If the range of k for which the quadratic equation polynomial f(x)=k x2-(2 k+3) x+6 is positive for exactly three negative integral values of x is (a, b], then a-4 b equals

Question

If the range of k for which the quadratic equation polynomial f(x)=k x2-(2 k+3) x+6 is positive for exactly three negative integral values of x is (a, b], then a-4 b equals (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/01843cc455423439a48a5e7dae3a9729-.png" /> Clearly, k <0 kx 2-2 kx -3 x +6=0 ( kx -3)( x -2)=0 ⇒ roots are (3/ k ), 2 (3/ k ) ∈[-4,-3) ⇒ k ∈(-1, (-3/4)] ∴ a -4 b =2

Q. If the range of $k$ for which the quadratic equation polynomial $f (x) = k x^{2} - (2 k + 3) x + 6$ is positive for exactly three negative integral values of $x$ is $(a, b]$ , then $a - 4 b$ equals

1

2

3

4

Clearly, $k < 0$
$k x^{2} - 2 k x - 3 x + 6 = 0$
$(k x - 3) (x - 2) = 0 \Rightarrow$ roots are $\frac{3}{k}, 2$
$\frac{3}{k} \in [- 4, - 3) \Rightarrow k \in (- 1, \frac{- 3}{4}]$
$∴ a - 4 b = 2$

Q. If the range of k for which the quadratic equation polynomial f(x)=kx2−(2k+3)x+6 is positive for exactly three negative integral values of x is (a,b], then a−4b equals

1

2

3

4

Clearly, k<0 kx2−2kx−3x+6=0 (kx−3)(x−2)=0⇒ roots are k3​,2 k3​∈[−4,−3)⇒k∈(−1,4−3​] ∴a−4b=2

Q. If the range of $k$ for which the quadratic equation polynomial $f (x) = k x^{2} - (2 k + 3) x + 6$ is positive for exactly three negative integral values of $x$ is $(a, b]$ , then $a - 4 b$ equals

Clearly, $k < 0$
$k x^{2} - 2 k x - 3 x + 6 = 0$
$(k x - 3) (x - 2) = 0 \Rightarrow$ roots are $\frac{3}{k}, 2$
$\frac{3}{k} \in [- 4, - 3) \Rightarrow k \in (- 1, \frac{- 3}{4}]$
$∴ a - 4 b = 2$