5²=(4-x)²+PR²
=> 25= 16-x²+PR²
=>???
Answers
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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
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