Prove that the points A(2, 4), B(2, 6) and
Question: Prove that the points $A(2,4), B(2,6)$ and $C(2+\sqrt{3}, 5)$ are the vertices of an equilateral triangle. Solution: The given points are $A(2,4), B(2,6)$ and $C(2+\sqrt{3}, 5)$. Now $A B=\sqrt{(2-2)^{2}+(4-6)^{2}}=\sqrt{(0)^{2}+(-2)^{2}}$ $=\sqrt{0+4}=2$ $B C=\sqrt{(2-2-\sqrt{3})^{2}+(6-5)^{2}}=\sqrt{(-\sqrt{3})^{2}+(1)^{2}}$ $=\sqrt{3+1}=2$ $A C=\sqrt{(2-2-\sqrt{3})^{2}+(4-5)^{2}}=\sqrt{(-\sqrt{3})^{2}+(-1)^{2}}$ $=\sqrt{3+1}=2$ Hence, the points $A(2,4), B(2,6)$ and $C(2+\sqrt{3}, 5...
Read More →1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...
Question: 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=1+2+3+4+5+\ldots+n=\frac{n(n+1)}{2}=\frac{n^{2}+n}{2}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n}\left(\frac{k^{2}+k}{2}\right)$ $\Rightarrow S_{n}=\frac{1}{2} \sum_{k=1}^{n}\left(k^{2}+k\right)$ $\Rightarrow S_{n}=\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\fra...
Read More →Find the value of k, if the points A(7, −2),
Question: Find the value of k, if the pointsA(7, 2),B(5, 1) andC(3, 2k) are collinear. Solution: The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$ If three points are collinear the area encompassed by them is equal to 0. The three given points are...
Read More →1.2.4 + 2.3.7 +3.4.10 + ...
Question: 1.2.4 + 2.3.7 +3.4.10 + ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=n(n+1)(3 n+1)=n\left(3 n^{2}+4 n+1\right)=\left(3 n^{3}+4 n^{2}+n\right)$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n} 3 k^{3}+\sum_{k=1}^{n} 4 k^{2}+\sum_{k=1}^{n} k$ $\Rightarrow S_{n}=3 \sum_{k=1}^{n} k^{3}+4 \sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$ $\Rightarrow S_{n}=\frac{3 n^{2}(...
Read More →Find the value of k if points A(k, 3), B(6, −2)
Question: Find the value of $k$ if points $\mathrm{A}(k, 3), \mathrm{B}(6,-2)$ and $\mathrm{C}(-3,4)$ are collinear. Solution: The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$ If three points are collinear the area encompassed by them is equal to...
Read More →Solve the following
Question: 1.2.5 + 2.3.6 + 3.4.7 + ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=n(n+1)(n+4)$ $=n\left(n^{2}+5 n+4\right)$ $=\left(n^{3}+5 n^{2}+4 n\right)$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} 5 k^{2}+\sum_{k=1}^{n} 4 k$ $\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$ $\Rightarrow S_{n}=\frac{n^{2}(...
Read More →Solve the following
Question: 1.2.5 + 2.3.6 + 3.4.7 + ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=n(n+1)(n+4)$ $=n\left(n^{2}+5 n+4\right)$ $=\left(n^{3}+5 n^{2}+4 n\right)$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} 5 k^{2}+\sum_{k=1}^{n} 4 k$ $\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$ $\Rightarrow S_{n}=\frac{n^{2}(...
Read More →Solve the following
Question: 22+ 42+ 62+ 82+ ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=(2 n)^{2}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $=\sum_{k=1}^{n}(2 k)^{2}$ $=\sum_{k=1}^{n} 4 k^{2}$ $=4 \sum_{k=1}^{n} k^{2}$ $=4 \frac{n(n+1)(2 n+1)}{6}$ $=\frac{2 n}{3}(n+1)(2 n+1)$...
Read More →Solve the following
Question: 22+ 42+ 62+ 82+ ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=(2 n)^{2}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $=\sum_{k=1}^{n}(2 k)^{2}$ $=\sum_{k=1}^{n} 4 k^{2}$ $=4 \sum_{k=1}^{n} k^{2}$ $=4 \frac{n(n+1)(2 n+1)}{6}$ $=\frac{2 n}{3}(n+1)(2 n+1)$...
Read More →Solve the following
Question: $1^{3}+3^{3}+5^{3}+7^{3}+\ldots$ Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $=\sum_{k=1}^{n}[2 k-1]^{3}$ $=\sum_{k=1}^{n}\left[8 k^{3}-1-6 k(2 k-1)\right]$ $=\sum_{k=1}^{n}\left[8 k^{3}-1-12 k^{2}+6 k\right]$ $=\sum_{k=1}^{n}\left[8 k^{3}-1-12 k^{2}+6 k\right]$ $=8 \sum_{k=1}^{n} k^{3}-\sum_{k=1}^{n} 1-12 \sum_{k=1}^{n} k^{2}+6 \sum_{k=1}^{n} k$ $=\...
