tạp chí toán học Mathematics 4 2016

FB .C OM/ EB OOK .SOS


FB .C OM/ EB OOK .SOS

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Vol. XXXIV

No. 4

April 2016

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64

CONTENTS
47

8 Maths Musing Problem Set - 160
10 Practice Paper - JEE Advanced

63

19 CBSE Board Class XII
Solved Paper 2016

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30 Math Musing Solutions
31 Practice Paper - JEE Advanced
40 Practice Paper - JEE Main

Mathematics Today
Chemistry Today
Physics For You
Biology Today

47 Olympiad Corner

61 Math Archives
63 You Ask We Answer

64 Practice Paper - AMU
72 Practice Paper - BITSAT

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MATHEMATICS TODAY | APRIL ‘16

7


M

aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths
Musing is to augment the chances of bright students seeking admission into IITs with additional study material.
During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new
pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our
readers. It is heartening that we receive solutions of Maths Musing problems from all over India.
Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope
that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.

JEE MAIN
1. If S is the sum of all the digits of the natural number
N = 1 + 11 + 111 + 1111 + ...... to 2011 terms, then
the sum of the digits of S is
(a) 17
(b) 18
(c) 19
(d) 21
2. If a, b, c are complex numbers such that |a| = 1,

| b | = 2 , | c | = 3 and |a + b + c| = 2, then
|bc + 2ca + 3ab| =
(a)

6

(b) 2 2

(c) 2 3

(d) 2 6

3. sec4θ – 8sec2θtanθ + 16tan2θ = 0 if θ =
π
5π
π
(a)
(b) π
(c)
(d)
12
12
3
6
4. A plane passes through the points (6, –4, 3) and
(0, 4, –3). he sum of its intercepts on the coordinate
axes is zero. Its distance from the origin is
3
6

12
18
(a)
(b)
(c)
(d)
7
7
7
7
5. A man goes in for an examination in which there
are four papers with a maximum of 20 marks. If N
is the number of ways of getting 40 marks on the
whole, then sum of the digits of N is
(a) 14
(b) 15
(c) 16
(d) 18

COMPREHENSION
A fair die is rolled four times. he probability
7. that each of the inal three rolls is atleast as large as
the roll preceding it is
5
5
(a)
(b)
9
18

7
7
(d)
36
72
8. that the list of outcomes contains exactly 3 distinct
numbers is
5
5
(a)
(b)
18
9
7
7
(d)
(c)
36
72
(c)

INTEGER MATCH
9. If a + b + c = 0, a3 + b3 + c3 = 3 and a5 + b5 + c5 = 10,
then a4 + b4 + c4 is
MATRIX MATCH
10.

Column I

(p)

8

(b) If the equation x2 + x – n = 0 has (q)
integer roots, the number of values of
n between 1 and 100 is

9

(c)

A circle is described on any focal chord (r)
2
of y = 20x as diameter. he locus of its
centre is a conic of latusrectum

10

(s)

11

(a)

4⎞
⎛
log 3 log1/ 5 ⎜ x 2 − ⎟
5⎠
⎝

> 1 then |x| ∈

⎛ 2 ⎞
,1⎟
(a) ⎜
⎝ 5 ⎠

⎛ 2 3 ⎞
,
(b) ⎜
⎟
⎝ 5 5⎠

⎛ 3 ⎞
(c) ⎜ 1,
⎟
5⎠
⎝

(d) (1, ∞)

⎛ n ⎞
⎛n⎞
If ⎜
⎟ = 165, ⎜ ⎟ = 330,
⎝ r − 1⎠
⎝r ⎠
⎛ n ⎞
⎟ = 462, then n is
⎜

⎝ r + 1⎠

JEE ADVANCED

1
6. If ⎛⎜ ⎞⎟
⎝2⎠

Column II

See Solution set of Maths Musing 159 on page no. 30

8

MATHEMATICS TODAY | APRIL ‘16



PAPER-1
SECTION - I
INTEGER ANSWER TYPE

1. In the expansion of (3–x/4 + 35x/4)n, the sum of the
binomial coeicients is 64 and the term with the
greatest binomial coeicient exceeds the third by
(n – 1) then ind the value of x.
2. Let p(x) be a polynomial of degree 4 having
p( x ) ⎞
⎛
extremum at x = 1, 2 and lim ⎜1 +

⎟ = 2. hen
⎝
x →0
x2 ⎠
the value of p(2) is
102 2n
3. he value of
−
C1 +
C2
81n 81n
81n
103 2n
102n
−
C3 + ..... +
is
81n
81n
4. Consider an equilateral triangle having vertices at
−i π ⎞
−5 π ⎞
iπ ⎞
⎛
⎛
⎛
2 2
2 6
2 6
⎟.

⎟ and C ⎜
points A ⎜
e ⎟, B⎜
e
e
⎜⎝ 3
⎟⎠ ⎜⎝ 3
⎜⎝ 3
⎟⎠
⎟⎠
2
2
If P(z) is any point on its incircle, then AP + BP +
CP2 =
1

10

2n

⎛ −1 ⎛ sin θ ⎞ ⎞
π
π
,
5. Let f (θ) = sin ⎜ tan ⎜
⎝ cos 2θ ⎟⎠ ⎟⎠ where − 4 ❁ θ ❁ 4 .
⎝
d
hen the value of
( f (θ)) is

d(tan θ)
6. If [x] stands for the greatest integer function, then
8

the value of

[x 2 ] dx

∫ [x 2 − 20x + 100] + [x 2 ] is
2

7. he maximum value of the function
f (x) = 2x3 – 15x2 + 36x – 48 on the set
A = {x | x2 + 20  9x} is
⎛ π⎞
−1
8. If sin x ∈ ⎜ 0, ⎟ , then ind the value of
⎝ 2⎠
10

MATHEMATICS TODAY | APRIL ‘16

⎡ cos−1(sin(cos −1 x )) + sin −1(cos(sin−1 x )) ⎤
tan ⎢
⎥
2
⎥⎦
⎢⎣
SECTION - II
ONE OR MORE THAN ONE CORRECT ANSWER TYPE

9. A man wants to divide 101 coins, a rupee each,
among his 3 sons with the condition that no one
receives more money than the combined total of
other two. he number of ways of doing this is :
(a)

103

C2 – 3 52C2

103

(b)

C2
3

103

C2
6
10. he solution of the diferential equation,
x(x2 + 3y2)dx + y(y2 + 3x2)dy = 0 is
(a) x4 + y4 + x2y2 = c (b) x4 + y4 + 3x2y2 = c
(c) x4 + y4 + 6x2y2 = c (d) x4 + y4 + 9x2y2 = c
(c) 1275

(d)

⎡ 4 2 2⎤
⎡ 1 −1 1 ⎤
⎥
⎢
11. Let A = ⎢ 2 1 −3⎥ and 10 B = ⎢⎢ −5 0 α ⎥⎥
⎢⎣ 1 −2 3 ⎥⎦
⎢⎣ 1 1 1 ⎥⎦
If B is the inverse of the matrix A then α is
(a) – 2 (b) – 1
(c) 2
(d) 5
✷
✷
12. Let the eccentricity of the hyperbola x − y = ✶
✷
✷
a
b
be reciprocal to that of the ellipse x2 + 4y2 = 4. If
the hyperbola passes through a focus of the ellipse,
then
✷ y✷
(a) the equation of the hyperbola is x − = ✶
✸

 

(b) a focus of the hyperbola is (2, 0)
(c) the eccentricity of the hyperbola is

✺
✸

(d) the equation of the hyperbola is x2 – 3y2 = 3



13. Equation of the circle of radius 5 which touches
x-axis and the line 3x = 4y is
(a) x2 + y2 – 30x – 10y + 225 = 0
(b) x2 + y2 + 30x + 10y + 225 = 0
(c) x2 + y2 + (10/3)x – 10y + 25/9 = 0
(d) x2 + y2 – (10/3)x + 10y + 25/9 = 0

SECTION - III
MATRIX MATCH TYPE

19. Match the columns.
Column I
(A)

14. If xy = yx ; x, y > 0, then dy/dx is
(a)
(c)

y( x log y − y )
x( y log x − x )

(b)

yx y −1 − y x log y
xy x −1 − x y log x

y 2 (log x − 1)
x 2 (log y − 1)

(d) none of these

15. If the derivative of an odd cubic polynomial vanishes
at two diferent values of ‘x’ then
(a) coeicient of x3 & x in the polynomial must be
same
(b) coeicient of x3 & x in the polynomial must be
of diferent sign
(c) the values of ‘x’where derivative vanishes are
closer to origin as compared to the respective
roots on either side of origin.
(d) the values of ‘x’ where derivative vanishes are
far from origin as compared to the respective
roots on either side of origin.
⎧x2 + 2 , x ✁ 0
⎪
16. If f ( x ) = ⎨ 3 , x = 0, then
⎪ x+2 , x > 0
⎩
(a) f (x) has a maximum at x = 0
(b) f (x) is strictly decreasing on the let of 0
(c) f ′(x) is strictly increasing on the let of 0
(d) f ′(x) is strictly increasing on the right of 0
17. Number of real roots of the equation cos7x + sin4 x = 1

in the interval (–π, π) is less than
(a) 3
(b) 4
(c) 5
(d) 6
18. If θ be the angle subtended at P(x1, y1) by the circle,
S ≡ x2 + y2 + 2gx + 2fy + c = 0, then
(a) cot θ =

2 S1
g2 + f 2 − c

⎡
⎤
S1
⎥
(b) θ = 2 cot −1 ⎢
⎢ g2 + f 2 − c ⎥
⎣
⎦
(c) tan(θ / 2) =

g2 + f 2 − c
S1

(d) none of these
12

MATHEMATICS TODAY | APRIL ‘16

10.8

∫

Column II
(P)
1/4

[ x ]dx equals

3.3

(B)

he point of maxima of (Q)
2

x 25 (1− x )75

52

in [0,1] is

(C)

sin[x ]
equals
x →0 [ x ]

(R)

1

(D)

4π

(S)

8

(T)

0

lim

∫

| cos x | dx equals

0

20. Match the columns.
Column I
(A) A line from the origin meets
the lines
x − 2 y −1 z +1
=
=

and
1
1
−2
x − (8 / 3) y + 3 z − 1
=
=
2
1
−1
at P and Q respectively. If
length PQ = d, then d ✂ is
(B) he values of x satisfying
tan➊✄(x + 3) – tan➊✄(x – 3)
⎛3⎞
= sin −1 ⎜ ⎟ are
⎝5⎠
(C) Non-zero vectors a, b and c

Column II
(P)

–4

(Q)

0

(R)

4

(S)

5

(T)

6

satisfy a ⋅ b = 0,
(b − a ) ⋅ (b + ❝ ) = ✵ and

2|b +c |= |b − a |.
If a = μb + 4c , then the
possible values of  are
(D) Let f be the function on [–,
] given by f (0) = 9 and
⎛ 9x ⎞
⎛x⎞
f ( x ) = sin ⎜ ⎟ sin ⎜ ⎟
⎝ 2 ⎠
⎝2⎠
for x 0. he value of
2
π

π

∫

f ( x )dx is

−π


PAPER-2
SECTION - I
INTEGER ANSWER TYPE

1. he value of c + 2 for which the area of the igure
bounded by the curve y = 8x2 – x5; the straight lines
16
x = 1 and x = c and x-axis is equal to , is
3
⎛
1 + x2 ⎞
2. he number of solutions of sin −1 ⎜
⎟
⎝ 2x ⎠
π
= sec( x − 1) is
2
3. If | z1 | = 2 , | z 2 | = 3 , | z 3 |= 4 and | 2z1 + 3z 2 + 4z 3 | = 4
then the expression | 8z 2z 3 + 27z 3z1 + 64z1z 2 | equals
24 × k. Find k
4. If the normals at the end points of a variable chord
PQ of the parabola y2 – 4y – 2x = 0 are perpendicular,
then the tangents at P and Q will intersect at
mx + n = 0. Find m + n.

