Let f: R arrow(0, (π/4)] be defined as f(x)= cot -1(x2+x+k), where k ∈ R. If f(x) is surjective function then 1+(1/ k )+(1/ k 2)+(1/ k 3)+ ldots ldots ldots ∞ terms is equal to

Question

Let f: R arrow(0, (π/4)] be defined as f(x)= cot -1(x2+x+k), where k ∈ R. If f(x) is surjective function then 1+(1/ k )+(1/ k 2)+(1/ k 3)+ ldots ldots ldots ∞ terms is equal to (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

Θ f ( x ) text is surjective ∴ 1 ≤ x 2+ x + k <∞ ∴ (- D /4 a )=1 ⇒ (-(1-4 k )/4)=1 ⇒ k =(5/4) ∴ text given sum =(1/1-(1 / k ))=(1/1-(4 / 5))=5

Q. Let $f : R \to (0, \frac{π}{4}]$ be defined as $f (x) = cot^{- 1} (x^{2} + x + k)$ , where $k \in R$ . If $f (x)$ is surjective function then $1 + \frac{1}{k} + \frac{1}{k ^{2}} + \frac{1}{k ^{3}} + \dots\dots\dots \infty$ terms is equal to

2

3

4

5

$Θ f (x) is surjective$
$∴ 1 \leq x^{2} + x + k < \infty$
$∴ \frac{- D}{4 a} = 1 \Rightarrow \frac{- ( 1 - 4 k )}{4} = 1 \Rightarrow k = \frac{5}{4}$
$∴ given sum = \frac{1}{1 - ( 1/ k )} = \frac{1}{1 - ( 4/5 )} = 5$

Q. Let f:R→(0,4π​] be defined as f(x)=cot−1(x2+x+k), where k∈R. If f(x) is surjective function then 1+k1​+k21​+k31​+………∞ terms is equal to

2

3

4

5

Θf(x) is surjective ∴1≤x2+x+k<∞ ∴4a−D​=1⇒4−(1−4k)​=1⇒k=45​ ∴ given sum =1−(1/k)1​=1−(4/5)1​=5

Q. Let $f : R \to (0, \frac{π}{4}]$ be defined as $f (x) = cot^{- 1} (x^{2} + x + k)$ , where $k \in R$ . If $f (x)$ is surjective function then $1 + \frac{1}{k} + \frac{1}{k ^{2}} + \frac{1}{k ^{3}} + \dots\dots\dots \infty$ terms is equal to

$Θ f (x) is surjective$
$∴ 1 \leq x^{2} + x + k < \infty$
$∴ \frac{- D}{4 a} = 1 \Rightarrow \frac{- ( 1 - 4 k )}{4} = 1 \Rightarrow k = \frac{5}{4}$
$∴ given sum = \frac{1}{1 - ( 1/ k )} = \frac{1}{1 - ( 4/5 )} = 5$