Q.12142/3
x sin x cos x dx (a 2 cos 2 x b 2 sin 2 x ) 2 [Ans.
[Sol.
[5]
1 x 1 b t a n t a n x C ] 2 2 2 2 2 2 ab a 2( b a ) a cos x b sin x 1
x s i n x co s x dx (a 2 cos 2 x b 2 sin 2 x ) 2
Integrating by parts taking ‘x’ as the t he Ist function and
sin x cos x as 2nd function (a 2 cos 2 x b 2 sin 2 x ) 2
For Integral of II put a2cos2x + b2sin2x = t (b2 – a2)sinx cosx dx =
1 dt 2
1 1 1 2 2 2 2 2 2 2 2 a cos x b sin x 2( b a ) 2( b a )
so I = x ·
=
x a 2 cos 2 x b 2 sin 2 x 2( b2 a 2 )
+
1 2( b 2 a 2 )
1 + 2 2 2 b (b a 2 )
·
1·
1 a 2 cos 2 x b 2 sin 2 x
sec 2 x dx a 2 b 2 tan 2 x
sec 2 x dx 2
a tan 2 x b
1 x 1 b 1 a tan x tan – 2 ( b 2 a 2 )(a 2 cos 2 x b 2 sin 2 x ) 2 b 2 (b 2 a 2 ) a a =
1 x 1 b tan x tan C ] a (a 2 cos 2 x b 2 sin 2 x) 2( b 2 a 2 ) ab 1
dx
ax 4
ln tan
[Sol.
Limit
x 0
= Limit ln tan ax x 0 4
sin bx bx bx
1 bx
1 tan tan ax = Limit ax . tan ax bx 4 4 x 0 bx ax
2a 1 tan 4 ax . tan 4 b ] Q.1053/3 The function f is defined by y = f (x). Where x = 2t – | t |, y = t² + t | t |, t R. Draw the graph of f for the interval 1 x 1. Also discuss its continuity & differentiability at x = 0. [REE ’91, 6] [Ans. f(x) = 2x² for 0 x 1 & f(x) = 0 for 1 x < 0, f is diff. & hence cont. at x = 0] = Limit x 0
[Sol.
x = 2t – | t | ; y = t2 + t | t | t if t 0 x= S 3t if t 0
y=
if t 0
2t 2
S 0
if t 0
for 0 < t < 1 we have x = t Therefore f is defined as y = f(x) =
2x2
S 0
for 0 x 1 for 1 x 0
The graph of f is as shown f is diffrentiable and continuous at x = 0.
]
Q.119/3 A tangent line is drawn to a circle of radius unity at the point A, and a segment AB is laid off whose length is equal to that of the arc AC. A straight line BC is drawn to intersect the extension of the diameter AO at the point P. Prove that : (1 cos ) (i) PA = (ii) Limit 0 PA = 3. sin Use of series expansion or L' Hospital's rule prohibited. [Sol.(i) From the figure, tan =
sin (1) 1 cos
Now in BPA tan =
(ii)
PA (1 cos ) PA = = tan sin
3 Limit (1 cos ) = Limit 1 cos Limit 0 sin 3 0 3 2 sin . 3 1 1 cos Now Limit = 0 2 2
sin Limit 3 t sin 3 t 3 t (3 sin t 4 sin 3 t ) Limit Limit Let l = 0 = t 0 = t 0 27 t 3 27 t 3 3 Limit t sin t + 4 t 0 27 9 t3 Limit 0
3 =6 sin
l =
9
+
(ii) =
4 27 1 .6=3] 2
8 4 = 9 27
l =
1 6
1 1 n is n 2 2 n 0 n
Q.6flcd The value of Lim (A) 1
[Sol.
