Let fx-A x-B- A- B - and y - fx passes through the points A- 2 A-B2 and 2 B-3-A-B2-1- If B1- B2- Bn- n - are different possible values of B- then the value of -r-1n Br is

Given : f(x)=Ax+B Putting nbsp;(A, 2 A B^2) and (2 B+3,(A+B)^2 1) in y =f(x), we get 2A B^2=A^2+B ⇒ A^2 +B^2 2A+B = 0⋯(1) (A+B)^2 1=2 A B+3 A+B ⇒ A^2+B^2 3 A ...

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Byju's Answer

Standard XIII

Mathematics

Relations between Roots and Coefficients : Higher Order Equations

Let fx=A x+B,...

Question

Let $f (x) = A x + B, A, B \in R$ and $y = f (x)$ passes through the points $(A, 2 A - B^{2})$ and $(2 B + 3, (A + B)^{2} - 1)$ . If $B_{1}, B_{2} \dots B_{n}, n \in N,$ are different possible value(s) of $B$ , then the value of $n \sum r = 1 B_{r}$ is

- \frac{9}{10}

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- \frac{18}{5}

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- \frac{9}{20}

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- \frac{9}{5}

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Solution

The correct option is D $- \frac{9}{5}$
Given : $f (x) = A x + B$
Putting $(A, 2 A - B^{2})$ and $(2 B + 3, (A + B)^{2} - 1)$ in $y = f (x)$ , we get
$2 A - B^{2} = A^{2} + B \Rightarrow A^{2} + B^{2} - 2 A + B = 0 \dots (1) (A + B)^{2} - 1 = 2 A B + 3 A + B \Rightarrow A^{2} + B^{2} - 3 A - B - 1 = 0 \dots (2)$
Subracting equation $(2)$ from $(1)$ , we get
$\Rightarrow A + 2 B + 1 = 0 \Rightarrow A = - (2 B + 1)$
Using equation $(1)$ , we get
$\Rightarrow (2 B + 1)^{2} + B^{2} + 2 (2 B + 1) + B = 0 \Rightarrow 5 B^{2} + 9 B + 3 = 0$
There are $2$ possible values $B_{1}, B_{2}$ , then
$2 \sum r = 1 B_{r} = B_{1} + B_{2} = - \frac{9}{5}$

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