MATHS Assertion (A) & Reason (B) MCQ

3. A is true; R is false

Solution:

Assertion Given limit $=\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-2^\text{x}}{\tan\text{x}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{(3^\text{x}-1)-(2^\text{x}-1)}{\text{x}}\times\frac{\text{x}}{\tan\text{x}}$

$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{2^\text{x}-1}{\text{x}}\Big)\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\tan\text{x}}$

$=(\log3-\log2)\times1$

$=\log\Big(\frac{3}{2}\Big)$

Reason $=\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\tan\text{x}}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}\times\frac{\text{x}}{\tan\text{x}}$

$=1\times1=1 $

4. A is false; R is true.

Soultion:

Assertion Given, $\lim\limits_\text{x}\rightarrow\pi\frac{\sin(\pi-\text{x)}}{\pi(\pi-\text{x)}}$

$\text{Let} \pi - \text{x} =\text{h}, \text{As } \text{x} \rightarrow \pi, \text{then} \text{ h} \rightarrow 0 $

$\therefore\lim\limits_{\text{x}\rightarrow\pi}\frac{\sin(\pi-\text{x})}{\pi(\pi-{\text{x})}\lim\limits_{\text{h}\rightarrow0}}\frac{\sin\text{h}}{\pi\text{h}}$

$\lim\limits_{\text{h}\rightarrow0}\frac{1}{\pi}\Big[\because\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$

Reason Given, $=\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{x}}{\pi-\text{x}}$

Put the limit directly, we get

$\frac{\cos0}{\pi-0}=\frac{1}{\pi}$

1. A is true, R is true; R is acorrect explanation of A.

Soultion:

Assertion Given, $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\sin\text{bx}}$

Multiplying and dividing by (ax) and (x), we get

$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}\times\frac{\text{bx}}{\sin\text{bx}}\times\frac{\text{ax}}{\text{bx}}$

$=1\times1\times\frac{\text{a}}{\text{b}}=\frac{\text{a}}{\text{b}}$

$\Big[\because\lim\limits_{\text{a}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{bx}}{\sin\text{bx}}=1\Big]$

4. A is false; R is true.

Solution:

Assertion Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\ell^{3+\text{x}}-\sin\text{x}-\ell^3}{\text{x}}$

$=\lim\limits_{\text{x}\rightarrow0}\Big[\frac{\ell^3\text{(e}^\text{x}-1)}{\text{x}}-\frac{\sin\text{x}}{\text{x}}\Big]=\ell^3-1$

Reason Given limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\tan4\text{x}}{\sin2\text{x}}$

$\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\tan4\text{x}}{4\text{x}}\Big)\Big|\Big|\Big(\frac{2\text{x}}{\sin2\text{x}}\Big)\Big|\Big|\Big(\frac{4\text{x}}{2\text{x}}\Big)$

$\lim\limits_{4\text{x}\rightarrow0}\frac{\tan4\text{x}}{4\text{x}}\times\frac{1}{\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}}\times\frac{4\text{x}}{2\text{x}}$

$=1\times1\times2=2 $

3. A is true; R is false

Solution:

Assertion Given, $\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$

Dividing each term by x, we get

$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\text{ax}}{\text{ax}}+\frac{\text{x}\cos\text{x}}{\text{x}}}{\frac{\text{b}\sin\text{x}}{\text{x}}}$ 

$=\lim\limits_{\text{x}\rightarrow0} \frac{\text{a}+\cos\text{x}}{\text{b}\Big|\frac{\sin\text{x}}{\text{x}}\Big|}$

$=\frac{\text{a}+\cos0}{\text{b}\times1}=\frac{\text{a}+1}{\text{b}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$

Reason $=\lim\limits_{\text{x}\rightarrow0}\text{x}\text{ sec}\text{ x}=0 \times\sec0=0 \times1=0$

1. A is true, R is true; R is acorrect explanation of A.

Soultion:

Assertion Given,

$\lim\limits_{\text{a}\rightarrow0}\frac{\sin\text{ax}}{\text{bx}}=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a})\sin\text{ax}}{\text{b}(\text{bx})}$

[dividing and multiplying by a]

$=\frac{\text{a}}{\text{b}}\times1=\frac{\text{a}}{\text{b}}\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}=1\Big]$

3. A is true; R is false

Solution:

Assertion Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\sqrt{1-\cos2\text{x}}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\sqrt{1-\cos2\text{x}}}$

$=\frac{1}{\sqrt{2}}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}.\frac{\text{x}}{\sin\text{x}}$

$=\frac{1}{\sqrt{2}}\times1\times=\frac{1}{\sqrt{2}}$

Reason Given, limit $=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2-1)}{(\text{x}-1)}$

$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\text{(x+1)}}{(\text{x}-1)}$

$=\lim\limits_{\text{x}\rightarrow1}(\text{x}+1)=2$

3. A is true; R is false

Solution:

Assertion Given,

$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{ax}^2+\text{bx+}\text{c}}{\text{cx}^2+\text{bx}+\text{a}}$

$=\frac{\text{a}\times(1)^2+\text{b}\times1+\text{c}}{\text{c}\times(1)^2+\text{b}\times1+\text{a}}$

$=\frac{\text{a}+\text{b}+\text{c}}{\text{c}+\text{b}+\text{c}}=1$

Reason $\lim\limits_{\text{x} \rightarrow -2}\frac{\frac{1}{\text{x}}+\frac{1}{2}}{\text{x}+2}$

