JB W8 DPP20

MATHEMATICS
TARGET : JEE (Main + Advanced) 2016

EST INF ORM AT IO

DPP DAILY PRACTICE PROBLEMS
Course : VIPUL (JB)

NO. 20
ANSWER KEY
DPP No. : 20 (JEE-ADVANCED)
1. (A) 2. (C) 3. (B) 4. (A) 5. (C) 6. (A)
7. (A) 9. 3
This DPP is to be discussed in the week (10-08-2015 to 15-08-2015)

DPP No. : 20 (JEE-ADVANCED)
Total Marks : 33 Max. Time : 36 min.
Single choice Objective ('1' negative marking) Q.1 to Q.7 (3 marks 3 min.) [21, 21]
Subjective Questions ('1' negative marking) Q.8 to Q.10 (4 marks 5 min.) [12, 15]
1. If pth term of an A.P. is q and qth term is p, its (p + q)th term is

;fn lekUrj Js.kh dk p ok in q rFkk qok in p gS rc (p + q) ok in gS
(A*) 0 (B) p + q (C) (p + q) (D) None of these buesa ls dksbZ
ugha
Sol. Let the A.P. be a, a + b, a + 2d, ...... then Tp = q a + (p 1)d = q ......(1)
Tq = p a + (q 1)d = p ......(2)
Subtracting (2) from (1), we get
(p 1 q + 1)d = q p d = 1 ( p q is desirable)
Substituting d = 1 in (1), we get a + (p 1)(1) = q a=p+q1
Hence Tp+q = a + (p + q 1)d = p + q 1 + (p + q 1)(1) = 0.
Hindi. ekuk fd lekUrj Js<+h a, a + b, a + 2d, ...... gS] rks Tp = q a + (p 1)d = q ......(1)
Tq = p a + (q 1)d = p ......(2)
(1) esa ls (2) dks ?kVkusa ij
(p 1 q + 1)d = q p d = 1 ( p q okaNuh; gS)
(1) esa d = 1 j[kus ij a + (p 1)(1) = q a=p+q1
vr% Tp+q = a + (p + q 1)d = p + q 1 + (p + q 1)(1) = 0.
2. The number of terms in the A.P.

a + b + .....+ c is
lekUrj Js.kh a + b + .....+ c esa inksa dh la[;k gS
ca
(A) c (B)
ba
b c 2a
(C*) (D) None of these buesa ls dksbZ ugha
ba
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Sol. Here, the first term = a,
C.D. = b a and last term = c.
If the number of terms = n, then c = Tn c = a + (n 1)(b a)
c a c a b c 2a
=n1 n= 1 =
ba ba ba
Hindi. ;gk izFke in= a,
lkoZvUrj = b a rFkk vafre in = c.
;fn inksa dh la[;k = n, rks c = Tn c = a + (n 1)(b a)
c a c a b c 2a
=n1 n= 1 =
ba ba ba
3. If P(x) = ax2 + bx + c and Q(x) = ax2 + dx + c, ac 0, then the equation P(x) . Q(x) = 0 has
(A) Exactly two real roots (B*) Atleast two real roots
(C) Exactly four real roots (D) No real roots
;fn P(x) = ax + bx + c vkSj Q(x) = ax + dx + c, ac 0 gks] rks lehdj.k P(x) . Q(x) = 0 j[krh gS&
2 2
(A) Bhd nks okLrfod ewy (B*) de ls de nks okLrfod ewy

(C) Bhd pkj okLrfod ewy (D) dksbZ okLrfod ewy ugha
Sol. D1 = b2 4ac
D2 = d2 + 4 ac
ac is either +ve or negative so at least one of D1 & D2 is +ve so atleast two real roots.
Hindi. D1 = b2 4ac
D2 = d2 + 4 ac
ac ;k rks /kukRed gksxk ;k _.kkRed vr% D1 vkSj D2 esa de ls de ,d /kukRed gksxk vr% de ls de nks ewy
okLrfod gksxsaA
4. If the difference between the corresponding roots of x 2 + ax + b = 0 and x2 + bx + a = 0 is same and a