5. If from a pack of 52 playing cards, one card is drawn
at random, the probability that it is either a king or
k
a queen is
. Find k.
13
6. Consider the set of eight vectors
^
^
^
❱ = { ☎ ✐ + b ❥ + ✆ ❦ : a , b , ✆ ∈{−1, 1}} . hree non-coplanar
vectors can be chosen from V in 2♣ ways. hen p is
7. he value of
⎛
1
1
1
4−
4−
6 + log 3 ⎜
⎜⎝ 3 2
3 2
3 2
2

4−

⎞
........⎟ is
⎟⎠

3 2
1

8. Let f : R  R and g : R  R be respectively given by
f (x) = |x| + 1 and g(x) = x2 + 1. Deine h : R  R by
⎧max{ f (x ), g (x )}, if x ≤ 0
h(x ) = ⎨
⎩ min{ f (x ), g (x )}, if x > 0
he number of points at which h(x) is not
diferentiable is
SECTION - II
ONE OR MORE THAN ONE CORRECT ANSWER TYPE
9. Let a = 2i − j + k, b = i + 2 j − k, c = i + j − 2k be
three vectors. A vector in the plane of b and c
whose projection on a is of magnitude 2 / 3 is
(b) 2i + 3 j + 3k
(a) 2i + 3 j − 3k
(c) −2i − j + 5k

(d) 2i + j + 5k

10. Which of the following is equivalent or connected
with f (x)?

⎧−1 if
⎪
(a) ⎨ 0 if
⎪ 1 if
⎩

⎧ x
if x ≠ 0
⎪
(b) ⎨| x |
⎪ 0 if x = 0
⎩

x>0
x =0
x<0

⎧ 1 if x > 0
⎧| x |
⎪
if x ≠ 0
⎪
(c) ⎨ x
(d) ⎨ 0 if x = 0
⎪−1 if x < 0
⎪⎩ 0 if x = 0
⎩
⎛ π⎤
11. Let f : R → ⎜ 0, ⎥ be a function deined by
⎝ 2⎦
–1 2
f (x) = cot (x + 4x + α2 – α),
then complete set of values of α for which f (x) is
onto, is
⎡1 − 17 1 + 17 ⎤
1 + 17

(a) ⎢
,
⎥ (b)
2
2 ⎥⎦
⎢⎣ 2
⎛
⎤
1 − 17 ⎤ ⎡1 + 17
(c) ⎜ −∞,
, ∞⎥
⎥∪⎢
⎜
2 ⎥⎦ ⎢⎣ 2
⎥⎦
⎝

(d)

1 − 17
2

{x}
{x}
+ cos
where a > 0 and {⋅}
a
a
denotes the fractional part of function. hen the set
of values of a for which f can attain its maximum

values is
⎛ 2⎤
4
(a) ⎛⎜ 0, ⎞⎟
(b) ⎜ 0, ⎥
π
⎝ π⎦
⎝
⎠

12. Let f (x) = sin

(c) (0, ∞)

(d) none of these.
2n

2n

13. he value of C0 + C1 + ..... + 2nCn is
(2n + 1)!
(a) 22n−1 + (2n − 1)! (b) 2n+1 +
1)!(n − 1)!
(
+
n
n !(n − 1)!
(c) 2n +

(2n)!

n !(n − 1)!
4♥

14. Let ❙♥ =

∑
✞

( −1)

✞ ( ✞ +1)
2

(d) 2n+2 +

(2n − 1)!
n !(n − 1)!

✝ 2 . hen S✟ can take value(s)

=1

(a) 1056 (b) 1088 (c) 1120
15. If

the

straight

line

(d) 1332

x1  y1  z
2
2
k

and

x1  y1  z
are coplanar, then the plane(s)
5
2
k

containing these two lines is (are)
(a) y + 2③ = –1
(b) y + ③ = –1
(c) y – ③ = –1
(d) y – 2③ = –1

MATHEMATICS TODAY | APRIL ‘16

13


✡ +✠
16. Let ✇ =
and P = {w☞ : n = 1, 2, 3, ...}. Further

☛
⎧
✍⎫
⎧
1⎫
H✌ = ⎨ z ∈ C ✿ ✎❡ z > ⎬ and H 2 = ⎨ z ∈ C ✿ ✎❡ z ✏ − ⎬ ,
☛
2⎭
⎩
⎭
⎩

where C is the set of all complex numbers.
If z1  P  H1, z2  P  H2 and O represents the
origin, then z1Oz2 =
π
(a)
2

(b)

π
6

2π
(c)
3

(d)

5π
6

SECTION - III
COMPREHENSION TYPE

Paragraph for Question No. 17 and 18
If u and v are two function of x, then
du
∫ u v dx = u ∫ vdx − ∫ dx ∫ v dx dx .
In applying the given rule, care has to be taken in the
selection of irst function and the second function.
Normally if both of the functions are directly integrable
then the irst function is chosen in such a way that the
derivative of the function thus obtained under integral
sign is easily integrable. Now integrate the following.

{

17.

}

∫ x cos x dx =
(a) x sin x + sin x + C (b) x sin x + cos x + C
(c) x cos x + sin x + C (d) none of these

18.

∫ loge | x | dx =

(a) log |x| –x + C
(c) x log |x| –x + C

(b) x log |x| + C
(d) none of these

Paragraph for Question No. 19 and 20
he solution of diferential equation is a relation
between the variables of the equation not containing
the derivatives, but satisfying the given diferential
equation. If y1 and y2 are two solutions of the diferential
dy
+ P( x ) ⋅ y = Q( x )
equation
dx
19. General solution of diferential equation is
(a) y = y1x
(b) y = y1 + c(y1 – y2)
(c) y1 = y + cy2
(d) none of these
20. αy1 + βy2 will also be a solution if
(a) α + β = 1
(b) α + β = –1
(c) α + β = 0
(d) α + β = 2
SOLUTIONS
PAPER-1
1. (0) : Given sum of the binomial coeicients in the
expansion of (3–x/4 + 35x/4)n = 64
hen putting 3–x/4 = 35x/4 = 1

14

MATHEMATICS TODAY | APRIL ‘16

∴ (1 + 1)n = 64
⇒ 2n = 64
∴ n=6
We know that middle term has the greatest binomial
coeicient, Here n = 6
⎛6 ⎞
∴ Middle term = ⎜ + 1⎟
⎝2 ⎠

th

term

= 4th term = T4
and given T4 = (n – 1) + T3
T3 + 1 = (n – 1) + T2 + 1
6
C3(3–x/4)3 (35x/4)3 = ( 6 – 1) + 6C2(3–x/4)4 (35x/4)2
6 . 5 . 4 −3x / 4 15x / 4
6 . 5 − x 5x / 2
⇒
⋅3
=5+
3
3 ⋅3
1. 2 . 3

1. 2
⇒ 20·33x = 5 + 15·33x/2
Let 33x/2 = t
....(i) (t > 0)
2
∴ 20t – 15t – 5 = 0
⇒ 4t2 – 3t – 1 = 0 ⇒ (4t + 1) (t – 1) = 0
1
∴ t≠−
(∵ t > 0)
4
3x/2
∴ t=1
from (i), 3
= 1 = 30
∴ 3x/2 = 0 ⇒ x = 0
2. (0) : Let p(x) = ax4 + bx3 + cx2 + dx + e
p(x) = 4ax3 + 3bx2 + 2cx + d
p(1) = 0, p(2) = 0
p( x) ⎞
⎛
lim ⎜1 + 2 ⎟ = 2
x→0 ⎝
x ⎠

⎛ x 2 + p( x) ⎞
⇒ lim ⎜
⎟⎠ = 2
x→0 ⎝
x2

So, p(0) = 0  e = 0
Again lim

x→0



(

2 x + p′ ( x)
=2
2x

)

p(0) = 0  d = 0

Again lim

x→0

( 2x + 2p′′(x) ) = 2

p(1) = 2  2c = 2  c = 1.
p(1) – p(2) = 0, gives on solution
a = 1/4, b = –1
p( x) =

x4

− x3 + x 2
4

Hence p(2) = 0

10 2n
102 2n
103 2n
C1 +
C2 −
C3 +
81n 81n
81n
81n
102n
..... +
81n
1 2n
2n
2n
2
[ C0 + C1(−10) + C2 (−10) +
=
81n
.... +2n Cn (−10)2n ]

3. (1) :

=

1
81n

1

−

{1 + (−10)}2n =

(−1)2n 92n
92n

=1


4. (5) : Given, A(z1 ) =
B(z2 ) =

f (4) = 2·43 – 15·42 + 36·4 – 48 = –16
f (5) = 2·53 – 15·52 + 36·5 – 48 = 7
hus the maximum value of f on [4, 5] is 7.

2i
3

i
2 ⎧⎪ 3
1 ⎫⎪
−i ⎬ = 1−
⎨

2
2
3
3 ⎩⎪
⎪⎭

i
2 ⎪⎧ 3 i ⎪⎫
− ⎬ = −1 −
⎨−
3
3 ⎪⎩ 2 2 ⎪⎭
1
Radius of incircle of ΔABC, i.e., r =
units
3
Hence, any point on incircle i.e., P(z) is
1
1
⎛ 1
⎞
cos α,
sin α ⎟ i.e.,
(cos α + i sin α)
⎜⎝
⎠
3
3
3
and C(z3 ) =

Solving for |AP|2 + |BP|2 + |CP|2, AP2 + BP2 + CP2 = 5
sin θ
⎛
⎞
⎛ sin θ ⎞
5. (1) : tan −1 ⎜
= sin −1
⎟
⎜⎝
2
⎝ cos 2θ ⎟⎠
sin θ + cos 2θ ⎠
π
π
⎛ sin θ ⎞
⎛ sin θ ⎞
= sin −1
= sin −1 ⎜
⎟⎠ as − ✑ θ ✑
⎜⎝
⎟
⎝
4
4
cos θ
cos2 θ ⎠
⎛ −1 ⎛ sin θ ⎞ ⎞
Thus f (θ) = sin ⎜ tan ⎜
⎟

⎝ cos2 θ ⎟⎠ ⎠
⎝
⎛
⎛ sin θ ⎞ ⎞ = sin(sin–1(tan)) = tan
= sin ⎜ sin −1 ⎜
⎝ cos θ ⎟⎠ ⎟⎠
⎝

hus d(f()) = d(tanθ)
8

6. (3) : I =

x 4 y 4 3(xy )2
+
+
+ c′ = 0
4
4
2
11. (d) : Since B is the inverse of A ∴ AB = I

[ x ]dx

2

I=

[ x 2 ]dx

....(i)

∫ [(x − 10)2 ] + [x 2 ]

2

8

Also, I =

[(10 − x 2 )]dx

....(ii)

∫ [x 2 ] + [(10 − x )2 ]

2

b

(∵ I =
8

∫

b

f (x )dx =

a

∫ f (a + b − x )dx )

a

(i) + (ii) gives 2 I = ∫ 1dx = 6
2

⇒ I = 3.
2

10. (c) : x(x2 + 3y2)dx + y(y2 + 3x2)dy = 0
⇒ x3dx + y3dy + 3xy(ydx + xdy) = 0
⇒ x3dx + y3dy + 3xy d(xy) = 0
⇒

2

∫ [x 2 − 20x + 100] + [x 2 ]

8

⎛ π⎞
8. (1) : As sin −1 x ∈ ⎜ 0, ⎟
⎝ 2⎠
⎛ π⎞
π
−1
−1
and cos x = − sin −1 x ⇒ cos x ∈ ⎜ 0, ⎟

⎝ 2⎠
2
1
⇒ sin(cos−1 x ) = cos(sin −1 x ) =
1 − x2
π
–1
–1
–1
hus, cos (sin(cos x)) + sin (cos(sin–1x)) =
2
π
⇒ Required value = tan = 1
4
9. (a, c) : Let the amount received by the sons be
` x, ` y and ` z respectively, then
x ≤ y + z = 101 – x
i.e. 2x ≤ 101
∴ x ≤ 50, y ≤ 50, z ≤ 50
x + y + z = 101
he corresponding multinomial is (1 + x + ...x50)3
∴ Coeicient of x101 in the expansion of
(1 + x + ... x50)3 = 1 × 103C101 – 3 · 52C50
= 103C2 – 3 · 52C2

7. (7) : As A = {x|x + 20  9x)}
= {x|x2 – 9x + 20  0)} = {x|(x – 4)(x – 5)  0)}
we have A = [4, 5]15
f (x) = 2x3 – 15x2 + 36x – 48
f (x) = 6x2 – 30x + 36 = 6(x – 2)(x – 3)

f has no critical points in [4, 5] as f ′(0)  0 in (4, 5)
and f ′(x) exists at all points.

