Maths Quiz on Binomial Theorem and Mathematical Induction for JEE Mains/Advance

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Maths Quiz on Binomial Theorem and Mathematical Induction
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JEE Maths Quiz on Binomial Theorem and Mathematical Induction

JEE Maths Quiz on Binomial Theorem and Mathematical Induction : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Binomial Theorem and Mathematical Induction

Quiz on Binomial Theorem and Mathematical Induction

 

Q1. In the polynomial (x − 1) (x − 2) (x − 3) . . . . . (x − 100), what is the coefficient of x99?
a)  5050
b) – 5050
c) 4040
d) – 4040

View Answer

b) – 5050

Solution:

(x − 1) (x − 2) (x − 3) . . . . . (x − 100)

Number of terms = 100

Coefficient of x99 = (x − 1) (x − 2) (x − 3) . . . . . (x − 100)

= (−1 −2 −3 −. . . . . . −100)

= − (1 + 2 + . . . . . +100)

= − [100 * 101 / 2

= – 5050

Q2.  If (1 + ax)= 1 + 8x + 24x2 +…., then the value of a and n are ________.
a) a = -2 and n = 4
b) a = 2 and n = 4
c) a = 4 and n = 6
d) a = 8 and n = 2

View Answer

b) a = 2 and n = 4

Q3. If the coefficient of x7 in (ax+ [1/bx])11 is equal to the coefficient of x−7 in (ax − 1/bx2)11, then ab = __________.
a) -1
b) 1
c) 2 
d) -2

View Answer

b) 1

Q4. In the expansion of (x + a)n, the sum of odd terms is P and sum of even terms is Q, then the value of (P− Q2) will be _________.
a) (x3 − a2)n
b) (x2 − a2)2
c) (x2 − a2)n
d) (xn − an)2

View Answer

c) (x2 − a2)n

Solution:

(x + a)n = xn + nC1xn−1 a + . . . . . = (xnC2xn−2a+ . . . . . . . + (nC1xn−1a + nC3xn−3a+ . . . . .) = P+Q

(x − a)= P − Q

As the terms are altered,

P− Q= (P + Q) (P − Q) = (x + a)(x − a)n

P− Q= (x2 − a2)n

Q5. (1 + x)− nx − 1 is divisible by (where n∈N)
a) 2x
b) x2
c) 2x3
d) All of these

View Answer

b) x2

(1 + x)= 1 + nx + ([n (n − 1)] / [2!]) * x+ ([n (n − 1) (n − 2)] / [3!]) * x+ . . . . .

(1 + x)− nx − 1 = x2 [([n (n − 1)] / [2!]) + ([n (n − 1) (n − 3)] / [3!]) * x + . . . . .]

From above it is clear that (1 + x)− nx − 1 is divisible by x2.

Trick: (1 + x)− nx − 1, put n = 2 and x = 3;

Then 4− 2 * 3 − 1 = 9 is not divisible by 6, 54 but divisible by 9, which is given by option (b) i.e., x= 9.

Q6. The approximate value of (1.0002) 3000 is __________.
a) 1.8
b) 1.6
c) 1.9
d) 1.7

View Answer

b) 1.6

Solution:

(1.0002) 3000 = (1 + 0.0002) 3000

= 1 + (3000) (0.0002) + ([(3000) (2999)] / [1.2]) * (0.0002)2

= 1 + (3000) (0.0002)

= 1.6

Q7. The digit in the unit place of the number (183!) + 3183 is _____.
a) 5
b) 7 
c) 9
d) 11

View Answer

b) 7 

We know that n! terminates in 0 for n³ at 5 and 34n terminator in 1, (because 3= 81)

Therefore, 3180 = (34)45 terminates in 1.

Also 3= 27 terminates in 7

3183 = 3180 * 33 terminates in 7.

183! + 3183 terminates in 7 i.e. the digit in the unit place = 7

Q8. If the three consecutive coefficients in the expansion of (1 + x)n are 28, 56 and 70, then the value of n is ______.
a) 2
b) 4
c) 6 
d) 8

View Answer

d) 8

Solution:

Let the three consecutive coefficients be nCr−1 = 28, nC= 56 and nCr+1 = 70, so that

nCnCr−1 = [n − r + 1] / [r] = 56 / 28 = 2 and nCr+1 / nC= [n − r] / [r + 1] = 70 / 56 = 5 / 4

This gives n + 1 = 3r and 4n − 5 = 9r

4n − 5 / n + 1 = 3

⇒ n = 8

 


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