Answer: Consider, e be the identity element in I+ with respect to * a * e = a = e * a, ∀ a ∈ I+ a * e = a and e * a = a, ∀ a ∈ I+ a + e = a and e + a = a, ∀ a ∈ I+ e = 0, ∀ a ∈ I+ Hence, 0 is the...
![\[\frac{{{x}^{2}}}{25}-\frac{{{y}^{2}}}{4}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-79b83cfd05141077d859b528a32f1b5d_l3.png "Rendered by QuickLaTeX.com")
Given Equation: \[\frac{{{x}^{2}}}{25}-\frac{{{y}^{2}}}{4}=1\] Comparing with the equation of hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] we get, a = 5 and b = 2 (v)...
Use king theorem of definite integral \[\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx\] \[y=\int _{0}^{\pi }\log (1+\cos (\pi -x))dx\] \[y=\int _{0}^{\pi }\log (1-\cos x)dx........(2)\] Adding...
Solution: $\left|\begin{array}{ccc} a & b & a x+b y \\ b & c & b x+c y \\ a x+b y & b x+c y & 0 \end{array}\right|$ $\begin{array}{l} \left.=\left(\frac{1}{x...
![\[\frac{2}{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-047bcea7c848d7bdc13d26f95e67a4a4_l3.png "Rendered by QuickLaTeX.com")
![\[5\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7dc8465919d0838925dbc6c99670258c_l3.png "Rendered by QuickLaTeX.com")
![\[10\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7e13914576c8f07fa7780b4b1aae684d_l3.png "Rendered by QuickLaTeX.com")
![\[8\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5ba6e07885b354f6c76c81fbd23c392a_l3.png "Rendered by QuickLaTeX.com")
![\[(4,1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1ae2e4bedacd7350453509fb457a0319_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \pm \mathbf{3},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-230cb6920566958d0c605f61419a7ed1_l3.png "Rendered by QuickLaTeX.com")
![\[\frac{3}{4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b506a58aabfe65c0a3f19affd6f9c8e7_l3.png "Rendered by QuickLaTeX.com")
![\[(6,4)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5d92f1e8ff96dfe442d91764fafe5723_l3.png "Rendered by QuickLaTeX.com")
Given Eccentricity = \[\frac{3}{4}\] We know that Eccentricity = c/a Therefore,c=\[\frac{3}{4}\]a
![\[\left( \mathbf{4},\text{ }\mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6e14255e5f2d2cbc940379e069bb5844_l3.png "Rendered by QuickLaTeX.com")
![\[(-1,4)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aeaab2e61affb979c28d3883772be7c5_l3.png "Rendered by QuickLaTeX.com")
Given: Center is at the origin and Major axis is along x – axis So, Equation of ellipse is of the form \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Given that ellipse passing...
![\[\left( \mathbf{0},\text{ }\pm \mathbf{4} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bb2be58eb8036aac914c2c677f788949_l3.png "Rendered by QuickLaTeX.com")
![\[e=\frac{4}{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c0fb8f8c87448e0d2f359f2d167e4c1a_l3.png "Rendered by QuickLaTeX.com")
Given: Coordinates of foci = \[\left( \mathbf{0},\text{ }\pm \mathbf{4} \right)\] …(i) We know that, Coordinates of foci = \[\left( 0,\text{ }\pm c \right)\] …(ii) The coordinates of the foci are...
![\[\left( \pm \mathbf{1},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-873e68902b5445bb221bebb969489511_l3.png "Rendered by QuickLaTeX.com")
Let the equation of the required ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Given: Coordinates of foci = \[\left( \pm 1,\text{ }0 \right)\] …(i) We know that,...
![\[\left( \pm \mathbf{2},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7120150033b6ca5ebc01c838e0e9b285_l3.png "Rendered by QuickLaTeX.com")
![\[\frac{1}{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eca8ad2eb4fe01867f7756c04c9a2882_l3.png "Rendered by QuickLaTeX.com")
Let the equation of the required ellipse be Given: Coordinates of foci = \[\left( \pm 2,\text{ }0 \right)\]…(iii) We know that, Coordinates of foci = \[\left( \pm c,\text{ }0 \right)\]…(iv) ∴ From...
