The solution of inequality $25 x^2-4 \leq 0$ is :
- ✓$\left[-\frac{2}{5}, \frac{2}{5}\right]$
- B$\left(-\frac{2}{5}, \frac{2}{5}\right)$
- C$\left(-\frac{2}{5}, \infty\right)$
- DNone of the above
Answer: A.
View full solution →29 questions across 5 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
M.C.Q (1 Marks)
10 Q→02True False[1 Marks ]
10 Q→03Fill In The Blanks[1 Marks ]
6 Q→042 Marks Questions
1 Q→05Match the following.
2 Q→One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Answer: A.
View full solution →Answer: B.
View full solution →Answer: B.
View full solution →Answer: B.
View full solution →Answer: C.
View full solution →| Part (a) | Part (b) |
| 1. If $\frac{2 x+4}{x-1} \geq 5$, then | (a) $x \in(4, \infty)$ |
| 2. If $3 x-7>x+1$, then | (b) $x \in\left(-\frac{1}{4}, \frac{5}{6}\right)$ |
| 3. If $\frac{5 x}{2}+\frac{3 x}{4} \geq \frac{39}{4}$, then | (c) $x \in(-\infty,-5) \cup(5, \infty)$ |
| 4. If $\frac{6 x-5}{4 x+1}<0$, then | (d) $x \in[3, \infty)$ |
| 5. If $\frac{x}{x-5}>\frac{1}{2}$, then | (e) $x \in(1,3]$ |
| Part (a) | Part (b) |
| 1. If $|x+2| \leq 9$, then | (a) $x \in(2,4) \cup(4,6)$ |
| 2. If $|x-1|>5$, then | (b) $x \in(10, \infty)$ |
| 3. If $-3 x+17<-13$, then | (c) $x \in[-11,7]$ |
| 4. If $\left|\frac{2}{x-4}\right|>1, x \neq 4$, then | (d) $x \in(-\infty, 3.9) \cup(4, \infty)$ |
| 5. If $2(2 x+3)-10<$ $6(x-2)$, then | (e) $x \in(-\infty,-4) \cup(6, \infty)$ |
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