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...

### Find the (v) length of the rectum of each of the following the hyperbola :

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...

### Find the length of tangent drawn to a circle with radius 8 cm form a point 17 cm away from the center of the circle

### Using properties of determinants prove that:

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...

### Find the eccentricity of an ellipse whose latus rectum is one half of its major axis.

### Find the eccentricity of an ellipse whose latus rectum is one half of its minor axis.

### Find the equation of an ellipse whose eccentricity is

, the latus rectum is

, and the center is at the origin.

### Find the equation of an ellipse, the lengths of whose major and mirror axes are

and

units respectively.

### Find the equation of the ellipse which passes through the point

and having its foci at

.

### Find the equation of the ellipse with eccentricity

, foci on the y-axis, center at the origin and passing through the point

.

Given Eccentricity = \[\frac{3}{4}\] We know that Eccentricity = c/a Therefore,c=\[\frac{3}{4}\]a

### Find the equation of the ellipse with center at the origin, the major axis on the x-axis and passing through the points

and

.

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...

### Find the equation of the ellipse whose foci are at

and

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...

### Find the equation of the ellipse whose foci are at

and e=1/2

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,...

### Find the equation of the ellipse whose foci are

and the eccentricity is

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...

### Find the equation of the ellipse the ends of whose major and minor axes are

and

respectively.

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...

### Find the equation of the ellipse whose vertices are the

and foci at

.

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{...

### Find the equation of the ellipse whose vertices are at

and foci at

.

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....

### Find the (v) length of the latus rectum of each of the following ellipses.

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...

### Find the (iii) coordinates of the foci, (iv) eccentricity

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...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}\] Divide by \[16\] to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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...

### Find the (v) length of the latus rectum of each of the following ellipses.

Answer :

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)...

### Find the (iii) coordinates of the foci, (iv) eccentricity

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...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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)...

### Find the (v) length of the latus rectum of each of the following ellipses.

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)...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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)...

### Find the (v) length of the latus rectum of each of the following ellipses.

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,...

### Find the (iii) coordinates of the foci, (iv) eccentricity

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,...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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, ...

### Find the (v) length of the latus rectum of each of the following ellipses.

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...

### Find the (iii) coordinates of the foci, (iv) eccentricity

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...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (v) length of the latus rectum of each of the following ellipses.

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...

### Find the (iii) coordinates of the foci, (iv) eccentricity

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...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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...

### Find the (v) length of the latus rectum of each of the following ellipses.

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)...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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)...

### Find the(v) length of the latus rectum of each of the following ellipses.

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)...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

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)...

### If (1 + i)z = (1 – i) then prove that

### A beam is supported at its ends by supports which are 12 m apart. Since the load is concentrated at its centre, there is a deflection of 3 cm at the centre, and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm?

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...

### A rod of length 15 cm moves with its ends always touching the coordinate axes. Find the equation of the locus of a point P on the rod, which is at a distance of 3 cm from the end in contact with the x-axis.

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...

### The towers of bridge, hung in the form of a parabola, have their tops 30 m above the roadway, and are 200 m apart. If the cable is 5 m above the roadway at the centre of the bridge, find the length of the vertical supporting cable, 30 m from the centre.

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...

### A parabolic reflector is 5 cm deep and its diameter is 20 cm. How far is its focus from the vertex?

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...

### The focus of a parabolic mirror is at a distance of 6 cm from its vertex. If the mirror is 20 cm deep, find its diameter.

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...

### Find the equation of the line passing through the intersection of the lines 3x – 4y + 1 = 0 and 5x + y – 1 = 0 and which cuts off equal intercepts from the axes.

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...

### Find the equation of the line through the intersection of the lines 2x + 3y – 2 = 0 and x – 2y + 1 = 0 and having x-intercept equal to 3.

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...

### Find the equation of the line through the intersection of the lines 2x – 3y + 1 = 0 and x + y – 2 = 0 and drawn parallel to y-axis.

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...

### Find the equation of the line through the intersection of the lines x – 7y + 5 = 0 and 3x + y – 7 = 0 and which is parallel to x-axis.

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...

### Find the equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and which is perpendicular to the line x + 2y + 1 = 0.

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...

### Find the equation of the line through the intersection of the lines 5x – 3y = 1 and 2x + 3y = 23 and which is perpendicular to the line 5x – 3y = 1.

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...

### Find the equation of the line drawn through the point of intersection of the lines x – y = 1 and 2x – 3y + 1 = 0 and which is parallel to the line 3x + 4y = 12.

