Question

If the roots of equation ax² + bx + 10 = 0 are not real and distinct, where a, b ∈ R and m and n are values of a and b, respectively, for which 5a + b is minimum, then the family of lines (4x + 2y + 3) + n(x − y − 1) = 0 are concurrent at

(A)  (1, −1)

(B)  (−1/6, −7/6)

(C)   (1, 1)

(D)  None of these

Answer 1

Correct option  (B) (−1/6, −7/6)

Let f(x) = ax² + bx + 10.

Since equation f(x) = 0 has no real and distinct roots, therefore, f(x) will have same sign for all real x. 

But f(0) = 10 > 0

Hence, f(x) ≥ 0 ∀ x ∈ R. This given

f(5) ≥ 0 ⇒ 5(5a + b) + 10 ≥ 0

⇒ 5a + b ≥ − 2 Minimum value of 5a + b = −2

According to question, 

5m + n = − 2

⇒ n = − 5m − 2

Given family of lines is

m (4x + 2y +3) − (5m + 2)(x − y − 1) = 0

⇒ 2(x − y − 1) + m (−x + 7y + 8) = 0

Clearly, this family of lines passes through the fixed point (-1/6 , - 7/6).