The number of integral values of $x$ for which $f (x) = 2 x^{2} - 20 x + 42$ is negative is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

$f (x) = 2 x^{2} - 20 x + 42$
Now,
$f (x) = 0 \Rightarrow 2 x^{2} - 20 x + 42 = 0 \Rightarrow x^{2} - 10 x + 21 = 0 \Rightarrow (x - 3) (x - 7) = 0 \Rightarrow x = 3, 7$

The graph of the $f (x)$ is

Therefore $f (x)$ is negative for $x \in (3, 7)$
Hence, the number of integral value is $3.$

Suggest Corrections

Similar questions

x

f (x) = 2 x^{2} - 20 x + 42

the set values of'a 'for which the $f (x) = a x^{2} + 2 x (1 - a) - 4$ is negative for exactly three integral value of $x$ , is

(k - 2) x^{2} + 8 x + k + 4 = 0

x

f (x) = \frac{2 x}{2 x^{2} + 3 x - 20}

x

y = x^{2} - 8 x + 12

Join BYJU'S Learning Program