MCQ 1011 Mark
$\lim \tan\text{x} = \text{x}\rightarrow \frac{\pi }{2}$
- A
$1$
- B
$0$
- C
$\frac{1}{\pi}$
- ✓
Answer$\text{L.H.L}.=\lim \tan\text{x}=+\infty \ \text{x}\rightarrow \Big(\frac{\pi}{2}\Big)^-$
$\text{R.H.L}.=\lim \tan\text{x}=-\infty \ \text{x}\rightarrow \Big(\frac{\pi}{2}\Big)^+$
Clearly left hand $ \text{limit} \neq$ right hand limit.
Hence given limit does not exist.
View full question & answer→MCQ 1021 Mark
Evaluate $\underset{\text{x}\, \rightarrow\,3}{\lim}\, (4\text{x}^2\, +\, 3)$
Answer$ =\displaystyle \lim _{\text{x}\rightarrow 3 }{ \left( 4{ \text{x} }^{ 2 }+3 \right) }$
$=4{ \left( 3 \right) }^{ 2 }+3$
$=36+3$
$=39$
View full question & answer→MCQ 1031 Mark
$\lim_\limits{\text{x}→0} \sin\text{x} (\sqrt{\text{x} + 1} - \sqrt{(1- \text{x})}$ is:
View full question & answer→MCQ 1041 Mark
The limit of $ \left[\frac{1}{\text{x}^2}+\frac{(2013)^\text{x}}{\text{e}^\text{x}-1}-\frac{1}{\text{e}^\text{x}-1}\right]\text{as} \text{ x}\rightarrow 0:$
AnswerCorrect option: A. Approaches $+\infty$
View full question & answer→MCQ 1051 Mark
lf $f(x) = 2x - 3, a = 2, l = 1f(x) = 2x - 3, a = 2, l = 1$ and $\epsilon = 0.001$ then $δ > 0$ satisfying $0<|x - a|< δ, ∣ f(x) - l ∣ < \epsilon $, is:
- A
$0.0050$
- ✓
$0.0005$
- C
$0.001$
- D
AnswerCorrect option: B. $0.0005$
$|f(x) - l|< 0.001 = \epsilon$
$\Rightarrow |2x - 3 - 1| (x) - l∣ < 0.001$
$\Rightarrow -0.001 < 2 x - 4 < 0.001$
$\Rightarrow -0.0005 < x - 2 < 0.0005$
$\Rightarrow ∣x - 2∣ < 0.0005$
$\Rightarrow ∣x - a∣ < 0.0005 = δ$
Hence, $δ = 0.0005 > 0$
View full question & answer→MCQ 1061 Mark
$\lim_\limits{\text{x} \rightarrow 2}\Bigg(\frac{\sqrt{1-\text{cos}{2(\text{x}-2)}}}{\text{x}-2}\Bigg):$
Answer$\lim_\limits{\text{t} \rightarrow 0}\frac{\sqrt{1-\cos2\text{t}}}{\text{t}}$
Clearly $\text{R.H.L}. = \sqrt{2}$
$\text{L.H.L.} = -\sqrt{2}$
Since $\text{R.H.L}. \neq \text{L.H.L.}$
So, limit does not exist.
Hence, option $A$ is correct.
View full question & answer→MCQ 1071 Mark
$\lim_\limits{\text{x} \rightarrow 1}(1+\sin\pi)π\text{x}:$
- ✓
$\pi$
- B
${\pi }^{ 2 }$
- C
${\pi }^{ 3 }$
- D
Answer$=\lim_\limits{\text{x} \rightarrow 1}(1+\sin\pi)π\text{x}$
$= (1+\sin \pi(1))$
$=\pi(1+0)$
$= \pi$
View full question & answer→MCQ 1081 Mark
$\lim_\limits{\text{x} \rightarrow \text{a}}\frac{\text{x}-\text{a}}{|\text{x}-\text{a}|}=$
AnswerUsing,
$\lim_\limits{\text{x} \rightarrow 0}|\text{x}|=-\text{x}$
$\lim_\limits{\text{x} \rightarrow 0}|\text{x}|=+\text{x}$
we get $\lim_\limits{\text{x} \rightarrow \text{a}}-\frac{\text{x}-\text{a}}{-(\text{x}-\text{a})}=-1$
$\lim_\limits{\text{x} \rightarrow \text{a}}+\frac{\text{x}-\text{a}}{-(\text{x}-\text{a})}=-1$
Since, $\text{LHL}$ is not equal to $\text{RHL}$,
hence the limit does not exist.