![\[\left( \pm \mathbf{4},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7e4c65ec2153cec645be0c25f93adce9_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2d6b922fc27485fd054fb582ccf195be_l3.png "Rendered by QuickLaTeX.com")
Given: Ends of Major Axis = \[\left( \pm \mathbf{4},\text{ }\mathbf{0} \right)\] and Ends of Minor Axis = \[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\] Here, we can see that the major axis is...
![\[(0,\pm 7)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-06930137911286949055c190e479f75e_l3.png "Rendered by QuickLaTeX.com")
![\[(0,\pm \sqrt{7})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1992b2f60ca3bf6ba810d7743190c38a_l3.png "Rendered by QuickLaTeX.com")
Given: Vertices = \[(0,\pm 4)\] …(i) The vertices are of the form = (0, ±a) …(ii) Hence, the major axis is along y – axis ∴ From eq. (i) and (ii), we get \[\begin{array}{*{35}{l}} a\text{ }=\text{...
![\[\left( \pm \mathbf{6},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2a03fb4e270050a2606c20b670ee0466_l3.png "Rendered by QuickLaTeX.com")
![\[(\pm \mathbf{4},~\mathbf{0})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-10ca5316ffee057110a312dedd039adb_l3.png "Rendered by QuickLaTeX.com")
Given: Vertices = \[\left( \pm \mathbf{6},\text{ }\mathbf{0} \right)\] …(i) The vertices are of the form = \[\left( \pm a,\text{ }0 \right)\] …(ii) Hence, the major axis is along x – axis ∴ From eq....
![\[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3badc0893d1a72abf2537d836a63e857_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
Given: \[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
![\[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c1bda8518d8de0566fb7bb8dae83e7d6_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}\] Divide by \[16\] to both the sides, we get...
Given: \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}\]…(i) Divide by \[18\] to both the sides, we get...
![\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1f0a66ca234bf1c2806a47a69c3bf010_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing eq. (i)...
Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing...
Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing eq. (i)...
Given: \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]…(i) Since, \[4\text{ }<\text{ }25\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii)...
![\[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cd71d027e4ee356834184fd92e4f618_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]…(i) Since, \[4\text{ }<\text{ }25\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii)...
![\[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c30dd6bfab0df4e91e525631f2ba2f4a_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since,...
Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since,...
Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since, ...
![\[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-862342dfce29b8b9c9392c90bd44304d_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...
![\[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b1c4d22d5b39a8859feb9ee71c371bb0_l3.png "Rendered by QuickLaTeX.com")
Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
![\[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aed3022a6a6b09b7823a798e1b61f3a4_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...
![\[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-07b8fa9753bba0be6e51668e79506671_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]…(i) Since, \[49\text{ }>\text{ }36\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] …(ii)...
Given: \[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]…(i) Since, \[49\text{ }>\text{ }36\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] …(ii)...
![\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d32e615a6517b76e65b04fd25ae8bf0a_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Since, \[25>9\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(ii)...
Given: \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Since, \[25>9\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(ii)...
Given, A beam is supported at its ends by supports which are 12 m apart. There is a deflection of 3 cm at the centre, and the deflected beam is in the shape of a parabola. EF are the...
Answer: Given, A rod of length 15 cm moves with its ends always touching the coordinate axes. A point P on the rod, which is at a distance of 3 cm from the end in contact with...
Answer: Given, Top of the towers are 30 m above the roadway and are 200 m apart. Cable is 5 m above the roadway at centre. A and B are the top of the towers. AE and BF are the...
Answer: Given, Parabolic reflector = 5 cm deep Diameter = 20 cm Reflector is 5 cm deep, OD = 5 cm Diameter of the mirror is 20 com, BC = 20 cm The equation...
Answer: Given, The focus of a parabolic mirror is at a distance of 6 cm from its vertex. And the mirror is 20 cm deep. O is the vertex A is the Focus OA = a = 6...
