R.d Sharma 2022 _mcqs for Class 9 Maths Chapter 2 - Exponents Of Real Numbers

Answer:

We have to find the value of. So,

The value of is 125

Hence the correct choice is

Answer:

Given

Here

By substituting in we get 

The value of is – 14

Hence the correct choice is .

Answer:

We have to find the product (say L) of the square root of x with the cube root of x is. So, 

$=x^{\frac{3+2}{6}}=x^{\frac{5}{6}}$

The product of the square root of x with the cube root of x is

Hence the correct alternative is

Answer:

We have to find he seventh root of x divided by the eighth root of x, so let it be L. So, 

The seventh root of x divided by the eighth root of x is

Hence the correct choice is .

Answer:

We have to find the value of . 

So,

The value of is

Hence the correct choice is .

Answer:

Disclaimer: In question in place of 29 it should be 19.

We have to find the value of ${\left\{{\left(23+2^2\right)}^{2/3}+(140-19{)}^{1/2}\right\}}^2$

$\Rightarrow {\left\{{\left(23+2^2\right)}^{2/3}+(140-19{)}^{1/2}\right\}}^2={20}^2=400$

Hence the correct answer is option (a).

Answer:

We have to simplify

So,

The value of is

Hence the correct choice is .

Answer:

We have to find the value of provided

So,

Equating the exponents we get

By substitute in we get 

The real value of is

Hence the correct choice is .

Answer:

We have to find the value ofprovided

So,

By equating the exponents we get

By substituting in we get 

The value of is

Hence the correct choice is

Answer:

We have to find the value ofif

Consider,

Multiply on both sides of powers we get 

By taking reciprocal on both sides we get,

Substituting in we get

By taking least common multiply we get 

Hence the correct choice is .

Answer:

We have to find the value of

So,

Hence the correct choice is .

Answer:

We have to find the value of

So, 

Also,

Hence the correct alternative is .

Answer:

We have to find the value of

So,

Since, is equal to ,,.

Hence the correct choice is

Answer:

We have to find the value of when a, b, c are positive real numbers.

So,

Taking square root as common we get 

$$\begin{gathered} \sqrt{a^{-1}b}\times \sqrt{b^{-1}c}\times \sqrt{c^{-1}a}=\sqrt{\frac{b}{a}\times \frac{c}{b}\times \frac{a}{c}} \\ \sqrt{a^{-1}b}\times \sqrt{b^{-1}c}\times \sqrt{c^{-1}a}=1 \end{gathered}$$

Hence the correct alternative is .

Answer:

We have to find value of provided

So,

Equating exponents of power we get

Hence the correct alternative is

Answer:

Find the value of

Hence the correct choice is .

Answer:

Find value of.

$\sqrt[5]{3125a^{10}{b}^5{c}^{10}}=5a^{2}b{c}^2$

Hence the correct choice is .

Answer:

Find the value of .

So,

Hence the correct choice is

Answer:

We have to find the value of if,

Substitute,into get,

Hence the correct choice is .

Answer:

We have to find the value of for

By using rational exponents

$7^{\frac{-1}{3}}=7^m$

Equating power of exponents we get

Hence the correct choice is .

Answer:

$\begin{array}{rcl}{(0.00243)}^{\frac{3}{5}}+{(0.0256)}^{\frac{3}{4}}& =& {\left(\frac{243}{100000}\right)}^{\frac{3}{5}}+{\left(\frac{256}{10000}\right)}^{\frac{3}{4}}\\ & =& {\left(243\times {10}^{-5}\right)}^{\frac{3}{5}}+{\left(256\times {10}^{-4}\right)}^{\frac{3}{4}}\\ & =& {\left(3^5\times {10}^{-5}\right)}^{\frac{3}{5}}+{\left(4^4\times {10}^{-4}\right)}^{\frac{3}{4}}\\ & =& 3^{5\times \frac{3}{5}}\times {10}^{-5\times \frac{3}{5}}+4^{4\times \frac{3}{4}}\times {10}^{-4\times \frac{3}{4}}\\ & =& 3^3\times {10}^{-3}+4^3\times {10}^{-3}\\ & =& 27\times {10}^{-3}+64\times {10}^{-3}\\ & =& 0.027+0.064\\ & =& 0.091\end{array}$

Hence, the correct answer is option (d).

Answer:

We have to find the value of. So,

By using law of rational exponents

we get

The value of is 4

Hence the correct choice is .

Answer:

We have to find the value of

Given that, therefore,

Hence the correct option is .