b, then
;fn lehdj.kksa x2 + ax + b = 0 vkSj x2 + bx + a = 0 ds laxr ewyksa dk vUrj leku gS rFkk a b rc
(A*) a + b + 4 = 0 (B) a + b 4 = 0 (C) a b 4 = 0 (D) a b + 4 = 0
Sol. Let are roots of x2 + ax + b = 0 and are roots of x2 + bx + a = 0
ekuk lehdj.k x2 + ax + b = 0 ds ewy gS rFkk lehdj.k x2 + bx + a = 0 ds ewy gS
so vr% + = a, = b
+ = b
= a
= a2 4b
= b2 4a
a2 4b = b2 4a
(a b) (a + b + 4) = 0
a b sovr% a + b + 4 = 0
5. The value of a for which the sum of the squares of the roots of the equation x2 (a 2) x a 1 = 0
assumes the least value is
a dk og eku] ftuds fy;s lehdj.k x2 (a 2) x a 1 = 0 ds ewyksa ds oxksZa dk ;ksxQy U;wure gks] gS
(A) 3 (B) 2 (C*) 1 (D) 0.
Sol. Let , be the roots, then
+ = a 2 and = (a + 1)
Now 2 +2 = ( + )2 2 = (a 2)2 + 2 (a +1) = a2 2a + 6 = (a 1)2 + 5
Which is least when a 1 = 0 i.e. when a = 1.
Hindi. ekuk , ewy gS] rks
+ = a 2 rFkk = (a + 1)
vc +2 = ( + )2 2= (a 2)2 + 2 (a +1) = a2 2a + 6 = (a 1)2 + 5.
2
tksfd U;wure gS tcfd a 1 = 0 vFkkZr~ tc a = 1
PAGE NO.-2
x2 bx m 1
6. If the roots of the equation are equal and of opposite sign, then the value of m will be
ax c m 1
x2 bx m 1
;fn lehdj.k ds ewy cjkcj rFkk foijhr fpUg ds gS] rc m dk eku gksxk
ax c m 1
a b ba ab ba
(A*) (B) (C) (D)
ab ab a b ba
Sol. (m + 1)x2 b(m + 1)x = ax(m 1) c(m 1)
x2(m + 1) x(bm + b + am a) + c(m 1) = 0
x2(m + 1) x[(a + b)m + (b a)] + c(m 1) = 0
for roots are of opposite sign and same magnitude coefficient of x = 0
ewy foijhr fpUg ds gS vkSj leku ifjek.k ds gS vr% x dk xq.kkad = 0
(b a) a b
m=
ab ab
7. The terms equidistance from a given term of an A.P. are multiplied together, then the difference of the
successive terms of the series so formed are in
(A*) A.P. (B) G.P. (C) H.P. (D) None of these
,d lekUrj Js.kh ds fn;s x;s in ls cjkcj nwjh ds inksa dks xq.kk fd;k tkrk gS rc bl Js.kh ds mkjkskj inksa dk
vUrj gksxk
(A*) lekUrj Js.kh esa (B) xq.kkskj Js.kh esa (C) gjkRed Js.kh esa (D) buesa ls dksbZ ugha
Sol. If the given term is a and the C.D. = d, then the series formed is
a2 d2, a2 4d2, a2 9d2 .........
3d2 , 5d2 , 7d2 ,...........A.P.
Hindi. ;fn fn;k x;k in a rFkk lkoZvUrj = d gks] rks Js.kh
a2 d2, a2 4d2, a2 9d2 .........
3d2 , 5d2 , 7d2 ,........... A.P.
8. Is 184 a term of the sequence 3, 7, 11, . . . . ?

D;k 184 vuqe 3, 7, 11, . . . . dk in gS ?
Sol. Clearly, the given sequence is an A.P. with first term a(= 3) and common difference d (=4)
Let the nth term of the given sequence be 184, then
an = 184
a + (n 1)d = 184
3 + (n 1) 4 = 184
1
4n = 185 n = 46
4
Since n is not a natural number. So 184 is not a term of the given sequence.
Hindi. Li"Vr;k nh x;h Js<+h lekUrj Js<+h gS ftldk izFke in a(= 3) rFkk lkoZvUrj d (=4) gS
ekuk fd fn;s x;s vuqe dk nok in 184 gS] rks
an = 184
a + (n 1)d = 184
3 + (n 1) 4 = 184
1
4n = 185 n = 46
4
D;ksafd n ,d izkr la[;k ugha gS vr% 184 fn;s x;s vuqe dk ,d in ugha gSA
1 3 1 an 10
9. If nth term of the series 3 , 2, 1 , 1 ,...... is , n N, then find the value of (a + b + c)
3 7 9 bn c
1 3 1 an 10
;fn Js.kh 3 , 2, 1 , 1 ,...... dk n oka in n N gks] rks (a + b + c) dk eku Kkr dhft;sA
3 7 9 bn c
Ans. 3
10 10 10 10 10 10 10 10 an 10
Sol. , 2, , ............. , , , Tn = =
3 7 9 3 5 7 9 2n 1 bn c
a = 0, b = 2, c = 1 a+b+c=3
PAGE NO.-3
10. If the 8th term of an A.P. is 31 and the 15th term is 16 more than the 11th term. Find the A.P.
;fn lekUrj Js<+h dk 8ok in 31 gS rFkk 15ok in] 11osa in ls 16 vf/kd gS rks] lekUrj Js<+h Kkr dhft;sA
Sol. Let a be the first term and d be the common difference of the A.P.
ekuk lekUrj Js<+h dk izFke in a gS rFkk lkoZvUrj d gS
We have ge tkurs gS,
a8 = 31 and rFkk a15 = 16 + a11
a + 7d = 31 and rFkk a + 14d = 16 + a + 10d
a + 7d = 31 and rFkk 4d = 16
a + 7d = 31 and rFkk d = 4
a + 7 4 = 31
a + 28 = 31
a=3
Hence the A.P. is bl izdkj lekUrj Js<+h gS a, a + d, a + 2d, a + 3d, . . . . . . .
i.e. vFkkZr 3, 7, 11, 15, 19, . . . . . .
PAGE NO.-4

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MATHEMATICS

TARGET : JEE (Main + Advanced) 2016

EST INF ORM AT IO

Course : VIPUL (JB)

This DPP is to be discussed in the week (10-08-2015 to 15-08-2015)

1. If pth term of an A.P. is q and qth term is p, its (p + q)th term is

2. The number of terms in the A.P.

(A) Bhd nks okLrfod ewy (B*) de ls de nks okLrfod ewy

4. If the difference between the corresponding roots of x 2 + ax + b = 0 and x2 + bx + a = 0 is same and a

tksfd U;wure gS tcfd a 1 = 0 vFkkZr~ tc a = 1

8. Is 184 a term of the sequence 3, 7, 11, . . . . ?

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