JEE  >  Maths 35 Years JEE Main & Advanced Past year Papers  >  Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem Download as PDF

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem


Test Description

27 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem for JEE 2022 is part of Maths 35 Years JEE Main & Advanced Past year Papers preparation. The Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem questions and answers have been prepared according to the JEE exam syllabus.The Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem below.
Solutions of Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem questions in English are available as part of our Maths 35 Years JEE Main & Advanced Past year Papers for JEE & Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem solutions in Hindi for Maths 35 Years JEE Main & Advanced Past year Papers course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem | 27 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Maths 35 Years JEE Main & Advanced Past year Papers for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 1

The coefficients of xp and xq in the expansion of (1+ x)p+q are[2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 1

 We have tp + 1 = p + qCp xp and tq + 1 = p+qCq xq
p + qCp = p + qCq.
[ Remember  nCr = nCn – r ]

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 2

If the sum of the coefficients in the expansion of ( a + b)n is 4096, then the greatest coeficient in the expansion is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 2

We have 2n = 4096 = 212 ⇒ n  = 12; the greatest coefff = coeff of middle term. So middle term

=   t7.; t7 = t6 + 1 ⇒ coeff of  

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 3

The positive integer just greater than (1 + 0.0001)10000 is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 3

(1 + 0.0001)10000   n = 10000

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 4

r and n are positive integers r > 1, n > 2 and coefficient of (r+2)th term and 3rth term in the expansion of (1 + x)2n are equal, then n equals [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 4

tr + 2 = 2nCr + 1 xr + 1;t3r = 2nC3r – 1 x3r – 1
Given 2nCr + 1 = 2nC3r – 1 ;
2nC2n – (r + 1) = 2nC3r – 1
⇒ 2n – r – 1 = 3r – 1
⇒ 2n = 4r
⇒ n = 2r

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 5

If  haing n radical signs then by methods of mathematical induction which is true [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 5

Then  = 7 + am < 7 + 7 < 14.

So by the prin ciple of mathematical induction an < 7 ∀ n.

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 6

If  x is positive, the first negative term in the expansion of (1 + x)27/5 is [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 6

For first negative term,  n - r + 1 < 0 ⇒ r >n+1

Therefore, first negative term  is T8 .

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 7

The number of integral terms in the expansion of is [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 7

Terms will be integral if  both are +ve

integer, which is so if r is  an integral multiple of 8. As 0 ≤ r ≤ 256
∴ r = 0,8,16,24,........256 , total 33 values.

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 8

Let S(K) = 1 + 3 + 5... + (2K - 1) =3+K 2 . Then which of the following is true [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 8

S(k) = 1+3+5+...+(2k – 1) = 3 + k2
S (1) : 1 = 3+ 1, which is not true
∵ S (1) is not true.
∴ P.M.I cannot be applied Let S(k) is true, i.e.

1 + 3 + 5.... + (2k - 1) =3+k
⇒ 1 + 3 + 5....+ (2k-1) + 2k +1
= 3+ k2 + 2k +1 = 3+ (k+1)2
∴ S (k ) ⇒ S (k+ 1)

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 9

The coefficient of the middle term in the binomial expansion in powers of x of (1 + αx)4 and of (1 - αx)6 is the same if a equals                            [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 9

The middle term in the expansion of  
(1 + αx)4 = T3 = 4C2 (αx)2 =6α2x2
The middle term in the expansion of  
(1 - αx)6 = T4 = 6C3 ( -αx)3 = -20α2x2
According to the question
2 = -20α3 ⇒ α =  -

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 10

The coefficient of xn in expansion of (1 + x) (1 – x)n  is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 10

Coeff of xn in (1 + x)(1-x )n
= Coeff of xn in (1 - x)n+ Coeff of xn -1 in (1 -x )n
= ( -1)n nCn + (-1)n -1 nCn-1
= ( -1)n 1 + (-1)n -1 n
= ( -1)n [1- n]

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 11

The value of     [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 11

We know ⇒ (50 C+50C3)
51C52 C3 + 53C3 + 54 C355C3
⇒ (51 C4 + 51C3) + 52 C3 + 53C3 + 54 C3+ 55C3
Proceeding in the same way, we get

⇒  54C4 + 55C3 = 56C4 .

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 12

IF  then which one of the following holds for all n ≥ 1, by the principle of mathematical induction [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 12

We observe that

  and we can prove by

induction that 

Now nA - ( n- 1)I = 

∴ nA - (n - 1) I = An

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 13

If the coefficient of x 7 in       equals the coefficient of x -7 in  , then a and b satisfy the relation [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 13

Tr +1 in the expansion

For the Coefficient of x7, we have

⇒ 22 – 3r = 7 ⇒ r = 5

∴ Coefficient of x11C5 (a)6 (b) -5 ...(1)

Again Tr + 1 in the expansion

11C(a)11- r ( -1)x (b)- r ( x)- 2r (x)11-r

For the Coefficient of x–7, we have

Now 11 – 3r = – 7 ⇒ 3r = 18  ⇒  r = 6

∴ Coefficient of  x- 7 11C6 ax 1 x (b)-6 
∴ Coefficient of x= Coefficient of x–7 
⇒  = 11C6 (a)x (b)-5  = 11C6 ax (b)-6 ⇒ ab = 1.