Answer:

We have to find the value of

Given

By equating the exponents we get 

Hence the correct alternative is .

Answer:

We have to find provided. So,

By raising both sides to the power

By substituting in we get

The value of is

Hence the correct choice is .

Answer:

For, we have to find the value of x.

So,

By raising both sides to the power we get

The value of is

Hence the correct alternative is

Answer:

Given.We have to find the value of

So,

The value of is

Hence the correct choice is

Answer:

We have to find the value of if

So,

Taking as common factor we get 

By equating powers of exponents we get 

By substituting in we get

Hence the correct choice is

Answer:

Simplify

${\left(256\right)}^{{-}^{\left(4^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}}={\left(256\right)}^{-{\left(2\right)}^{\left(-3\right)}}$
 

Hence the correct choice is .

Answer:

We have to find the value of provided

So,

By cross multiplication we get 

By equating exponents we get 

And 

Hence the correct choice is

Answer:

Find the value of

So,

Hence the correct statement is.

Answer:

We have to find provided

So,

Substitute in to get

Hence the value of is

The correct choice is

Answer:

We have to find the value ofprovided

So,

Equating the power of exponents we get

The value of is 

Hence the correct alternative is

Answer:

We have to find the value ofprovided

So,

Equating the power of exponents we get

The value of  is 

Hence the correct alternative is

Answer:

Given that, a, b, c are positive integers such that $a^{b^c}=256$.

Now,
$256={\left(16\right)}^2={\left(4^2\right)}^2$

Thus, the values of a, b, c can be 4, 2 and 2.

Therefore,
$\begin{array}{rcl}abc& =& 4\times 2\times 2\\ & =& 16\end{array}$

Hence, the correct answer is option (b).

Answer:

Given that, a, b, c are positive integers such that $a^{b^c}=6561$.

Now,
$6561={\left(81\right)}^2={\left(9^2\right)}^2$

Thus, the values of a, b, c can be 9, 2 and 2.

Therefore,
$\begin{array}{rcl}abc& =& 9\times 2\times 2\\ & =& 36\end{array}$

Hence, the correct answer is option (b).

Answer:

Given that, 2^x = 3^y = 6^z.

Let 2^x = 3^y = 6^z = k.
Thus, $k^{\frac{1}{x}}=2,k^{\frac{1}{y}}=3 \mathrm{and} k^{\frac{1}{z}}=6$.

Now,
$$\begin{gathered} 2\times 3=6 \\ \Rightarrow k^{\frac{1}{x}}\times k^{\frac{1}{y}}=k^{\frac{1}{z}} \\ \Rightarrow k^{\frac{1}{x}+\frac{1}{y}}=k^{\frac{1}{z}} \left(\because m^a·m^b=m^{ab}\right) \\ \Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z} \end{gathered}$$

Thus,
$\begin{array}{rcl}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}& =& \frac{1}{z}+\frac{1}{z}\\ & =& \frac{2}{z}\end{array}$

Hence, the correct answer is option (c).

Answer:

Given:  x^y = y^z = z^x and xz = y²

$$\begin{gathered} \mathrm{Let} x^y=y^z=z^x=k \\ \Rightarrow x^y=k \\ \Rightarrow x={\left(k\right)}^{\frac{1}{y}} .....\left(1\right) \\ \mathrm{Also}, y^z=k \\ \Rightarrow y={\left(k\right)}^{\frac{1}{z}} .....\left(2\right) \\ \mathrm{And} z^x=k \\ \Rightarrow z={\left(k\right)}^{\frac{1}{x}} .....\left(3\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Since} xz=y^2 .....\left(4\right) \\ \mathrm{Substituting} \left(1\right),\left(2\right) \mathrm{and}\left(3\right) \mathrm{in} \left(4\right) \\ {\left(k\right)}^{\frac{1}{y}}·{\left(k\right)}^{\frac{1}{x}}={\left[{\left(k\right)}^{\frac{1}{z}}\right]}^2 \\ \Rightarrow {\left(k\right)}^{\frac{1}{y}+\frac{1}{x}}={\left(k\right)}^{\frac{2}{z}} \left[ \because a^m·a^n=a^{m+n} and {\left(a^m\right)}^n=a^{mn}\right] \\ \Rightarrow {\left(k\right)}^{\frac{x+y}{yx}}={\left(k\right)}^{\frac{2}{z}} \\ \mathrm{Bases} \mathrm{are} \mathrm{same} \mathrm{so} \mathrm{equate} \mathrm{the} \mathrm{power} \\ \Rightarrow \frac{x+y}{yx}=\frac{2}{z} \\ \Rightarrow z\left(x+y\right)=2yx \\ \Rightarrow z=\frac{2xy}{x+y} \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