View full question & answer→MCQ 1091 Mark
$\lim_\limits{\text{x} \rightarrow 0}\frac{\sin|\text{x}|}{\text{x}}$ is equal to:
Answer$=\lim_\limits{\text{x} \rightarrow 0}\frac{\sin|\text{x}|}{\text{x}}$
$\text{ LHL} =-1,\text{RHL}=1$
Limit does not exist.
View full question & answer→MCQ 1101 Mark
If $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}},$ then $f'(1)$ is:
- ✓
$\frac{5}{4}$
- B
$\frac{4}{5}$
- C
$1$
- D
$0$
AnswerCorrect option: A. $\frac{5}{4}$
$\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$
$=\frac{1}{2}\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}$
$=\frac{1}{2}\text{x}^{\frac{1}{2}}-\text{2x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to $x,$ we get
$\text{f}'(\text{x})=\frac{1}{2}\times\frac{1}{2}\text{x}^{\frac{1}{2}-1}-2\times\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}\ [\text{f}(\text{x})=\text{x}^\text{n}$
$\Rightarrow\text{f}'(\text{x})=\text{nx}^{\text{n}-1}]$
$\Rightarrow\text{f}'(\text{x})=\frac{1}{4}\text{x}^{-\frac{1}{2}}+\text{x}^{-\frac{3}{2}}$
$\therefore\text{f}'(\text{1})=\frac{1}{4}\times1+1=\frac{5}{4}$
View full question & answer→MCQ 1111 Mark
Mark the correct alternative in each of the following : If $\text{f(x)}=\text{x}\sin\text{x},$ then $\text{f}'\Big(\frac{\text{x}}{2}\Big)=$
- A
$0$
- ✓
$1$
- C
$-1$
- D
$\frac{1}{2}$
Answer$\text{f(x)}=\text{x}\sin\text{x}$
Differentiating both sides with respect to $x,$ we get
$\text{f}'\text{(x)}=\text{x}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\sin\text{x}\times\frac{\text{d}}{\text{dx}}\text{(x)} ($Product rule$)$
$=\text{x}\times\cos\text{x}+\sin\text{x}\times1$
$=\text{x}\cos\text{x}+\sin\text{x}$
Putting $\text{x}=\frac\pi{2},$ we get
$=\frac{\pi}{2}\times0+1$
$=1$
Hence, the correct answer is option $(b)$
View full question & answer→MCQ 1121 Mark
Evaluate the following limit : $ \displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin^2 3\text{x}}{\text{x}^2}$
Answer$\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin^2 3\text{x}}{\text{x}^2}$
$=\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{x}} \times\frac{\sin 3\text{x}}{\text{x}}$
$= \displaystyle\lim_{\text{x} \rightarrow 0} 3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg) \times3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg)$
$=3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}\times3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}$
$ = 3\times3$
$=9$
View full question & answer→MCQ 1131 Mark
Choose the correct answer. If $\text{f}(\text{x})=\text{x}-[\text{x}],\in\text{R}$ then $\text{f}'\big(\frac{1}{2}\big)$ is equal to:
- A
$\frac{3}{2}$
- ✓
$1$
- C
$0$
- D
$-1$
AnswerGiven $f(x) = x - [x]$
we have ti first check for differentiability of $f(x)$ at $\text{x}=\frac{1}{2}$
$\therefore \text{Lf}'\Big(\frac{1}{2}\Big)=\text{LHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}-\text{h}\big)-\big[\frac{1}{2}-\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}-\text{h}-0-\frac{1}{2}+0}{-\text{h}}$
$=\frac{-\text{h}}{-\text{h}}$
$=1$
$\therefore \text{Rf}'\Big(\frac{1}{2}\Big)=\text{RHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}+\text{h}\big)-\big[\frac{1}{2}+\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}+\text{h}-1-\frac{1}{2}+1}{\text{h}}$
