FROM BHANU KUMAR
SRI CHAITANYA EDUCATIONAL INSTITUTIONS
JUNIOR INTER ( INTERMEDIATE )
JUNIOR INTER MATHEMATICS- IB (MAX MARKS 75)
STUDENT HAS TO ANSWER ALL 10 VSAQ (2MARKS EACH) QUESTIONS,
5 OUT OF 7 SAQ( 4 MARKS EACH) QUESTIONS AND
5 OUT OF 7 LAQ (7 MARKS EACH) QUESTIONS
MATHEMATICS - 1B
BLUE PRINT
S.NO
TOPIC NAME
NUMBER OF
VSAQ
(2M)
SAQ (4M) LAQ (7M)
TOTA
L
1
LOCUS
1
4
2
CHANGE OF AXES
1
4
3
STRAIGHT LINES
4
PAIR OF STRAIGHT LINES
5
3D GEOMENTRY
1
6
PLANES
1
2
7
LIMITS & CONTINUITY
3
6
8
DIFFERENTIATIONS
1
9
TANGENT & NORMALS
1
10
MAXIMA & MINIMA
11
RATE OF CHANGE
12
ERRORS & APROXIMATIONS
13
PARTIAL DIFFERENTIATION
TOTAL NO. OF QUESTIONS
2
1
2
1
13
2
14
1
9
1
17
1
9
1
7
1
4
1
2
1
10
7
4
7
95
Maths - I Year IB - Part I
DIFFERENTIATION
LAQ
1. If x y  e x y then S.T
Sol :
x y  e
dy
log x

dx 1  log x 2
(Mar 08, 07, Sep 00)
x y 
Apply log on both sides
log e x y  loge
x y 
y.log x   x  y .log e  log e  1
y.log x  x  y
 y  y log x  x
 y 1  log x   x
 y
dy

dx

x
1  log x
d
1  log x
dx
2
1  log x 
1  log x .1 x.
1
1  log x  x. 
 x
1  log x
2

1  log x 1
1  log x 
2
dy
log x

dx 1  log x 2
2. Find the derivative of x tan x  sin x 
cos x
Sol: Let y  x tan x  sin x 
cos x
u  x tan x , v  sin x 
cos x
dy
d
du dv
 u  v  

dx dx
dx dx
w.r.t ‘x’ (Mar 08, 07, May 06)
u  x tan x Apply log on both sides
log u = tan x.log x
diff . w.r.to ' x '
d
d
log u   tan x.log x
dx
dx
d
dv
du
uv   u  v
dx
dx
dx
1 du
d
d
 tan x log x   log x  tan x
u dx
dx
dx
1 du
d
d
 tan x log x   log x  tan x
u dx
dx
dx
 tan x

du
 x tan x 
 sec 2 x.log x  ----- (1)
 x

dx
v  sin x
cos x
Apply log on both sides

log v  log sin x
cos x

log v  cos x.log sin x 
diff . w.r.to ' x '
d
d
log v    cos x.log sin x 
dx
dx
d
dv
du
uv   u  v
dx
dx
dx
1 dv
1 d sin x 
 cos x.
 log sin x. sin x
v dx
sin x dx
dv
cos x
 sin x   cot x cos x  sin x.log sin x  ------ (2)
dx
From (1) and (2)
 tan x

dy
cos x
 x tan x 
 sec 2 x.log x   sin x cos x.cot x  sin x.log sin x 
 x

dx
1 y 2
dy
3.
If 1 x  1 y  a  x  y  then S.T

(Mar 08, 05)
dx
1 x 2
SOL:
Put x  sin a and y  sin b
2
2
1 x 2  1 y 2  a  x  y 
 1 sin 2 a  1 sin 2 b  a sin a  sin b 
cos a  cos b  a sin a  sin b 

 a  b   a  b 
 a  b   a  b 
2 cos 
 cos 
  a 2 cos 
 sin 

 2   2 
 2   2 

 a  b 
cos 

 a  b 
 2 
 a  cot 
  a

 a  b 

sin 

 2 
cot a  b   2a  a  b  cot 1 2a 
sin1 x  sin 1 y  cot 1 2a
diff . w.r.to ' x '
d
d
sin1 x  sin 1 y   cot 1 2a 

