Putting $\frac{{{x}^{2}}}{\left( {{x}^{4}}-{{x}^{2}}-12 \right)}=\frac{t}{{{t}^{2}}-t-12}=\frac{t}{(t-4)(t+3)}$ $=\frac{A}{t-4}+\frac{B}{t+3}.....(1)$ Where $t={{x}^{2}}$ $A(t+3)+B(t-4)=t$ Now put...
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CBSE Study Material
Put $t=cosx$ $dt=-sinxdx$ $\frac{-dt}{\sin x}=dx$ $I=\int \frac{-dt}{{{\sin }^{2}}x(1+2t)}$ $=\int \frac{dt}{(1-{{\cos }^{2}}x)(1+2t)}=\int \frac{dt}{(1-{{t}^{2}})(1+2t)}$ Putting,...
Evaluate:
Put $t=sinx$ $dt=cosx$ $I=\int \frac{\sin x\cos x}{{{\cos }^{2}}x(1-\sin x)}dx$ $=\int \frac{tdt}{(1-{{\sin }^{2}}x)(1-t)}=\int \frac{tdt}{(1-{{t}^{2}})(1-t)}$ Putting...
![\[I=\int \frac{1}{\sin x{{\cos }^{2}}x}dx\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0e1477669fd34335f69cbb5cf2b35c18_l3.png "Rendered by QuickLaTeX.com")
$=\int \left( \frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x{{\cos }^{2}}x} \right)dx$ $=\int \frac{{{\sin }^{2}}x}{\sin x{{\cos }^{2}}x}dx+\int \frac{{{\cos }^{2}}x}{\sin x{{\cos }^{2}}x}dx$ $=\int...
Differentiate 
with respect to 
As per the given question,
Put $t=sinx$ $dt=cosxdx$ $I=\int \frac{dt}{(1-{{\sin }^{2}}x)(5-4t)}=\int \frac{dt}{(1-{{t}^{2}})(5-4t)}$ $\frac{1}{(1-{{t}^{2}})(5-4t)}=\frac{1}{(1-t)(1+t)(5-4t)}$ Putting,...
If x and y are acute such that 
, prove that 
Answer: Given: $\sin x=\frac{1}{\sqrt{5}}\,and\,\sin y=\frac{1}{\sqrt{10}}$ Now we will calculate value of cos x and cosy Sin(x + y) = sinx.cosy + cosx.siny
Differentiate 
with respect to 
, if 
As per the given question,
Put $t=cosx$ $\frac{dt}{-\sin x}=dx$ $I=\int \frac{dt}{-\frac{\sin x}{\sin x(3+2t)}}$ $=-\int \frac{dt}{{{\sin }^{2}}x(3+2t)}=-\int \frac{dt}{(1-{{\cos }^{2}}x)(3+2t)}$ $=-\int...
Evaluate:
Put $t={{x}^{6}}$ $dt=6{{x}^{5}}dx$ \[\int \frac{dt}{\frac{\left( 6{{x}^{5}} \right)}{x\left( t+1 \right)}}=\frac{1}{6}\int \frac{dt}{{{x}^{6}}(t+1)}=\frac{1}{6}\int \frac{dt}{t(t+1)}\]...
Put $t={{x}^{2}}$ $dt=5{{x}^{4}}dx$ Putting $\frac{1}{t\left( t+1 \right)}=\frac{A}{t}+\frac{B}{t+1}......(1)$ $A(t+1)+Bt=1$ Now put $t+1=0$ $t=-1$ $A(0)+B(-1)=1$ $B=-1$ Now put $t=0$...
Differentiate 
with respect to 
, if 
As per the given question,
Putting 
${{e}^{x}}=t+1$ $dt={{e}^{x}}dx$ $\frac{dt}{{{e}^{x}}}=dx$ $\frac{dt}{t+1}=dx$ Put, $\frac{1}{(1-t){{t}^{2}}}=\frac{A}{t+1}+\frac{Bt+C}{{{t}^{2}}}$ $A\left( {{t}^{2}} \right)+(Bt+C)(t+1)=1$ put...
Putting $\frac{{{x}^{2}}+1}{({{x}^{2}}+4)({{x}^{2}}+25)}=\frac{t+1}{\left( t+4 \right)\left( t+25 \right)}$ $=\frac{A}{t+4}+\frac{B}{t+25}.....(1)$ Where $t={{x}^{2}}$ $(A+B)t+(25A+4B)=t+1$...
Put $\frac{1}{\left( {{x}^{2}}+2 \right)\left( {{x}^{2}}+4 \right)}=\frac{1}{(t+2)(t+4)}=\frac{A}{t+2}+\frac{B}{t+4}......(1)$ $A(t+4)+B(t+2)=1$ Put $t+4=0$ $t=-4$ $A(0)+B(-4+2)=1$ $B=-\frac{1}{2}$...