(B*) 2
(C) 4
(D) none
1 1 1 1 1 1 1 1 1 1 1 ....... 1 2 n 2 2 n 1 2 2 2 2 2 2 2 2 Lim = Lim = 2 Ans. ] n 1 n 1 1 1 2 2
SUBJECTIVE:
Q.7153/2 A triangle has side lengths 18, 24 and 30. Find the area of the triangle whose vertices are the incentre, circumcentre and centroid of the triangle. [Ans. 3 units] [Sol. radius of the incircle, (18 · 24) 2 18 · 24 r = = 2 (18 24 30) = = 6 s 72 Hence coordinates of incentre (6, 6)
18 , 24 = (6, 8) 3 3
centroid is =
Hence D formed by I, G and circumcentre is
16 6 1 16 6 1 A = 9 12 1 = 3 6 0 = 3 sq. units 26 8 1 20 2 0
8 57 Q.8275/5 Find an upper triangular matrix A such that A3 = 0 27 [Sol.
a b Let A = 0 c be the required matrix.
]
[2]
(elements below the leading diagonals are all zero)
a 3 a 2 b abc b 2 a 2 ab bc 3 0 c3 c 2 and A = 0
Then,
A2 =
8 57 A3 = 0 27
a 3 a 2 b abc b 2 0 = c3
8 57 0 27
a3 = 8, c3 = 27 and a2 b + abc + bc2 = – 57 a = 2 , c = 3 and b = – 3 Thus , a = 2 , c = 3 and b = – 3 ]
ax 4
ln tan
Q.9 68/3 Evaluate : Limit x 0
sin bx
( b 0) .
Use of series expansion and L’Hospital’s rule is not allowed.
[ Ans:
2a ] b
M ATH EM ATIC S
BANSAL CLASSES Tar g et IIT JEE 200 7 CLASS : XIII (XYZ)
D a i ly Pra c t ic e Pro b l e m s
DATE : 28-29//08/2006
TIME : 60 Min.
DPP. NO.-20
Only one is correct:
Q.1
[Sol.
The number of permutation of the letters AAAAB BB C in which the A's appear together in a block of four letters or the B's appear in a block of 3 letters, is (A*) 44 (B) 50 (C) 60 (D) none n(A B) = n(A) + n(B) – n(A B) where A = A's together and B = B's together 5! n(A) = 3! = 20
now
A A A A B B B C
6! n(B) = 4 ! = 30
BBB
A A A A C
n(A B) = 3! = 6 AAAA n(A B) = 50 – 6 = 44 Ans. ] Q.2
If {x} denotes the fractional part function then the number x =
(A) [Sol.
C
BBB
1 2
(B) 0
(C*) –
3 2 2
32 2 2 2
1 2
simplifies to
(D) none
( 3 1) 2( 2 1) ( 3 1) 2 2 ( 3 1) 2 2 1 = = = – Ans. ] ( 4 2 3 ) 2(3 2 2 ) ( 3 1) 2 2( 2 1) 2 2 4 2 2 32
tan kx for x 0 x Q.3cont Let f (x) = . If f (x) is continuous at x = 0 then the number of values of k is 3x 2k 2 for x 0 (A) none (B) 1 [Hint: LHL = k, RHL = 2k 2 = f (0)
(C*) 2 k = 2k 2
(D) more than 2 k = 0 and k = 1/2 ]
1 1 1 Lim Q.4limit Find y2 x y 2 x y 2 1 (A) 0 (B) ln x (C*) – 2 x xx 2y 1 1 1 Lim · Lim [Hint: y2 = – y2 = – 2 Ans. ] x ( x y 2) y 2 x (x y 2) x
(D) does not exist
Q.5s&p Let p(x) be the cubic polynomial 7x3 – 4x2 + K. Suppose the three roots of p(x) form an arithmetic progression. Then the value of K, is 4 16 16 (B) (C) 21 147 441 Let the roots be p – q, p and p + q. Then 3p = 4/7, 3p2 – q 2 = 0, and p(p2 – q 2) = – K/7. Solving yields (A) [Sol.
3
4 K = 7p(2p2); K = 14 ; 21
K=
128 Ans. ] 1323
(D*)
128 1323
cot = 2 1 = 2.414... 8 100x = 241.4... greatest integer = 241 Ans.
