y = sec x on [0, π] − {π/2} and y = sec⁻¹x[0, π] − {π/2} पर y = sec x और y = sec⁻¹x[0, π] − {π/2} par y = sec x aur y = sec⁻¹x

NCERT Fig 2.4 · Class 12 Maths · Chapter 2 · interactive diagram

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Parts and what they mean

y = \sec x on [0, \pi] - \{\pi/2\} ([0, \pi] - \{\pi/2\} पर y = \sec x) — Two pieces: from 1 upwards on [0, \pi/2) and from -\infty up to -1 on (\pi/2, \pi]. Range \mathbb{R} - (-1, 1).
y = \sec^{-1} x (principal branch) (y = \sec^{-1} x (मुख्य शाखा)) — Domain \mathbb{R} - (-1, 1) (no graph between −1 and 1), range [0, \pi] - \{\pi/2\}.
Asymptote x = \pi/2 (अनंतस्पर्शी x = \pi/2) — \cos\frac{\pi}{2} = 0, so \sec\frac{\pi}{2} is not defined.
Asymptote y = \pi/2 (अनंतस्पर्शी y = \pi/2) — That is why \pi/2 is removed from the range of \sec^{-1}.
Mirror line y = x (दर्पण रेखा y = x) — Fold the paper along this line: the trig graph lands exactly on the inverse graph. A point (a, b) goes to (b, a).
Full graph (dashed, faint) (पूरा ग्राफ़ (धुँधला, डैश)) — The trig function on its natural domain — it is not one-one there.
A point and its mirror (एक बिंदु और उसका प्रतिबिंब) — (a, b) on sec ↔ (b, a) on \sec^{-1}.
(\pi/3, 2) on y = \sec x (y = \sec x पर (\pi/3, 2)) — Because \sec\frac{\pi}{3} = \frac{1}{\cos(\pi/3)} = 2.
(2, \pi/3) on y = \sec^{-1}x (y = \sec^{-1}x पर (2, \pi/3)) — So \sec^{-1}2 = \frac{\pi}{3}.
End points (filled) (अंत बिंदु (भरे हुए)) — \sec^{-1}(1) = 0 and \sec^{-1}(-1) = \pi are included.
x-axis (x-अक्ष) — The horizontal number line; inputs x are read here.
y-axis (y-अक्ष) — The vertical number line; outputs y = f(x) are read here.
Origin O(0, 0) (मूल बिंदु O(0, 0)) — Where the two axes cross.

Original diagram drawn by Simple Learning Hub, based on NCERT Fig 2.4. Report a mistake