A GoodProblem for JEE

Algebra Level 4

If the value of 2 [ ( 10 1 ) + ( 10 3 ) ] + ( 10 5 ) 2 \left [ { 10 \choose 1} + { 10 \choose 3 } \right ] + { 10 \choose 5} is in the form of 10 ! × 2 a b ! \frac {10! \times 2^a}{b!} for positive integers a , b a,b .

Expression the value of x 3 − 5 x 2 + 33 x − 10 x^3 - 5x^2 + 33x - 10 in terms of a a and b b when x = 2 + 5 i x = 2+5i

a b \frac{a}{b} a+b b-a ab

Tanishq Varshney by Tanishq Varshney , 23, India

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2 solutions

Pranjal Jain
Jan 30, 2015

( 1 + x ) 10 = ( 10 0 ) + ( 10 1 ) x + ( 10 2 ) x 2 + ( 10 3 ) x 3 + . . . + ( 10 10 ) x 10 (1+x)^{10}=\dbinom{10}{0}+\dbinom{10}{1}x+\dbinom{10}{2}x^2+\dbinom{10}{3}x^3+...+\dbinom{10}{10}x^{10}

Substitute x = 1 x=1 ,

2 10 = ( 10 0 ) + ( 10 1 ) + ( 10 2 ) + ( 10 3 ) + . . . + ( 10 10 ) 2^{10}=\dbinom{10}{0}+\dbinom{10}{1}+\dbinom{10}{2}+\dbinom{10}{3}+...+\dbinom{10}{10}

Substitute x = − 1 x=-1 ,

0 = ( 10 0 ) − ( 10 1 ) + ( 10 2 ) − ( 10 3 ) + . . . + ( 10 10 ) 0=\dbinom{10}{0}-\dbinom{10}{1}+\dbinom{10}{2}-\dbinom{10}{3}+...+\dbinom{10}{10}

Subtract 2nd equation from 1st,

2 10 = 2 [ ( 10 1 ) + ( 10 3 ) + . . . + ( 10 9 ) ] 2 9 = ( 10 1 ) + ( 10 3 ) + . . . + ( 10 9 ) 2 9 = 2 [ ( 10 1 ) + ( 10 3 ) ] + ( 10 5 ) 2^{10}=2\left [\dbinom{10}{1}+\dbinom{10}{3}+...+\dbinom{10}{9}\right ]\\2^9=\dbinom{10}{1}+\dbinom{10}{3}+...+\dbinom{10}{9}\\2^9=2\left [\dbinom{10}{1}+\dbinom{10}{3}\right ]+\dbinom{10}{5}

So a = 9 , b = 10 a=9,b=10

x = 2 + 5 i x − 2 = 5 i x 2 − 4 x + 4 = − 25 x 2 − 4 x + 29 = 0 x=2+5i\\x-2=5i\\x^2-4x+4=-25\\x^2-4x+29=0

x 3 − 5 x 2 + 33 x − 10 = x ( x 2 − 4 x + 29 ) − x 2 + 4 x − 10 = − ( x 2 − 4 x + 29 ) + 19 = 19 = a + b x^3-5x^2+33x-10\\=x(x^2-4x+29)-x^2+4x-10\\=-(x^2-4x+29)+19\\=19=a+b

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Gautam Sharma
Jan 30, 2015

Dont calculate 1st part only calculate 2nd part which is equal to 19 and then see options you will see only one fits well because when we find out that value of second part is 19 then a b , a b , b − a ab,\frac { a }{ b } ,b-a cannot be equal to 19 because

a , b a , b will not be greater than equal to 10. so only possibility is a + b a+b

0 Helpful 0 Interesting 0 Brilliant 0 Confused

How can you say a b , a b , b − a ab,\frac{a}{b},b-a cannot be 19? And btw I knew the result of first part before solving. Just proved it here. No extra time used.

Pranjal Jain - 6 years, 4 months ago

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I m not against solving it, thats good because other will learn from it . I was saying that checking options will also do the same.

And if b is greater than 10 then 1 st part will not remain integer.And it is easy to figure out that b ≠ \neq 1. So

a b , a b , b − a ab,\frac { a }{ b } ,b-a ≠ \neq 19 And forgive me if i hurt you.

Gautam Sharma - 6 years, 4 months ago

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