]
Q.8182/6 Sum the series, cot 1(2a 1 + a) + cot 1 (2a1 + 3a) + cot 1(2a 1 + 6a) + cot1(2a1 + 10a) + ...... + to ' n ' terms. Also find the sum of infinite terms, (a > 0). ( n 1) a 2 n ( n 1) nth term of 1 + 3 + 6 + 10 +........ is = 2
[Ans. Sn = tan 1
[Sol.
n (n 1) a Tn = cot 1 2 a 1 2
tan 1
a ; S = 2 2
tan 1
4 n (n 1)a 2 = cot 1 2 a
(n 1)a na a 2a 2 2 2 1 = tan1 = tan1 2 = tan (n 1)a na 4 n(n 1)a 1 1 na (n 1)a 2 · 2 2 2 a na = tan –1 ( n 1) – tan –1 2 2 Put n = 1 , 2 , 3 ,........ , n we have Sn = tan 1
( n 1) a 2
tan 1
a ; 2
S =
2
tan 1
a a = cot 1 ] 2 2
a a = cot 1 ] 2 2
dy dy dt dy 1 dy · = · · sec t = = dx dt dx dt cos t dt d dy dt d 2 y · sec t · = = dt dt dx dx 2
d 2 y dy d 2 y dy 2 sec t · 2 sec t · tan t sec t = sec t 2 + dt sec2t · tan t dt dt dt
(1) becomes 2 d 2 y dy 2 sec t sec t · tan t 2 – sin t · sec t dy + y = 0 cos t 2 dt dt dt dy dy d 2 y + tan t – tan t + y = 0 dt dt dt 2
d 2 y + y = 0 ] dt 2
Q.6 11/1 Find the real solutions to the system of equations log10(2000xy) – log10x · log10y = 4 log10(2yz) – log10y · log10z = 1 and log10(zx) – log10z · log10x = 0 [Ans. x = 1, y = 5, z = 1 or x = 100, y = 20, z = 100] [Q.11, Ex-2, log] [Sol. From (1) 3 + log10(2xy) – log10x · log10y = 4 ....(1) log10(xy) – log10x · log10y = 1 – log10(2) from (2) log10(yz) – log10y · log10z = 1 – log10(2) hence log x + log y – log x · log y = log y + log z – log y· log z log x (1– log y) = log z (1 – log y) (log x – log z)(1 – log y) = 0 hence either, log x = log z or log10y = 1 y = 10 but y = 10 does not satisfy equation (1) hence rejected. log x = log z from (3) (log10x)2 = 2(log10x) log10x [log10x – 2] = 0 x = 1 or x = 100 if x = z = 1 then y = 5 x = z = 100 then y = 20 Ans. ] 44
cos n
1 Q.735/6 Let x = n44
sin n
find the greatest integer that does not exceed 100x.
[Ans. 241]
n 1
[Sol.
sin 22 · cos 22.5 cos1 cos 2 ........ cos 44 sin 1 2 x= = sin 22 sin 1 sin 2 ......... sin 44 · sin 22.5 sin 1 2
(using the formula of sum of cos series S =
sin n 2 (n 1) sin n 2 (n 1) cos sin , for sine series S = ) sin 2 2 sin 2 2
13 3 15 26 3 65
Q.3239/5 Find the value of the determinant
[ Sol. D =
3 13 3 5 2 13 3 5 13
5
=
5 10 . 5
3 2 5 1 2 = 5 15 5 3 5 15
2 5 5 3 5
1 5. 3 . 5 5 3
2 5 5 15
2 5 3
1 2 5
[REE’92, 6] [Ans. 5 3 2 5 3 ]
1 1 2 5 1 2 13 2 5 2 5 5 15 5 Vanishes
Use C2 C2 – C1 ]
Q.4129/2 Using algebraic geometry prove that in an isosceles triangle the sum of the distances from any point of the base to the lateral sides is constant. [4] (You may assume origin to be the middle point of the base of the isosceles triangle) [Sol. Equation of the sides AC and AB as :
x y x y = 1 and – = 1 a b a b and P(K, 0) where 0 < K < a and a > 0; b > 0
PN =
K 1 a
1 1 a 2 b2
K a
a b 2
(P and O are on the same side of AC)
2
ab bK PN = ; a 2 b2
Hence
similarly PM =
ab1
K 1 ab a a b 2
PN + PM =
2
2ab a 2 b2
Kb ba a b 2
2
(P and O are on the same side of BC)
which is constant
Q.5 8/4 If 'y' is a twice differentiable function of
]
x, transform the equation, (1 – x2)
d 2 y d y – x + y = 0 by 2 d x d x
means of the transformation, x = sin t, in terms of the independent variable 't '. [Sol.