Given: 6^x – y = 36 and 3^x + y = 729

$$\begin{gathered} \mathrm{Since} 6^{x-y}=36 \\ \Rightarrow 6^{x-y}=6^2 \\ \Rightarrow x-y=2 .....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{And} 3^{x+y}=729 \\ \Rightarrow 3^{x+y}=3^6 \\ \Rightarrow x+y=6 .....\left(2\right) \end{gathered}$$

$\begin{array}{rcl}\therefore x^2-y^2& =& \left(x+y\right)\left(x-y\right)\\ & =& 6\times 2 \left[\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right]\\ & =& 12\end{array}$

Hence, the correct answer is option (a).

Answer:

Converting all the numbers to base 3 for comparing.
3¹⁹⁸ is already a number with base 3.

$\begin{array}{rcl}{27}^{64}& =& {\left(3^3\right)}^{64}\\ & =& 3^{3\times 64}\\ & =& 3^{192}\end{array}$

$\begin{array}{rcl}{9}^{100}& =& {\left(3^2\right)}^{100}\\ & =& 3^{2\times 100}\\ & =& 3^{200}\end{array}$

$\begin{array}{rcl}{81}^{49}& =& {\left(3^4\right)}^{49}\\ & =& 3^{4\times 49}\\ & =& 3^{196}\end{array}$

So, $3^{192}<3^{196}<3^{198}<3^{200}$
⇒  27⁶⁴ <  81⁴⁹ <  3¹⁹⁸ < 9¹⁰⁰
Thus, the greatest among 3¹⁹⁸, 27⁶⁴, 9¹⁰⁰ and 81⁴⁹ is 9¹⁰⁰.

Hence, the correct answer is option (a).

 

Answer:

Given: $\sqrt[x]{3}\times \sqrt[y]{5}=10125$

$$\begin{gathered} \sqrt[x]{3}\times \sqrt[y]{5}=10125 \\ \Rightarrow {\left(3\right)}^{\frac{1}{x}}\times {\left(5\right)}^{\frac{1}{y}}=3^4\times 5^3 \\ \mathrm{Comparing} \mathrm{the} \mathrm{powers} \mathrm{of} \mathrm{same} \mathrm{bases} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \Rightarrow \frac{1}{x}=4 \mathrm{and} \frac{1}{y}=3 \\ \Rightarrow x= \frac{1}{4} \mathrm{and} y=\frac{1}{3} \\ \mathrm{Now}, xy=\frac{1}{4}\times \frac{1}{3} \\ \Rightarrow xy=\frac{1}{12} \\ \Rightarrow 12xy=1 \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

Given
Option (a) :
Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong. 

Option (b) : 

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong.

Option (c) : 

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong. 

Option (d) : 

Left hand side: 

Right Hand side:

Left hand side is equal to right hand side 

The statement is true.

Hence the correct choice is .

Answer:

Statement-2 (Reason): ${\left({\left(a^m\right)}^n\right)}^s=a^{mns},a>0.$

$\begin{array}{rcl}{\left({\left(a^m\right)}^n\right)}^s& =& {\left(a^{m\times n}\right)}^s \left[\because {\left(p^x\right)}^y=p^{xy}\right]\\ & =& {\left(a^{mn}\right)}^s\\ & =& \left(a^{mn\times s}\right) \left[\because {\left(p^x\right)}^y=p^{xy}\right]\\ & =& a^{mns}\end{array}$

Thus, Statement-2 is true.

Statement-1 (Assertion): ${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^{-\frac{1}{3}}.$

$\begin{array}{rcl}{\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}& =& {\left[{\left\{{\left(7^{-2}\right)}^{-2}\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}\\ & =& {\left(7\right)}^{\left(-2\right)\times \left(-2\right)\times \left(-\frac{1}{3}\right)\times \left(\frac{1}{4}\right)}\\ & =& {\left(7\right)}^{-\frac{1}{3}}\end{array}$

Thus, Statement-1is true.
Also, Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): If aⁿ = k, then $a=k^{\frac{1}{n}}$.
aⁿ = k 
⇒ $a=k^{\frac{1}{n}}$

Thus, Statement-2 is true.