$=\lim\limits_{\text{x}\rightarrow-2}\frac{(2+\text{x)}}{2\text{x}(\text{x}+2)}=\lim\limits_{\text{x}\rightarrow-2}\frac{1}{2\text{x}}$

$=\frac{1}{2(-2)}=-\frac{1}{4}$

4. A is false; R is true.

Solution:

Assertion Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{3^{2+\text{x}}-9}{\text{x}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{3^2(3^\text{x}-1)}{\text{x}}=9\log3$

Reason Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^{\sin\text{x}}-1}{\sin\text{x}}$

$=\text{Let}\text{ y}=\sin\text{x}$

$\text{then,}\text{ y}\rightarrow0\text{ as}\text{ x}\rightarrow0$

$\therefore\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^{\sin\text{x}}-1}{\sin\text{x}}$

$=\lim\limits_{\text{y}\rightarrow0}\frac{\text{a}^\text{y}-1}{\text{y}}=\log\text{a}$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion Given, limit $=\lim\limits_\text{x}\rightarrow\frac{\ell^\text{x}-\ell^-{\text{x}}}{\text{x}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{\ell^{2\text{x}}-1}{\text{x}\ell^\text{x}}$

$=\lim\limits_{2\text{x}\rightarrow0}\frac{\ell^{2\text{x}}-1}{2\text{x}}\times\lim\limits_{2\text{x}\rightarrow0}\frac{2}{\ell^\text{x}}$

$=1\times\frac{2}{1}=2$

4. A is false; R is true.

Soultion:

Assertion $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$

Dividing each term by x, we get

$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\text{ax}}{\text{x}}+\frac{\text{bx}}{\text{x}}}{\frac{\text{ax}}{\text{x}}+\frac{\sin\text{bx}}{\text{x}}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\text{a}\sin\text{ax}}{\text{ax}}+\text{b}}{\text{a}+\frac{\text{b}\sin\text{bx}}{\text{bx}}}$

$=\frac{\text{a}\times1+\text{b}}{\text{a}+\text{b}\times1}=\frac{\text{a}+\text{b}}{\text{a}+\text{b}}=1$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$

Reason  $\lim\limits_{\text{x}\rightarrow1}\ (5\text{x} + 5\text{x} +1) $

$5(1)^\circ + 5(1)41=54+541=11$

4. A is false; R is true.

Solution:

Assertion Given limit $=\lim\limits_{\text{a}\rightarrow0}\Big(\frac{\ell^\text{x}-\ell^{\sin\text{x}}}{\text{x}-\sin\text{x}}\Big)$

$=\lim\limits_{\text{a}\rightarrow0}\ell^{\sin\text{x}}\Big(\frac{\ell^{\text{x}-\sin\text{x}}-1}{\text{x}-\sin\text{x}}\Big)=\ell^{\sin0}\times1=1$

Reason Given limit $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3^{2\text{x}}-2^{3\text{x}}}{\text{x}}\Big)$

$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{(3^{2\text{x}}-1)-(2^{3\text{x}}-1}{\text{x}}\Big)$

$=2\times\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3^{2\text{x}}-1}{2\text{x}}\Big)-3\times\lim\limits_{\text{x}\rightarrow0}\Big(\frac{2^{3\text{x}}-1}{3\text{x}}\Big)$

$= 2\log 3-3\log 2 $

$=\log 9-\log8 $

$=\log\Big(\frac{9}{8}\Big)$

3. A is true; R is false

Solution:

Assertion Given limit 

$=\lim\limits_{\text{x}\rightarrow0}\frac{\ell^{\tan\text{x}}-1}{\tan\text{x}}\times\frac{\tan\text{x}}{\text{x}}$

$=1\times1=1$

Reason Given limit

$\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\ell^{4\text{x}}-1}{\text{x}}\Big)$

$\lim\limits_{4\text{x}\rightarrow0}4\Big(\frac{\ell^{4\text{x}}-1}{\text{x}}\Big)=4$

3. A is true; R is false

Solution:

Assertion Given limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}-1}{\log_\text{e}\text{x}}$

$\text{put } \text{x}=1+\text{h} \text{ as}\text{ x}\rightarrow1,\text{h}\rightarrow0 $

$\therefore\lim\limits_{\text{h}\rightarrow0}\frac{1+\text{h}-1}{\log_\text{e}(1+\text{h)}}=\frac{1}{\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h)}}{\text{h}}}=1$

Reason Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\log(\sin\text{x+1)}}{\text{x}}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{\log(\sin\text{x+1)}}{\sin\text{x}}\times\frac{\sin\text{x}}{\text{x}}=1$

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Assertion Given,
$\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^2\times\Big(\frac{\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\Big)\times4$

$=\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$

$=\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{1-\cos\text{x00}}=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{2\sin^2\frac{\text{x}}{2}}$

$\Big[\because1-\cos2\text{x}=2\sin^2\text{x}\text{ and }1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\Big]$

Multiplying and dividing by x2 and then multiplying by $\frac{4}{4}$ in the numerater. we get

$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin^2\text{x}}{\text{x}^2}\times\frac{4\times\frac{\text{x}^2}{4}}{\sin^2\frac{\text{x}}{2}}$

$=1\times1\times4=4 $