Question

if one root of the polynomial f(x)=3^3+11x+p is reciprocal of the other then find value of p​

Answer 1

Step-by-step explanation:

Since \cos(x)cos(x) is a periodic function, the inverse function \cos^{-1}(x)cos

−1

(x) is only defined for -1\leq x\leq1−1≤x≤1 .

So, let's find the value in the domain by using the properties of \cos(x)cos(x) .

\hookrightarrow\cos(\dfrac{7\pi}{6})=\cos(2\pi-\dfrac{7\pi}{6})↪cos(

6

7π

)=cos(2π−

6

7π

)

\hookrightarrow\cos(\dfrac{7\pi}{6})=\cos(\dfrac{5\pi}{6})↪cos(

6

7π

)=cos(

6

5π

)

We know that the inverse function is a kind of function that assigns a domain value for the range.

So,

\hookrightarrow f^{-1}(f(x))=x↪f

−1

(f(x))=x

So,

\large\hookrightarrow\red{\boxed{\red{\bold{\cos^{-1}(\cos(\dfrac{7\pi}{6}))=\red{\underline{\dfrac{5\pi}{6}}}}}}}↪

cos

−1

(cos(

6

7π

))=

6

5π

This is our answer.

\arcsin(x),\ \arccos(x),\ \arctan(x)arcsin(x), arccos(x), arctan(x) refers to \sin^{-1}(x),\ \cos^{-1}(x),\ \tan^{-1}(x)sin

−1

(x), cos

−1

(x), tan

−1

(x) .

For the restricted range of \sin(x),\ \cos(x),\ \tan(x)sin(x), cos(x), tan(x) , view \text{[Attachment 1, 2, 3]}[Attachment 1, 2, 3] .

The domains of \arcsin(x)arcsin(x) and \arccos(x)arccos(x) are,

\large\hookrightarrow\red{\boxed{\red{\bold{-1\leq x\leq1}}}}↪

−1≤x≤1