⎡ 1 −1 1 ⎤
⎥ 1
⎢
⎢ 2 1 −3⎥ × 10
⎢⎣ 1 1 1 ⎥⎦

⎡ 4 2 2 ⎤ ⎡1 0 0⎤
⎥
⎥ ⎢
⎢
⎢ −5 0 α ⎥ = ⎢ 0 1 0 ⎥
⎢⎣ 1 −2 3 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦

⎡ 1 −1 1 ⎤ ⎡ 4 2 2 ⎤ ⎡10 0 0 ⎤
⎥
⎥ ⎢
⎥ ⎢
⎢
i.e., ⎢ 2 1 −3⎥ ⎢ −5 0 α ⎥ = ⎢ 0 10 0 ⎥
⎢⎣ 1 1 1 ⎥⎦ ⎢⎣ 1 −2 3 ⎥⎦ ⎢⎣ 0 0 10 ⎥⎦
Comparing (1, 3)th entry on both sides we get,
∴ 2 – α + 3 = 0 ⇒ α = 5.
12. (b, d)
13. (a, b, c, d) : Since, the circle touches x-axis and its
radius is 5, y-coordinate of the centre is 5 or –5.
Circle also touches 3x – 4y = 0.
Case I : When centre is (h, 5), then

(3h − 20)
❂ ✒ 5 or 3h – 20 = ±25 or h = 15 or –5/3
5
so centre is (15, 5) or (–5/3, 5)
MATHEMATICS TODAY | APRIL ‘16

15


Case II : When centre is (h, –5), then
(3h + 20)
✓ ✔ 5 or h = 5/3 or –15
5
so centre is (5/3, –5) or (–15, –5)
Hence, circles are
2
5⎞
⎛
2
(x – 15)2 + (y – 5)2 = 25; ⎜ x + ⎟ + ( y − 5) = 25;
⎝
⎠
3
2
5⎞
⎛
2
(x
+
15)2 + (y + 5)2 = 25,

⎜⎝ x − ⎟⎠ + ( y + 5) = 25;
3
which are given in (a), (b), (c) and (d).
14. (a, b, c) : Taking log on both sides, we get
y log x = x log y
xdy
y
dy
log x + = log y +
Diferentiating,
ydx
x
dx
dy log y − y / x y( x log y − y )
=
∴
=
dx log x − x / y x( y log x − x )
=
Also,

yx

y 2 (log x − 1)
x 2 (log y − 1)
y −1

− y x log y

x −1

[using (i)]
=

...(i)

...(ii)

( y / x ) ⋅ x y − y x log y
x

y

= dy/dx
... (iv)
(ii), (iii) and (iv) shows that (a), (b) and (c) all are
correct.
15. (b, c) : Let f (x) = ax3 + bx2 + cx + d
f (–x) = –ax3 + bx2 – cx + d
Since f(x) is odd
∴ ax3 + bx2 + cx + d ≡ ax3 – bx2 + cx – d
d = 0, b = 0
∴ f (x) = ax3 + cx
f ′(x) = 3ax2 + c = 0 or 3ax2 = –c
c
∴ x2 = − > 0
3a
a and c must be of diferent signs.
−c
Non-zero roots of f(x) = 0 are ✔

a
−c
∴ ✔
are closer to origin than the roots.
3a
⎧ x 2 ✰ 2, x ✕ 0
⎪
x=0
16. (a, b, c) : f ( x ) = ⎨ 3,
⎪ x + 2, x > 0
⎩
f (0) = 3, lim− f ( x ) = 2, lim+ f ( x ) = 2
x →0

16

x →0

MATHEMATICS TODAY | APRIL ‘16

∴
∴
∴
∴

f(x) has a maximum at x = 0
f ′(x) = 2x, x < 0
f ′(x) < 0 for x < 0
f(x) is decreasing on the let of 0
f ′′(x) = 2, x < 0

f ′′(x) > 0, x < 0
f ′(x) is increasing on the let of 0.

17. (b, c, d) : cos7x + sin4x = 1
cos7x ≤ cos2x and sin4x ≤ sin2x
∴ cos7x + sin4x ≤ cos2x + sin2x = 1
he equality holds only if
cos7x = cos2x and sin4x = sin2x
i.e. cos2x(1 – cos5x) = 0 and sin2x(1 – sin2x) = 0
i.e. cosx = 0 or cosx = 1 and
sinx = 0 or sinx = ±1
∴ cosx = 0 or cosx = 1.
i.e. x = ±π/2 and x = 0 ∴ here are 3 solutions.

...(iii)

xy
− x log x ( x / y ) ⋅ y − x log x
log y − y / x
=
[∵ xy = yx]
log x − x / y
y

∴

18. (b, c) : We have tan(θ / 2) =

r
S1

=

g2 + f 2 − c
S1

⎧ g2 + f 2 − c ⎫
⎪
⎪
or θ = 2 tan −1 ⎨
⎬
S1
⎪⎭
⎪⎩
⎫
⎧
S1
⎪
⎪
= 2 cot −1 ⎨
⎬.
2
2
⎪⎩ g + f − c ⎪⎭
19. (A)  (Q); (B)  (P); (C)  (R); (D)  (S)
20. (A)  (T); (B)  (P, R); (C)  (Q); (D)  (R)
PAPER-2
1

1. (1) : For c ✕ 1; ∫ (8 x 2 − x 5 )dx =

⇒

c
3

16
3

8 1 8c
c 6 16
− −
+
=
3 6
3
6
3

⎡ 8 c 3 ⎤ 16 8 ✖ 17
− + =
⇒ c3 ⎢ − + ⎥ =
⎢⎣ 3 6 ⎥⎦ 3 3 6 6
⇒ c = –1
Again, for c ≥ 1, none of the values of c satisfy the
c

16
2
5
required condition that ∫ (8 x − x )dx =

3
1
∴ c+2=1
⎛ 1 + x2 ⎞ π
2. (1) : sin −1 ⎜
⎟ = sec(x − 1)
⎜ 2x ⎟ 2
⎝
⎠


If λ + μ = 2, we have
r = 2i + ( λ + 2) j − (2 + μ)k

1 + x 2 ≤ 1 i.e. 1 + x 2 ≤ | 2x |
2x
∴ x = 1, –1
π
π
Let x = 1 then sin −1 1 = sec 0 =
2
2
π
−1
Let x = –1 then sin (−1) = sec(−2)
2

does not hold good. ∴ only solution is x = 1.
⎛ 8 27 64 ⎞
3. (4) : | 8z 2z 3 + 27z 3z1 + 64z1z 2 | = z1z 2z 3 ⎜ + + ⎟

⎝ z1 z 2 z 3 ⎠
=| z1 || z 2 || z 3 |
= 2⋅3⋅ 4 ⋅

8z1
2

+

| z1 |

27z 2
2

| z2 |

+

64z 3
| z 3 |2

8z1 27z 2 64z 3
+
+
4
9
16

= 24 ⋅ 2z1 + 3z 2 + 4z 3

(∵| z |=| z |)

= 24 | 2z1 + 3z 2 + 4z 3 |= 24 | 2z1 + 3z 2 + 4z 3 | = 24 × 4

⇒ k=4
4. (7) : Since the normals are perpendicular
∴ the tangent will also be perpendicular to each other
∴ they will intersect on the directrix.
Equation of parabola is
y2 – 4y + 4 = 2x + 4 or (y – 2)2 = 2(x + 2)
∴ Equation of the directrix is
x + 2 = –1/2. i.e. 2x + 5 = 0.
∴ m+n=7

∴ For λ = μ = 1, r = 2i + 3 j − 3k which is given
in (a)
For (b), λ + 2 = 3 and 2 + μ = –3
λ = 1, μ = –5, λ + μ ≠ 2
(b) is false.
For (d), λ + 2 = 1, 2 + μ = 5
⇒ λ = –1, μ = 3
But (λ + μ) ≠ 2. ∴ (d) is false.
If λ + μ = –2, only possibility is (c),
so, 2λ + μ = –1 and λ + 2μ = –5
⇒ –2 + λ = –1, –2 + μ = –5
⇒ λ = 1, μ = –3 ⇒ λ + μ = –2.
∴ (c) is correct.
10. (b, c, d) : By deinition of signum function
⎧ 1, if x > 0
⎪

sgn (x) = ⎨ 0, if x = 0 ∴ (d) is correct
⎪−1, if x ✙ 0
⎩
∴ (d) is connected with f (x) = sgn (x) also signum
function can be deined by the choices (b) and (c) as

5. (2) : Number of ways of selecting one card out of
52 cards n(S) = 52C1 = 52
n(E) = number of selection of a card which is either
a king or a queen = 4C1 + 4C1 = 4 + 4 = 8
Required probability = 8/52 = 2/13
So, k = 2
6. (5)

7. (4)

8. (3)

9. (a, c) : Let r = λb + μc
(a vector in the plane of b and c )
= ( λ + μ) i + (2 λ + μ) j − ( λ + 2μ) k
Length of projection of r on a =
∴
∴

2( λ + μ) − (2 λ + μ) − ( λ + 2μ)
6

r ⋅a
=

a
✗✘

2
3

2
3

–λ – μ = ± 2 or, λ + μ = ±2
MATHEMATICS TODAY | APRIL ‘16

17


| x | ⎪⎧ 1, if x > 0
x
=
or
|x|
x ⎨⎪⎩−1, if x ✚ 0
⎧ |x| x
⎪ x = | x | = 1, if
⎪⎪
0,
if
∴ sgn (x) = ⎨
⎪|x| x
⎪ =
= −1, if

⎪⎩ x | x |

x>0
x =0
x<0

11. (b, d) : Clearly x2 + 4x + α2 – α ≥ 0 ∀ x ∈R and must
take all values of the interval [0, ∞)
⇒ D=0
i.e. 16 – 4(α2 – α) = 0 ⇒ α2 – α = 4
1 ➧ 17
.
⇒ α=
2
{x} π ⎞
12. (a, b) : f (x) = 2 sin ⎛⎜
+ ⎟
⎝ a 4⎠
π⎞
⎛
π⎞
⎛
{x} = ⎜ 2nπ + ⎟ a ∴ 0 ✚ ⎜ 2nπ ✛ ⎟ a ✚ 1
4⎠
4⎠
⎝
⎝
4
∴ 0✚a✚
∴ 0✚a✚ 4 .