Statement-1 (Assertion): If a^x = b^y = c^z = abc, then xy + yz + zx = xyz.

$$\begin{gathered} a^x=b^y=c^z=abc .....\left(1\right) \\ \mathrm{As}, a^x=b^y \\ \Rightarrow b=a^{\frac{x}{y}} \\ \mathrm{and} a^x=c^z \\ \Rightarrow c=a^{\frac{x}{z}} .....\left(2\right) \\ \mathrm{From} \left(1\right), \\ {a}^x=abc \\ \Rightarrow a^x=a\times a^{\frac{x}{y}}\times a^{\frac{x}{z}} \\ \Rightarrow a^x=a^{1+\frac{x}{y}+\frac{x}{z}} \end{gathered}$$

As bases are same equate the powers.

$$\begin{gathered} \Rightarrow x=1+\frac{x}{y}+\frac{x}{z} \\ \Rightarrow x=\frac{yz+xz+xy}{yz} \\ \Rightarrow xyz=yz+xz+xy \end{gathered}$$

Thus, Statement-1 is true.
Also, Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): $\sqrt{a\sqrt{a\sqrt{a.....}}}n \mathrm{terms}=a^{\frac{2^n-1}{2^n}}$.
$\mathrm{Let} x=\sqrt{a\sqrt{a\sqrt{a.....}}}n \mathrm{terms}$
Then,

$$\begin{gathered} x=a^{\frac{1}{2}}\sqrt{a^{\frac{1}{2}}\sqrt{a^{\frac{1}{2}}...n \mathrm{terms}}} \\ \Rightarrow x=a^{\frac{1}{2}}\times {\left(a^{\frac{1}{2}}\right)}^{\frac{1}{2}}\times \sqrt{{\left(a^{\frac{1}{2}}\right)}^{\frac{1}{2}...n \mathrm{terms}}} \\ \Rightarrow x=a^{\frac{1}{2}}\times \left(a^{\frac{1}{2^2}}\right)\times ...\times \left(a^{\frac{1}{2^n}}\right) \\ \Rightarrow x=a^{\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}} .....\left(1\right) \end{gathered}$$

Now, let $S_n=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}$         .....(2)
Multiplying (2) on both sides by $\frac{1}{2}$, we get
$\frac{1}{2}{S}_n=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}}$         .....(3)

Subtracting (3) from (2), we get
$$\begin{gathered} S_n-\frac{1}{2}{S}_n=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}}\right) \\ \Rightarrow S_n\left(1-\frac{1}{2}\right)=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{n+1}} \\ \Rightarrow S_n\left(\frac{1}{2}\right)=\frac{1}{2}-\frac{1}{2^{n+1}} \\ \Rightarrow S_n=\left(\frac{1}{2}-\frac{1}{2^{n+1}}\right)\times 2 \\ \Rightarrow S_n=\frac{1}{2}\left(1-\frac{1}{2^n}\right)\times 2 \\ \Rightarrow S_n=\left(1-\frac{1}{2^n}\right) \\ \Rightarrow S_n=\frac{2^n-1}{2^n} .....\left(4\right) \end{gathered}$$
Substituting (4) in (1), we get
$$\begin{gathered} x=a^{\frac{1}{2}+\frac{1}{2^2}...+\frac{1}{2^n}} \\ \Rightarrow x=a^{\frac{2^n-1}{2^n}} \end{gathered}$$

Thus, Statement-2 is true.


Statement-1 (Assertion): $\sqrt{7\sqrt{7\sqrt{7\sqrt{7}}}}=\sqrt[16]{7^{15}}$.

$\mathrm{Let} x=\sqrt{7\sqrt{7\sqrt{7\sqrt{7}}}}$

Using statement 2 for n = 4 and a = 7, we get

$\sqrt{a\sqrt{a\sqrt{a.....}}}n \mathrm{terms}=a^{\frac{2^n-1}{2^n}}$
$\begin{array}{rcl}& \Rightarrow & \sqrt{7\sqrt{7\sqrt{7\sqrt{7}}}}=7^{\frac{2^4-1}{2^4}}\\ & =& 7^{\frac{15}{16}}\\ & =& {\left(7^{15}\right)}^{\frac{1}{16}}\\ & =& \sqrt[16]{7^{15}}\end{array}$