$=\frac{\text{h}}{\text{h}}$
$=1$
Since, $\text{LHD = RHD}$
$\text{f}'\big(\frac{1}{2}\big)=1$
View full question & answer→MCQ 1141 Mark
Differentiate with respect to $x^4+ 3x^2− 2x:$
- ✓
$4x^3+ 6x − 2$
- B
$4x^3+ 6x − 3$
- C
$4x^4+ 6x − 2$
- D
AnswerCorrect option: A. $4x^3+ 6x − 2$
View full question & answer→MCQ 1151 Mark
What is the value of $ \lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}^2\sec\text{x}}{\sin\text{x}}?$
Answer$\lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}^2\sec\text{x}}{\sin\text{x}}\times\ \lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}}{\cos\text{x}} $$$
$= 1 \times 0$
$= 0$
View full question & answer→MCQ 1161 Mark
If $\lim\limits_{\text{x}\to \text{a}}\frac{\text{x}^5-\text{a}^5}{\text{x - a}}=80$ then the value of $aa$ is:
AnswerGiven,$ \lim\limits_{\text{x}\to \text{a}}\frac{\text{x}^5-\text{a}^5}{\text{x - a}}=80$
or, $ 5\text{a}^4=80 [$Using direct formula$]$ or, $\text{a}^4=16$ or, $\text{a}=2.$
View full question & answer→MCQ 1171 Mark
if $\text{f(x)} = \begin{vmatrix} \cos \text{x}& \text{x} & 1\\ 2\sin \text{x} & \text{x}^{2} & 2\text{x}\ \\ \tan \text{x} & \text{x} & 1\end{vmatrix},$ then $\displaystyle \lim_{\text{x}\rightarrow 0} \dfrac {\text{f}(\text{x})}{\text{x}}.$
AnswerCorrect option: A. Exists and is equal to $-2$
View full question & answer→MCQ 1181 Mark
Evaluate : $\displaystyle\lim_{\text{x}\rightarrow 2} \dfrac{\text{x}^2-4}{\text{x}+3}:$
AnswerUsing direct substitution, we obtain,$ =\displaystyle\lim_{\text{x}\rightarrow 2} \dfrac{\text{x}^2-4}{\text{x}+3}$
$ =\dfrac{4-4}{2+3}$
$=0$
View full question & answer→MCQ 1191 Mark
$\lim_\limits{\text{x} \rightarrow 0}\frac{2\text{x}^2+3\text{x}+4}{2}=$
AnswerAs their is not any $x$ term in the denominator,
we can directly substitute the value of $x$ as $0.$
Thus, we have $\lim_\limits{\text{x} \rightarrow 0}\frac{2\text{x}^2+3\text{x}+4}{2}$
$=\frac { 2.0+3.0+4 }{ 2 }$
$=\frac{4}{2}$
$=2$
View full question & answer→MCQ 1201 Mark
Evaluate: $\displaystyle \lim_{\text{x} \rightarrow 0}{\left( \frac{\text{a}^\text{x} + \text{b}^\text{x} + \text{c}^\text{x}}{3} \right)^{\frac{2}{\text{x}}}}$
AnswerCorrect option: D. $(\text{abc})^{\frac{2}{3}}$
$ \displaystyle \lim _{ \text{x} \rightarrow 0 }{ { \left( \frac { { \text{a} }^{ \text{x} }+{ \text{b} }^{ \text{x} }+{ \text{c} }^{ \text{x} } }{ 3 } \right) }^{ \frac{ 2 }{ \text{x} } } } ={ \left( \frac { 3 }{ 3 } \right) }^{\frac { 2 }{ 0 } }$
$ =1\infty \text{ form}={ \text{e} }^{ \displaystyle \lim _{ \text{x}\rightarrow 0 }{ { \left( \cfrac { { \text{a} }^{ \text{x} }+{ \text{b} }^{ \text{x} }+{ \text{c} }^{ \text{x} } }{ 3 } -1 \right) }^{\frac{ 2 }{ \text{x} } } } }$
$ =\text{e}\frac{2}{3}(\log \text{a}+\log \text{b}+\log \text{c})$
$= (\text{abc})^{\frac{2}{3}}$
View full question & answer→MCQ 1211 Mark
What is the number of critical points of $\text{f}(\text{x}) = \frac{|\text{x}^2 - 1|}{ \text{x}^2} ?$
Answer$f(x)$ is not differentiable at $x = 1$ and $x = -1$
And $x = 0$ is not a critical point not in the domain.
Therefore $1$ and $-1$ are critical points.