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...

### Find the equation of the line drawn through the point of intersection of the lines x + y = 9 and 2x – 3y + 7 = 0 and whose slope is −???? . ????

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...

### Find the equation of the line drawn through the point of intersection of the lines x – y = 7 and 2x + y = 2 and passing through the origin.

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...

### Find the equation of the line drawn through the point of intersection of the lines x – 2y + 3 = 0 and 2x – 3y + 4 = 0 and passing through the point (4, -5).

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...

### Transform the equation 2×2 + y2 – 4x + 4y = 0 to parallel axes when the origin is shifted to the point (1, -2).

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 –...

### Find what the given equation becomes when the origin is shifted to the point (1, 1). xy – x – y + 1 = 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 xy – x – y + 1 = 0...

### Find what the given equation becomes when the origin is shifted to the point (1, 1)

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...

### Find what the given equation becomes when the origin is shifted to the point (1, 1). xy – y2 – x + y = 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 xy – y2 – x + y = 0...

### Find what the given equation becomes when the origin is shifted to the point (1, 1)

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...

### At what point must the origin be shifted, if the coordinates of a point (-4,2) become (3, -2)?

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...

### If the origin is shifted to the point (2, -1) by a translation of the axes, the coordinates of a point become (-3, 5). Find the origin coordinates of the point.

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...

### If the origin is shifted to the point (0, -2) by a translation of the axes, the coordinates of a point become (3, 2). Find the original coordinates of the point.

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...

### If the origin is shifted to the point (-3, -2) by a translation of the axes, find the new coordinates of the point (3, -5).

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...

### If the origin is shifted to the point (1, 2) by a translation of the axes, find the new coordinates of the point (3, -4).

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...

### Find the equations of the medians of a triangle whose sides are given by the equations 3x + 2y + 6 = 0, 2x – 5y + 4 = 0 and x -3y – 6 = 0.

### Find the equation of the perpendicular drawn from the point P(-2, 3) to the line x– 4y + 7 = 0. Also, find the coordinates of the foot of the perpendicular.

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...

### Find the equation of the perpendicular drawn from the origin to the line 4x – 3y + 5 = 0. Also, find the coordinates of the foot of the perpendicular.

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...

### Find the area of the triangle, the equations of whose sides are y = x, y = 2x and y – 3x = 4.

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...

### Find the area of the triangle formed by the lines x = 0, y = 1 and 2x + y = 2.

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...

### Find the area of the triangle formed by the lines x + y = 6, x – 3y = 2 and 5x – 3y + 2 = 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...

### Find the image of the point P(1, 2) in the line x – 3y + 4 = 0.

### Find the value of k so that the lines 3x – y – 2 = 0, 5x + ky – 3 = 0 and 2x + y – 3 = 0 are concurrent.

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...

### Show that the lines 3x – 4y + 5 = 0, 7x – 8y + 5 = 0 and 4x + 5y = 45 are concurrent. Also find their point of intersection.

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...

### Show that the lines x + 7y = 23 and 5x + 2y = a 16 intersect at the point (2, 3).

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 +...

### Find the points of intersection of the lines 4x + 3y = 5 and x = 2y – 7.

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...

### The perpendicular distance of a line from the origin is 5 units, and its slope is -1. Find the equation of the line.

### Find the distance between the parallel lines p(x + y) = q = 0 and p(x + y) – r =0

### Find the distance between the parallel lines y = mx + c and y = mx + d

### Find the distance between the parallel lines 8x + 15y – 36 = 0 and 8x + 15y + 32 = 0.

### Find the distance between the parallel lines 4x – 3y + 5 = 0 and 4x – 3y + 7 = 0

### A vertex of a square is at the origin and its one side lies along the line 3x – 4y – 10 = 0. Find the area of the square.

### Find all the points on the line x + y = 4 that lie at a unit distance from the line 4x+3y=10.

### What are the points on the x-axis whose perpendicular distance from the line is 4 units?

### The points A(2, 3), B(4, -1) and C(-1, 2) are the vertices of ΔABC. Find the length of the perpendicular from C on AB and hence find the area of ΔABC

### Show that the length of the perpendicular from the point (7, 0) to the line 5x + 12y – 9 = 0 is double the length of perpendicular to it from the point (2, 1)

### Find the values of k for which the length of the perpendicular from the point (4, 1) on the line 3x – 4y + k = 0 is 2 units