Find a vector 
which is perpendicular to both 
and 
, and is such that 
.Solution: $\begin{array}{l} \vec{a}=4 \hat{\imath}+5 \hat{\jmath}-\hat{k} \\ \vec{b}=\hat{\imath}-4 \hat{\jmath}+5 \hat{k} \end{array}$ $\vec{c}=3 \hat{\imath}+\hat{\jmath}-\hat{k}$ Let...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 3x – 4y + 1 = 0 …(i) 5x + y – 1 = 0 …(ii) Now, we find the point of...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 2x + 3y – 2 = 0 …(i) x – 2y + 1 = 0 …(ii) Now, we find the point of...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 2x – 3y + 1 = 0 …(i) x + y – 2 = 0 …(ii) Now, we find the point of...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – 7y + 5 = 0 …(i) 3x + y – 7 = 0 …(ii) Now, we find the point of...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 2x – 3y = 0 …(i) 4x – 5y = 2 …(ii) Now, we find the point of intersection...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. 5x – 3y = 1 …(i) 2x + 3y = 23 …(ii) Now, we find the point of intersection...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – y = 1 …(i) 2x – 3y + 1 = 0 …(ii) Now, we find the point of intersection...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x + y = 9 …(i) 2x – 3y + 7 = 0 …(ii) Now, we find the point of intersection...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – y = 7 …(i) 2x + y = 2 …(ii) Now, we find the point of intersection of...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. x – 2y + 3 = 0 …(i) 2x – 3y + 4 = 0 …(ii) Now, we find the point of...
Answer : Let the new origin be (h, k) = (1, -2) Then, the transformation formula become: x = X + 1 and y = Y + (-2) = Y – 2 Substituting the value of x and y in the given equation, we get 2x2 + y2 –...
Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get xy – x – y + 1 = 0...
x2 – y2 – 2x + 2y = 0 Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get...
Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get xy – y2 – x + y = 0...
x2 + xy – 3x – y + 2 = 0 Answer : Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we...
Answer : Let (h, k) be the point to which the origin is shifted. Then, x = -4, y = 2, X = 3 and Y = -2 ∴ x = X + h and y = Y + k ⇒ -4 = 3 + h and 2 = -2 + k ⇒ h = -7 and k = 4 Hence, the origin must...
Answer : Let the new origin be (h, k) = (2, -1) and (x, y) = (-3, 5) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ -3 = X + 2 and...
Answer : Let the new origin be (h, k) = (0, -2) and (x, y) = (3, 2) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ 3 = X + 0 and 2...
Answer : Let the new origin be (h, k) = (-3, -2) and (x, y) = (3, -5) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ 3 = X – 3 and...
Answer : Let the new origin be (h, k) = (1, 2) and (x, y) = (3, -4) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k ⇒ 3 = X + 1 and...
Let the equation of line AB be x – 4y + 7 = 0 and point C be (-2, 3) CD is perpendicular to the line AB, and we need to find: Equation of Perpendicular drawn from point C Coordinates of D Let the...
Let the equation of line AB be 4x – 3y + 5 = 0 and point C be (0, 0) CD is perpendicular to the line AB, and we need to find: Equation of Perpendicular drawn from point C Coordinates of D Let the...
Answer : The given equations are y = x …(i) y = 2x …(ii) and y – 3x = 4 …(iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC From eq. (i) and (ii), we get x = 0...
Answer : The given equations are x = 0 …(i) y = 1 …(ii) and 2x + y = 2 …(iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC From eq. (i) and (ii), we get x = 0...
The given equations are x + y = 6 …(i) x – 3y = 2 …(ii) and 5x – 3y + 2 = 0 or 5x – 3y = -2 …(iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC Firstly, we...
Answer : Given that 3x – y – 2 = 0, 5x + ky – 3 = 0 and 2x + y – 3 = 0 are concurrent We know that, The lines a1x + b1y + c1 = 0, a1x + b1y + c1 = 0 and a1x + b1y + c1 = 0 are concurrent if It is...
Answer : Given: 3x – 4y + 5 = 0, 7x – 8y + 5 = 0 and 4x + 5y = 45 or 4x + 5y – 45 = 0 To show: Given lines are concurrent The lines a1x + b1y + c1 = 0, a1x + b1y + c1 = 0 and a1x + b1y + c1 = 0 are...
