If the range of ^{-1} hx + ^{-1} hx is [a,b), then

Correct Answer is : a=log⁡(1+2),b=∞a=(1+{2}), b=∞a=log(1+2),b=∞

Correct Answer is : a=log⁡(1+2),b=∞a=(1+{2}), b=∞a=log(1+2),b=∞

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TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 26 of 160

Marks: +1, -0

If the range of

\sec^{-1} hx + \csc^{-1} hx

[a,b)

, then

Solution:

Range of

\operatorname{sech}^{-1} x = \operatorname{cosech}^{-1} x

[a, b]

\because \operatorname{sech}^{-1} x = \cosh^{-1}\left(\frac{1}{x}\right) = \ln\left(\frac{1}{x} + \sqrt{\frac{1}{x^{2}}-1}\right)

\therefore

Domain

x \in (0,1]

. . . (i)

\operatorname{cosech}^{-1}(x) = \sinh^{-1}\left(\frac{1}{x}\right) = \ln\left(\frac{1}{x} + \sqrt{\frac{1}{x^{2}}+1}\right)

\therefore

Domain

x \in \mathbb{R} - \{0\}

. . . (ii)

\therefore

Domain of

\operatorname{sech}^{-1}(x) + \operatorname{cosech}^{-1}(x)

is Eqs. (i) and (ii)

\Rightarrow x \in (0,1]

\therefore f(0) = \operatorname{sech}^{-1}(0) + \operatorname{cosech}^{-1}(0) = \infty

f(l) = \operatorname{sech}^{-1}(l) + \operatorname{cosech}^{-1}(l) = 0 + \ln(1+\sqrt{2})

\therefore

Range

= [(\ln(1+\sqrt{2}), \infty)

Hence,

a = \ln(1+\sqrt{2}), b = \infty

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