Read More →Show that the points A(5, 2), B(2, − 2) and C(− 2, t) are the vertices of a right triangle with
Question: Show that the points $A(5,2), B(2,-2)$ and $C(-2, t)$ are the vertices of a right triangle with $\angle B=90^{\circ}$, then find the value of $t$. Solution: $\because \angle B=90^{\circ}$ $\therefore A C^{2}=A B^{2}+B C^{2}$ $\Rightarrow(5+2)^{2}+(2-t)^{2}=(5-2)^{2}+(2+2)^{2}+(2+2)^{2}+(-2-t)^{2}$ $\Rightarrow(7)^{2}+(t-2)^{2}=(3)^{2}+(4)^{2}+(4)^{2}+(t+2)^{2}$ $\Rightarrow 49+t^{2}-4 t+4=9+16+16+t^{2}+4 t+4$ $\Rightarrow 8-4 t=4 t$ $\Rightarrow 8 t=8$ $\Rightarrow t=1$ Hence,t= 1....
Read More →If (x, y) be on the line joining the two points
Question: If (x, y) be on the line joining the two points (1, 3) and (4, 2) , prove that x + y + 2= 0. Solution: Since the point (x,y) lie on the line joining the points (1, 3) and (4, 2); the area of triangle formed by these points is 0. That is, $\Delta=\frac{1}{2}\{x(-3-2)+1(2-y)-4(y+3)\}=0$ $-5 x+2-y-4 y-12=0$ $-5 x-5 y-10=0$ $x+y+2=0$ Thus, the result is proved....
Read More →Show that the points A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.
Question: Show that the pointsA(3, 0),B(6, 4) andC( 1, 3) are the vertices of an isosceles right triangle. Solution: The given points areA(3, 0),B(6, 4) andC( 1, 3). Now $A B=\sqrt{(3-6)^{2}+(0-4)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}$ $=\sqrt{9+16}=\sqrt{25}=5$ $B C=\sqrt{(6+1)^{2}+(4-3)^{2}}=\sqrt{(7)^{2}+(1)^{2}}$ $=\sqrt{49+1}=\sqrt{50}=5 \sqrt{2}$ $A C=\sqrt{(3+1)^{2}+(0-3)^{2}}=\sqrt{(4)^{2}+(-3)^{2}}$ $=\sqrt{16+9}=\sqrt{25}=5$ $\because A B=A C$ and $A B^{2}+A C^{2}=B C^{2}$ Therefore,A(3, 0),B(...
Read More →If three points (x1, y1) (x2, y2), (x3, y3) lie on the same line, prove that
Question: If three points (x1, y1) (x2, y2), (x3, y3) lie on the same line, prove that $\frac{y_{2}-y_{3}}{x_{2} x_{3}}+\frac{y_{3}-y_{1}}{x_{3} x_{1}}+\frac{y_{1}-y_{2}}{x_{1} x_{2}}=0$ Solution: GIVEN: If three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ lie on the same line TO PROVE: $\frac{\left(y_{2}-y_{3}\right)}{x_{2} x_{3}}+\frac{\left(y_{3}-y_{1}\right)}{x_{3} x_{1}}+\frac{\left(y_{1}-y_{2}\right)}{x_{1} x_{2}}=0$ PROOF: We know that three...
Read More →Prove that the points A(7, 10), B(−2, 5) and C(3, −4)
Question: Prove that the pointsA(7, 10),B(2, 5) andC(3, 4) are the vertices of an isosceles right triangle. Solution: The given points areA(7, 10),B(2, 5) andC(3, 4). $A B=\sqrt{(-2-7)^{2}+(5-10)^{2}}=\sqrt{(-9)^{2}+(-5)^{2}}=\sqrt{81+25}=\sqrt{106}$ $B C=\sqrt{(3-(-2))^{2}+(-4-5)^{2}}=\sqrt{(5)^{2}+(-9)^{2}}=\sqrt{25+81}=\sqrt{106}$ $A C=\sqrt{(3-7)^{2}+(-4-10)^{2}}=\sqrt{(-4)^{2}+(-14)^{2}}=\sqrt{16+196}=\sqrt{212}$ Since, AB and BC are equal, they form the vertices of an isosceles triangle. A...
Read More →Prove that the points (a, b), (a1, b1) and
Question: Prove that the points (a, b), (a1, b1) and (a a1, b b1) are collinear if ab1= a1b. Solution: The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$ If three points are collinear the area encompassed by them is equal to 0. The three given poin...