dx
dx
1
1 x
2
1
1 x 2

dy
0
1 y dx

1 y 2
dy
dy


dx
1 y 2 dx
1 x 2
1
2
1
4. If x y  y x  ab then Show that
 yx y1  y x log y 
dy
 (Mar 2003)
  y
 x log x  xy x1 
dx
SOL: Given x y  y x  ab
Let u = xy and v = yx
u + v = ab
du dv
diff . w.r.to ' x '
  0 -----------(2)
dx dx
u = xy take log on both sides
 log u  log x y
 log u  y log x
diff . w.r.to ' x '
1 du
1
dy
.  y.  log x.
u dx
x
dx
1 du
1
dy
.  y.  log x.
u dx
x
dx

du
dy 
 x y  y.x1  log x. 

dx
dx 
du
dy
 y. x y1  x y log x
dx
dx
x
v = y take log on both sides
log v = log yx
 log v  x.log y
diff . w.r.to ' x '
d
d
log v    x.log y 
dx
dx
1 dv
1 dy
.  x.
 log y
v dx
y dx


dv
dy
 y x  xy 1  log y 


dx
dx
dv
dy
 xy x1  y x log y
dx
dx
From (2) yx y1  x y log x
dy
dy
 xy x1  y x log y  0
dx
dx
dy  y
x log x  xy x1     xy y1  y x log y 


dx
 xy y1  y x log y 
dy


 y
x1 

dx
 x log x  xy 
5. Find the derivative
2
3
5
6
dy
1  2 x  1  3x 
of each of the functions y 
dx
1  6 x  1  7 x 
3
4
.
6
7
(2010)
SOL: take log on both sides and diff. w.r.to x
SAQ
1)
Find the derivative of
Sol : Let
x  1, with respect to x using first principle
f  x   x 1
f  x  h  x  h  1
f 1  x   Lt
h 0
f  x  h  f  x 
x  h  1  x 1
 Lt
h 0
h
h
 x  h  1   x  1
1/ 2
 Lt
h 0
h
1/ 2
 x  h 1   x  1

1/ 2
1/ 2
 x  h 1   x  1
1/ 2
1/ 2
(JUNE2005)

2.
1
 x  1   x 1
1/ 2
1/ 2

1
2 x 1
Find the derivative of cos ax with respect to x using first principle (MAR2009)
Let f (x) = cos ax
f(x + h) = cos  ax  ah
f 1  x   Lt
h 0
f  x  h  f  x
h
 ax  ah  ax 
 ax  ah  ax 
2 sin 
 .sin 


cos ax  ah  cos ax




2
2
 Lt
 Lt
h 0
h 0
h
h
 2ax  a 

 Lt sin nq  n
 2sin 
.

 2  2  q0 q

= a.sin ax
3.
Find the derivative of sin 2x with respect to x using first principle (2010, 2002)
Let f(x) = sin 2x
f  x  h  sin 2 x  2h
f  x  h  f  x
sin  2 x  2h  sin 2 x
 Lt
h 0
h 0
h
h
 2 x  2h  2 x 
 2 x  2h  2 x 
2 cos 
 .sin 





2
2
 Lt
h 0
h
 4 x  2h 
 4x 
sin h
 2. Lt cos 
. Lt
 2.cos    2.cos 2 x



 2 
h 0
 2  h 0 h
f 1  x   Lt
4.
Find the derivative of tan 2x with respect to x using first principle (MAR2005)
SOL: let f  x   tan 2 x and f  x  h  tan 2 x  2h
f 1  x   Lt
h 0
f  x  h  f  x
h
sin  2 x  2h sin 2 x

tan 2 x  2h  tan 2 x
cos 2 x  2h cos 2 x
 Lt
 Lt
h 0
h 0
h
h
sin  2 x  2h.cos 2 x  cos 2 x  2h.sin 2 x
h 0
h.cos  2 x  2h .cos 2 x
 Lt
sin 2 x  2h  2 x 
h 0 h.cos 2 x  2 h .cos 2 x
 Lt
sin 2h
1
1
 Lt
.
h 0
h 0 cos  2 x  2h  cos 2 x
h
 Lt
1
1
1