(8n + 1)π
π
2n
2n
2n
13. (a) : C0 + C1 + .... + Cn
= 2nC0 + 2nC1 + .... + 2nC2n – (2nCn + 1 + .... + 2nC2n)
= 22n – (2nCn – 1 + 2nCn – 2 + ... + 2nC0)
= 22n – (2nC0 + 2nC1 + ... + 2nCn – 1 + 2nCn – 2nCn)
∴ 2(2nC0 + 2nC1 + .... 2nCn) = 22n + 2nCn
∴ 2nC0 + 2nC1 + .... + 2nCn
(2n − 1)!
2n{(2n − 1)!}
= 22n – 1 +
= 22n – 1 + n !(n − 1)! .
2n(n !){(n − 1)!}

⇒

14. (a, d)

15. (b, c)

16. (c,d )

∫ x cos x dx
Let I = ∫ x cos x dx

{ }

18. (c) : I = ∫ loge | x | dx = ∫ log e | x | . 1 dx
I

II

1
⇒ I = log | x | ⋅ x − ∫ ⋅ xdx = x log | x | − ∫ 1 ⋅ dx
x
⇒ I = x log | x | − x + C
18

20. (a) : Given y = αy1 + βy2 is also a solution
d
⎛ dy
⎞
∴
(αy1 + βy2 ) + β ⎜ 2 + p( x ) ⋅ y2 ⎟ = Q( x )
⎝ dx
⎠
dx
⎛ dy
⎞
⎛ dy
⎞
⇒ α ⎜ 1 + p(x ) y1 ⎟ + β ⎜ 2 + p(x ) ⋅ y2 ⎟ = Q(x )
⎝ dx
⎠
⎝ dx
⎠
̈̈

Now taking x as irst function and cos x as second
function, apply method of integration by parts.
⎤
⎡ d
I = x ∫ cos x dx − ∫ ⎢
(x ) ∫ cos xdx ⎥ dx
⎦
⎣ dx
⇒ I = x sin x − ∫1 ⋅ sin x dx ∴ I = x sin x + cos x + C

)

d
( y − y1 ) + p( x ) ⋅ ( y − y1 ) = 0
...(iv)
dx
From (ii) and (iii)
d
...(v)
( y − y2 ) + p( x ) ⋅ ( y1 − y2 ) = 0
dx 1
From (iv) and (v)
d
( y − y1 )
y − y1
dx
=
d
y1 − y2

( y − y2 )
dx 1
d
d
( y − y1 )
( y1 − y2 )
= dx
⇒ dx
y − y1
y1 − y2
On integrating we get
log(y – y1) = log(y1 – y2) + logc ⇒ y = y1 + c(y1 – y2)

⇒ αQ(x) + β Q(x) = Q(x) ⇒ α + β = 1.

17. (b) :

(

19. (b) : As y1, y2 are the solutions of the diferential
equation
dy
...(i)
∴
+ p( x ) ⋅ y = Q( x )
dx
dy1
+ p( x ) ⋅ y1 = Q( x )
...(ii)
dx

dy2
...(iii)
+ p( x ) ⋅ y2 = Q( x )
dx
From (i) and (ii)
⎛ dy dy1 ⎞
⎜⎝ −
⎟ + p( x ) ⋅ ( y − y1 ) = 0
dx dx ⎠

MATHEMATICS TODAY | APRIL ‘16

Solution Sender of Maths Musing
SET-159
1. Gouri Sankar Adhikary

West Bengal

2. Khokon Kumar Nandi

West Bengal

3. Gajula Ravinder

Telangana

4. V. Damodhar Reddy

Telangana

5. Samrat Gupta

West Bengal
SET-158

1. S. Ahamed Thawfeeq

Kerala


Units



VSA(1 mark)

SA(4 marks)

VBQ(4 marks)

LA(6 marks)

Total

Relations and Functions

---

---

---

6(1)

6(1)

Inverse Trigonometric Functions

---

4(1)

---

---

4(1)

Matrices

2(2)

---

---

---

2(2)

Determinants

1(1)

---

4(1)

6(1)

11(3)

Continuity and Diferentiability

---

8(2)

---

---

8(2)

Application of Derivatives

---

4(1)

---

6(1)

10(2)

Integrals

---

12(3)

---

---

12(3)

Application of Integrals

---

---

---

6(1)

6(1)

Diferential Equations

---

8(2)

---

---

8(2)

Vector Algebra

2(2)

4(1)

---

---

6(3)

hree Dimensional Geometry

1(1)

4(1)

---

6(1)

11(3)

Linear Programming

---

---

---

6(1)

6(1)

Probability

---

4(1)

---

6(1)

10(2)

6(6)

48(12)

4(1)

42(7)

100(26)

Total

Time Allowed : 3 hours

Maximum Marks : 100
GENERAL INSTRUCTIONS

(i)
(ii)
(iii)
(iv)
(v)
(vi)

All questions are compulsory.
Please check that this question paper contains 26 questions.
Questions 1-6 in Section A are very short-answer type questions carrying 1 mark each.
Questions 7-19 in Section B are long-answer I type questions carrying 4 marks each.
Questions 20-26 in Section C are long-answer II type questions carrying 6 marks each.
Please write down the serial number of the question before attempting it.

SECTION - A

1

1

1
1
1 + cos θ

1. he two vectors j + k and 3i − j + 4k represent the
two sides AB and AC, respectively of a ΔABC. Find
the length of the median through A.

3. Find the maximum value of 1 1 + sin θ

2. Find the vector equation of a plane which is at a
distance of 5 units from the origin and its normal
vector is 2i − 3 j + 6k.

4. If A is a square matrix such that A2 = I, then ind the
simpliied value of (A – I)3 + (A + I)3 – 7A.

1

1

MATHEMATICS TODAY | APRIL ‘16

19


⎡ 0 2b −2⎤
5. Matrix A = ⎢⎢ 3 1 3 ⎥⎥ is given to be symmetric,
⎢⎣3a 3 −1⎥⎦

ind values of a and b.

SECTION - B

7. Find the general solution of the following
−1
dy
diferential equation : (1 + y 2 ) + (x − e tan y ) = 0

dx

8. Show that the vectors a , b and c are coplanar if
a + b , b + c and c + a are coplanar.
9. Find the vector and cartesian equations of the line
through the point (1, 2, –4) and perpendicular to
the two lines
r = (8i − 19 j + 10k) + λ(3i − 16 j + 7k) and
r = (15i + 29 j + 5k ) + μ(3i + 8 j − 5k ).

OR
A and B throw a pair of dice alternately. A wins the
game if he gets a total of 7 and B wins the game if he
gets a total of 10. If A starts the game, then ind the

probability that B wins.
1
5

1
7

1
3

1
8

11. Prove that : tan−1 + tan−1 + tan−1 + tan−1 =

π
4

OR
x) = tan–1 (2 cosec x)

12. he monthly incomes of Aryan and Babban are in
the ratio 3 : 4 and their monthly expenditures are in
the ratio 5 : 7. If each saves `15,000 per month, ind
their monthly incomes using matrix method. his
problem relects which value?
13. If x = a sin 2t(1 + cos 2t) and y = b cos 2t(1 – cos 2t),
ind the values of
20

d2 y
dx 2

−

2

y
1 ⎛ dy ⎞
⎜⎝ ⎟⎠ − = 0.
y dx
x

π
π
dy
at t = and t = .
3
4
dx

MATHEMATICS TODAY | APRIL ‘16

⎧1 − sin3 x
, if x ✜ π / 2
⎪
⎪ 3 cos2 x
⎪
f (x ) = ⎨ p,
if x = π / 2

⎪ q(1 − sin x )
⎪
, if x > π / 2
⎪⎩ (π − 2 x )2

is continuous at x = π/2.
15. Show that the equation of normal at any point t on
the curve x = 3 cost – cos3t and y = 3 sint – sin3t is
4(y cos3t – x sin3t)= 3 sin 4t.
16. Find

(3 sin θ − 2)cos θ

∫ 5 − cos2 θ − 4 sin θ dθ.
OR
π

10. hree persons A, B and C apply for a job of Manager
in a Private Company. Chances of their selection
(A, B and C) are in the ratio 1 : 2 : 4. he probabilities
that A, B and C can introduce changes to improve
proits of the company are 0.8, 0.5 and 0.3
respectively. If the change does not take place, ind
the probability that it is due to the appointment of C.

Solve for x : 2

If y = xx, prove that

14. Find the values of p and q, for which

6. Find the position vector of a point which divides
the join of points with position vectors a − 2b and
2a + b externally in the ratio 2 : 1.

tan–1(cos

OR

⎛π
4

⎞

Evaluate ∫ e2 x ⋅ sin ⎜⎝ + x ⎟⎠ dx.
0

17. Find

∫

x
a3 − x 3
2

18. Evaluate

∫

dx.

x 3 − x dx.

−1

19. Find the particular solution of the diferential
equation (1 – y2)(1 + logx)dx + 2xy dy = 0, given
that y = 0 when x = 1.
SECTION - C

20. Find the coordinate of the point P where the line
through A(3, –4, –5) and B(2, –3, 1) crosses the
plane passing through three points L(2, 2, 1),
M(3, 0, 1) and N(4, –1, 0). Also, ind the ratio in
which P divides the line segment AB.
21. An urn contains 3 white and 6 red balls. Four balls
are drawn one by one with replacement from the
urn. Find the probability distribution of the number
of red balls drawn. Also ind mean and variance of
the distribution.
22. A manufacturer produces two products A and B.
Both the products are processed on two diferent
machines. he available capacity of irst machine
is 12 hours and that of second machine is 9 hours


per day. Each unit of product A requires 3 hours on
both machines and each unit of product B requires
2 hours on irst machine and 1 hour on second
machine. Each unit of product A is sold at ` 7 proit

and that of B at a proit of `4. Find the production
level per day for maximum proit graphically.
23. Let f : N → N be a function deined as
f(x) = 9x2 + 6x – 5. Show that f : N → S, where S
is the range of f, is invertible. Find the inverse of f
and hence ind f –1(43) and f –1(163).
yz − x 2

zx − y 2

24. Prove that zx − y 2 xy − z 2
xy − z 2

yz − x 2

xy − z 2
yz − x 2 is divisible by
zx − y 2

(x + y + z), and hence ind the quotient.
OR
Using elementary transformations, ind the inverse
⎡8 4 3 ⎤
⎢
⎥
of the matrix A = ⎢2 1 1⎥ and use it to solve the
⎢⎣1 2 2⎥
⎦

following system of linear equations:

8x + 4y + 3z = 19
2x + y + z = 5
x + 2y + 2z = 7
25. Show that the altitude of the right circular cone of
maximum volume that can be inscribed in a sphere

...
∴

2

5⎞
⎛ 3⎞
2 ⎛
So, length of AD = ⎜⎝ 0 − ⎟⎠ + (0) + ⎜⎝ 0 − ⎟⎠
2
2

9 25
34 units
+
=
4 4
2
2. We know that perpendicular distance of a point
with position vector a from the plane r ⋅ n = d is
a ⋅n − d
n
Here, a = 0i + 0 j + 0k , n = 2i − 3 j + 6k
According to question,

0−d
=5
4 + 9 + 36
⇒ d = 35
∴ Required equation of plane is
r ⋅ (2i − 3 j + 6k ) = 35
1
1
1
1
3. Let Δ = 1 1 + sin θ
1
1
1 + cos θ
Applying R2 → R2 – R1 and R3 → R3 – R1, we get
1
1
1
Δ = 0 sin θ
0 = sin θ cos θ
0
0
cos θ

terms of volume of the sphere.