Differentiate 
with respect to 
, if 
As per the given question,
Put $\frac{17}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}=\frac{A}{2x+1}+\frac{Bx+C}{{{x}^{2}}+4}.......(1)$ $A\left( {{x}^{2}}+4 \right)+(Bx+C)(2x+1)=17$ Put $2x+1=0$ $x=-\frac{1}{2}$ $A\left(...
Put $\frac{1}{({{x}^{2}}+1){{(x+1)}^{2}}}=\frac{A}{x+1}+\frac{B}{{{(x+1)}^{2}}}+\frac{Cx+D}{{{x}^{2}}+1}.....(1)$ $A(x+1)({{x}^{2}}+1)+B({{x}^{2}}+1)+(Cx+D){{(x+1)}^{2}}=1$ Put $x+1=0$ $x=-1$...
If θ and Φ lie in the first quadrant such that 
, find the values of (iii) tan (θ – Φ)
Answer: (iii)We will first find out the Values of tanθ and tanΦ
Differentiate 
with respect to 
, if 
As per the given question,
If θ and Φ lie in the first quadrant such that , find the values of (i) sin (θ – Φ ) (ii) cos (θ – Φ)
Answer: Given: $\sin \theta =\frac{8}{17}\,and\,\cos \phi =\frac{12}{13}$
Differentiate 
with respect to 
, if 
As per the given question,
Differentiate 
with respect to 
, if
As per the given question,
Put $\frac{1}{{{x}^{3}}-1}=\frac{1}{(x+1)({{x}^{2}}-x+1)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}-x+1}.....(1)$ $A\left( {{x}^{2}}-x+1 \right)+(bx+C)(x+1)=1$ Now putting $x+1=0$ $x=-1$ $A(1+1+1)+C(0)=1$...
Prove that:
Answer: Using cos(90° + θ) = - sinθ(I quadrant cosx is positive cosec( - θ) = - cosecθ tan(270° - θ) = tan(180° + 90° - θ) = tan(90° - θ) = cotθ (III quadrant tanx is positive) Similarily sin(270° +...
Put $\frac{1}{{{x}^{3}}-1}=\frac{1}{\left( x-1 \right)({{x}^{2}}+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1}.....(1)$ $A\left( {{x}^{2}}+x+1 \right)+(Bx+C)(x-1)=1$ Now putting, $x-1=0$ $x=1$...
Differentiate 
with respect to 
As per the given question,
Put $t={{x}^{2}}$ $dt=2xdx$ Now putting, $\frac{{{x}^{2}}}{\left( {{x}^{4}}-1 \right)}=\frac{t}{(t-1)(t+1)}=\frac{A}{t-1}+\frac{B}{t+1}......(1)$ $A(t+1)+B(t-1)=t$ Putting, $t+1=0$ $t=-1$...
Differentiate 
with respect to 
, if 
As per the given question,
Differentiate with respect to 
, if
As per the given question,
Prove that:
Answer: Using sin(90° + θ) = cosθ and sin( - θ) = sinθ,tan(90° + θ) = - cotθ Sin(180° + θ) = - sinθ(III quadrant sinx is negative)
Put $t={{x}^{2}}$ $dt=2xdx$ Now putting, $\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}.......(1)$ $A(t+3)+B(t+1)=1$ Putting $t+3=0$ $x=-3$ $A(0)+B(-3+1)=1$ $B=-\frac{1}{2}$ Putting $t+1=0$...
Now putting, $\frac{3x+5}{\left( {{x}^{3}}-{{x}^{2}}+x-1 \right)}=\frac{A}{x-1}+\frac{Bx-C}{\left( {{x}^{2}}+1 \right)}......(1)$ $A\left( {{x}^{2}}+1 \right)+(Bx+C)(x-1)=3x+5$ Putting, $x-1=09$...
Differentiate with respect to , if
As per the given question,
Now putting, $\frac{8}{(x+2)({{x}^{2}}+4)}=\frac{A}{x+2}+\frac{Bx+C}{({{x}^{2}}+4)}.....(1)$ $A\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)(x+2)=8$ Putting, $x+2=0$ $x=-2$ $A(4+4)+0=8$ $A=1$ By...
Now putting, $\frac{5{{x}^{2}}18x+17}{{{(x-1)}^{2}}(2x-3)}=\frac{A}{(2x-3)}+\frac{B}{x-1}+\frac{C}{{{(x-1)}^{2}}}......(1)$ $A{{\left( x-1 \right)}^{2}}+B(2x-3)(x-1)+C(2x-3)=5{{x}^{2}}-18x+17$...
Now putting, $\frac{5x+8}{{{x}^{2}}(3x+8)}=\frac{A}{(3x+8)}+\frac{Bx+C}{{{x}^{2}}}....(1)$ $A{{x}^{2}}+\left( Bx+C \right)\left( 3x+8 \right)=5x+8$ Putting, $3x+8=0$ $x=-\frac{8}{3}$ $A\left(...