### Prove that the product of the lengths of perpendiculars drawn from the points

### Find the length of the perpendicular from the origin to each of the following lines : (i) 7x + 24y = 50 (ii) 4x + 3y = 9 (iii) x = 4

### Find the distance of the point (4, 2) from the line joining the points (4, 1) and (2, 3)

### Find the distance of the point (2, 3) from the line y = 4.

### Find the distance of the point (-4, 3) from the line 4(x + 5) = 3(y – 6).

### Find the distance of the point (-2, 3) from the line 12x = 5y + 13.

### Find the distance of the point (3, -5) from the line 3x – 4y = 27

### Reduce each of the following equations to normal form :

### Reduce the equation to the normal form x cos ???? + y sin ???? = p, and hence find the values of ???? and p.

### Reduce the equation to the normal form x cos ???? + y sin ???? = p, and hence find the values of ???? and p.

### Reduce the equation x + y – √2 = 0 to the normal form x cos ???? + y sin ???? = p, and hence find the values of ???? and p.

### Find the inclination of the line:

### Reduce the equation 5x – 12y = 60 to intercepts form. Hence, find the length of the portion of the line intercepted between the axes

### Reduce the equation 3x – 4y + 12 = 0 to intercepts form. Hence, find the length of the portion of the line intercepted between the axes

### Reduce the equation y + 5 = 0 to slope-intercept form, and hence find the slope and the y-intercept of the line.

### Reduce the equation 5x + 7y – 35 = 0 to slope-intercept form, and hence find the slope and the y-intercept of the line

### Reduce the equation 2x – 3y – 5 = 0 to slope-intercept form, and find from it the slope and y-intercept.

### Find the equation of the line which is at a distance of 3 units from the origin such that tan ???? = ???? ,where ???? is the acute angle which this perpendicular makes ???????? with t

### The length of the perpendicular segment from the origin to a line is 2 units and the inclination of this perpendicular is ???? such that sin ???? = ????/ ???? and ???? is acute.

### Find the equation of the line for which p = 4 and ???? = 1800

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...

### Find the equation of the line for which p = 2 and ???? = 3000

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...

### Find the equation of the line for which p = 3 and ???? = 2250

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...

### Find the equation of the line for which p = 8 and ???? = 1500

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...

### Find the equation of the line for which p = 5 and ???? = 1350

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...

### Find the equation of the line for which p = 3 and ???? = 450

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...

### If the straight line ????/a + ???? /????=1 find the values of a and b.

### A straight line passes through the point (5, -2) and the portion of the line intercepted between the axes is divided at this point in the ratio 2 : 3. Find the equation of the line.

### Find the equation of the line whose portion intercepted between the coordinate axes is divided at the point (5, 6) in the ratio 3 : 1.

### Find the equation of the line whose portion intercepted between the axes is bisected at the point (3, -2).

### Find the equation of the line which passes through the point (22, -6) and whose intercept on the x-axis exceeds the intercept on the y-axis by 5.

### Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes, whose sum is 9.

### Find the equation of the line which passes through the point (3, -5) and cuts off intercepts on the axes which are equal in magnitude but opposite in sign.

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...

### Find the equation of the line and cuts off equal intercepts on the coordinate axes and passes through the point (4,7).

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...

### Find the equation of the line which cuts off intercepts 4 and -6 on the x-axis and y-axis respectively.

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...

### Find the equation of the line which cuts off intercepts -3 and 5 on the x-axis and y-axis respectively.

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...

### A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : 2. Find the equation of the line.

### Find the equation of the line passing through ( – 3, 5) and perpendicular to the line through the points (2, 5) and ( – 3, 6).

### Find the equation of the line whose y – intercept is – 3 and which is perpendicular to the line joining the points ( – 2, 3) and (4, – 5).

### Find the equation of the line which is perpendicular to the line 3x + 2y = 8 and passes through the midpoint of the line joining the points (6, 4) and (4, – 2).

### Find the equation of the line that has x – intercept – 3 and which is perpendicular to the line 3x + 5y = 4

### Find the equation of the line passing through the point (2, 4) and perpendicular to the x – axis.

### Find the equation of the line passing through the point (2, 3) and perpendicular to the line 4x + 3y = 10

### Find the equation of the line passing through the point (0, 3) and perpendicular to the line x – 2y + 5 = 0

### Find the equation of the line which is parallel to the line 2x – 3y = 8 and whose y – intercept is 5 units.

### Find the equation of the line through the point ( – 1, 5) and making an intercept of – 2 on the y – axis.

### Find the equation of the bisectors of the angles between the coordinate axes

### Find the equation of the line cutting off an intercept – 2 from the y – axis and equally inclined to the axes.

### Find the equation of the line whose inclination is ???????? and which makes an ???? interc

### Find the equation of the line which makes an angle of 300 with the positive direction of the x – axis and cuts off an intercept of 4 units with the negative direction of the y – axis.

