F- 1-3- is a function defined by fx-x-x-x2-1- Let the range of f be a1-a2- b1-b2- If fundamental period of a1a2b1b2x-32 is p-q- where gcdp-q-1- then the value of p-q is

f(x)=xx^2 +1, amp; 1 x 2 2xx^2 +1, amp; 2≤ x 3 Consider y=xx^2 +1 dydx=1(x^2 +1) 2x· x(x^2+1)^2 ⇒dydx=1 x^2(x^2+1)^2 0 ∀ x ∈ (1,3) ⇒ y is strictly decrea ...

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Standard XII

Mathematics

How to Find the Inverse of a Function

f: 1,3→ℝ is a...

Question

$f : (1, 3) \to R$ is a function defined by $f (x) = \frac{x [x]}{x^{2} + 1} .$ Let the range of $f$ be $(a_{1}, a_{2}) \cup (b_{1}, b_{2}] .$ If fundamental period of $cot (\frac{a_{1} a_{2}}{b_{1} b_{2}} x + \frac{3}{2})$ is $\frac{p π}{q},$ where $gcd (p, q) = 1,$ then the value of $p + q$ is

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Solution

$f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{x}{x^{2} + 1}, & 1 < x < 2 \frac{2 x}{x^{2} + 1}, & 2 \leq x < 3 \end{matrix}$

Consider $y = \frac{x}{x^{2} + 1}$
$\frac{d y}{d x} = \frac{1 (x^{2} + 1) - 2 x \cdot x}{(x^{2} + 1)^{2}}$
$\Rightarrow \frac{d y}{d x} = \frac{1 - x^{2}}{(x^{2} + 1)^{2}} < 0 \forall x \in (1, 3)$
$\Rightarrow y$ is strictly decreasing in its domain.
$∴$ Range of $f$ is $(\frac{2}{5}, \frac{1}{2}) \cup (\frac{3}{5}, \frac{4}{5}]$

Now, $cot (\frac{a_{1} a_{2}}{b_{1} b_{2}} x + \frac{3}{2}) = cot (\frac{5}{12} x + \frac{3}{2})$
$∴$ Fundamental period of $cot (\frac{5}{12} x + \frac{3}{2})$ is $\frac{π}{5 / 12} = \frac{12 π}{5}$

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f:(1,3)→R is a function defined by f(x)=x[x]x2+1. Let the range of f be (a1,a2)∪(b1,b2]. If fundamental period of cot(a1a2b1b2x+32) is pπq, where gcd(p,q)=1, then the value of p+q is

$f : (1, 3) \to R$ is a function defined by $f (x) = \frac{x [x]}{x^{2} + 1} .$ Let the range of $f$ be $(a_{1}, a_{2}) \cup (b_{1}, b_{2}] .$ If fundamental period of $cot (\frac{a_{1} a_{2}}{b_{1} b_{2}} x + \frac{3}{2})$ is $\frac{p π}{q},$ where $gcd (p, q) = 1,$ then the value of $p + q$ is