Answer : Suppose the given two lines intersect at a point P(2, 3). Then, (2, 3) satisfies each of the given equations. So, taking equation x + 7y = 23 Substituting x = 2 and y = 3 Lhs = x + 7y = 2 +...
Answer : Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations. ∴ 4x + 3y = 5 or 4x + 3y – 5 = 0 …(i) and x = 2y – 7 or x – 2y + 7 = 0...
Answer : Given: p = 4 and ???? = 1800 Here p is the perpendicular that makes an angle ???? with positive direction of x-axis , hence the equation of the straight line is given by: Formula used: x...
Answer : Given: p = 2 and ???? = 3000 Here p is the perpendicular that makes an angle ???? with positive direction of x-axis , hence the equation of the straight line is given by: Formula used: X...
Answer : Given: p = 3 and ???? = 2250 Here p is the perpendicular that makes an angle ???? with positive direction of x-axis , hence the equation of the straight line is given by: Formula used: x...
Answer : Given: p = 8 and ???? = 1500 Here p is the perpendicular that makes an angle ???? with positive direction of x-axis , hence the equation of the straight line is given by: Formula used: x...
Answer : Given: p = 5 and ???? = 1350 Here p is the perpendicular that makes an angle ???? with positive direction of x-axis , hence the equation of the straight line is given by: Formula used: x...
Answer : To Find:The equation of the line. Given: p = 3 and ???? = 450 Here p is the perpendicular that makes an angle ???? with positive direction of x-axis , hence the equation of the straight...
Answer : To Find: The equation of the line passing through (3, -5) and cuts off intercepts on the axes which are equal in magnitude but opposite in sign. Given : Let a and b be two intercepts of...
Answer : To Find: The equation of the line with equal intercepts on the coordinate axes and that passes through the point (4,7). Given : Let a and b be two intercepts of x-axis and y-axis...
Answer : To Find:The equation of the line with intercepts 4 and -6 on the x-axis and y- axis respectively. Given : Let a and b be the intercepts on x-axis and y-axis respectively. Then,x-intercept...
Answer : To Find: The equation of a line with intercepts -3 and 5 on the x-axis and y- axis respectively. Given :Let a and b be the intercepts on x-axis and y-axis respectively. Then, the...
, where 
is ![[-\infty, 0]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-94a61bc9af7b9f196d46b44726815c11_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(D) is correct. $f(x)=a x$, where $a>0$ Case 1 : When $x<0$, then ax lies between $(0,1)$ Case 2 : When $x \geq 0$, then $a x \geq 1$ Union of above two cases, gives us the...
. Then, range 
![[-1,1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bd9ec9b7fdf72f6f954becf26cf7129a_l3.png "Rendered by QuickLaTeX.com")
![(0,1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3243e3012721250a044fef79f51a0c2f_l3.png "Rendered by QuickLaTeX.com")
Solution: Option (B) is correct. $\mathrm{f}(\mathrm{x})=\frac{x^{2}}{\left(1+x^{2}\right)}$ The range of $f(x)$ can be found out by putting $f(x)=y$ $\begin{array}{l}...
. Then, range 
![(-\infty, 1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2eb654b4016c2cc85fa136277a9d7f78_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $f(x)=\frac{1}{\left(1-x^{2}\right)}$ The range of $f(x)$ can be found out by putting $f(x)=y$ $\begin{array}{l} \mathrm{y}=\frac{1}{\left(1-x^{2}\right)} \\...
. Then, 
![(-\infty,-1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17e48fe12e81cf52057d1a1ecdfc8269_l3.png "Rendered by QuickLaTeX.com")
Solution: Option (B) is correct. $f(x)=\log (1-x)+\sqrt{x^{2}-1}$ Solving inequality, $\log (1-x) \geq 0$ $\Rightarrow 1-x \geq \mathrm{e}^{0} \quad \begin{array}{c}\text { (Log taken to the...
. Then, dom (f) and range (f) are respectively 
and 
and and Solution: Option(A) is correct. $f(x)=x^{3}$ $f(x)$ can assume any value, therefore domain of $f(x)$ is $R$ Range of the function can be positive or negative Real numbers, as the cube of any number...