2
 2.sec 2 2 x
cos 2 x cos 2 x
cos 2 2 x
Find the derivative of sec 3x with respect to x using first principle(MAR2008)
 2
5.
let f  x   sec3x
f  x  h  sec 3 x  3h
f 1  x   Lt
h 0
f  x  h  f  x
h
sec 3 x  h  sec3 x
 Lt
h 0
h 0
h
 Lt
 Lt
h 0
1
1

cos 3 x  3h cos 3x
h
cos 3x  cos 3x  3h
cos 3x  3h.cos 3 x.h
 3 x  3x  3h 
 3x  3 x  3h 

2sin 
.sin






2
2
 Lt
h 0
h.cos 3 x  3h.cos 3 x
 6 x  3h 
 2 Lt sin 
. Lt
 2  h0
h 0
6.
 3h 
sin 

 2 
h
 6 x   3 
2 sin  . 
 2   2 
sin 3 x
1

 3.

cos 3x.cos 3x
cos 3 x cos 3x
 3.sec3 x.tan 3 x
Find the derivative of cos2 x with respect to x using first principle
let f  x  cos 2 x
f  x  h  cos 2  x  h
cos 2  x  h  cos 2 x
h 0
h
f 1  x   Lt
cos 2  x  h  cos 2 x
sin  x  x  h.sin  x  x  h
 Lt
 Lt
h 0
h 0
h
h
sin h
 sin 2 x 1
h 0
h
 Lt sin  2 x  h. Lt
h 0
  sin 2x
7.
If x  tan e y find
dy
(MAR2005,JUNE2003)
dx
Sol : tan e y  x
 e y  tan 1 x
d y
d
e  tan1 x
dx
dx
 dy 
1
e y .

 dx  1  x 2
dy
e y

dx 1  x 2
8.
Find
dy
if x  a cos 3 t , y  a sin 3 t
dx
Sol : x  a cos3 t
diff . w.r.to ' t '
diff . w.r.to ' t '
dx d
  a cos3 t 
dt dt
d
 a 3cos 2 t  cos t 
dt
dy d
 a sin 3 t
dt dt
d
= a 3sin 2 t  sin t
dt
 3a cos 2 t  sin t 
 3a sin 2 t.cos t
dy dy / dt
3a sin 2 t.cos t


  tan t
dx dx / dt 3a cos2 A.sin A
dy
Find
if x  a cos t  t sin t  , y  a sin t  t cos t  (MAR2008, EP2002)
dx

9.
y  sin t
Sol : x  a cos t  t sin t 
diff . w.r.to ' t '
dx d
 a cos t  t sin t 
dt dt
d

dx
d
 a  cos t  t sin t 
 dt

dt
dt
 a  sin t  t cos t  sin t   at cos t
y  a sin t  t cos t 
diff . w.r.to ' t '
dy d
 a sin t  t cos t 
dt dt
d

d
 a  sin t  t cos t 
 dt

dt
 a cos t  t sin t  cos t 
 a cos t  t sin t  cos t   at sin t

10.
dy dy / dt at sin t


 tan t
dx dx / dt at cos t
Find
 2 x 
dy
if y  tan 1 
(MAR2004, SEP92,98)
1 x 2 
dx
 2 x 
Sol : Let y  tan 1 
1 x 2 
put x  tan q  q  tan 1 x
 2 tan q 
y  tan 1 
1 tan 2 q 
y  tan 1  tan 2q 
y  2q
y  2 tan 1 x
diff . w.r.to ' x '
dy
d
2
 2 tan 1 x 
dx dx
1 x2
11.
 1  x 2  1 x 2 

 w.r.to.x ( MAR2010,2004, 06, MAY97)
Find the derivatives of tan 1 
 1  x 2  1 x 2 
 1  x 2  1 x 2 