### If A(1, 4), B(2, – 3) and C( – 1, – 2) are the vertices of a ΔABC, find the equation of (i) the median through A (ii) the altitude through A (iii) the perpendicular bisector of BC

### the midpoints of the sides BC, CA and AB of a ΔABC are D(2, 1), B( – 5, 7) and P( – 5, – 5) respectively. Find the equations of the sides of ΔABC.

### If A(4, 3), B(0, 0) and C(2, 3) are the vertices of a ΔABC, find the equation of the bisector of ∠A.

### . Find the equations of the altitudes of a ΔABC, whose vertices are A(2, – 2), B(1, 1) and C( – 1, 0).

### Find the slope and the equation of the line passing through the points: ( – 1, 1) and (2, – 4)

### Find the equation of a line whose inclination with the x – axis is 1500 and which passes through the point (3, – 5).

### Find the equation of a line which is equidistant from the lines y = 8 and y = – 2.

### . Using slopes, find the value of x for which the points A(5, 1), B(1, -1) and C(x, 4) are collinear.

### Without using Pythagora’s theorem, show that the points A(1, 2), B(4, 5) and C(6, 3) are the vertices of a right-angled triangle.

### Find the distance between the points:

(i) A(2, -3) and B(-6, 3)

(ii) C(-1, -1) and D(8, 11)

(iii) P(-8, -3) and Q(-2, -5)

(iv) R(a + b, a – b) and S(a – b, a + b)

### Mark (√) against the correct answer in the following: The range of , where is

A.

B.

C.

D.

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...

### Mark (√) against the correct answer in the following: Let . Then, range

A.

B.

C.

D.

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}...

### Mark (√) against the correct answer in the following: Let . Then, range

A.

B.

C.

D. none of these

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)} \\...

### Mark (√) against the correct answer in the following: Let . Then,

A.

B.

C.

D.

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...

### Mark (√) against the correct answer in the following: Let . Then, dom (f) and range (f) are respectively

A. and

B. and

C. R and R +

D. 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...

### Mark (√) against the correct answer in the following: Let . Then, dom (f) and range (f) are respectively.

A. and

B. and

C. and

D. 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...

### Mark (√) against the correct answer in the following: Let . Then,

A.

B.

C.

D. None of these

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...

### Mark (√) against the correct answer in the following: Let . Then,

A.

B.

C.

D. none of these

Solution: Option(C) is correct. $\mathrm{f}(\mathrm{x})=\sqrt{\cos x}$ As per graph of $\sqrt{\cos x}$ the domain is $\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$

### Mark (√) against the correct answer in the following: Let . Then, ?

A.

B.

C.

D. None of these

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...

### Mark (√) against the correct answer in the following: Let . Then,

A.

B.

C.

D.

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...

### Mark (√) against the correct answer in the following: Let Then,

A.

B.

C.

D. none of these

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...

### Mark (√) against the correct answer in the following: Let . Then, ?

A.

B.

C.

D.

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}...

### Mark (√) against the correct answer in the following: Let . Then, ?

A.

B.

C.

D.

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 }...

### Mark (√) against the correct answer in the following: Let . Then, dom (f) – ?

A.

B.

C.

D.

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...

### Mark (√) against the correct answer in the following: Let . Then, dom

A.

B.

C.

D.

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 \\...

### Mark (√) against the correct answer in the following: If f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)} then (g o f) = ?

A. {(3, 1), (1, 3), (3, 4)}

B. {(1, 3), (3, 1), (4, 3)}

C. {(3, 4), (4, 3), (1, 3)}

D. {(2, 5), (5, 2), (1, 5)}

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 $\mathrm{g}$,...

### Mark (√) against the correct answer in the following: If and then

A. 0

B. 1

C.

D.

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 $h$,...

### Mark (√) against the correct answer in the following: If and then

A.

B.

C.

D.

Solution: Option(B) is correct. $f(x)=8 x^{3}$ $g(x)=x^{1 / 3}$ According to the combination of $\mathrm{f}$ and $\mathrm{g}$, $\operatorname{gof}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{x}))$...

### Mark (√) against the correct answer in the following: If then

A.

B.

C.

D. None of these

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}...