. Then, dom (f) and range (f) are respectively. and and and and 
Solution: Option(C) is correct. $f(x)=x_{2}$ $f(x)$ can assume any value, therefore domain of $f(x)$ is $R$ Range of the function can only be positive Real numbers, as the square of any number is...
. Then, 
![[1,2]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ab425e31cf7af5fb3d46646ff538fdcd_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(C) is correct. $f(x)=\sqrt{\log \left(2 x-x^{2}\right)}$ For $f(x)$ to be defined $2 x-x^2$ should be positive. Solving inequality, (Log taken to the opposite side of the equation...
. Then, ![\left[0, \frac{\pi}{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cdf7f72b896a3c8ef3745557ca17a3c9_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{3 \pi}{2}, 2 \pi\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aa7b802c7af63a3c3f09db8b0292cb35_l3.png "Rendered by QuickLaTeX.com")
![\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-be8b8bc855076ca9e77d0af23204c861_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(C) is correct. $\mathrm{f}(\mathrm{x})=\sqrt{\cos x}$ As per graph of $\sqrt{\cos x}$ the domain is $\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$
. Then, 
? 
![\left[0, \frac{2}{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b52292076dc04c74fe44cded15544d8b_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{-2}{3}, \frac{2}{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c6d1d6173855bcc8e2e22c311f052282_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $\begin{array}{l} \mathrm{f}(\mathrm{x})=\cos ^{-1}(3 \mathrm{x}-1) \end{array}$ Domain for function $\cos ^{-1} \mathrm{x}$ is $[-1,1]$ and range is $[0, \pi]$ When...
. Then, ![\left[\frac{-1}{2}, \frac{1}{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2b2127e7c856ce7280a6fc7257d27da7_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d03f7933dddf8dbceee8c92c85707a25_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-36caf65bb0d2f985fd702c950c7c7653_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $f(x)=\cos ^{-1}2 x$ Domain for function $\cos ^1 \mathrm{x}$ is $[-1,1]$ and range is $[0, \pi]$ When a function is multiplied by an integer, the domain of the...
Then, 
![[-1,1]-\{0\}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0d140ee7fc9eafbc15a3b47a6ad32f55_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $f(x)=\frac{\sin ^{-1} x}{x}$ The domain of the function is defined for $\mathrm{x} \neq 0$ domain of $\sin ^{-1} x$ is $[-1,1]$ So, domain of...
. Then, 
? 
Solution: Option(D) is correct. $\mathrm{f}(\mathrm{x})=\frac{x}{\left(x^{2}-1\right)}$ The domain of the function is defined for $\begin{array}{l} \mathrm{x}^{2}-1 \neq 0 \\ \Rightarrow \mathrm{x}...
. Then, ? 
![(-\infty,-1] \cup(1, \infty)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f70651ff938d8644c2a7b1e4ab8dcdbc_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(C) is correct. $\mathrm{f}(\mathrm{x})=e^{\sqrt{x^{2}-1}} \cdot \log (x-1)$ The domain of the function is defined for $\begin{array}{l} \mathrm{x}-1>0 \quad \text { and }...
. Then, dom (f) – ? 
![[1,4]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b466cd4c540c856fc9d44d24b88d173d_l3.png "Rendered by QuickLaTeX.com")
![(-\infty, 4]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7342c3266939896b38339bca55408448_l3.png "Rendered by QuickLaTeX.com")
![(-\infty, 1] \cup(4, \infty)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3cacb249d82a72cc2e90b22ea1598fb5_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(D) is correct. $\mathrm{f}(\mathrm{x})=\sqrt{\frac{x-1}{x+4}}$ The domain of the function can be defined for $\sqrt{\frac{x-1}{x+4}} \geq 0$ $\begin{array}{l} \Rightarrow...
. Then, dom ![[-3,3]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9cbb1b08c8e67cf43d5548e7d359de56_l3.png "Rendered by QuickLaTeX.com")
![[-\infty,-3]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-697587bce5e8d6c46d3797bdabaae044_l3.png "Rendered by QuickLaTeX.com")
![(-\infty,-3] \cup(4, \infty)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4747a453581bf4e2adb77e958e00ac9_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(A) is correct. $\mathrm{f}(\mathrm{x})=\sqrt{9-x^{2}}$ The domain of the function can be defined for $\sqrt{9-x^{2}} \geq 0$ $\begin{array}{l} \Rightarrow \sqrt{9-x^{2}} \geq 0 \\...