Differentiate 
with respect to 
As per the given question,
Now putting, $\frac{3x+1}{(x+2){{(x-2)}^{2}}}=\frac{A}{(x+2)}+\frac{B}{(x-2)}+\frac{C}{{{(x-2)}^{2}}}......(1)$ $A{{\left( x-2 \right)}^{2}}+B\left( x+2 \right)(x-2)+C(x+2)=3x+1$ Putting, $x-2=0$...
Now putting, $\frac{2x}{{{(2x+1)}^{2}}}=\frac{A}{(2x+1)}+\frac{B}{{{(2x+1)}^{2}}}....(1)$ $A(2x+1)+B=2x$ Putting $2x+1=0$ $x=\frac{-1}{2}$ $A(0)+B=-1$ By equating the coefficient of x, $2A=2$ $A=1$...
Differentiate with respect to , if
As per the given question,
Now putting, $\frac{{{x}^{2}}+x+1}{(x+1)({{x}^{2}}+1)}=\frac{A}{(x+2)}+\frac{Bx+c}{({{x}^{2}}+1)}$ $A\left( {{x}^{2}}+1 \right)+(Bx+C)(x+2)={{x}^{2}}+x+1$...
Prove that
Answer:
Now putting, $\frac{{{x}^{2}}+1}{(x-3){{(x-1)}^{2}}}=\frac{A}{(x-3)}+\frac{B}{(x-1)}+\frac{C}{{{(x-1)}^{2}}}......(1)$ $A{{\left( x-1 \right)}^{2}}+B\left( x-3 \right)(x-1)+C\left( x-3...
Now putting, $\frac{{{x}^{2}}+1}{(x+3){{(x-1)}^{2}}}=\frac{A}{(x+3)}+\frac{B}{(x-1)}+\frac{C}{{{(x-1)}^{2}}}....(1)$ $A{{\left( x-1 \right)}^{2}}+B\left( x+3 \right)(x-1)+C(x+3)={{x}^{2}}+1$ Now put...
Differentiate 
with respect to 
.
As per the given question, ......(i) ........(ii)
Differentiate with respect to 
, if 
As per the given question,
Differentiate with respect to , if 
As per the given question,
Differentiate 
with respect to , if 
As per the given question,
Differentiate 
with respect to 
, if 
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Differentiate with respect to , if 
As per the given question,
Differentiate with respect to , if 
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if 
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, if 
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Differentiate 
with respect to 
As per the given question,
Differentiate 
with respect to 
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with respect to 
As per the given question,
If 
and 
, find 
when 
As per the given question,
If 
find 
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If 
, find at 
.
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If 
, find
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If 
and 
find the value of at 
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If 
and 
, show that at 
.
As per the given question,
Show that the points A(1, 1, 1), B(-2, 4, 1), C(1, -5, 5) and D(2, 2, 5) are the vertices of a square.
Answer: (x1,y1,z1) = (1, 1, 1) (x2,y2,z2) = (-2, 4, 1) (x3,y3,z3) = (1, 5, 5) (x4,y4,z4) = (2, 2, 5) $\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} -...
Show that the points A(0, 1, 2), B(2, -1, 3) and C(1, -3, 1) are the vertices of an isosceles right-angled triangle.
Answer: (x1,y1,z1) = (0, 1, 2) (x2,y2,z2) = (2, -1, 3) (x3,y3,z3) = (1, -3, 1) $\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ =>...
Show that the points A(4, 6, -5), B(0, 2, 3) and C(-4, -4, -1) from the vertices of an isosceles triangle.
Answer: (x1,y1,z1) = (4, 6, -3) (x2,y2,z2) = (0, 2, 3) (x3,y3,z3) = (-4, -4, -1) $\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ =>...
Show that the points A(1, -1, -5), B(3, 1,3) and C(9, 1, -3) are the vertices of an equilateral triangle.
Answer: (x1,y1,z1) = (1, -1, -5) (x2,y2,z2) = (3, 1,3) (x3,y3,z3) = (9, 1, -3) $\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ =>...
Find the distance between the points : (i) R(1, -3, 4) and S(4, -2, -3) (ii) C(9, -12, -8) and the origin
Answers: (i) R(1, -3, 4) and S(4, -2, -3) (x1,y1,z1) = (1, -3, 4) (x2,y2,z2) = (4, -2, -3) $\begin{array}{l} D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ D =...
If 
and 
find , at .
As per the given question,
Find the distance between the points : (i) A(5, 1, 2) and B(4, 6, -1) (ii) P(1, -1, 3) and Q(2, 3, -5)
Answers: (i) A(5, 1, 2) and B(4, 6, -1) (x1,y1,z1) = (5, 1, 2) (x2,y2,z2)= (4, 6, -1) $\begin{array}{l} D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ D = \sqrt...
Find 
, if 
As per the given question,
If 
and 
, find
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If 
and 
, find
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If 
and 
are connected parametrically by the equation, without eliminating the parameter, find . 
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If 
and 
, prove that 
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and 
, prove that 
.
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and 
, prove that 
at 
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, prove that 
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If and are connected parametrically by the equation, without eliminating the parameter, find . 