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 14

If x is so small that x3 and higher powers of x may be neglected, then      may be approximated as         [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 14

∵ xand higher powers of x may be neglected

 

(as x3 and higher powers of x can be neglected)

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 15

If the expansion in powers of x of the function   is a0 + a1 x + a2 x2+ a3x3 ...... then an is [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 15

(1 - ax) -1 (1- bx)-1
= (1 + ax + a 2x+ ...)(1 + bx + b 2x2+ ...)
∴ Coefficient of xn
xn = bn + abn -1 + a2bn - 2 + ....... + an-1b+an

{which is a G.P. with  

∴ Its sum is  = 

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 16

For natural numbers  m, n if  (1 – y)m (1 + y)n = 1 + a1y + a2y2 + ....... and a1 = a2 = 10, then (m, n)  is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 16

(1 - y) m (1+y)n

 

= 1 + (n-m) +

∴ a1 = n -m = 10

and   

So, n – m = 10 and (m -n)2 - (m +n)= 20
⇒ m +n= 80
∴  m = 35, n = 45

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 17

In the binomial expansion of (a – b)n, n ≥ 5, the sum of 5th and 6th terms is zero, then a/b equals

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 17

Tr + 1 = (–1)rnCr (a)n – r. (b)r is an expansion of (a – b)n
∴ 5th term = t5 = t4+1
= (–1)4. nC4 (a)n–4.(b)4 = nC4 . an–4 . b
6th term = t= t5+1 = (–1)5 nC5 (a)n–5 (b)
Given t5 + t6 = 0
∴ nC. an–4 . b4 + (– nC5 . an–5 . b5) = 0

or,  

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 18

The sum of the series is              [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 18

We know that,  (1 + x)20 20C0 + 20C1x + 20C2 x2 + ...... 20C10 x10 + ..... 20C20 x20
Put x = –1,  (0) = 20C0 – 20C1 + 20C2 – 20C3 + ...... + 20C10 – 20C11 .... + 20C20
⇒ 0 = 2[20C0 – 20C1 + 20C2 – 20C3
+ ..... – 20C9] + 20C10
 20C10 = 2[20C– 20C20C2 – 20C3
+ ...... – 20C9 + 20C10]
⇒ 20C– 20C1 + 20C2 – 20C+ .... + 20C10 =  20C10

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 19

Statement -1 :                                    [2008]

Statement-2 :  

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 19

We have

 

= nx (1+ x) n–1 + (1+ x)n = RHS
∴ Statement 2 is correct.
Putting x = 1, we get

 

∴ Statement 1 is also true and statement 2 is a correct explanation for statement 1.

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 20

The remainder left out when 82n – (62)2n +1 is divided by 9 is :[2009]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 20

(8)2n – (62) 2n + 1 =  (64) – (62)2n + 1
=  (63 + 1)n – (63 – 1)2n + 1

= 63 ×  +........]+ 1

–  63 × 

⇒ 63 × some integral value + 2
⇒  82n – (62)2n+1 when divided by 9 leaves 2 as the remainder.

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 21

Let  

Statement-1 : S3 = 55 × 29
Statement-2: S1 = 90 × 28 and S2 = 10 × 28 .           [2010]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 21

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 22

The coefficient of x7 in the expansion of (1– x – x2 + x3 )6 is [2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 22

 

(1 – x – x+ x3)6 = [(1– x) – x(1 – x)]
= (1– x)6 (1 – x2)
= (1 – 6x + 15x– 20x3 + 15x4 – 6x5 + x6)
× (1 – 6x2 + 15x– 20x+ 15x8 – 6x10 + x12)
Coefficient of x7 = (– 6) (– 20) +  (– 20)(15) + (– 6) (–6)                              
= – 144

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 23

If n is a positive integer , then  is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 23

x Some integer  ∴   irrational number

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 24

The term independent of x in expansion of   is                     [JEE M 2013]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 24

Given expression can be written as

= (x1/3 – x–1/2)10

General term  = Tr+1​= 10Cr (x1/3)10–r(–x–1/2)r

Term will be independent of x when

⇒ r = 4

So, required term = T5 = 10C4 = 210

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 25

If the coefficents of x3 and x4 in the expansion of (1 + ax + bx2 ) (1-2 x )18 in powers of x are both zero, then (a, b) is equal to: [JEE M 2014]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 25

Consider (1 + ax + bx2) (1 – 2x)18
= (1 + ax + bx2) [18C0 18C1 (2x) + 18C2(2x)218C3(2x)3 + 18C4(2x)4 –.......]

Coeff of x3 = 18C3 (–2)3 + a. (–2)2. 18C2 + b (–2). 18C1 = 0 Coeff. of x3 = – 18C3.8 + a × 4. 18C2– 2b × 18 = 0

= –51 × 16 × 8 + a × 36 × 17 – 36b = 0
= –34 × 16 + 51a – 3b = 0
= 51a – 3b = 34 × 16 = 544
= 51a – 3b = 544 ....(i)
Only option number (b) satisfies the equation number (i)

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 26

The sum  of coefficients of integral power of x in the binomial expansion   is:    [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 26

...(1)

...(2)

Adding equation (1) and (2)

Putting x = 1, we get above as  

Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 27

If the number of terms in th e expan sion of   x ≠ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is : [JEE M 2016]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - Question 27

Total number of terms = n+2C2 = 28 (n + 2) (n + 1) = 56
x = 6
Sum of coefficients = (1 – 2 + 4)n = 36 = 729

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem Page
In this test you can find the Exam questions for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem solved & explained in the simplest way possible. Besides giving Questions and answers for Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem, EduRev gives you an ample number of Online tests for practice