Sol : Let y  tan 1 
 1  x 2  1 x 2 
put x 2  cos 2q  2q  cos1  x 2 
1
 q  .cos1  x 2 
2
2
2


 1  cos 2q  1 cos 2q 
1  2 cos q  2 sin q 




y  tan 
= y  tan 
 1  cos 2q  1 cos 2q 
 2 cos 2 q  2 sin 2 q 
1
 p
 cos q  sin q 

y  tan 1 
= y  tan 1  tan   q 


 cos q  sin q 
  4

p
p 1
 q  y   cos1 x 2
4
4 2
diff . w.r.to ' x '
= y

dy
d p 1
1
1
   cos1 x 2   
2x =

dx dx  4 2
2
1 x 4
12.
Find the derivative of tan 1
Sol : Let y  tan 1
y  tan 1
x
1 x 4
1 cos x
w.r .to ' x ' (JUN2002)
1  cos x
1 cos x
1  cos x
2 sin 2 x / 2
2 cos 2 x / 2
y  tan 1  tan x / 2
y x/2
diff . w.r.to ' x '
dy
d
  x / 2
dx dx
dy 1

dx 2
13.
2
dy sin  a  y 
Find the derivative of sin y = x sin (a + y) then S.T

dx
sin a
Sol : Given sin y  x sin a  y 
 x
sin y
sin a  y 
Diff. w. r. to y
dx
d  sin y 
 
dy dy  sin  a  y 




sin a  y  d sin y  sin y d sin  a  y 

dy
dy

sin 2 a  y 
sin  a  y .cos y  sin y.cos a  y 
sin 2  a  y 
dx sin  a  y  y 

dy
sin 2 a  y 
2
dy sin  a  y 

dx
sin a
15.
 x 2  x  2 
Find the derivative of log  2

 x  x  2 
 x 2  x  x 
 = log  x 2  x  2   log  x 2  x  2 
Let y  log  2
 x  x  2 
Diff. w. r. to x
2 x 1
dy
2 x 1
 2
 2
dx x  x  2 x  x  2
2
2
dy  2 x 1 x  x  2  2 x 1 x  x  2

dx
 x 2  x  2 x 2  x  2

2 1 x 2 
 x 2  2
2
 x2
=
2 1 x 2 
x4  3x2  4
3D GEOMETRY
LAQ
1)
If the relations between the direction cosines of any two lines are given by l  m  n  0 ,
l 2  m 2  n 2  0, find the angle between the lines (2007, 2004)
Sol : Let a1, b1, c1 and a2, b2, c2 are the d.r’s of the two lines satisfying the equations
l  m  n  0 ------ (1) and l 2  m 2  n 2  0 ------ (2)
From (1), l = – m – n = –(m + n)
Substituting in (2)
m  n   m 2  n 2  0
2
m2  n 2  2mn  m 2  n 2  0
2m 2  2mn  0
2m  m  n   0
m = 0 or m = –n
case I :- m = 0, substituting in (1)
l  n  0  l  n
l : m : n  n : o : n
D.R’s of the first line l1 are
a1 , b1 , c1  1, 0,1
case II :- m = – n, substituting in –(1)
l = 0, l : m : n  0 : n : n
D.R’s of the second line l2 are
a2 , b2 , c2  0,1, 1
Suppose q be the acute angle between the two lines
cos q 
cos q 
2.
a1a2  b1b2  c1c2
a12  b12  c12 a22  b22  c22
0  0 1
2. 2
 1/ 2  cos p / 3  q  60o
If the relations between the direction cosines of any two lines are 3l  m  5n  0,
6mn  5lm  2l n  0 find the angle between the lines
(MARCH 2010 , 2009, 2006)
Sol :- Let a1, b1, c1 and a2, b2, c2 are the d.r’s of the two lines satisfying the equations
3l  m  5n  0  (1) and 6mn  5lm  2nl  0  (2)
from (1) m = –3l – 5n
substituting in (2)
6n 3l  5n  5l 3l  5n  2nl  0
18nl  30n 2 15l 2  25nl  2nl  0
15l 2  45nl  30n 2  0
15 l 2  3nl  2n 2   0
l 2  3nl  2n 2  0
l 2  2nl  nl  2n 2  0
l l  2nl   n l  2n  0
l  2nl  n  0
l  2n, l  n
Case – I :- If l  n
substituting in (1) 3n  m  5n  0
m  2n  0, m  2n
l : m : n = –n : –2n : n
DR’s of the first line l1 are
a1 , b1 , c1  1, 2,1
Case – II :- If l  2n, substituting in
 6 n  m  5n  0
m  n  0, m  n
– (1)
l : m : n  2n : n : n  2,1,1
D.R’s of the second line l2 are
a2 , b2 , c2  2,1,1
Suppose q be the acute angle between the two lines
cos q 