Find the intervals in which f(x) = sin 3x – cos 3x,
0 < x < π, is strictly increasing or strictly decreasing.
26. Using integration ind the area of the region
{(x, y) : x2 + y2 ≤ 2ax, y2 ≥ ax, x, y ≥ 0}

SOLUTIONS

1. Take A to be as origin (0, 0, 0).
∴ Coordinates of B are (0, 1, 1)
and coordinates of C are (3, –1, 4).

2

=

4r
. Also ind maximum volume in
of radius r is
3

OR

D is the mid point of BC.
⎛3 5⎞
Coordinates of D are ⎜ , 0, ⎟
⎝2 2⎠

∴

Maximum value of Δ is

1
.
2

4. Given, A2 = I
Consider, (A – I)3 + (A + I)3 – 7A
= A3 – I3 – 3A2I + 3AI2 + A3 + I3 + 3A2I + 3AI2 – 7A
= 2A3 + 6AI2 – 7A = 2AA2 + 6AI – 7A
= 2AI + 6A – 7A = 2A + 6A – 7A = A
⎡ 0 2b −2⎤
⎢
⎥
5. Given, A = ⎢ 3 1 3 ⎥
⎢⎣3a 3 −1⎥⎦
... A is symmetric.
∴ A′ = A
⎡ 0 3 3a ⎤ ⎡ 0 2b −2⎤
⇒ ⎢ 2b 1 3 ⎥ = ⎢ 3 1 3 ⎥
⎢
⎥ ⎢
⎥
⎢⎣ −2 3 −1⎥⎦ ⎢⎣3a 3 −1⎥⎦
MATHEMATICS TODAY | APRIL ‘16

21


On comparing, we get
3
−2
and b =
a=
2
3

6. Required position vector =
=

2 ⋅ (2a + b ) − 1(a − 2b )
2 −1

4a + 2b − a + 2b
= 3a + 4b
1

−1
dy
7. We have, (1 + y 2 ) + (x − e tan y ) = 0
dx
2
tan −1 y dy
) = −(1 + y )
⇒ (x − e
dx
−1

⇒
⇒

dx x − e tan y
=
dy
−(1 + y 2 )
−1
1

e tan y
dx
+
⋅x =
dy 1 + y 2
1 + y2

his is a homogeneous linear diferential equation.
I.F. = e
∴

∫

dy
1+ y 2

−1
= e tan y

Solution is
xe

tan −1 y

=∫
=∫

⇒
⇒

x ⋅e
x=

tan −1 y

e

−1
(e tan y )2

1 + y2
−1
e 2 tan y

1 + y2

dy + c

dy + c

−1
e 2 tan y
=
+ c1
2

tan −1 y

2

−1
+ c1e − tan y

8. Since, a + b , b + c and c + a are coplanar.
∴ (a + b ) ⋅ ⎡⎣(b ✢ c ) ✣ (c ✢ a )⎤⎦ = 0
⇒ (a + b ) ⋅ ❬b ✣ c ✢ b ✣ a ✢ c ✣ c ✢ c ✣ a ✤ = 0
⇒

(a + b ) ⋅ (b ✣ c ✢ b ✣ a ✢ c ✣ a ) = 0 ⎡⎣∵ c ✣ c = 0⎤⎦

⇒

a ⋅ (b ✣ c ) ✢ a ⋅ (b ✣ a ) ✢ a ⋅ (c ✣ a ) ✢ b ⋅ (b ✣ c )

⇒

+ b ⋅ (b ✣ a ) ✢ b ⋅ (c ✣ a ) ✥ 0
2 ⎡⎣a ⋅ (b ✣ c )⎤⎦ = 0 ⇒ a ⋅ (b ✣ c ) ✥ 0
a , b and c are coplanar.

⇒

9. he required line is perpendicular to the lines
which are parallel to vectors b1 = 3i − 16 j + 7k and
b2 = 3i + 8 j − 5k
So, it is parallel to the vector b ✥ b1 ✣ b2
22

MATHEMATICS TODAY | APRIL ‘16

i
j
k
Now, b ✥ b1 ✣ b2 = 3 −16 7 = 24i + 36 j + 72k
3 8 −5
So, vector equation of the line passing through
the point (1, 2, – 4) and parallel to the vector
b = 24i + 36 j + 72k is
r = (i + 2 j − 4k) + s(24i + 36 j + 72k )
Cartesian equation is,
xi + y j + zk = i + 2 j − 4k + s(24i + 36 j + 72k)
x −1 y − 2 z + 4
x −1 y − 2 z + 4
=
=
=
=
⇒
⇒
2
3
6
24
36
72
10. Let I = Event that improvement changes take place.
1
Probability of selection of A, P(A) =
7
2

Probability of selection of B, P(B) =
7
4
Probability of selection of C, P(C) =
7
Probability that A introduce changes to improve,
P(I/A) = 0.8
Probability that B introduce changes to improve,
P(I/B) = 0.5
Probability that C introduce changes to improve,
P(I/C) = 0.3
Probability that A does not introduce changes,
P ( I / A) = 1 − 0.8 = 0.2
Probability that B does not introduce changes,
P ( I / B ) = 1 − 0.5 = 0.5
Probability that C does not introduce changes
P ( I / C ) = 1 − 0.3 = 0.7
So, required probability,
P (C / I ) =

P (C )P ( I / C )
P ( A)P ( I / A) + P (B)P ( I / B) + P (C )P ( I / C )

4
✣ 0. 7
7
= 0. 7
=
1
2

4
✣ 0. 2 ✢ ✣ 0. 5 + ✣ 0. 7
7
7
7
OR
Total outcomes = 36
Favourable outcomes for A to win are
{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}
6 1
=
∴ Probability of A to win, P(A) =
36 6


1 5
Probability of A to lose, P ( A) = 1 − =
6 6
Fovourable outcomes for B to win are
{(4, 6), (6, 4), (5, 5)}
3
1
=
∴ Probability of B to win, P(B) =
36 12
1 11
Probability of B to lose, P (B ) = 1 − =
12 12
Required probability
= P ( A)P (B) + P ( A)P (B )P ( A)P (B)

⇒
⇒
⇒
∴

5 1 5 11 5 1 5 11 5 11 5 ★
= ✦ ✧ ✦ ✦ ✦ ✧ ✦ ✦ ✦ ✦ ✦ + ....
6 12 6 12 6 12 6 12 6 12 6 12
5
5 / 72
=
5 11 17
1− ✩
6 12

1
1
1
1
11. L.H.S. = tan −1 + tan −1 + tan −1 + tan −1
5
7
3
8
⎛ 1 1 ⎞
⎛ 1 1 ⎞
+
+
⎟

⎜
⎟
⎜
= tan −1 ⎜ 5 7 ⎟ + tan −1 ⎜ 3 8 ⎟
1 1
1 1
⎜1− ⋅ ⎟
⎜1− ⋅ ⎟
⎝ 3 8⎠
⎝ 5 7⎠

⎛ 6 11 ⎞
+
π
⎟
⎜
⎛ 325 ⎞
−1
= tan −1 ⎜ 17 23 ⎟ = tan −1 ⎜
⎟⎠ = tan (1) =
⎝
6 11
4
325
⎜1− ⋅ ⎟
⎝ 17 23 ⎠
OR
Given equation is
2 tan–1(cos x) = tan–1 (2 cosec x)
...(i)

Let 2 tan–1(cos x) = θ ⇒ cos x = tan (θ/2)
...(ii)
Now,
2
2
2
...(iii)
=
=
2 cosec x =
2
2
sin x
1 − cos x
1 − tan (θ / 2)
From (ii) and (iii), given equation becomes
2
2
⎡
⎤
θ = tan −1 ⎢
⎥ ⇒ tan θ =
θ
θ
⎢ 1 − tan2 ⎥
1 − tan2
⎢⎣
⎥
2
2⎦

⇒

2 tan(θ / 2)
2

1 − tan (θ / 2)

=

2
1 − tan2 (θ / 2)

tan(θ / 2) = 1 − tan2 (θ / 2)

θ
⎤
⎡
⎢⎣∵ From(iv), tan 2 > 0⎥⎦

[From (ii)]
π 7 π 9π
,
, ....
4 4 4

✪✫ ,

12. Let the monthly income of Aryan be ` 3x and that
of Babban be ` 4x

Also, let monthly expenditure of Aryan be ` 5y and
that of Babban be ` 7y
According to question,
3x – 5y = 15000
4x – 7y = 15000
hese equations can be rewritten as
⎡ 3 −5⎤ ⎡ x ⎤ ⎡15000⎤
⎢ 4 −7 ⎥ ⎢ y ⎥ = ⎢15000⎥
⎣
⎦⎣ ⎦ ⎣
⎦
⎛ 4⎞
Applying R2 → R2 + ⎜ − ⎟ R1 , we get
⎝ 3⎠

⎛ 12 ⎞
⎛ 11 ⎞
= tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟
⎝ 34 ⎠
⎝ 23 ⎠

⇒

θ
1
tan =
2
2
1
cos x =

2
π
x ✪ 2nπ ✫ , n ∈ Z
4

But for equation (i) to be satisied cosec x and cos x
must have same sign.
∴ x lies in 1st quadrant.
⇒ x = π/4

+ P ( A)P (B )P ( A)P (B )P ( A)P (B) + ....

=

tan2 (θ / 2) = 1 − tan2 (θ / 2) ⇒ 2 tan2 (θ / 2) = 1

...(iv)

⎡3 −5 ⎤ ⎡ x ⎤ ⎡15000 ⎤
⎢0 −1 / 3⎥ ⎢ y ⎥ = ⎢ −5000⎥
⎣
⎦⎣ ⎦ ⎣
⎦
−1
y = −5000 ⇒ y = 15000
3
3x – 5y = 15000
⇒ 3x = 15000 + 75000 ⇒ x = 30000
So, monthly income of Aryan = 3 × 30000 = `90000
Monthly income of Babban = 4 × 30000 = `120000

So,

13. x = a sin2t (1 + cos2t)
y = b cos2t (1 – cos2t)
dx
Now,
= 2a cos 2t (1 + cos 2t ) + a sin 2t (−2 sin 2t )
dt
= 2a cos2t + 2a[cos2 2t – sin2 2t]
= 2a cos2t + 2a cos 4t
dy
and
= −2b sin 2t (1 − cos 2t ) + b cos 2t (2 sin 2t )
dt
= –2b sin 2t + 4b (sin2t cos2t)
= –2b sin2t + 2b sin4t
So, dy = dy / dt = 2b(sin 4t − sin 2t )
dx dx / dt 2a(cos 4t + cos 2t )
MATHEMATICS TODAY | APRIL ‘16

23


b ⎡ sin π − sin(π / 2) ⎤
⎡ dy ⎤
= ⎢
⎥
⎢⎣ dx ⎥⎦
at t = π/ 4 a ⎣ cos π + cos( π / 2) ⎦

⎛π
⎞
f (x ) = lim f ⎜ + h ⎟
⎝
⎠
2
h→0
x → π /2
⎡
⎛π
⎞⎤
q ⎢1 − sin ⎜ + h ⎟ ⎥
⎝
⎠⎦
q(1 − cos h)
2
= lim
= lim ⎣
2
h→0 ⎡
h→0
4h 2
⎛π
⎞⎤
⎢ π − 2 ⎜⎝ 2 + h ⎟⎠ ⎥
⎦
⎣

Now,

b ⎡ 0 −1 ⎤ b
= ⎢
=
a ⎣ −1 + 0 ⎥⎦ a
b ⎡ sin(4 π / 3) − sin(2 π / 3) ⎤
⎡ dy ⎤
= ⎢
⎢⎣ dx ⎥⎦
⎥
at t = π/3 a ⎣ cos( 4 π / 3) + cos(2 π / 3)) ⎦

