For the following questions, match the items in Column-I to one or more items in Column-II. Column I Column II A) If C k + C k + 1 + C k + 2 + C k + 3 + C k + 4 8 for P) 1 k = 2 , 3 , 4 ,

Question

For the following questions, match the items in Column-I to one or more items in Column-II.

Column I Column II

A) If $C_{k} + C_{k + 1} + C_{k + 2} + C_{k + 3} + C_{k + 4} > 8$ for P) 1

$k = 2, 3, 4, 4$ , then the quadratic equations whose roots

are $α, β$ and $α^{k}, β^{k}$ have $m$ common roots, then $m =$

B) If the number of solutions of the equation Q) 2

$2 x^{2} - 5 x + 3 + (x - 1) = 0$ is $n$ , then $n =$

C) If the constant term of the quadratic expression R) 0

$\sum_{k = 1}^{n} (\frac{x - x _{k}}{x _{k}}) (\frac{x - x _{k}}{x})$ as $n \to \infty$ is $p$ , then $p =$

D) The equations $x^{2} + 4 a^{2} = 1 - 4 a x$ and S) -1

$x^{2} + 4 b^{2} = 1 - 4 b x$ have only one root in common,

then the value of $a - b$ is

T) -2

TestHub · Accepted Answer

Given 8 C k + 2 + 2 8 C k + 3 + 8 C k + 4 > 10 C 4 ⇒ ( 8 C k + 2 + 8 C k + 3 ) + ( 8 C k + 3 + 8 C k + 4 ) > 10 C 4 ⇒ 9 C k + 3 + 9 C k + 4 > 10 C 4 ⇒ 10 C k + 4 > 10 C 4 only 10 C 5 > 10 C 4 ⇒ k + 4 = 5 ⇒ k = 1 ∴ α k = α and β k = β Hence, the quadratic equation having roots α and β and α k and β k are identical and have both roots common. ∴ m = 2 . (B) For 1 ≤ x < 2 3 or 2 3 ≤ x < ∞ , x − 1 > 0 . Therefore, no solution is possible. For x ≤ 1 , the given equation is ( 2 x 2 − 5 x + 3 ) + x − 1 = 0 . ∴ 2 x 2 − 4 x + 2 = 0 ⇒ x 2 − 2 x + 1 = 0 ⇒ x = 1 . ∴ The equation has only one solution. ∴ n = 1 . (C) Constant term C = 1 ⋅ 2 1 + 2 ⋅ 3 1 + ……… . + n ( n + 1 ) 1 ∴ ∴ p = 1 t n = n ( n + 1 ) 1 = n 1 − n + 1 1 C = x = 1 ∑ n t n = 1 − n + 1 1 D) ( x + 2 a ) 2 = 1 ( x + 2 b ) 2 = 1 x = ± 1 − 2 a , x = ± 1 − 2 b 1 − 2 a = − 1 − 2 b ⇒ b − a = − 1 − 1 − 2 a = 1 − 2 b ⇒ b − a = 1 ⇒ ∣ a − b ∣ = 1

Mathematics - Binomial Theorem Question with Solution | TestHub

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