a1a2  b1b2  c1c2
a12  b12  c12 a22  b22  c22
2  2 1
6. 6
 1/ 6
q  cos1 1/ 6
3.
A line makes angles a, b , g , d with the four diagonals of a cube S.T
4
( 2008, 2005)
3
Sol :- Consider a unit cube OALB; CNPM as shown in the figure
Y
cos 2 a  cos 2 b  cos 2 g  cos 2 d 
 0,1,0 B
M
L
1,1,0 
1,1,1P
 0,1,1
A1,0,0X
O
 0,0,0 
Z
C 0,0,1
N
1, 0,1
Let one of the vertices of the cube be the origin O and the co-ordinate axes be along three edges
OA, OB, OC passing through the origin. The co-ordinates of the vertices of the cube with
respect to the frame of reference OABC are as shown in figure. The diagonals of the cube are
OP, CL, AM, BN
 1 1 1 
D.R’s of OP = (1, 1, 1) ; D.C’s of OP =  ,
, 
 3 3 3 
 1 1 1
D.R’s of CL = (1, 1, –1) ; D.C’s of CL =  ,
, 
 3 3 3 
D.R’s of AM = (–1, 1, 1);
 1 1 1 
D.C’s of AM =  ,
, 
 3 3 3 
D.R’s of BN = (1, –1, 1) ;
 1 1 1 
D.C’s of BN =  ,
, 
 3 3 3 
Let the direction cosines of the given ray be (l, m, n).
If this ray makes the angles with the four diagonals of the cube then
l
m
n
l  m n
cos a 



3
3
3
3
cos b 
l
m
n
l  mn



3
3
3
3
cos g 
l
m
n
l  m  n



3
3
3
3
cos d 
l
m
n
l m  n



3
3
3
3
 cos 2 a  cos 2 b  cos 2 g  cos 2 d 
l  m  n
2
3

l  m  n
2

4 l 2  m 2  n 2 
3
3

l  m  n
2

3
l  m  n
2

3
4
 l 2  m2  n2  1
3
SAQ
1.
If two vertices of a triangle are (3, –9, 11), (–2, 5, 7) and the centroid is (–3, 0, 3) then find the
third vertex
Sol: A(3, –9, 11), B(–2, 5,7), C(x, y, z)
Given centroid = (–3, 0, 3)
Centroid le of ABC is
 x  x2  x3 y1  y2  y3 z1  z 2  z3 
G   1
,
,



3
3
3
 3  2  x 9  5  y 11  7  z 
 
,
,
  3, 0,3


3
3
3
1 x
4  y
18  z
 3,
 0,
3
3
3
3
1 + x = –9
4  y  0
18  z  9
y4
x  9 1
z  9 18
x  10
z  9
Third vertex C(–10, 4, –9)
2.
Find the centroid of the tetrahedron formed by the points (3, 2, –4), (5, 4, –6),
(9, 8, –10), (3, 4, 10)
Sol : Let A(3, 2 –4), B(5, 4, –6), C(9, 8, –10), D(3, 4, 10)
 x1 x2  x3  x4 y1  y2  y3  y4 z1  z2  z3  z4 