OR
xx

We have, y =
⇒ y = ex log x
Diferentiating w.r.t. x, we get
dy
1
⎞
x log x ⎛
✬e
⎜⎝ x ✭ + log x ⎟⎠
dx
x
⇒

dy
= x x (1 + log x )
dx

⇒

dy
1 ⎛ dy ⎞
= y(1 + log x ) ⇒ (1 + log x ) = ⋅ ⎜ ⎟ ...(i)
dx
y ⎝ dx ⎠

Again diferentiating w.r.t. x, we get

dx
⇒
⇒

2

= (1 + log x ) ⋅

dy
1
✮y✭
dx
x
2

d2 y

y
1 ⎛ dy ⎞

= ⎜ ⎟ +
2
⎝
⎠
y dx
x
dx

d2 y
dx 2

y
1 ⎛ dy ⎞
⎜ ⎟ − = 0 . Hence proved
y ⎝ dx ⎠
x

14. ✳✳✳ f(x) is continuous at π/2.
⇒

[From (i)]

2

−

lim

x → π /2 −

f (x ) = lim

x → π /2 +

f (x ) = f (π / 2)

⎛π
⎞
f (x ) = lim f ⎜ − h ⎟
⎝
⎠
−
2
→
h
0
x → π /2
⎛π
⎞
1 − sin3 ⎜ − h ⎟
⎝2
⎠
1 − cos3 h
= lim
= lim
⎛π
⎞ h→0 3 sin2 h
h→0
3 cos2 ⎜ − h ⎟
⎝2

⎠

Now, lim

(1 − cos h)(1 + cos2 h + cos h)
3(1 − cos h)(1 + cos h)
h→0

= lim

2

= lim

h→0

24

(1 + cos h + cos h) 1 + 1 + 1 1
=
=
3(1 + cos h)
3(1 + 1) 2
MATHEMATICS TODAY | APRIL ‘16

+

h
2 sin2
q

2 ✬q✭2✬q
✬ ✭ lim
and f (π / 2) = p
4 4 8
4 h → 0 h2
4
✭
4
Hence, from (i)
q
1
1
= p = ⇒ p = and q = 4
2
8
2
15. x = 3 cost – cos3t and y = 3 sint – sin3 t
Now, dx = −3 sin t + 3 cos2 t sin t = −3 sin t (1 − cos2 t )
dt
= –3 sint sin2t = –3sin3t
dy
Also,
= 3 cos t − 3 sin2 t cos t = 3 cos t (1 − sin2 t )
dt
= 3 cost cos2t = 3 cos3t

⎡− 3
3⎤
−
⎥

⎢
b
3b
2 ⎥ = b ⎡− 3 ⎤
= ⎢ 2
⎢
⎥ =
1
1
−
a⎢
a
⎥ a ⎣ −1 ⎦
−
⎢⎣ 2 2 ⎥⎦

d2 y

lim

...(i)

dy dy / dt 3 cos3 t
cos3 t
=
=
=−
dx dx / dt −3 sin3 t
sin3 t
−1 sin3 t

=
Slope of normal =
dy cos3 t
dx
Required equation of normal is

So,

y − (3 sin t − sin3 t ) =

sin3 t ⎡
3
⎣ x − (3 cos t − cos t )⎤⎦
cos3 t

⇒ y cos3t – 3 sint cos3t + sin3t cos3t
= x sin3t – 3 cost sin3t + sin3t cos3t
3
⇒ y cos t – x sin3t = 3 sint cost (cos2t – sin2t)
3 sin 2t ⋅ cos 2t
3
3
⇒ y cos t − x sin t =
2
3 sin 4t
3
3
⇒ y cos t − x sin t = ⋅
2 2
3

3
⇒ 4(y cos t – x sin t) = 3 sin4t. Hence proved
16. Let I = ∫

(3 sin θ − 2)cos θ

dθ
5 − cos2 θ − 4 sin θ
cos θ
sin θ cos θ
= 3∫
dθ − 2 ∫
dθ
2
4 + sin θ − 4 sin θ
4 + sin2 θ − 4 sin θ
= 3I1 – 2I2 (say)
Now, I1 = ∫

sin θ cos θ

4 + sin2 θ − 4 sin θ

dθ


Put sin2θ = t ⇒ 2 sinθ cosθ dθ = dt
dt
dt
1

1
=
∴ I1 = ∫
2 4 + t − 4 t 2 ∫ ( t − 2)2
Put
⇒
∴

⎡ 1 ⎛ −1 5π/2 1 π/2 ⎞
= e − π /2 ⎢ ⎜
−
e
e ⎟
⎠
2
⎣2 ⎝ 2
1 ⎛ 1 5 π /2 1 π /2 ⎞ ⎤ I
− ⎜−
−
e
e ⎟⎥ −
⎠⎦ 4
4⎝
2
2
⇒ I + 1 I = − 1 ⎣⎡e 2 π + 1⎦⎤ + 1 (e 2 π + 1)
4
2 2
4 2

t −2=u⇒ t =u+2
1
dt = du ⇒ dt = 2(u + 2)du
2 t
du
du
(u + 2)
I1 = ∫
du = ∫
+ 2∫
2
u
u
u2

2
2
+ c1 = log( t − 2) −
+ c1
u
t −2
2
= log(sin θ − 2) −
+c
sin θ − 2 1
= log u −

I2 = ∫

cos θ

4 + sin2 θ − 4 sin θ

(e 2 π + 1) ⎡ 1 ⎤
1
5
I=
− 1⎥ = −
(e 2 π + 1)
⎢
4
4 2
2 2 ⎣2 ⎦
−1
I=
(1 + e 2 π )
5 2

⇒

dθ

Put sinθ = t ⇒ cosθ dθ = dt
dt
dt
=∫
∴ I2 = ∫
4 + t 2 − 4t
(t − 2)2
−1
−1

+c
+ c2 =
=
t −2
sinθ − 2 2
6
2
∴ I = 3 log(sin θ − 2) −
+
+ c,
sin θ − 2 sin θ − 2
where c = 3c1 – 2c2
4
= 3 log(sin θ − 2) −
+c
sin θ − 2
OR

⇒

dx
a3 − x 3
Put x3/2 = t
3 1/2
x dx = dt
⇒
2
2
dt
dt

2
∴ I= ∫
= ∫
3
a3 − t 2 3 (a3/2 )2 − t 2
2⎡
⎛ t ⎞⎤
2 ⎡ −1 ⎛ x 3/2 ⎞ ⎤
+
= ⎢sin −1 ⎜
c
⎢sin ⎜
=
⎟⎥ + c
⎝ a3/2 ⎟⎠ ⎥⎦
3⎣
⎝ a3/2 ⎠ ⎥⎦
3 ⎢⎣
2
⎛x⎞
= sin −1 ⎜ ⎟
⎝a⎠
3
2

18. Let I =

∫ |x

⎛π

⎞
Let I = ∫ e 2 x ⋅ sin ⎜ + x ⎟ dx
⎝4
⎠
π
π
+ x = t ⇒ x = t − ⇒ dx = dt
4
4
π
5π
When x = 0, t = and when x = π, t =
4
4

Put

I=

∫

e

sin t dt = e − π/2

=e

− π /2

5 π/ 4

⎡⎛
2t
⎢ ⎜ sin t e
⎢⎝
2
⎣

− x | dx

∫

I=
0

=

∫ (x

e 2t sin t dt

5 π/ 4 5 π/ 4
⎤
⎞
e 2t ⎥
−
t
dt
cos
⎟⎠

∫
2 ⎥
π/ 4
π/ 4
⎦

⎡1 ⎛
5π
π⎞
= e − π/2 ⎢ ⎜ e 5π/2 sin − e π/2 sin ⎟
⎝
2
4
4⎠
⎣
5 π/ 4 5 π/ 4
⎤
⎛ e 2t
⎞
e 2t
− ∫
− ⎜ cos t ⎟
sin t dt ⎥
⎝ 4
⎠ π/ 4
⎥
4
π/ 4
⎦

x ∈(0, 1)
2

3
3
3
∫ | x − x | dx + ∫ | x − x | dx + ∫ | x − x | dx

−1
3

x ∈(−1, 0) ∪ [1, 2)
1

0

∴

−1

π/ 4

π/ 4

+c

−1

0

∴

3

3/2

⎧⎪ x 3 − x ,
| x 3 − x |= ⎨
3
⎪⎩−(x − x ),

π

5 π/ 4 2 ⎛⎜ t − π ⎞⎟
⎝ 4⎠

x

17. Let I = ∫

1

0

2

1

− x )dx + ∫ −(x 3 − x )dx + ∫ (x 3 − x )dx
1

0

0

1

2

⎛ −x 4 x2 ⎞ ⎛ x 4 x2 ⎞
⎛ x 4 x2 ⎞
+ ⎟ +⎜ − ⎟
=⎜ − ⎟ +⎜
⎝ 4
2 ⎠ −1 ⎝ 4
2 ⎠0 ⎝ 4
2 ⎠1

⎡ ⎛ 1 1 ⎞⎤ ⎛ 1 1 ⎞ ⎡
⎛ 1 1 ⎞⎤
= ⎢0 − ⎜ − ⎟ ⎥ + ⎜ − + − 0 ⎟ + ⎢(4 − 2) − ⎜ − ⎟ ⎥
⎝
⎠
⎝ 4 2 ⎠⎦
⎝
⎠
4 2 ⎦
4 2
⎣
⎣

1 11
1 1
= + +2+ =
4 4
4 4
19. (1 – y2)(1 + log x) dx + 2xy dy = 0
⇒ (1 – y2)(1 + log x) dx = – 2xy dy
(1 + log x )
2y
dx = −
dy
⇒
x
1 − y2
MATHEMATICS TODAY | APRIL ‘16

25


On integrating both sides, we get
(1 + log x )
2y
∫ x dx = − ∫ 1 − y 2 dy
(1 + log x )2
= log | 1 − y 2 | +c
2
Now, y = 0 when x = 1
⇒

(1 + log 1)2

1
= log(1) + c ⇒ c =
⇒
2
2
2
(1 + log x )
1
= log | 1 − y 2 | +
∴
2
2
2
2
⇒ (1 + log x) = 2 log |1 – y | + 1
20. General equation of a plane passing through
L(2, 2, 1) is a(x – 2) + b(y – 2) + c(z – 1) = 0 ...(i)
It will pass through M(3, 0, 1) and N(4, –1, 0) if
a(3 – 2) + b(0 – 2) + c(1 – 1) = 0
⇒ a – 2b + 0⋅c = 0
...(ii)
a(4 – 2) + b(–1– 2) + c(0 – 1) = 0
⇒ 2a – 3b – c = 0
...(iii)
Solving (ii) and (iii), we get
a b c
a
b
c
=

=
⇒ = = =λ
2 + 0 0 + 1 −3 + 4
2 1 1
⇒ a = 2λ, b = λ, c = λ
∴ From (i), we get
2λ(x – 2) + λ(y – 2) + λ(z – 1) = 0
⇒ 2x + y + z – 7 = 0
...(*)
Equation of line passing through A(3, –4, –5) and
x −3 y +4 z +5
=
=
B(2, –3, 1) is
2 − 3 −3 + 4 1 + 5
x −3 y +4 z +5
=
=
= r (say )
⇒
1
6
−1
⇒ x = –r + 3, y = r – 4, z = 6r – 5
Any point on the line AB is P(–r + 3, r – 4, 6r – 5)
... It lies on the plane (*)
∴ 2(–r + 3) + (r – 4) + (6r – 5) – 7 = 0
⇒ 5r = 10 ⇒ r = 2
So, coordinates of point P are (1, –2, 7).
Let P divides the line segment AB in k : 1