2 d y d y (1 – x2) 2 – x + y = 0 d x d x
x = sin t ;
dx = cos t dt
....(1)
d 2 y [ Ans. 2 + y = 0 ] dt
M ATH EM ATIC S
BANSAL CLASSES Tar g et IIT JEE 200 7 CLASS : XIII (XYZ)
DATE : 25-26//08/2006
TIME : 60 Min.
(b) ( p 3 6 p) sin p dp
Q.1115(b) Integrate: [Sol.
D a i ly Pra c t ic e Pro b l e m s DPP. NO.-19
[Ans. (b) (3 sin p – p cos p)p2 + C]
3 I = ( p 6 p) sin p dp I
II
2 = – (p3 + 6p) cos p + (3 p 6) cos p dp I
II
= – (p3 + 6p) cos p + (3p 2 + 6) sin p – 6 p sin p dp
= – (p3 + 6p) cos p + (3p2 + 6) sin p – 6 [ p cos p cos p dp] (p3 +
(3p 2 +
=– 6p) cos p + 6) sin p + 6p cos p – 6 sin p + C = 3p2 sin p – (p3 + 6p) cos p + 6p cos p + C = 3p2 sin p – p3 cos p + C = (3 sin p – p cos p) p2 + C ] x
for continuity in [0, ]. Plot its graph and state the n 1 4 sin 2 x nature of discontinuity and jump of discontinuity if applicable.
Q.2 38/3 Examine the function f (x) = Limit
n
[Sol.
5 , 4 sin2x = 1 6 6 x x f (x) = Limit n 1 1n 2 5 for < x < , 4 sin2x > 1 6 6 x 0 f (x) = Limit n n 1 (greater than1) for x =
or
5 < x < , 4 sin2x < 1 6 6 x x f (x) = Limit n 1 (less than1) n
for 0 < x <
or
x
for 0 x
or
6 Hence f (x) = x for x or 5 6 6 2 0 for x 5 6
5 x 6 and graph is as shown above.
6
From the graph it is clear that f (x) is continuous everywhere in [0,] except at removable discontinuity of finite type. Jump at x =
6
it is
6
units and at x =
6
and
5 and has 6
5 5 it is units. ] 6 6
1
t 1 · 2 2 1 1 2 2 1 1 1 tan = · 2 · I = tan = 2I = · Ans.] tan 6 3 6 3 2 3 3 3 3 3 0 Q.6
Find the general solution of the equation , 2 + tan x · cot
[Sol.
2 tan x 2 · cot x 2 1 tan 2 ( x 2) tan x 2 2+ = 0 1 tan 2 x 2 + 2 tan x 2
1 1 tan 2 ( x 2) 2+ + = 0 1 tan 2 ( x 2) 2
2 4 1 tan
x + 4 + 2
x = y 2 y2 + 4y + 4 = 0 (y + 2)2 = 0 x 1 – tan2 = – 1 2 x x tan2 = 3 = tan2 2 3
x x + cot x · tan = 0 2 2
and 2 2 x or 2
1 tan 2 x = 0 2
1 – tan2
x = n ± 3 2
y=–2
x = 2n ±
2 , n N] 3 1
0
Q.720/Ex-2, def Let , be the distinct positive roots of the equation tan x = 2x then evaluate (sin x · sin x ) dx , independent of and . 1
[Sol.