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As per the given question,
In which octant does each of the given points lie? (i) (-1, -6, 5) (ii) (4, 6, 8)
Answers: (i) (-1, -6, 5) lies in octant III (ii) (4, 6, 8) lies in octant...
In which octant does each of the given points lie? (i) (-6, 5, -1) (ii) (4, -3, -2)
Answers: (i) (-6, 5, -1) lies in octant VI (ii) (4, -3, -2) lies in octant...
In which octant does each of the given points lie? (i) (-4, -1, -6) (ii) (2, 3, -4)
Answers: (i) (-4, -1, -6) lies in octant VII (ii) (2, 3, -4) lies in octant...
In which plane does the point (4, -3, 0) lie?
The x, y, z coordinates of the point are 4, -3, 0. As the distance of point along the z-axis is 0, the plane in which the point lies is the xy-plane.
If a point lies on yz-plane then what is its x-coordinate?
The x-coordinate is the distance of a point from the origin parallel or along the x- axis. To measure the x coordinate, you must move either to the left of the origin or to its right. In case of a...
If a point lies on the z-axis, then find its x-coordinate and y-coordinate.
The X and y coordinates of a point are its distance from the origin along or parallel to the horizontal x-axis and y-axis. To measure the x and y coordinates, you must move either to the left of the...
Prove that:
Answer:
Find the equation of the parabola with vertex at the origin, passing through the point P(5, 2) and symmetric with respect to the y-axis.
Answer: The equation of a parabola with vertex at the origin and symmetric about the y-axis is x2 = 4ay The point P(5,2) passes through above parabola, 52 =...
Find the equation of the parabola with vertex at the origin and focus F(0, 5).
Answer: Vertex : A (0,0) Focus F(0,5) is of the form F(0,a) Vertex A(0,0) and Focus F(0,a), The equation of parabola is x2 = 4ay a = 5 The equation of parabola is...
Find the equation of the parabola with focus F(0, -3) and directrix y = 3.
Answer: Given, The equation of directrix is, y = 3 y - 3 = 0 Above equation is of the form, y - a = 0 Focus of the parabola F(0,-3) is of the form F(0,-a) a = 3...
Find the equation of the parabola with focus F(4, 0) and directrix x = -4.
Answer: Given, Equation of directrix, x = -4 x + 4 = 0 The equation is of the form, x + a = 0 Focus of the parabola F(4,0) is of the form F(a,0) a = 4 For...
Find the equation of the parabola with vertex at the origin and focus at F(-2, 0).
Answer : Vertex : A (0,0) focus F(-2,0) is of the form F(-a,0) For Vertex A(0,0) and Focus F(-a,0), The equation of parabola is y2 = - 4ax a = 2 The equation of...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, 3x2 = -16y x2 = -16/3 y Comparing the given equation with parabola having an equation, x2 = 4ay 4a = 16/2 a = 4/3 Focus: F(0, -a) = F(0, -4/3) Vertex: A(0,...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, x2 = -18y Comparing given equation with parabola having equation, x2 = -4ay 4a = 18 a = 9/2 Focus: F(0, -a) =F(0, -9/2) Vertex: A(0, 0) =A(0, 0) Equation...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, x2 = - 8y Comparing given equation with parabola having equation, x2 = - 4ay 4a = 8 a = 2 Focus : F(0,-a) = F(0,-2) Vertex : A(0,0) = A(0,0) Equation of...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, 3x2 = 8y x2 = 8/3 y Comparing the given equation with parabola having an equation, x2 = 4ay 4a = 8/3 a = 2/3 Focus : F(0, a) = F(0, 2/3) Vertex :...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, x2 = 10y Comparing given equation with parabola having equation, x2 = 4ay 4a = 10 a = 5 Focus : F(0,a) = F(0,2.5) Vertex : A(0,0) = A(0,0) Equation of the...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, x2 = 16y Comparing given equation with parabola having equation, x2 = 4ay 4a = 16 a = 4 Focus : F(0,a) = F(0,4) Vertex : A(0,0) = A(0,0) Equation of the...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, 5y2 = -16x y2 = -16/5 x Comparing the given equation with parabola having an equation, y2 = - 4ax 4???? = 16/5 ???? = 4/5 Focus : F(-a,0) = F(-4/5, 0)...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, y2 = -6x Comparing given equation with parabola having equation, y2 = - 4ax 4a = 6 a = 2/3 Focus: F(-a, 0) = F(-2/3, 0) Vertex: A(0, 0) =A(0, 0)...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : 
Answer: Given, y2 = -8x Comparing given equation with parabola having equation, y2 = - 4ax 4a = 8 a = 2 Focus : F(-a,0) = F(-2,0) Vertex : A(0,0) = A(0,0) Equation...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: 
Answer: Given, 3y2 = 8x y2 = 8/3 x Comparing the given equation with parabola having equation, y2 = 4ax 4a = 8/3 a = 2/3 Focus : F(a, 0) = F(2/3, 0) Vertex :...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: 
Answer: Given, y2 = 10x Comparing given equation with parabola having equation, y2 = 4ax 4a = 10 a =2.5 Focus : F(a,0) = F(2.5,0) Vertex : A(0,0) = A(0,0) Equation of the...