Solution: Option(B) is correct. $\begin{array}{l} \mathrm{f}=\{(1,2),(3,5),(4,1)\} \\ \mathrm{g}=\{(2,3),(5,1),(1,3)\} \end{array}$ According to the combination of $\mathrm{f}$ and $\mathrm{g}$,...
and 
then 
Solution: Option(A) is correct. $\begin{array}{l} f(x)=x ^2 \\ g(x)=\tan x \\ h(x)=\log x \end{array}$ According to the combination of $f, g$ and $h$,...
and 
then 
Solution: Option(B) is correct. $f(x)=8 x^{3}$ $g(x)=x^{1 / 3}$ According to the combination of $\mathrm{f}$ and $\mathrm{g}$, $\operatorname{gof}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{x}))$...
then 
Solution: Option(D) is correct. $f(x)=x^2-3 x+2$ According to the combination of $\mathrm{f}$ and $\mathrm{f}$, $\operatorname{fof}(x)=f(f(x))$ Therefore, fof $(x)=f(f(x))$ $\begin{array}{l}...
![f(x)=\sqrt[3]{3-x^{3}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6c33645c75bd40d87b10258556b03f43_l3.png "Rendered by QuickLaTeX.com")
then (f of) 
Solution: Option(B) is correct. $f(x) \sqrt[3]{3-x^{3}}$ According to the combination of $f$ and $f$, $\text { fof }(x)=f(f(x))$ Therefore, fof $(x)=f(f(x))$...
then (f of of) 
Solution: Option(C) is correct. $\mathrm{f}(\mathrm{x})=\frac{1}{(1-x)}$ According to the combination of $\mathrm{f}$ and $\mathrm{f}$, fofof $(x)=f(f(f(x)))$ Therefore, fof $(x)=f(f(f(x))$...
then 
Solution: Option(C) is correct. $\begin{array}{l} \mathrm{f}\left(\mathrm{x}+\frac{1}{x}\right)=\left(\mathrm{x} ^2+\frac{1}{x^{2}}\right) \\ \Rightarrow...
and 
then 
Solution: Option(C) is correct. $\begin{array}{l} \mathrm{f}(\mathrm{x})=(\mathrm{x}^2-1) \\ \mathrm{g}(\mathrm{x})=(2 \mathrm{x}+3) \end{array}$ According to the combination of $\mathrm{f}$ and...
then 
Solution: Option(A) is correct. $\mathrm{f}(\mathrm{x})=\frac{(4 x+3)}{(6 x-4)}, \mathrm{x} \neq \frac{2}{3}$ According to the combination of $\mathrm{f}$ and $\mathrm{f}$,...
(y) = ? 
Solution: Option(B) is correct. $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{X}: \mathrm{f}(\mathrm{x})=4 \mathrm{x} 2+12 \mathrm{x}+15$ We need to find $\mathrm{f}-1$, Suppose...
. Then 
Solution: Option(A) is correct. $\text { f: } R-\left\{-\frac{4}{3}\right\} \rightarrow-\left\{\frac{4}{3}\right\}: f(x)=\frac{4 x}{(3 x+4)}$ We need to find $\mathrm{f}-1$ Suppose $f(x)=y$...
(y) = ? 
Solution: Option(C) is correct. $\mathrm{f}: \mathrm{Q} \rightarrow \mathrm{Q}: \mathrm{f}(\mathrm{x})=(2 \mathrm{x}+3)$ We need to find $\mathrm{f}-1$ Suppose $\mathrm{f}(\mathrm{x})=\mathrm{y}$...
Solution: Option(B) is correct. One-One Function Suppose $\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{q}_{1}, \mathrm{q}_{2}$ be two arbitrary elements in $\mathrm{R}$ + Therefore, $f\left(p_{1},...
