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Quadratic Equation and Inequalities question

2018 · 15 Apr · Shift 2 · Q29
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Quadratic Equation and Inequalities question

2018 · 15 Apr · Shift 2 · Q29

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If f(x) is a quadratic expression such that f (1) + f (2) = 0, and −-− 1 is a root of f (x) = 0, then the other root of f(x) = 0 is :
  1. A
    −58-{5 \over 8}−85​
  2. B
    −85-{8 \over 5}−58​
  3. C
    58{5 \over 8}85​
  4. D
    85{8 \over 5}58​
View written solutionFree

Correct answer: D

Let the quadratic be f(x)=a(x+1)(x−r),f(x)=a(x+1)(x-r),f(x)=a(x+1)(x−r), where one root is −1-1−1 and the other root is rrr.

We are given: f(1)+f(2)=0.f(1)+f(2)=0.f(1)+f(2)=0.

Step 1: Compute f(1)f(1)f(1) and f(2)f(2)f(2)

Using f(x)=a(x+1)(x−r),f(x)=a(x+1)(x-r),f(x)=a(x+1)(x−r), we get

f(1)=a(1+1)(1−r)=2a(1−r),f(1)=a(1+1)(1-r)=2a(1-r),f(1)=a(1+1)(1−r)=2a(1−r),

and

f(2)=a(2+1)(2−r)=3a(2−r).f(2)=a(2+1)(2-r)=3a(2-r).f(2)=a(2+1)(2−r)=3a(2−r).

So, f(1)+f(2)=2a(1−r)+3a(2−r)=0.f(1)+f(2)=2a(1-r)+3a(2-r)=0.f(1)+f(2)=2a(1−r)+3a(2−r)=0.

Step 2: Simplify

Factor out aaa: a[2(1−r)+3(2−r)]=0.a[2(1-r)+3(2-r)]=0.a[2(1−r)+3(2−r)]=0.

Since f(x)f(x)f(x) is quadratic, a≠0a\neq 0a=0. Hence, 2(1−r)+3(2−r)=0.2(1-r)+3(2-r)=0.2(1−r)+3(2−r)=0.

Expand: 2−2r+6−3r=0,2-2r+6-3r=0,2−2r+6−3r=0, 8−5r=0.8-5r=0.8−5r=0.

Thus, r=85.r=\frac{8}{5}.r=58​.

Step 3: Identify the other root

Therefore, the other root is 85.\boxed{\frac{8}{5}}.58​​.

So the correct option is D.

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