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: 
Answer: Given, y2= 12x Comparing given equation with parabola having equation, y2 = 4ax 4a = 12 a = 3 Focus : F(a,0) = F(3,0) Vertex : A(0,0) = A(0,0) Equation of...
Prove that:
Answer:
Prove that:
Answer: (II quadrant tanx negative) - tan45° = - 1
Find the equation of the parabola, which is symmetric about the y-axis and passes through the point P(2, -3).
Answer: The equation of a parabola with vertex at the origin and symmetric about the y-axis is x2 = 4ay The point P(2,-3) passes through above parabola, 22...
Prove that
Answer: (ii)cot105° - tan105° = cot(180° - 75°) - tan(180° - 75°) (II quadrant tanx is negative and cotx as well) = - cot75° - ( - tan75°) = tan75° - cot75°
Find when: 
and 
We have, $y=2 a t$ $\frac{d y}{d t}=2 a \frac{d}{d t}(t)=2 a(1)=2 a \\ \text { also } x=a t^{2} \\ \frac{d x}{d x}=a \frac{d}{d t}\left(t^{2}\right)=a(2 t)=2 a t \\ \text { now } \frac{d y}{d...
Prove that: tan15° + cot15° = 4
Answer: (iii) tan15° + cot15° = First, we will calculate tan15°,
Prove that:
Answer: (i) sin75° = sin(90° - 15°) .…….(using sin(A - B) = sinAcosB - cosAsinB) = sin90°cos15° - cos90°sin15° = 1.cos15° - 0.sin15° = cos15° Cos15° = cos(45° - 30°) …………(using cos(A - B) = cosAcosB...
Prove that:
Answer:
Prove that:
(i) cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x = cos x Answer: (i) cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x = sin((n + 2)x + (n + 1)x)(using cos(A - B) = cosAcosB + sinAsinB) =...
If ( – 1, 3) and (????, β) are the extremities of the diameter of the circle x2 + y2 – 6x + 5y – 7 = 0, find the coordinates (????, β).
Answer: Given, x2 + y2 – 6x + 5y – 7 = 0 Centre (3, -5/2) ( - 1, 3) & (????, β) are the 2 extremities of the diameter Using mid - point formula, $\begin{array}{l} \frac{{\alpha - 1}}{2} = 3\\...
Show that the quadrilateral formed by the straight lines x – y = 0, 3x + 2y = 5, x – y = 10 and 2x + 3y = 0 is cyclic and hence find the equation of the circle.
Answer: Slope of CD = 1 AB||CD Slope of BD = AC = - 1 AC||B They form a rectangle with all sides = 900 The quadrilateral is cyclic as sum of opposite angles...
Find the equation of the circle circumscribing the triangle formed by the lines x + y = 6, 2x + y = 4 and x + 2y = 5.
Answer: The required circle equation is, Using Laplace Expansion, 27(x2 + y2) - 459x - 513y + 1350 = 0 x2 + y2 - 17x - 19 + 50 =...
Find the equation of a circle passing through the origin and intercepting lengths a and b on the axes.
Answer: AD = b units and AE = a units. D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre. The general equation of a circle: (x - h)2 + (y - k)2 = r2...
Find the equation of the circle which passes through the points A(1, 1) and B(2, 2) and whose radius is 1. Show that there are two such circles.
Answer: The general equation of a circle is, (x - h)2 + (y - k)2 = r2 …(i) (h, k) is the centre and r is the radius. Putting A(1, 1) in (i) (1 - h)2 + (1 - k)2 = 12 ...
Prove that the centres of the three circles 
, 
and 
are collinear.
Answer: Given, x2 + y2 – 4x – 6y – 12 = 0 Centre ( - g1, - f1) = (2, 3) x2 + y2 + 2x + 4y – 5 = 0 Centre ( - g2, - f2) = ( - 1, - 2) x2 + y2 – 10x – 16y + 7 = 0 Centre ( - g3, - f3) = (5, 8) ...
Find the equation of the circle concentric with the circle 
and of double its area.
Answer: Two or more circles are said to be concentric If they have the same centre and different radii. Given, x2 + y2 - 6x + 12y + 15 = 0 Radius r = The...
Find the equation of the circle concentric with the circle 
and which touches the y-axis.
Answer: The general equation of the circle is, x2 + y2 + 2gx + 2fy + c = 0 Radius, r = $\begin{array}{l} r = \sqrt {{{(2)}^2} + {{(3)}^2} - ( - 3)} \\ r =...
Find the equation of the circle which passes through the points (1, 3) and (2, – 1), and has its centre on the line 2x + y – 4 = 0.