Thus, Statement-1 is true.
Also, Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): $\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}.....\infty =x, x>0.$
$$\begin{gathered} \mathrm{Let} y =\sqrt{x\sqrt{x\sqrt{x\sqrt{x...\infty }}}} .....\left(1\right) \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides} \\ \Rightarrow y^2=x\sqrt{x\sqrt{x\sqrt{x\sqrt{x...\infty }}}} \\ \Rightarrow y^2-xy=0 \left[\because \mathrm{From} \left(1\right)\right] \\ \Rightarrow y\left(y-x\right)=0 \\ \Rightarrow y=0 \mathrm{or} \left(y-x\right)=0 \\ \Rightarrow y=0 \mathrm{or} y=x \\ \Rightarrow \sqrt{x\sqrt{x\sqrt{x\sqrt{x...\infty }}}} =x \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): $\sqrt{5\sqrt{5\sqrt{5\sqrt{5}}}}.....\infty =5\sqrt{5}.$

Now, according to Statement-2:$\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}.....\infty =x, x>0.$ 

$\sqrt{5\sqrt{5\sqrt{5\sqrt{5}}}}.....\infty =5$

Thus, Statement-1 is false.

Hence, the correct answer is option (d).

Answer:

Statement-2 (Reason): $\sqrt{x+\sqrt{x+\sqrt{x+}}}.....\infty =x, x>0.$
$$\begin{gathered} \mathrm{Let} y=\sqrt{x+\sqrt{x+\sqrt{x+}}}.....\infty \\ \mathrm{Squaring} \mathrm{both} \mathrm{the} \mathrm{sides} \\ \Rightarrow y^2=x+\sqrt{x+\sqrt{x+}}.....\infty \\ \Rightarrow y^2=x+y \left[\because \mathrm{From}\left(1\right)\right] \\ \Rightarrow y^2-y-x=0 \\ \mathrm{Now}, \mathrm{at}\mathit{ }y=x \\ \Rightarrow x^2-x-x=x^2\ne 0 \left[\because x>0\right] \end{gathered}$$

So, y = x does not satisfy the equation.
Thus, Statement-2 is false.

Statement-1 (Assertion): $\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+}}}}.....\infty =3.$
$$\begin{gathered} \mathrm{Let} y=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+}}}}.....\infty .....\left(1\right) \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides} \\ \Rightarrow y^2=6+\sqrt{6+\sqrt{6+\sqrt{6+}}}.....\infty \\ \Rightarrow y^2=6+y \left[\because \mathrm{From} \left(1\right)\right] \\ \Rightarrow y^2-y-6=0 \\ \Rightarrow y^2-3y+2y-6=0 \\ \Rightarrow y\left(y-3\right)+2\left(y-3\right)=0 \\ \Rightarrow \left(y+2\right)\left(y-3\right)=0 \\ \Rightarrow \left(y+2\right)=0 \mathrm{or} \left(y-3\right)=0 \\ \Rightarrow y=-2 \mathrm{or} y=3 \end{gathered}$$

Now as y is a square root and it is a real number.
∴  y must be positive, i.e., $\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+}}}}.....\infty$ must be positive.
⇒ y = 3                   .....(2)

From (1) and (2), we get
$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+}}}}.....\infty =3.$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): If m, n, p are rational numbers and a is any positive real number, then ${\left({\left(a^m\right)}^n\right)}^p=a^{mnp}.$

$\begin{array}{rcl}{\left({\left(a^m\right)}^n\right)}^p& =& {\left(a^{m\times n}\right)}^p \left[\because {\left(q^x\right)}^y=q^{xy}\right]\\ & =& {\left(a^{mn}\right)}^p\\ & =& \left(a^{mn\times p}\right) \left[\because {\left(q^x\right)}^y=q^{xy}\right]\\ & =& a^{mnp}\end{array}$

Thus, Statement-2 is true.


Statement-1 (Assertion): If m, n are positive integers, then for any positive real number a, ${\left\{\sqrt[m]{\sqrt[n]{a}}\right\}}^{mn}=a.$
$\begin{array}{rcl}{\left\{\sqrt[m]{\sqrt[n]{a}}\right\}}^{mn}& =& {\left\{{\left(\sqrt[n]{a}\right)}^{\frac{1}{m}}\right\}}^{mn} \left(\because \sqrt[x]{q}=q^{\frac{1}{x}}\right)\\ & =& {\left\{{\left(a^{\frac{1}{n}}\right)}^{\frac{1}{m}}\right\}}^{mn} \left(\because \sqrt[x]{q}=q^{\frac{1}{x}}\right) \\ & =& a^{\frac{1}{n}\times \frac{1}{m}\times mn} \left(\mathrm{From} \mathrm{Statement}‐1\right)\\ & =& a^{\frac{mn}{nm}}\\ & =& a^1\\ & =& a\\ & & \end{array}$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).