srinight objective type questions: 1. $$ \sqrt{9 x^{2}}= $$ a) $$ \pm 3 x $$ b) $$ 3 x $$ c) $$ -3 x $$ d) $$ 9 x $$ 2. $$ (x-3)^{2} $$ is equila to b) $$ x^{2}+6 x+9 $$ c) $$ x^{2}-6 x-9 $$ d) $$ x^{2}+6 x-9 $$ 3. A trinomial is a perfect square if it is the square of a) monomial b) binomial c) trinomial d) all these 4. $$ (a+b+c)^{2} $$ is equal to a) $$ a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a $$ b) $$ a^{2}+b^{2}+c^{2}-2 a b-2 b c-2 c a $$ c) $$ a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a $$ d) $$ a^{2}+b^{2}+c^{2}-2 a b-2 b c+2 c a $$ ane or more than one correct answer type questions : $$ \sqrt{16 x^{2} y^{4}}= $$ a) $$ 4 x y^{2} $$ b) $$ -\left(4 x y^{2}\right) $$ c) $$ \frac{4 x y^{3}}{y} $$ d) $$ -4 x y \times(-y) $$ eger type questions : $$ \sqrt{16 a^{2}}=k a, \text { then } k= $$ $$ \qquad $$

Question

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Let's solve each of the objective type questions step by step.

### Question 1
**Problem:** $$ \sqrt{9 x^{2}}= $$

**Solution:**
The square root of a squared term can be expressed as:
$$
\sqrt{a^2} = \pm a
$$
Thus,
$$
\sqrt{9 x^{2}} = \sqrt{(3x)^{2}} = \pm 3x
$$

**Answer:** a) $$ \pm 3 x $$

---

### Question 2
**Problem:** $$ (x-3)^{2} $$ is equal to

**Solution:**
To expand $$ (x-3)^{2} $$:
$$
(x-3)^{2} = x^{2} - 2 \cdot 3 \cdot x + 3^{2} = x^{2} - 6x + 9
$$

**Answer:** None of the provided options are correct. The correct expansion is $$ x^{2} - 6x + 9 $$.

---

### Question 3
**Problem:** A trinomial is a perfect square if it is the square of

**Solution:**
A trinomial can be a perfect square if it is the square of a binomial. For example, $$ (a+b)^{2} = a^{2} + 2ab + b^{2} $$ is a perfect square trinomial.

**Answer:** b) binomial

---

### Question 4
**Problem:** $$ (a+b+c)^{2} $$ is equal to

**Solution:**
To expand $$ (a+b+c)^{2} $$:
$$
(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
$$

**Answer:** a) $$ a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a $$

---

### One or More Than One Correct Answer Type Question
**Problem:** $$\sqrt{16 x^{2} y^{4}}=$$

**Solution:**
We can simplify this expression:
$$
\sqrt{16 x^{2} y^{4}} = \sqrt{16} \cdot \sqrt{x^{2}} \cdot \sqrt{y^{4}} = 4 \cdot x \cdot y^{2} = 4xy^{2}
$$

Now let's check the options:
- a) $$ 4 x y^{2} $$ is correct.
- b) $$ -\left(4 x y^{2}\right) $$ is also correct since the square root can be positive or negative.
- c) $$ \frac{4 x y^{3}}{y} = 4 x y^{2} $$ is correct.
- d) $$ -4 x y \times(-y) = 4 x y^{2} $$ is also correct.

**Answer:** a), b), c), d) (all are correct)

---

### Integer Type Question
**Problem:** $$\sqrt{16 a^{2}}=k a, \text{ then } k=$$

**Solution:**
We can simplify:
$$
\sqrt{16 a^{2}} = \sqrt{(4a)^{2}} = \pm 4a
$$
Thus, $$ k = \pm 4 $$.

**Answer:** $$ k = 4 $$ or $$ k = -4 $$ (depending on the context, typically $$ k = 4 $$ is taken as the principal square root).

---

This concludes the solutions to the questions provided. If you have any further questions or need additional explanations, feel free to ask!
**Solutions to the Objective Type Questions:** 1. $$ \sqrt{9 x^{2}} = \pm 3 x $$ **Answer:** a) $$ \pm 3 x $$ 2. $$ (x-3)^{2} = x^{2} - 6x + 9 $$ **Answer:** None of the provided options are correct. The correct expansion is $$ x^{2} - 6x + 9 $$. 3. A trinomial is a perfect square if it is the square of **Answer:** b) binomial 4. $$ (a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca $$ **Answer:** a) $$ a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a $$ **One or More Than One Correct Answer Type Question:** $$ \sqrt{16 x^{2} y^{4}} = 4xy^{2} $$ **Answer:** a), b), c), d) (all are correct) **Integer Type Question:** $$ \sqrt{16 a^{2}} = 4a \quad \text{or} \quad -4a $$ **Answer:** $$ k = 4 $$ or $$ k = -4 $$

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