Answer: The equation of a circle: x2 + y2 + 2gx + 2fy + c = 0…(i) Putting (1, 3) & (2, - 1) in (i) 2g + 6f + c = - 10..(ii) 4g - 2f + c = - 5..(iii) The centre lies on the given straight line, (...
Prove that (i) sin(50° + θ)cos(20° + θ) – cos(50° + θ)sin(20° + θ) = 1/2 (ii) cos(70° + θ)cos(10° + θ) + sin(70° + θ)sin(10° + θ) = 1/2
Answer: (i) We have: sin(50° + θ)cos(20° + θ) - cos(50° + θ)sin(20° + θ) = sin(50° + θ - (20° + θ))(using sin(A - B) = sinAcosB - cosAsinB) = sin(50° + θ - 20° - θ) = sin30° = 1/2 (ii) We have:...
Show that the points A(1, 0), B(2, – 7), c(8, 1) and D(9, – 6) all lie on the same circle. Find the equation of this circle, its centre and radius.
Answer: The general equation of a circle: (x - h)2 + (y - k)2 = r2 …(i) (h, k) is the centre and r is the radius. Putting (1, 0) in (i) (1 - h)2 + (0 - k)2 = r2 h2 + k2 +...
Find the equation of the circle concentric with the circle 
and passing through the point P(5, 4).
Answer: Two or more circles are said to be concentric If they have the same centre and different radii. Given, x2 + y2 + 4x + 6y + 11 = 0 The concentric circle...
Find the equation of the circle which is circumscribed about the triangle whose vertices are A( – 2, 3), b(5, 2) and C(6, – 1). Find the centre and radius of this circle.
Answer: The general equation of a circle: (x - h)2 + (y - k)2 = r2 ...(i) (h, k) is the centre r is the radius Putting A( - 2, 3), B(5, 2) and c(6, - 1) in the equation, h2 + k2 + 4h - 6k + 13 = r2...
Find the equation of the circle passing through the points (20, 3), (19, 8) and (2, – 9) Also, find the centre and radius.
Answer: The required circle equation, Using Laplace Expansion, 102(x2 + y2) - 1428x - 612y - 11322 = 0 x2 + y2 - 14x -6y - 11 = 0 The equation with centre = (7, 3) Radius...
Find the equation of the circle passing through the points (i) (0, 0), (5, 0) and (3, 3) (ii) (1, 2), (3, – 4) and (5, – 6). Also, find the centre and radius
Answers: (i) The required circle equation, Using Laplace Expansion, 15(x2 + y2) - 75x - 15y = 0 x2 + y2 - 5x - y =0 The equation with centre = (2.5, 0.5) Radius = (ii) The...
Show that the equation 
does not represent a circle.
Answer: Radius = The radius is negative which is not possible x2 + y2 - 3x + 3y + 10 = 0 does not represent a circle.
Show that the equation 
represents a point circle. Also, find its centre.
Answer: The general equation of a circle is, x2 + y2 + 2gx + 2fy + c = 0 c, g, f are constants. x2 + y2 + 2x + 10y + 26 = 0 The equation represents a circle with 2g = 2 ⇒g = 1, 2f = 10 ⇒f = 5 and c...
Show that the equation 
represents a circle. Find its centre and radius.
Answer: The general equation of a conic is, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 a, b, c, f, g, h are constants For a circle, a = b and h = 0. The equation is, x2 + y2 + 2gx + 2fy + c = 0 ...
Prove that (v) cos130°cos40° + sin130°sin40° = 0
(v) cos130°cos40° + sin130°sin40° = cos(130° - 40°) (using cos(A - B) = cosAcosB + sinAsinB) = cos90° = 0
Show that the equation 
represents a circle. Find its centre and radius.
Answer: The general equation of a conic is, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 a, b, c, f, g, h are constants For a circle, a = b and h = 0. The equation is, x2 + y2 + 2gx + 2fy + c = 0 x2 + y2 +...
Show that the equation 
represents a circle. Find its centre and radius.
Answer: The general equation of a conic is, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 a, b, c, f, g, h are constants For a circle, a = b and h = 0. The equation is, x2 + y2 + 2gx + 2fy + c = 0 x2 + y2 –...
Prove that
Answer: (iii) cos75°cos15° + sin75°sin15° = cos(75° - 15°) (using cos(A - B) = cosAcosB + sinAsinB) = cos60° = 1/2 (iv) sin40°cos20° + cos40°sin20° = sin(40° + 20°) (using sin(A + B) = sinAcosB +...
Prove that
Answer: (i) sin80°cos20° - cos80°sin20° = sin(80° - 20°) (using sin(A - B) = sinAcosB - cosAsinB) = sin60° = $\frac{\sqrt{3}}{2}$ (ii)cos45°cos15° - sin45°sin15° = cos(45° + 15°) (using cos(A + B) =...