,
,



4
4
4
Centroid of the tetrahedron = 
 3  5  9  3 2  4  8  4 4  6  40  10   20 18 10   9 5 
 
,
,
  , ,
  5, , 

  4 4 4   2 2 
4
4
4
3.
Find the centroid of the triangle formed by the points (2, 1, 4), (3, –1, 2), (5, 0, 6)
Sol : Let
A 2,1, 4 B 3, 1, 2 C 5, 0, 6
10

 centroid   , 0, 4
 3

4.
Find the coordinates of the point at which yz – plane intersects the line segment joining the
points (–2, 3, 7) an (6, –1, 2)
Sol : Let A (–2, 3, 7), B(6, –1, 2)
yz – plane divides AB in the ratio
x1 : x2  2 : 6
 2 : 6  1: 3
1 6  32 11  33 12  37

Required point = 
,
,
1 3
1 3
1  3 

 6  6 1  9 2  21 

0, 2, 23 
 
,
,


 4
4
4  
4 
Note : xy  plane divides  z1 : z2
zy  plane divides  x1 : x2
5.
Find the coordinates of the point (1, –5, 3) in the new system when the origin is shifted to (–4, 3,
9)
Sol : Let  x, y, z   1, 5, 3 ,  h, k , l   4,3,9
x  x  h, y  y  k , z  z  l
x  5, y  8 z  6
  x, y, z   5, 8, 6
6.
S.T the point (1, 2, 3), (7, 0, 1) (–2, 3, 4) are collinear
Sol : Let A(1, 2, 3), B(7, 0, 1), C(–2, 3, 4)
AC  9  1  1  11,
AB  36  4  4  44,  2 11,
BC  81  9  9  99,  3 11,
 AC  AB  11  2 11  3 11  BC ,
Given points are collinear
If 3, 2, 1 ,  4,1,1 , 6, 2,5 are three vertices and 4, 2, 2 is the centroid of a tetrahedron, find
the fourth vertex of the tetrahedron
Sol: Let A(3, 2, –1), B(4, 1, 1), C(6, 2, 5),
D(x, y, z) centroid = (4, 2, 2)
 3  4  6  x 2  1  2  y 1  1  5  z 
 
,
,
  4, 2, 2


4
4
4
7.
x 13
 4,
4
x + 13 = 16
x=3
5 y
 2,
4
5+y=8
y=3
  x, y, z   3,3,3
LIMITS and CONTINUTY VSAQ
1)Show that Lt 
x 2
Sol:
Lt 
x 2
x2
 1
x2
x2
x2
 Lt
x  2 x  2 h x  2
 Lt
2h2
2h2
 Lt
h
h
 Lt
 1
h0 h
h
h 0
h 0
5 z
 2,
4
5+z=8
z=3
a x 1
2) Evaluate Lt x
x 0 b  1
Sol:
ax 1
Lt
a x  1 x0 x
Lt x

x 0 b  1
bx 1
Lt
x0
x
log e a

 logb a
log e b
 sin 2 x

3) Is f(x) defined by  x
 1
 sin 2 x

Sol: f(x)=  x
 1
if
x0
if
x0
if
x0
if
x0
continuous at 0?
sin 2 x
x 0
x
Given f(0)=1 and Lt f ( x) = Lt
x 0
sin 2 x
 sin 2 x 
= Lt 
2  2 Lt
 2.1  2

x 0
2 x 0
2x
 2x 
Since Lt f ( x )  f (0) f(x) is not continuous at x=0
x 0
 cos ax  cos bx
if
2

x
4)Show that f(x) = 
 1 b2  a 2 
if
 2
Sol: Given f(0)=
x0
where a and b are real constants, is continues at ‘0’
x0
1 2
b  a2 

