Question

5²=(4-x)²+PR²
=> 25= 16-x²+PR²
=>???​

Answer 1

Step-by-step explanation:

Given,

f(x)=4x

2

−16x+k=0,k∈R

and 1<α<2 and 2<β<3

As a=4>0,

(i) f(1)>0

⇒4−16+k>0

k>12

(ii) f(3)>0

⇒36−48+k>0

k>12

(iii) Discriminant, D>0

⇒256−16k>0

k<16

(iv) f(2)<0

16−32+k<0

⇒k<16

The integral value of k=13,14,15.

The no. of integral values are 3.

The correct answer is option