⎛ 3 ✯ 1 ✱ 2 ✯ k 1 ✯ (−4) ✱ k ✯ (−3) −5 ✯ 1 ✱ k ✯ 1 ⎞
∴ ⎜
,
,
⎟⎠
⎝ k +1
k +1
k +1
= (1, –2, 7)
3 + 2k
⇒
= 1 ⇒ 3 + 2k = k + 1 ⇒ k = – 2
k +1
Hence, P divides the line segment AB in the ratio
2 : 1 externally.
26

MATHEMATICS TODAY | APRIL ‘16

21. Number of white balls = 3
Number of red balls = 6
Total number of balls = 9
Let X be the random variable denoting the number
of red balls drawn.
∴ X can take values 0, 1, 2, 3, 4
P(X = 0) = Probability of getting no red ball in four
draws
1 1 1 1 1
= ✯ ✯ ✯ =
3 3 3 3 81

P(X = 1) = Probability of getting one red ball in four
draws
⎡1 1 1 2 ⎤ 8
=4✯⎢ ✯ ✯ ✯ ⎥ =
⎣ 3 3 3 3 ⎦ 81
P(X = 2) = Probability of getting two red balls in
four draws
⎛ 1 1 2 2 ⎞ 24
✲6✯⎜ ✯ ✯ ✯ ⎟ =
⎝ 3 3 3 3 ⎠ 81
P(X = 3) = Probability of getting three red balls in
four draws
⎛ 1 2 2 2 ⎞ 32
=4✯⎜ ✯ ✯ ✯ ⎟ =
⎝ 3 3 3 3 ⎠ 81
P(X = 4) = Probability of getting four red balls in
four draws
2 2 2 2 16
= ✯ ✯ ✯ =
3 3 3 3 81
he probability distribution of X is
0
1/81

X
P(X)
Mean =

1
8/81

2
24/81
1

3
32/81
8

24

∑ XP( X ) ✲ 0 ✯ 81 + 1 ✯ 81 + 2 ✯ 81 +
3✯

=

4
16/81

216 8
=
81 3

16
32
✱4✯
81
81

Variance = ∑ X 2 P ( X ) − (Mean)2

✲0✯

=

1
8
24
32
16 ⎛ 8 ⎞
+ 1 ✯ + 4 ✯ + 9 ✯ + 16 ✯ − ⎜ ⎟
81
81
81
81
81 ⎝ 3 ⎠

2

648 64 8
− =
81 9 9

22. he given information can be represented in the
tabular form as below:


Machines

Time required to Max. machine
produce

hours
available
A
B

First
Machine

3

2

12

Second
Machine

3

1

9

Proit (in `)

7

4

Let the manufacturer produces x units of product A

and y units of product B.
∴ 3x + 2y ≤ 12 and 3x + y ≤ 9
Let Z denote the total proit.
∴ Z = 7x + 4y
Clearly x ≥ 0 and y ≥ 0
Above L.P.P. can be stated mathematically as:
Maximize Z = 7x + 4y
subject to 3x + 2y ≤ 12,
3x + y ≤ 9 and x, y ≥ 0
To solve graphically, we convert the inequations
into equations
3x + 2y = 12, 3x + y = 9, x = 0, y = 0
he line 3x + 2y = 12 meets the coordinate axes at
A(4, 0) and B(0, 6). Similarly 3x + y = 9, meets the
coordinate axes at C(3, 0) and D(0, 9)

Coordinates of the corner points of the feasible
region are O(0, 0), C(3, 0), E(2, 3), B(0, 6)
Values of the objective function at corner points of
the feasible region
Points
O(0, 0)
C(3, 0)
E(2, 3)
B(0, 6)

Z
Z
Z
Z

Value of objective function
Z = 7x + 4y
=0
= 21
= 14 + 12 = 26 (Max.)
= 24

∴ Z is maximum at x = 2, y = 3
For maximum proit he should manufacture 2 units
of product A and 3 units of product B.

23. We have, f : N → S, f(x) = 9x2 + 6x – 5
Consider, f(x1) = f(x2)
⇒ 9x12 + 6x1 – 5 = 9x22 + 6x2 – 5
⇒ 9(x12 – x22) + 6(x1 – x2) = 0
⇒ (x1 – x2)[9x1 + 9x2 + 6] = 0
⇒ x1 = x2
[... x1, x2 ∈N]
⇒ f is one-one.
Let y ∈S be arbitrary number.
Consider, y = f(x)
⇒ y = 9x2 + 6x – 5 ⇒ y = (3x + 1)2 – 6
y + 6 −1
y + 6 = 3x + 1 ⇒ x =
∈N
⇒
3
⇒ x = f –1(y)
Since, f is one-one and onto.

So, f is invertible.
y + 6 −1
x + 6 −1
Also f −1 ( y ) =
or f −1 (x ) =
3
3
49
−
1
7
−
1
=
=2
Now, f −1 (43) =
3
3
and f −1 (163) =

169 − 1 13 − 1
=
=4
3
3

yz − x 2

zx − y 2

xy − z 2

24. Let Δ = zx − y 2

xy − z 2

yz − x 2

xy − z 2 yz − x 2 zx − y 2
Applying C1 → C1 + C2 + C3
−(x 2 + y 2 + z 2 − xy − yz − zx ) zx − y 2

xy − z 2

Δ = −(x 2 + y 2 + z 2 − xy − yz − zx ) xy − z 2

yz − x 2

−(x 2 + y 2 + z 2 − xy − yz − zx ) yz − x 2 zx − y 2
C1
Applying C1 →
2
2
2
−(x + y + z − xy − yz − zx )
Δ = −(x 2 + y 2 + z 2 − xy − yz − zx )
1 zx − y 2

xy − z 2

1 xy − z 2

yz − x 2

1 yz − x 2 zx − y 2
Applying R1 → R1 – R3, R2 → R2 – R3
Δ = −(x 2 + y 2 + z 2 − xy − yz − zx )

0 (x − y )(x + y + z ) ( y − z )(x + y + z )
0 (x − z )(x + y + z ) ( y − x )(x + y + z )
yz − x 2
zx − y 2
R2
R1
Applying R1 →
, R2 →
(x + y + z )
(x + y + z )
1

MATHEMATICS TODAY | APRIL ‘16

27


Δ = −(x 2 + y 2 + z 2 − xy − yz − xz )(x + y + z )2

⇒

0

0

x−y
x−z

y−z
y−x

1

yz − x 2

zx − y 2

Δ = −(x + y + z )(x 3 + y 3 + z 3 − 3xyz )
y−z
0 x−y
y−x
0 x−z
yz − x 2

1

Applying R1 → R1 – R2

Δ = −(x + y + z )(x 3 + y 3 + z 3 − 3xyz )
0 z−y
0 x−z

1

Expanding along C1, we get

yz − x 2

zx − y 2

x−z
y−x
zx − y 2

Δ = −(x + y + z )(x 3 + y 3 + z 3 − 3xyz )
[0 − 0 + {(z − y )( y − x ) − (x − z )2 }]
= (x + y + z )(x 3 + y 3 + z 3 − 3xyz )
2

2

2

(x + y + z − xy − yz − zx )
Hence, Δ is divisible by (x + y + z) and quotient is
(x3 + y3 + z3 – 3xyz)(x2 + y2 + z2 – xy – yz – zx)
OR
⎡8 4 3 ⎤
A = ⎢⎢2 1 1⎥⎥
⎢⎣1 2 2⎥⎦
Since, AA–1 = I
⎡8 4 3 ⎤
⎡1 0 0 ⎤
⎢2 1 1⎥ A−1 = ⎢0 1 0⎥

⇒ ⎢
⎥
⎢
⎥
⎢⎣1 2 2⎥⎦
⎢⎣0 0 1⎥⎦
Applying R1 ↔ R3
⎡1 2 2 ⎤
⎡0 0 1 ⎤
⎢2 1 1⎥ A−1 = ⎢0 1 0⎥
⎢
⎥
⎢
⎥
⎢⎣8 4 3⎥⎦
⎢⎣1 0 0⎥⎦
Applying R2 → R2 – 2R1 and R3 → R3 – 8R1
2 ⎤
⎡1 2
⎡0 0
⎢0 −3 −3 ⎥ A−1 = ⎢0 1
⎢
⎥
⎢
⎢⎣0 −12 −13⎥⎦
⎢⎣1 0
Applying R3 → – R3 + 4R2
⎡1 2 2 ⎤
⎡0 0
⎢0 −3 −3⎥ A−1 = ⎢ 0 1

⎢
⎥
⎢
⎢⎣0 0 1 ⎥⎦
⎢⎣ −1 4
28

1⎤
−2⎥⎥
−8⎥⎦
1⎤
−2⎥⎥
0 ⎥⎦

MATHEMATICS TODAY | APRIL ‘16

⎛ 1⎞
Applying R2 → ⎜ − ⎟ R2 – R3 and R1 → R1 – 2R3
⎝ 3⎠
1 ⎤
−8
⎡1 2 0 ⎤
⎡2
⎢0 1 0⎥ A−1 = ⎢ 1 −13 / 3 2 / 3⎥
⎢
⎥
⎢
⎥
⎢⎣0 0 1⎥⎦
⎢⎣ −1

4
0 ⎥⎦
Applying R1 → R1 – 2R2
2 / 3 −1 / 3⎤
⎡1 0 0 ⎤
⎡0
⎢0 1 0⎥ A−1 = ⎢ 1 −13 / 3 2 / 3 ⎥
⎢
⎥
⎢
⎥
⎢⎣0 0 1⎥⎦
⎢⎣ −1
4
0 ⎥⎦
2 / 3 −1 / 3⎤
⎡0
−1 ⎢
So, A = 1 −13 / 3 2 / 3 ⎥
⎢
⎥
⎢⎣ −1
4
0 ⎥⎦

...(i)

he given system of linear equations can be written
as AX = B, where
⎡8 4 3⎤

⎡x ⎤
⎡19⎤
⎢
⎥
⎢
⎥
A = ⎢2 1 1⎥ , X = ⎢ y ⎥ , B = ⎢⎢ 5 ⎥⎥
⎢⎣1 2 2⎥⎦
⎢⎣ z ⎥⎦
⎢⎣ 7 ⎥⎦
∴

he solution of above equation is X = A–1B
2 / 3 −1 / 3⎤ ⎡19⎤
⎡0
⎢
[From (i)]
X = ⎢ 1 −13 / 3 2 / 3 ⎥⎥ ⎢⎢ 5 ⎥⎥
⎢⎣ −1
4
0 ⎥⎦ ⎢⎣ 7 ⎥⎦
⎡ 0 + 10 / 3 − 7 / 3 ⎤ ⎡1⎤
= ⎢⎢19 − 65 / 3 + 14 / 3⎥⎥ = ⎢⎢2⎥⎥ ⇒ x = 1, y = 2, z = 1
⎢⎣ −19 + 20 + 0 ⎥⎦ ⎢⎣1⎥⎦

25. Let ABC be a cone of maximum volume inscribed

in the sphere.
Let OD = x
2

2





⇒ BD = r − x

and AD = AO + OD = r + x

= altitude of cone.