2
 bx  ax   bx  ax 
2sin 
 sin 

cos ax  cos bx
2   2 

Lt f ( x) = Lt
= Lt
x 0
x 0
x 0
x2
x2
ba
ba 
2sin x 
 .sin x 

2 
2 


= Lt
x 0
x2
ba
ba
sin x 
sin x 
 ba

2  

2  ba


= 2 Lt
.
.
 . Lt

x 0
 b  a   2  x 0  b  a   2 
x
x


 2 
 2 
ba ba 1 2
2
= 2.1. 
 .1. 
 = b  a 
2
2
2

 

Since Lt f ( x )  f (0) f(x) is continuous at x=0
x 0
IMPARTENT LONG ANSWER QUESTIONS
1.
Find the centroid of the triangle formed by the lines 12 x 2  20 xy  7 y 2  0 and
2x  3 y  4  0
2.
The product of the perpendiculars from a, b  to the pair of lines ax 2  2hxy  by 2  0 is
aa 2  2hab  bb 2
 a  b  4h 2
2
3
Show that the area of the triangle formed by the lines ax 2  2hxy  by 2  0 and
lx  my  n  0 is
n 2 h 2  ab
Sq. units
am 2  2hklm  bl 2
4.
Show that the lines  x  2a   3 y 2  0, x  a form an equilateral triangle and find the area of
triangle
5
Prove that the lines represented by the equations x 2  4 xy  y 2  0, x  y  3 form an
equilateral triangle and find the area of triangle
6
If S  ax 2 2hxy  by 2  2 gx  2 fy  c  0 represents a pair of parallel lines then S.T
2
h 2  ab, af 2  bg 2 . also the distance between the parallel lines  2
g 2  ac
f 2  bc
2
a a  b 
b a  b 
7
Find the angle between the lines joining the origin to the points of intersection of the curve
x 2  2 xy  y 2  2 x  2 y  5  0 and the line 3 x  y  1  0
8
Find the condition for the chord lx  my  1 of the circle x 2  y 2  a 2 (whose centre is the
origin) to subtend a right angle at the origin
Find the values of k, if the lines joining the origin to the points of intersection of the curve
2 x 2  2 xy  3 y 2  2 x  y 1  0 and the lines x  2 y  k are mutually perpendicular
9
10
Show that the lines joining the origin to the points of intersection of the curve
x 2  xy  y 2  3 x  3 y  2  0 and the straight line x  y  2  0 are mutually perpendicular
11
12
13
Show that the pairs of lines 3 x 2  8 xy  3 y 2  0 and 3 x 2  8 xy  3 y 2  0  2 x  4 y 1  0
form a square
Show that the equation 2x2 – 13xy – 7y2 + x + 23y – 6 = 0 represents a pair of straight lines. Also
find the angle between the co-ordinates of the point of intersection of the lines
If p, q are the lengths of the perpendicular’s drawn from the origin to the tangent and normal
drawn at any point on x 2/ 3  y 2 / 3  a 2 / 3 respectively, then show that 4 p 2  q 2  a 2
14
Show that the curved surface of a cylinder inscribed in a sphere of radius R is maximum then the
height of the cylinder is 2 R.
15
Show that the semi-vertical angle of the right circular cone of maximum volume and of given
slant height is tan 1 2
16
If the tangent at any point on the curve x 2/ 3  y 2 / 3  a 2 / 3 Intersects the coordinate axes in A, B
then show that the length AB - is constant.
17
The tangent at any point ‘P’ on the curve x m y n  a mn cuts the coordinate axes in A and B.
Show that AP : PB is constant
18
If x y  y x  a b then S .T
19
20
 
 yx y1  y x log y 
dy

  y
 x log x  xy x1 
dx
1 y 2
dy
If 1 x  1 y  a  x  y  then S.T

dx
1 x 2
Find the orthocentre of the triangle whose sides are given by 7x + y – 10 = 0, x  2 y  5  0 and
x y2 0
2
2
21
Find the circumcentre of the triangle whose sides are given by x + y + 2 = 0,
5x – y – 2 = 0 and x – 2y + 5 = 0
22
If a, b  is the centroid of the triangle formed by lines ax 2  2hxy  by 2  0 and
a
b
2
lx  my  n  0 then prove that


2
bl  hm am  hl 3am  2hlm  bl 2 