Then, 
is Solution: Option(D) is correct. $f:\mathrm{N} \rightarrow \mathrm{N}: \mathrm{f}(\mathrm{x})=$ $f: N \rightarrow N: f(x)=\left\{\begin{array}{l}\frac{1}{2}(n+1) \text {, when } n \text { is odd } \\...
and 
. Then 
is Solution: Option(B) is correct. f: $\mathrm{A} \rightarrow \mathrm{B}: \mathrm{f}(\mathrm{x})=\frac{(x-2)}{(x-3)}$ Where, $\mathrm{A}=\mathrm{R}-\{3\}$ and $\mathrm{B}=\mathrm{R}-\{1\}$ One-One...
Solution: Option() is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}+$ Therefore, $f(p)=f(q)$ $\begin{array}{l}...
Solution: Option (C) is correct. One-one function $\cos x$ graph cuts y axis repeatedly, hence it is many-one. Onto function Range of $f(x)$ is $[-1,1]$ Co-domain is $\mathrm{R}$ So here, Range of...
is Solution: Option(D) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}+$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
is Solution: Option(B) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
is Solution: Option(D) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
+ x + 1 is Solution: Option (B) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{N}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
Solution: Option (B) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{N}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow 2 \mathrm{p}=2...
The correct answer is a) number of charge carriers can change with temperature T b) time interval between two successive collisions can depend on T
Answer : Given: g = {(1, 2), (2, 5), (3, 8), (4, 10), (5, 12), (6, 12)} We know that, A function ‘f’ from set A to set B is a correspondence (rule) which associates elements of set A to elements of...
Solution: A function is stated as the relation between the two sets, where there is exactly one element in set B, for every element of set A. A function is represented as f: A → B, which means ‘f’...
Solution: $A=\{1,2,3\}$ and $\bar{R}=\{(1,1),(2,2),(3,3),(1,2),(2,3)\}$ (Given) $\mathrm{R}$ is reflexive if $\mathrm{a} \in \mathrm{A}$ and $(\mathrm{a}, \mathrm{a}) \in \mathrm{R}$ Here,...
for all 
Show that R satisfies none of reflexivity, symmetry and transitivity.Solution: $\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=\mathrm{b}_{2}\right\}$ for all $\mathrm{a}, \mathrm{b} \in \mathrm{N}$ (As given) Non-Reflexivity: Assume $a$ be an arbitrary...
Solution: $\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{d}(\mathrm{A}, \mathrm{B})<2$ units $\}$, where $\mathrm{d}(\mathrm{A}, \mathrm{B})$ is the distance between the points $\mathrm{A}$ and...
be the set of all real numbers and let 
and 
Show that 
is an equivalence relation on .Solution: $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{S}$ and $\mathrm{a}=\pm \mathrm{b}\} .$ (Given) If $\mathrm{R}$ is Reflexive, Symmetric and Transitive, then...
on 
, defined by 
is an equivalent relation.Solution: If $R$ is Reflexive, Symmetric and Transitive, then $R$ is an equivalence relation. Reflexivity: Suppose $a$ and $\mathrm{b}$ be an arbitrary element of $\mathrm{N} \times \mathrm{N}$...
and 
is divisible by 5
Show that is an equivalence relation on 
.Solution: $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{Z}$ and $(\mathrm{a}-\mathrm{b})$ is divisible by 5$\}$ (As given) If $R$ is Reflexive, Symmetric and Transitive,...
be the set of all triangles in a plane. Show that the relation 
is an equivalence relation on 
.Solution: Suppose $\mathrm{R}=\left\{\left(\Delta 1, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ be a relation defined on A. (As given) If $\mathrm{R}$ is Reflexive, Symmetric and...
Solution: Relation: Suppose $P$ and $Q$ are two sets. Therefore, a relation $R$ from $P$ to $Q$ is a subset of $P \times Q$. Therefore, $\mathrm{R}$ is a relation to $\mathrm{P}$ to $\mathrm{Q}...
Solution: A $= (1, 2, 3, 4, 5, 6)$ and $R = {(a, b): a, b \in \text{A and b} = a + 1}$ (As given) Therefore, R = {(1,2), (2,3), (3,4), (4,5), (5,6)} (i) Non−reflexive: If $x \in A$ and $(x, x) \in...