1 1 1 cos( )x cos( )x dx 2 sin x · sin x dx = I= 2 20 0
1
1 sin( ) x sin( ) x 1 sin( ) sin( ) = 2 0 = 2 2 tan 2 tan
2( ) tan tan 2( ) tan tan
now
and
sin( ) sin( ) 2( – ) = and 2( + ) = cos cos cos cos
and
....(1)
sin( ) sin( ) and we get 2( ) 2( ) I = (cos · cos ) – (cos · cos ) = 0 Ans. ] substituting the value of
Q.8
x
x
Find the differentiable function which satisfies the equation f (x) = – f ( t ) tan t dt tan( t x ) dt
where x , 2 2
0
[Ans. cos x – 1]
0
[to be put in T/S DE]
Alternatively:
f (x) = x cos x – sin x; f ' (x) = – x sin x
f '(x) = – x sin x + cos x – sin x g (x) = x sin x + cos x; g ' (x) = x cos x g ' (x) = x cos x + sin x – sin x f (x) · g ' (x) + g (x) · f ' (x) = – x2 I=–
d [f ( x )·g( x )] dx dx = f ( x )g( x )
F' ( x ) dx F(x )
[ f (x) g(x) = F (x)]
ln F( x ) + C ln f ( x ) · g ( x ) + C
]
Q.4160/6 In a ABC, given sin A : sin B : sin C = 4 : 5 : 6 and cos A : cos B : cos C = x : y : z. Find the ordered pair that (x, y) that satisfies this extended proportion. [Sol.
sin A sin B sin C = = a b c
a = 4k, b = 5k, c = 6k
52 6 2 4 2 1 cos A = = ; 2 ·5· 6 4
hence
cos A cos B cos C 14 9 16 18
4 2 6 2 52 4 2 52 6 2 9 1 cos B = = ; cos C = = 2 ·5· 6 2· 4·5 16 8
dividing by 16 cos A cos B cos C 4 9 2 x = 4 and y = 9 Ans. ]
sin 1 x dx 2 x x 1 0
1
Q.527(Ex-2, Def) [Sol.
[Ans.
2 6 3
]
put x = sin2 2
0
I=
2
0
I=
sin 2d = sin 4 sin 2 1
( 2) sin 2d sin 4 sin 2 1
2
0
( 2) sin 2d = cos 4 cos2 1 ....(2)
(1) + (2) 2I =
2
2
0
sin 2 d sin 4 sin 2 1
put sin2 = t
1
dt 1 = 2 2 = t t 1 20 0
dt 2 3 t 1 2 2
2
2
0
( 2) sin 2d (1 sin 2 ) 2 (1 sin 2 ) 1
M ATH EM ATIC S
BANSAL CLASSES Tar g et IIT JEE 200 7 CLASS : XIII (XYZ)
D a i ly Pra c t ic e Pro b l e m s
DATE : 23-24//08/2006
TIME : 60 Min.
DPP. NO.-18
b
Q.1
a
2 12 Find the value of a and b where a < b, for which the integral ( 24 2 x x ) dx has the largest
[Ans. a = – 6, b = 4, I = 12]
value.
Q.2
1 sin x cos x y = x Solve the differential eqaution: y' + x . e cos x e cos x
[Sol.
dy + dx
ex C [Ans. y = ] 1 e x cos x
e x (sin x cos x ) 1 y = 1 e x cos x e x cos x
I.F. = e
e x (sin x cos x ) 1 e x cos x
= excos x – 1
x 2x x e cos x 1 e cos x e dx = dx = y(excos x – 1) = e x cos x 1 ex cos x = – ex + C y(1 – excos x) = ex + C Ans. ]
e x (e x cos x 1) 1 ex cos x dx
x2 Q.330/Ex-1, def Integrate : dx ( x cos x sin x )(x sin x cos x ) [Sol. put x = tan dx = sec2 d
[Illustration]
[Ans. ln
x sin x cos x ] x cos x sin x
tan 2 sec2 d [(tan cos(tan ) sin(tan )][tan sin(tan ) cos(tan )]
tan 2 sec 2 cos 2 d = ] [sin cos(tan ) cos sin(tan )][sin sin(tan ) cos cos(tan )]
(sec 2 1) d = = – [(sec( tan )] cos( tan )
(sec 2 1) d sec(tan ) · cos(tan )
put tan – = y
dy =–2 = – 2 cosec 2 y dy 2 sin y cos y put 2y = t
=–
1 cos t t 2 cosec t dt = – ln(cosec t – cot t) = – ln = – ln tan = – ln (tan y) 2 sin t 2
= – ln [tan(tan – = ln
)]= – ln[tan(x – tan –1x)]=– ln
x sin x cos x Ans. x cos x sin x
tan x x 1 x tan x 1 tan x x = ln tan x x