Let V be the volume of cone
1
1
⇒ V = π(BD)2 ( AD) = π(r 2 − x 2 )(r + x )
3
3
dV 1 ⎡ 2
2
⇒
= π ⎣(r − x ) + (r + x )(−2 x )⎤⎦
dx 3
π
= ⎡⎣r 2 − 3x 2 − 2rx ⎤⎦
3
d 2V π
and

= ✴ −6 x − 2r ✹
dx 2 3
dV
=0
For maximum or minimum,
dx
⇒ r2 – 3x2 – 2rx = 0 ⇒ r2 – 3rx + rx – 3x2 = 0


⇒ r(r – 3x) + x(r – 3x) = 0 ⇒ (r – 3x)(r + x) = 0
⇒ r = 3x
[... r + x ≠ 0]
r
⇒ x=
3
⎛ d 2V ⎞
⎤
π ⎡ ⎛r ⎞
Also, ⎜
=
−6 ⎜ ⎟ − 2r ⎥
⎟
⎝ dx 2 ⎠ r 3 ⎢⎣ ⎝ 3 ⎠
⎦
x=

3

−4
π

= ✻ −2r − 2r ✼ =
rπ ✽ 0
3
3
⇒ V is maximum when x =

r
3

r 4r
and altitude of cone = AD = r + x = r + =
3 3
r
Also, maximum volume of cone when x =
3
1 ⎛ 2 r2 ⎞ ⎛ r ⎞ π ⎛ 8 2 ⎞ ⎛ 4 ⎞
= π ⎜r − ⎟ ⎜r + ⎟ = ⎜ r ⎟ ⎜ r ⎟
3 ⎝
9 ⎠⎝ 3⎠
3 ⎝9 ⎠⎝3 ⎠
8 ⎛4
⎞ 8 (Volume of sphere)
= ⎜ πr 3 ⎟ =
⎠ 27
27 ⎝ 3
OR
f(x) = sin 3x – cos 3x ⇒ f ′(x) = 3 cos 3x + 3 sin 3x
f ′(x)= 0 ⇒ 3 cos 3x = –3 sin 3x
⇒ cos 3x = – sin 3x ⇒ tan 3x = –1
3π

11π
7π
which gives 3x =
or
or
4
4
4
π
11π
7π
⇒ x = or
or
[... 0 ✽ x ✽ π]
4
12
12
π
7π
11π
and x =
divides
he points x = , x =
4
12
12
the interval into four disjoint intervals, namely
⎛ π ⎞ ⎛ π 7 π ⎞ ⎛ 7 π 11π ⎞ ⎛ 11π ⎞
, π⎟
⎜⎝ 0, ⎟⎠ , ⎜⎝ , ⎟⎠ , ⎜⎝ ,

⎟,⎜
4
4 12
12 12 ⎠ ⎝ 12 ⎠
⇒ f ′(x) > 0 in ⎛⎜ 0, π ⎞⎟
⎝ 4⎠

or f is strictly increasing in ⎛⎜ 0, π ⎞⎟
⎝ 4⎠
⎛ π 7π ⎞
⇒ f ′(x) < 0 in ⎜⎝ , ⎟⎠
4 12
⎛ π 7π ⎞
or f is strictly decreasing in ⎜⎝ , ⎟⎠
4 12
⎛ 7 π 11π ⎞
⎟
⇒ f ′(x) > 0 in ⎜⎝ ,
12 12 ⎠
⎛ 7 π 11π ⎞
or f is strictly increasing in ⎜⎝ ,
⎟
12 12 ⎠
11π ⎞
⇒ f ′(x) < 0 in ⎛⎜
,π
⎝ 12 ⎟⎠
⎛ 11π ⎞
or f is strictly decreasing in ⎜
,π

⎝ 12 ⎟⎠

⇒ f is strictly increasing in the intervals
⎛ π ⎞ ⎛ 7 π 11π ⎞
⎜⎝ 0, ⎟⎠ ∪ ⎜⎝ ,
⎟
4
12 12 ⎠
and f is strictly decreasing in the intervals
⎛ π 7 π ⎞ ⎛ 11π ⎞
, π⎟
⎜⎝ , ⎟⎠ ∪ ⎜⎝
4 12
12 ⎠
26. Let R = {(x, y) : x2 + y2 ≤ 2ax, y2 ≥ ax, x, y ≥ 0}
= {(x, y) : x2 + y2 ≤ 2ax} ∩ {(x, y) : y2 ≥ ax}
∩ {(x, y) : x ≥ 0, y ≥ 0}
⇒ R = R1 ∩ R2 ∩ R3
where R1 = {(x, y) : x2 + y2 ≤ 2ax},
R2 = {(x, y) : y2 ≥ ax}
R3 = {(x, y) : x ≥ 0, y ≥ 0}
Region R1 : (x – a)2 + y2 = a2 represents a circle with
centre at (a, 0 ) and radius a.
Region R2 : y2 = ax represents a parabola with vertex
at (0, 0) and its axis along x-axis.
Region R3 : x ≥ 0, y ≥ 0 represents the irst
quadrant.
⇒ R = R1 ∩ R2 ∩ R3 is the shaded portion in the
igure.
Since, given curves are x2 + y2 = 2ax and y2 = ax

So, point of intersection of the curves are (0, 0) and
(a, a)

a

∴

Required area, A = ∫

(

)

a2 − (x − a)2 − ax dx

0

⎡1
= ⎢ (x − a) a2 − (x − a)2 +
⎣2
a
⎤
a2 −1 ⎛ x − a ⎞ 2
3/2
sin ⎜
ax ⎥
−
⎝ a ⎟⎠ 3
2
⎦0

⎧ ⎛ 2 a 3/2 ⎞ ⎛ 1 2 −1
⎞⎫
A = ⎨⎜ −
a ⎟ − ⎜ a sin ( −1) ⎟ ⎬
⎝
⎠⎭
⎝
⎠
3
2
⎩
⎧ 2a2 a2 ⎛ π ⎞ ⎫ ⎛ 2a2 a2 π ⎞
+
= ⎨−
− ⎜ − ⎟⎬ = ⎜ −
⎟
2 ⎝ 2 ⎠⎭ ⎝ 3
4 ⎠
⎩ 3
⎛π 2⎞
⇒ A = ⎜ − ⎟ a2 sq. units
⎝ 4 3⎠
MATHEMATICS TODAY | APRIL ‘16

̈̈
29


+ve

b/2

–ve

b
2
1
c⎛b⎞ b
Area
=
∴ Smax = ⎜ ⎟ × sin A
( of ΔABC )
2
b⎝2⎠ 2
Hence, S is maximum when y =

SOLUTION SET-159

10 s −1

1. (b) : M = ∑
10

∑ ( 2s − 2r )

s =1 r =0

= ∑[s ⋅ 2 − (1 + 2 + 2 + ... + 2
2

s

s =1

2

s −1

10

)] = ∑ ( (s − 1)2 + 1)

8

s

s =1

= 4 (1 + 2⋅2 + 3⋅2 + ... + 9⋅2 ) +10
= 14 + 8⋅211, summing A.G.P. = 14 + 214 = 16398
2. (d) : H, T, X stand for head, tail, head or tail. he
required sequence of atleast 8 consecutive heads is
8HXXXX, T8HXXX, XT8HXX, XXT8HX, XXXT8H.
1
4
3
Probability = + =
8
9
2

2
28
10
1
1 10 ⎛ 1
2
1 ⎞
m
= ∑⎜ −
+
3. (b) :
=∑
⎟
n r =1 r (r + 1)(r + 2) 2 r =1 ⎝ r r + 1 r + 2 ⎠
1 1 2 1⎤
65
1⎡
1−1+ + − + ⎥ =
 m + n = 329
⎢
2 11 11 12 ⎦
264
2⎣
1
1
4. (c) : α + β = –a, αβ = –
⇒ α2 + β2 = a2 +
2
2a
a2

1
1
1
⇒ α 4 + β4 = a 4 + 2 +
−
= a4 + 2 +
4
4
a
2a 4
2a
2
1 ⎞
⎛
= 2 + ⎜ a2 −
⎟ + 2 ≥2+ 2
2 ⋅ a2 ⎠
⎝
1 t 3 1⋅ 3 t 5
5. (b, c) : sin–1 t = t + ⋅ +
+ ...
2 3 2⋅4 5
=

π /2 ⎛

⎞
1 c 3 cos3 x 1 3 c 5 cos5 x
c
cos

x
+
+ ⋅ ⋅
+ ... ⎟ dx
⎜
I= ∫ ⎜
⎟
2
3
2 4
5
⎠
0 ⎝
3
5
c
c
= c + + + ...  a1 + a2 + a3 = 1 + 9 + 25 = 35
9 25
6. (a) : AF || ED and AE || FD
Now, in ABC and EDC,
DEC = BAC, ACB is common.
⇒ ABC ~ EDC
c
b− y x
Now,
= ⇒ x = (b − y )
b
b
c

Now, S = Area of parallelogram AFDE = 2 (Area of AEF)
A
⎞ c
⎛1
⇒ S = 2 ⎜ xy sin A ⎟ = ( b − y ) y sin A
b
⎠
⎝2
y
dS ⎛ c
⎞
= ⎜ sin A ⎟ ( b − 2 y )
F
E
dy ⎝ b
⎠
dS
x
Sign scheme of
,
dy
B
C
D

30

MATHEMATICS TODAY | APRIL ‘16

7. (b) : he plane contains the point C(1, 2, 3) and

A(0, –1, 1) and is parallel to the line with d.r.’s –2, 2, 1
y +1 z −1
∴
2
1 = 0  x + 5y – 8z + 13 = 0
3
2
15
5
he distance of D(2, 0, 0) from the plane is
=
2
90
8. (d) : he foot of perpendicular from C on the line is
⎛ −4 1 5 ⎞
⎜ , , ⎟
⎝ 3 3 3⎠
⎛ −11 −4 1 ⎞
,
, ⎟
 he image of C in the line is F ⎜
3 3⎠
⎝ 3
x
−2
1

he distance of the point D from F = 34
⎛ 15 ⎞
9. (4) : P ⎜ α, α ⎟ lies on 15x = 8y

8 ⎠
⎝
⎛ 3 ⎞
Q ⎜ β, β ⎟ lies on 3x = 10y
⎝ 10 ⎠
he midpoint of PQ is (8, 6)
32
80
15
3
∴ α + β = 16, α + β = 12. Solving, α = , β =
7
7
8
10
60 m
⎛ 32 60 ⎞
⎛ 80 24 ⎞
P = ⎜ , ⎟ , Q = ⎜ , ⎟ ⇒ PQ = = , m − 8n = 4
7 n
⎝ 7 7 ⎠
⎝ 7 7 ⎠
10. (a) → (q); (b) → (t); (c) → (p); (d) → (r)
(a) x, y, z are in A.P.  x + z = 2y, even
∴ Both x and z are even or both are odd
⎛5⎞ ⎛5⎞
∴ he number of triples is ⎜ ⎟ + ⎜ ⎟ = 20
⎝2⎠ ⎝2⎠
⎛ 10 − 3 + 1 ⎞ ⎛ 8 ⎞
(b) he number of triples is ⎜

⎟ = ⎜ ⎟ = 56
3
⎠ ⎝ 3⎠
⎝
(c) 1 4 7 10
2 5 8
3 6 9
x3 + y3 is divisible by 3 if x + y is divisible by 3.
∴ x and y belong to 3rd row or one from 1st row and
the second from 2nd row.
⎛ 3⎞ ⎛ 4 ⎞⎛3⎞
∴ he number of subsets is ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ = 15
⎝2⎠ ⎝ 1 ⎠⎝1⎠
2
2
(d) x – y = (x + y) (x – y)
⎛4⎞

⎛ 3⎞ ⎛ 4 ⎞⎛ 3⎞

∴ he number of subsets is ⎜ ⎟ + 2 ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ = 24
⎝ 2 ⎠ ⎝2⎠ ⎝ 1 ⎠⎝1⎠
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