### Mark (√) against the correct answer in the following: If then (f of)

A.

B.

C.

D. None of these

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 $(x)=f(f(x))$...

### Mark (√) against the correct answer in the following: If then (f of of)

A.

B.

C.

D. None of these

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 $(x)=f(f(f(x))$...

### Mark (√) against the correct answer in the following: If then

A.

B.

C.

D. None of these

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...

### Mark (√) against the correct answer in the following: If and then

A.

B.

C.

D. None of these

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...

### Mark (√) against the correct answer in the following: If then

A.

B.

C.

D. None of these

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 $\mathrm{f}$,...

### Mark (√) against the correct answer in the following: Let f : N → X : f(x) = 4×2 + 12x + 15. Then, (y) = ?

A.

B.

C.

D. None of these

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...

### Mark (√) against the correct answer in the following:

Let . Then

A.

B.

C.

D. None of these

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 $f(x)=y$...

### Mark (√) against the correct answer in the following: Let f : Q → Q : f(x) = (2x + 3). Then, (y) = ?

A. (2y – 3)

B.

C.

D. none of these

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 $\mathrm{f}(\mathrm{x})=\mathrm{y}$...

### Mark (√) against the correct answer in the following:

Let A and B be two non – empty sets and let

f : (A × B) → (B × A) : f(a, b) = (b, a). Then, f is

A. one – one and into

B. one – one and onto

C. many – one and into

D. many – one and onto

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},...

### Mark (√) against the correct answer in the following: Let Then, is

A. one – one and into

B. one – one and onto

C. many – one and into

D. many – one and onto

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 } \\...

### Mark (√) against the correct answer in the following:

Let and . Then is

A. one – one and into

B. one – one and onto

C. many – one and into

D. many – one and onto

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...

### Mark (√) against the correct answer in the following:

f : C → R : f(z) = |z| is

A. one – one and into

B. one – one and onto

C. many – one and into

D. many – one and onto

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}...

### Mark (√) against the correct answer in the following:

f : R → R : f(x) = cos x is

A. one – one and into

B. one – one and onto

C. many – one and into

D. many – one and onto

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...

### Mark (√) against the correct answer in the following:

f : R + → R + : f(x) = is

A. many – one and into

B. many – one and onto

C. one – one and into

D. one – one and onto

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...

### Mark (√) against the correct answer in the following:

f : R → R : f(x) = is

A. one – one and onto

B. one – one and into

C. many – one and onto

D. many – one and into

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...

### Mark (√) against the correct answer in the following:

f : R → R : f(x) = is

A. one – one and onto

B. one – one and into

C. many – one and onto

D. many – one and into

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...

### Mark (√) against the correct answer in the following:

f : N → N : f(x) = + x + 1 is

A. one – one and onto

B. one – one and into

C. many – one and onto

D. many – one and into

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...

### Mark (√) against the correct answer in the following:

f : N → N : f(x) = 2x is

A. one – one and onto

B. one – one and into

C. many – one and onto

D. many – one and into

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...

### Temperature dependence of resistivity ρ(T) of semiconductors, insulators, and metals is significantly based on the following factors:

a) number of charge carriers can change with temperature T

b) time interval between two successive collisions can depend on T

c) length of material can be a function of T

d) mass of carriers is a function of T

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

### Define a function. What do you mean by the domain and range of a function? Give examples.

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’...

### Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Show that R is reflexive but neither symmetric nor transitive.

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,...

### Let 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...

### Let S be the set of all points in a plane and let R be a relation in S defined by R = {(A, B) : d(A, B) < 2 units}, where d(A, B) is the distance between the points A and B. Show that R is reflexive and symmetric but not transitive.

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...

### Let 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...

### Show that the relation 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 $\mathrm{N} \times \mathrm{N}$...

### Let 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,...

### Let 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...

### Define a relation on a set. What do you mean by the domain and range of a relation? Give an example.

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}...

### Let A = {1, 2, 3, 4, 5, 6) and let R = {(a, b) : a, b ∈ A and b = a + 1}. Show that R is

(i) not reflexive,

(ii) not symmetric

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...

### On the set S of all real numbers, define a relation R = {(a, b) : a ≤ b}. Show that R is

(i) reflexive

(ii) transitive

Solution: (i) Reflexivity: Suppose $\mathrm{p}$ is an arbitrary element of $\mathrm{S}$. So now, $\mathrm{p} \leq \mathrm{p}$ $\Rightarrow(p, p) \in R$ Therefore, $\mathrm{R}$ is reflexive. (ii)...