Find the values of all trigonometric functions of 135 deg
Answer:
Find the value of (ix) cos (495֯ )
(ix)cos495° = cos(360° + 135°) …………(using cos(360° + x) = cosx) = cos135° = cos(180° - 45°) ………….(using cos(180° - x) = - cosx) = - cos45° = - 1//2
Find the value of (vii) cot ( – 315֯ ) (viii) sin ( – 1230֯ )
Answer: vii) $ \cot \left( -{{315}^{\circ }} \right)=\frac{1}{\tan \left( -{{315}^{\circ }} \right)} $ $ \Rightarrow \frac{1}{-\tan \left( {{315}^{\circ }} \right)}=\frac{1}{-\tan \left(...
Find the value of (v) cosec ( – 690֯ ) (vi) tan (225֯ )
Answer:
Find the value of (iii) tan ( – 120֯ ) (iv) sec ( – 420֯ )
Answer: (iii) tan( - 120°) = - tan12 …….(tan( - x) = tanx) = - tan(180° - 60°) ……. (in II quadrant tanx is negative) = - ( - tan60°) = tan60°
Find the value of (i) cos 840֯(ii) sin 870֯
Answer: (i) Cos840° = Cos(2.360° + 120°) …………(using Cos(2ϖ + x) = Cosx) = Cos(120°) = Cos(180° - 60°) = - Cos60° ……………(using Cos(ϖ - x) = - Cosx) = - 1/2 (ii) sin870° = sin(2.360° + 150°)...
Find the probability that a leap year selected at random w ill contain 53 Sundays.
A game consists o f tossing a 1 rupee coin three times, and noting Its outcomes each time. Find the probability o f getting (I) 3 heads, (II) at least 2 tails.
All kings, queens, and aces are removed from a pack o f 52 cards. The remaining cards are well-shuffled and then a card Is drawn from I t Find the probability that the drawn card Is
(I) a black face card,
(II)a red face card.
All red face cards are removed from a pack o f playing cards. The remaining cards are well-shuffled and then a card Is drawn at random from them. Find the probability that the drawn card Is
(I) a red card,
(II) a face card,
(III)a card of clubs.
What Is the probability that an ordinary year has 53 Mondays?
A card Is drawn at random from a well-shuffled pack o f 52 cards. Find the probability that the card was drawn Is neither a red card nor a queen.
5 cards the ten, Jack, queen, king and ace o f diamonds are well shuffled with their faces downward. One card Is then picked up at random. (a) What Is the probability that the drawn card Is the queen? (b) If the queen Is drawn and put aside and a second card Is drawn, find the probability that the second card Is (I) an ace, (II) a queen.
A letter Is chosen at random from the letter o f the word ‘ASSOCIATION’. Find the probability that the chosen letter Is a (I) vowel (II) consonant (III) S
Two dice are rolled once. Find the probability o f getting such numbers on 2 dice whose product Is a perfect square.
A die Is rolled twice. Find the probability that _9_ _ 3 12 4
(I) 5 w ill not come up either time,
(II) 5 w ill come up exactly one time,
(III) 5 w ill come up both the times.
A group consists o f 12 persons, o f which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group Is selected at random. Assuming that each person Is equally likely to be selected, find the probability o f selecting a person who Is
(I) extremely patient,
(II) extremely kind o r honest. Which o f the above values did you prefer more?
A carton consists o f 100 shirts o f which 88 are good and 8 have minor defects. Rohlt, a trader, w ill only accept the shirts which are good. But, Kamal, and another trader w ill only reject the shirts which have major defects. 1 shirt Is drawn at random from the carton. What Is the probability that It Is acceptable to
(I)Rohlt,
(II) Kamal?
A Jar contains 54 marbles, each o f which some are blue, some are green and some are white. The probability o f selecting a blue marble at random Is and the probability o f selecting a green marble at random Is | . How many white marbles does the Jar contain?
The sides of a rectangle are given by the equations x = – 2, x = 4, y = – 2 and y = 5. Find the equation of the circle drawn on the diagonal of this rectangle as its diameter.
Answer: The intersection points in clockwise fashion are:( - 2, 5), (4, 5), (4, - 2), ( - 2, -2). The equation of a circle passing through the coordinates of the end points of diameters is (x - x1)...
A Jar contains 24 marbles. Some o f these are green others are blue. If a marble Is drawn at random from the Jar, the probability that It Is green Is | . Find the number o f blue marbles In the Jar.
Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(p, q) and B(r, s)
Answer: The equation of a circle passing through the coordinates of the end points of diameters is (x - x1) (x - x2) + (y - y1)(y - y2) = 0 Substituting the values:(x1, y1) = (p, q) & (x2, y2) =...
Find the equation of the circle, the coordinates of the end points of one of whose diameters are A( – 2, – 3) and B( – 3, 5)
Answer: The equation of a circle passing through the coordinates of the end points of diameters is (x - x1) (x - x2) + (y - y1)(y - y2) = 0 Substituting the values:(x1, y1) = ( - 2, - 3) & (x2,...
Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(5, – 3) and B(2, – 4)
Answer: The equation of a circle passing through the coordinates of the end points of diameters is (x - x1) (x - x2) + (y - y1)(y - y2) = 0 Substituting the values:(x1, y1) = (5, - 3) & (x2, y2)...
Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(3, 2) and B(2, 5)
Answer: The equation of a circle passing through the coordinates of the end points of diameters is (x - x1) (x - x2) + (y - y1)(y - y2) = 0 Substituting the values (x1, y1) = (3, 2) & (x2, y2) =...
If two diameters of a circle lie along the lines x – y = 9 and x – 2y = 7, and the area of the circle is 38.5 sq cm, find the equation of the circle.
Answer : The point of intersection of two diameters is the centre of the circle. The point of intersection of two diameters x – y = 9 and x – 2y = 7 is (11, 2). ∴...
A bag contains 18 balls out o f which x balls are red.
(I)If one ball Is drawn at random from the bag, what Is the probability that It Is not red?
(II) If two more red balls are put In the bag, the probability o f drawing a red ball w ill be | times the probability o f drawing a red ball In the firs t case. Find the value o f x.
Find the equation of the circle passing through the point ( – 1, – 3) and having its centre at the point of intersection of the lines x – 2y = 4 and 2x + 5y + 1 = 0.
Answer: The intersection of the lines: x – 2y = 4 and 2x + 5y + 1 = 0. is (2, - 1) The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is...
Find the equation of the circle whose centre is (2, – 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4.
Answer: The intersection of the lines: 3x + 2y = 11 and 2x + 3y = 4 Is (5, - 2) The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the...
Find the equation of the circle of radius 5 cm, whose centre lies on the y – axis and which passes through the point (3, 2).
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. The centre lies on Y - axis, ∴ it’s X - coordinate...
Find the equation of the circle whose centre is (2, – 5) and which passes through the point (3, 2).
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. (h, k) = (2, - 5) For determining the equation of...
Find the centre and radius of each of the following circles : 
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Comparing the given equation of circle with...
Find the centre and radius of each of the following circles : 
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Comparing the given equation of circle with...
Find the centre and radius of each of the following circles : 
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Comparing the given equation of circle with...
Find the centre and radius of each of the following circles : 
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Comparing the given equation of circle with...
Find the equation of a circle with Centre at the origin and radius 4
Answer: The general form of the equation of a circle is: (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle...
Find the equation of a circle with Centre ( – a, – b) and radius 
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle...
Find the equation of a circle with Centre (a cos ????, a sin ????) and radius a
Answer: The general form of the equation of a circle is: (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle...
Find the equation of a circle with Centre (a, a) and radius √2
Answer: The general form of the equation of a circle is (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle...
The probability o f selecting a red ball at random from a Jar that contains only red, blue and orange balls Is j . The probability of selecting a blue ball at random from the same Jar Is j .If the Jar contains 10 orange balls, find the total number o f balls In the Jar.
Find the equation of a circle with Centre ( – 3, – 2) and radius 6
Answer: The general form of the equation of a circle is: (x - h)2 + (y - k)2 = r2 (h, k) is the center of the circle. r is the radius of the circle. Substituting the center and radius of the circle...
Find the equation of a circle with Centre (2, 4) and radius 5
Answer: The general form of the equation of a circle is: (x - h)2 + (y - k)2 = r2 (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle...
A piggy bank contains hundred 50-p coins, seventy Rs. 1 coin, fifty Rs. 2 coins and th irty Rs. 5 coins. If It Is equally likely that one of the coins will fall out when the blank Is turned upside down, what Is the probability that the coin(I) will bea R s. 1 coin? (II) will not be a Rs. 5 coin (III) will be 50-p or a Rs. 2 coin?
A box contains 80 discs, which are numbered from 1 to 80. If one disc Is drawn at random from the box, find the probability that It bears a perfect square number.
Tickets numbered 2 ,3 ,4 , 5……………100,101 are placed In a box and mix thoroughly. One ticket Is drawn at random from the box. Find the probability that the number on the ticket Is(III) a number which Is a perfect square (Iv) a prime number less than 40.
Tickets numbered 2 ,3 ,4 , 5……………100,101 are placed In a box and mix thoroughly. One ticket Is drawn at random from the box. Find the probability that the number on the ticket Is
(I)an even number
(II)a number less than 16
Cards marked with numbers 1,3, 5……………..101 are placed In a bag and mixed thoroughly. A card Is drawn at random from the bag. Find the probability that the number on the drawn card Is
(I)less than 19,
(II) a prime number less than 20.
A box contains cards bearing numbers 6 to 70. If one card Is drawn at random from the box, find the probability that It bearsA box contains cards bearing numbers 6 to 70. If one card Is drawn at random from the box, find the probability that It bearsA box contains cards bearing numbers 6 to 70. If one card Is drawn at random from the box, find the probability that It bears
(III) an odd number less than 30,
(Iv) a composite number between 50 and 70.
Prove that:
Answer: Taking LHS: We know that: Putting the values, we get = 4 = RHS ∴ LHS = RHS