Let $\text{f(x)}=\sqrt{1+\text{x}^2}$ then.
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$\text{f(xy)} = \text{f(x)}.\text{f(y)}$
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$\text{f(xy)} \geq \text{f(x)}.\text{f(y)}$
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$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$
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None of these
71 questions across 7 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
M.C.Q (1 Marks)
12 Q→02True False[1 Marks ]
5 Q→03Fill In The Blanks[1 Marks ]
1 Q→041 Marks Question
27 Q→052 Marks Questions
18 Q→063 Marks Question
6 Q→075 Marks Questions
2 Q→One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Let $\text{f(x)}=\sqrt{1+\text{x}^2}$ then.
$\text{f(xy)} = \text{f(x)}.\text{f(y)}$
$\text{f(xy)} \geq \text{f(x)}.\text{f(y)}$
$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$
None of these
Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is.
The domain and range of real function f defined by $\text{f(x)}=\sqrt{\text{x}-1}$ is given by.
Domain $= [1, \infty),$ Range $= [0, \infty)$
Domain $= [1, \infty),$ Range $= [0, \infty)$
Domain $= [1, \infty),$ Range $= [0, \infty)$
Domain $= [1, \infty),$ Range $= [0, \infty)$
The domain of the function f defined by $\text{f(x)}=\sqrt{4-\text{x}}+\frac{1}{\sqrt{\text{x}^2-1}}$ is equal to.
$(–\infty, –1) \cup (1, 4]$
$(–\infty, –1] \cup (1, 4]$
$(–\infty, –1) \cup [1, 4]$
$(–\infty, –1) \cup [1, 4)$
Domain of $\sqrt{\text{a}^2-\text{x}^2}(\text{a}>0)$ is.
| Column I | Column II | ||
| (a) | $\text{f}-\text{g}$ | (i) | $\Big\{\Big(2, \frac{4}{5}\Big), \Big(8, \frac{-1}{4}\Big), \Big(10, \frac{-3}{13}\Big)\Big\}$ |
| (b) | $\text{f}+\text{g}$ | (ii) | $\{(2, 20), (8, -4), (10, -39)\}$ |
| (c) | $\text{f}\times\text{g}$ | (iii) | $\{(2, 1), (8, -5), (10, -16)\}$ |
| (d) | $\frac{\text{f}}{\text{g}}$ | (iv) | $\{(2, 9), (8, 3), (10, 10)\}$ |
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