Question : If $p^2-4 p-1=0$, then the value of $p^2+3 p+\frac{1}{p^2}-\frac{3}{p}$ is:
Option 1: 40
Option 2: 50
Option 3: 65
Option 4: 30
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Correct Answer: 30
Solution :
Given: $p^2-4 p-1=0$
⇒ $p-\frac{1}{p}=4$
Squaring both sides, we get:
$⇒(p-\frac{1}{p})^2=(4)^2$
$⇒ (p^2+\frac{1}{p^2}-2×p×\frac{1}{p})=16$
$⇒ (p^2+\frac{1}{p^2})=16+2$
$⇒ (p^2+\frac{1}{p^2})=18$
Now, $p^2+3 p+\frac{1}{p^2}-\frac{3}{p}$
$=p^2+\frac{1}{p^2}+3 p-\frac{3}{p}$
$=p^2+\frac{1}{p^2}+3(p-\frac{1}{p})$
Putting the values, we get:
$=18+3×4$
$=30$
Hence, the correct answer is 30.
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