Read More →Using the distance formula, show that the given points are collinear:
Question: Using the distance formula, show that the given points are collinear:(i) (1, 1), (5, 2) and (9, 5) (ii) (6, 9), (0, 1) and (6, 7)(iii) (1, 1), (2, 3) and (8, 11) (iv) (2, 5), (0, 1) and (2, 3) Solution: (i)LetA(1, 1),B(5, 2) andC(9, 5) be the given points. Then $A B=\sqrt{(5-1)^{2}+(2+1)^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5$ units $B C=\sqrt{(9-5)^{2}+(5-2)^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5$ units $A C=\sqrt{(9-1)^{2}+(5+1)^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{100}=10$ units $\therefore A B+...
Read More →For what value of a the point (a, 1), (1, −1), and
Question: For what value ofathe point (a, 1), (1, 1), and (11, 4) are collinear? Solution: The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$ If three points are collinear the area encompassed by them is equal to 0. The three given points areA(a,1)...
Read More →The true statement amongst the following
Question: The true statement amongst the following is :Both $\Delta \mathrm{S}$ and $\mathrm{S}$ are functions of temperature.Both $S$ and $\Delta S$ are not functions of temperature.$\mathrm{S}$ is not a function of temperature but $\Delta \mathrm{S}$ is a function of temperature.$S$ is a function of temperature but $\Delta S$ is not a function of temperature.Correct Option: 1 Solution: A system at higher temperature has greater entropy (randomness). $\Delta \mathrm{S}$ is related with $q$ and ...
Read More →If enthalpy of atomisation for
Question: If enthalpy of atomisation for $\mathrm{Br}_{2}(\mathrm{I})$ is $x \mathrm{~kJ} / \mathrm{mol}$ and bond enthalpy for $\mathrm{Br}_{2}$ is $y \mathrm{~kJ} / \mathrm{mol}$, the relation between them:is $x=y$does not existis $xy$is $xy$Correct Option: , 3 Solution: $\Delta \mathrm{H}_{\text {atomisation }}=\Delta \mathrm{H}_{\text {vap }}+$ Bond energy Hence $xy$...
Read More →Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given
Question: Four points $\mathrm{A}(6,3), \mathrm{B}(-3,5), \mathrm{C}(4,-2)$ and $\mathrm{D}(\mathrm{x}, 3 \mathrm{x})$ are given in such a way that $\frac{\Delta D B C}{\Delta A B C}=\frac{1}{2}$, find $x$. Solution: GIVEN: four points $A(6,3), B(-3,5) C(4,-2)$ and $D(x, 3 x)$ such that $\frac{\Delta D B C}{\Delta A B C}=\frac{1}{2}$ TO FIND: the value ofx PROOF: We know area of the triangles formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3}, y_{3}\righ...
Read More →If the point (x, y) is equidistant from the points (a + b, b − a) and (a − b, a + b), prove that bx = ay.
Question: (i) If the point (x,y) is equidistant from the points (a+b,ba) and (ab, a + b), prove thatbx=ay.(ii) If the distances ofP(x, y) fromA(5, 1) andB(1, 5) are equal then prove that 3x= 2y. Solution: (i) As per the question, we have $\sqrt{(x-a-b)^{2}+(y-b+a)^{2}}=\sqrt{(x-a+b)^{2}+(y-a-b)^{2}}$ $\Rightarrow(x-a-b)^{2}+(y-b+a)^{2}=(x-a+b)^{2}+(y-a-b)^{2} \quad$ (Squaring both sides) $\Rightarrow x^{2}+(a+b)^{2}-2 x(a+b)+y^{2}+(a-b)^{2}-2 y(a-b)=x^{2}+(a-b)^{2}-2 x(a-b)+y^{2}+(a+b)^{2}-2 y(a...
Read More →Solve the following
Question: If $a \neq b \neq c$, prove that the points $\left(a, a^{2}\right),\left(b, b^{2}\right),\left(c, c^{2}\right)$ can never be collinear. Solution: GIVEN: If $a \neq b \neq c$ TO PROVE: that the points $\left(a, a^{2}\right),\left(b, b^{2}\right),\left(c, c^{2}\right)$, can never be collinear. PROOF: We know three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ are collinear when $\frac{1}{2} \| x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_...
Read More →At constant volume,
Question: At constant volume, $4 \mathrm{~mol}$ of an ideal gas when heated from $300 \mathrm{~K}$ to $500 \mathrm{~K}$ changes its internal energy by $5000 \mathrm{~J}$. The molar heat capacity at constant volume is __________________ . Solution: $(6.25) \quad \Delta U=n C_{v} \Delta T$ $5000=4 \times C_{v}(500-300)$ $\mathrm{C}_{\mathrm{v}}=6.25 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$...
Read More →The magnitude of work done by a gas that undergoes a reversible expansion
Question: The magnitude of work done by a gas that undergoes a reversible expansion along the path $A B C$ shown in the figure is ______________ . Solution: (48.00) Work done is given by the area under the trapezium. $\therefore|\mathrm{w}|=\frac{1}{2}(6+10) \times 6=48 \mathrm{~J}$...
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