Let f(x)=x3-6 sin θ ⋅ x2+(2 sin 2 θ+q) x-1, q, θ ∈ R be a cubic polynomial in x. If exaclty one tangent can be drawn to the graph of y=f(x) which is parallel to the line y=x, then if q is smallest, then the value of .( d / dx )( f -1( x ))| x =9 is

Question

Let f(x)=x3-6 sin θ ⋅ x2+(2 sin 2 θ+q) x-1, q, θ ∈ R be a cubic polynomial in x. If exaclty one tangent can be drawn to the graph of y=f(x) which is parallel to the line y=x, then if q is smallest, then the value of .( d / dx )( f -1( x ))| x =9 is (A) (1/2) (B) (1/13) (C) (3/17) (D) (2/9)

Tardigrade Admin · Accepted Answer

Smallest value of q =1 when sin θ=0 Now, f ( x )= x 3+ x -1 f (2)=9 ⇒ f -1(9)=2 .( d / dx )( f -1( x ))| x =9=(1/ f prime(2))=(1/13) Θ f prime( x )=3 x 2+1

$\frac{1}{2}$

$\frac{1}{13}$

$\frac{3}{17}$

$\frac{2}{9}$

Smallest value of $q = 1$ when $sin θ = 0$
Now,
$f (x) = x^{3} + x - 1$
$f (2) = 9 \Rightarrow f^{- 1} (9) = 2$
$\frac{d}{d x} (f^{- 1} (x)) ∣ ∣_{x = 9} = \frac{1}{f ^{'} ( 2 )} = \frac{1}{13} {Θ f^{'} (x) = 3 x^{2} + 1}$

Q. Let f(x)=x3−6sinθ⋅x2+(2sin2θ+q)x−1,q,θ∈R be a cubic polynomial in x. If exaclty one tangent can be drawn to the graph of y=f(x) which is parallel to the line y=x, then if q is smallest, then the value of dxd​(f−1(x))∣∣​x=9​ is

21​

131​

173​

92​

Smallest value of q=1 when sinθ=0 Now, f(x)=x3+x−1 f(2)=9⇒f−1(9)=2 dxd​(f−1(x))∣∣​x=9​=f′(2)1​=131​{Θf′(x)=3x2+1}

$\frac{1}{2}$

$\frac{1}{13}$

$\frac{3}{17}$

$\frac{2}{9}$

Smallest value of $q = 1$ when $sin θ = 0$
Now,
$f (x) = x^{3} + x - 1$
$f (2) = 9 \Rightarrow f^{- 1} (9) = 2$
$\frac{d}{d x} (f^{- 1} (x)) ∣ ∣_{x = 9} = \frac{1}{f ^{'} ( 2 )} = \frac{1}{13} {Θ f^{'} (x) = 3 x^{2} + 1}$