometry of $\mathrm{AB}_{4}$ is :Square pyramidalTetrahedralRectangular planarSquare planarCorrect Option: 1 Solution: For $\mathrm{AB}_{4}$ compound possible geometry are Structure with $s p^{3} d^{2}$ hybridisation is polar due to lone pair moment while in other possibilities molecules is non-polar. Square pyramidal can be polar due to lone pair moment as the bond pair moments will get cancelled out....
Read More →Particle A of mass
Question: Particle $A$ of mass $m_{A}=\frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B}=\frac{m}{3}$. If both particles move along the $x$-axis after the collision, the change $\Delta \lambda$ in de-Broglie wavelength of particle $A$, in terms of its de-Broglie wavelength $\left(\lambda_{0}\right)$ before collision is :(1) $\Delta \lambda=\frac{3}{2} \lambda_{0}$(2) $\Delta \lambda=\frac{5}{2} \lambda_{0}$(3) $\...
Read More →If f(x)
Question: If $f(x)=\left\{\begin{array}{cl}\frac{1}{|x|} ;|x| \geq 1 \\ a x^{2}+b ;|x|1\end{array}\right.$ is differentiable at every point of the domain, then the values of a and b are respectively:(1) $\frac{1}{2}, \frac{1}{2}$(2) $\frac{1}{2},-\frac{3}{2}$(3) $\frac{5}{2},-\frac{3}{2}$(4) $-\frac{1}{2}, \frac{3}{2}$Correct Option: , 4 Solution: $f(x)=\left\{\begin{array}{cc}\frac{1}{|x|}, |x| \geq 1 \\ a x^{2}+b, |x|1\end{array}\right.$ at $x=1$ function must be continuous So, $1=a+b \ldots$ ...
Read More →Solve this
Question: Let $\bar{x}$ be the mean of $x_{1}, x_{2}, \ldots, x_{n}$ and $\bar{y}$ be the mean of $y_{1}, y_{2}, \ldots, y_{n}$. If $\bar{z}$ is the mean of $x_{1}, x_{2}, \ldots, x_{n}, y_{1}, y_{2}, \ldots, y_{n}$, then $\bar{z}=?$ (a) $(\bar{x}+\bar{y})$ (b) $\frac{1}{2}(\bar{x}+\bar{y})$ (c) $\frac{1}{n}(\bar{x}+\bar{y})$ (d) $\frac{1}{2 n}(\bar{x}+\bar{y})$ Solution: (b) $\frac{1}{2}(\bar{x}+\bar{y})$ $\bar{z}=\frac{\left(x_{1}+x_{2}+\ldots+x_{n}\right)+\left(y_{1}+y_{2}+\ldots+y_{n}\right)...
Read More →Match List-I with List-II.
Question: Match List-I with List-II. Choose the correct answer from the options given below:$(\mathrm{a})-(\mathrm{iii}),(\mathrm{b})-(\mathrm{iv}),(\mathrm{c})-(\mathrm{i}),(\mathrm{d})-(\mathrm{iii})$(a) $-($ i $),($ b $)-($ ii $),($ c $)-($ iii $),($ d $)-($ iv $)$$(\mathrm{a})-(\mathrm{ii}),(\mathrm{b})-(\mathrm{i}),(\mathrm{c})-(\mathrm{i} v),(\mathrm{d})-(\mathrm{iii})$$(\mathrm{a})-(\mathrm{i} v),(\mathrm{b})-(\mathrm{iii}),(\mathrm{c})-(\mathrm{ii}),(\mathrm{d})-(\mathrm{i})$Correct Opti...
Read More →If the function
Question: If the function $f(x)=\frac{\cos (\sin x)-\cos x}{x^{4}}$ is continuous at each point in its domain and $\mathrm{f}(0)=\frac{1}{\mathrm{k}}$, then $\mathrm{k}$ is______. Solution: $\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}=f(0)$ $\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{\sin x+x}{2}\right)^{\sin \left(\frac{x-\sin x}{2}\right)}}{x^{4}}=\frac{1}{K}$ $\Rightarrow \lim _{x \rightarrow 0} 2\left(\frac{\sin x+x}{2 x}\right)\left(\frac{x-\sin x}{2 x^{3}}\...
Read More →Let f
Question: Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $f(x)=\left\{\begin{array}{rr}x+a, x0 \\ |x-1|, x \geq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{rl}x+1, x0 0 \\ (x-1)^{2}+b, x \geq 0 \end{array}\right.$ where $\mathrm{a}, \mathrm{b}$ are non-negative real numbers. If (gof) $(\mathrm{x})$ is continuous for all $\mathrm{x} \in \mathrm{R}$, then $\mathrm{a}+\mathrm{b}$ is equal to________. Solution: $g[f(x)]=\...
Read More →A block of mass 1.9 kg is at rest at the edge of a table,
Question: A block of mass $1.9 \mathrm{~kg}$ is at rest at the edge of a table, of height $1 \mathrm{~m}$. A bullet of mass $0.1 \mathrm{~kg}$ collides with the block and sticks to it. If the velocity of the bullet is $20 \mathrm{~m} / \mathrm{s}$ in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$. Assume there is no rotational motion and losss of energy after the collision ...
Read More →The mean of 100 observations is 50.
Question: The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be(a) 50.5(b) 51(c) 51.5(d) 52 Solution: (b) 51Mean of 100 observations = 50 Sum of 100 observations $=100 \times 50=5000$ It is given that one of the observations, 50, is replaced by 150. New sum = (5000-50 + 150) = 5100And, Resulting mean $=\frac{5100}{100}=51$...
Read More →Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Question: Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Dipole-dipole interactions are the only non-covalent interactions, resulting in hydrogen bond formation Reason $\mathrm{R}$ : Fluorine is the most electronegative element and hydrogen bonds in HF are symmetrical In the light of the above statements, choose the most appropriate answer from the options given below :$\mathrm{A}$ is false but $\mathrm{R}$ is trueBoth $A$ and...
Read More →Let alpha in R be such that the function
Question: Let $\alpha \in \mathrm{R}$ be such that the function $f(x)= \begin{cases}\frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, x \neq 0 \\ \alpha, x=0\end{cases}$ is continuous at $x=0$, where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $\mathrm{x}$. Then:(1) $\alpha=\frac{\pi}{\sqrt{2}}$(2) $\alpha=0$(3) no such $\alpha$ exists(4) $\alpha=\frac{\pi}{4}$Correct Option: , 3 Solution: $\operatorname{Lim}_{\mathrm{x} \rightarrow 0^{+}} \mathr...
Read More →The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively.
Question: The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is(a) 64.86(b) 65.31(c) 64.91(d) 64.61 Solution: (c) 64.91Mean of 100 items = 64 Sum of 100 items $=64 \times 100=6400$ Correct sum $=(6400+36+90-26-9)=6491$ Correct mean $=\frac{6491}{100}=64.91$...
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