Thus, there are $2$ critical points.
View full question & answer→MCQ 1221 Mark
Derivative of the function $f(x) = (x - 1) (x - 2)$ is:
- A
$2x + 3$
- B
$3x - 2$
- C
$3x + 2$
- ✓
$2x - 3$
AnswerCorrect option: D. $2x - 3$
View full question & answer→MCQ 1231 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}$ is equal to:
- A
$1$
- ✓
$\frac{\text{m}}{\text{n}}$
- C
$\frac{-\text{m}}{\text{n}}$
- D
$\text{m}^{2}\text{n}^{2}$
AnswerCorrect option: B. $\frac{\text{m}}{\text{n}}$
Given $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}=\lim\limits_{\text{x} \rightarrow 1}\frac{\frac{\text{x}^{\text{m}-(1)^\text{m}}}{\text{x}-1}}{\frac{\text{x}^{\text{n}-(1)^{\text{n}}}}{\text{x}-1}}$
$=\frac{\text{m}(1)^{\text{m}-1}}{\text{n}(1)^{\text{n}-1}}=\frac{\text{m}}{\text{n}}$
$=\frac{\text{m}}{\text{n}}$
View full question & answer→MCQ 1241 Mark
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{t}^\text{x}-\text{e}\sin\text{x}}{2(\text{x}-\sin\text{x)}}=$
- ✓
$-\frac{1}{2}$
- B
$\frac{1}{2}$
- C
$1$
- D
$\frac{3}{2}$
AnswerCorrect option: A. $-\frac{1}{2}$
View full question & answer→MCQ 1251 Mark
Evaluate : $ \displaystyle \underset{\text{x}\rightarrow 2}{\lim} \text{x}^2-5\text{x}+6$
Answer$ = \displaystyle \underset{\text{x}\rightarrow 2}{\lim} \text{x}^2-5\text{x}+6$
$ ={ 2 }^{ 2 }-5\times 2+6$
$= 4 - 10 + 6$
$= 0$
View full question & answer→MCQ 1261 Mark
If $\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}},$ then $\frac{\text{dy}}{\text{dx}}=$
- ✓
$-\frac{4\text{x}}{(\text{x}^2-1)^2}$
- B
$-\frac{4\text{x}}{\text{x}^2-1}$
- C
$\frac{1-\text{x}^2}{\text{4x}}$
- D
$\frac{4\text{x}}{\text{x}^2-1}$
AnswerCorrect option: A. $-\frac{4\text{x}}{(\text{x}^2-1)^2}$
$\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}}$
$=\frac{\text{x}^2+1}{\text{x}^2-1}$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2} ($Quotient rule$)$
$=\frac{(\text{x}^2-1)(\text{2x}+0)-(\text{x}^2+1)(\text{2x}-0)}{(\text{x}^2-1)^2}$
$=\frac{\text{2x}^3-\text{2x}-\text{2x}^3-\text{2x}}{(\text{x}^2-1)^2}$
$=\frac{-\text{4x}}{(\text{x}^2-1)^2}$
View full question & answer→MCQ 1271 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos4\theta}{1-\cos6\theta}$ is equal to:
- ✓
$\frac{4}{9}$
- B
$\frac{1}{2}$
- C
$\frac{-1}{2}$
- D
$-1$
AnswerCorrect option: A. $\frac{4}{9}$
Given $\lim\limits_{\theta \rightarrow 0}\frac{1-\cos4\theta}{1-\cos6\theta}=\lim\limits_{\theta \rightarrow 0}\frac{2\sin^{2}2\theta}{2\sin^{2}3\theta}$
$=\lim\limits_{\theta \rightarrow 0}\frac{\sin^{2}2\theta}{\sin^{2}3\theta}=\lim\limits_{\theta \rightarrow 0}\Big[\frac{\sin2\theta}{\sin3\theta}\Big]^{2}$
$=\lim\limits_{\theta \rightarrow 0}\bigg[\frac{\frac{\sin2\theta}{2\theta}\times2\theta}{\frac{\sin3\theta}{2\theta}\times3\theta}\bigg]$
$=\Big[\frac{2\theta}{2\theta}\Big]^{2}$
$=\Big(\frac{2}{3}\Big)^{2}$
$=\frac{4}{9}$
View full question & answer→MCQ 1281 Mark
The value of $ \lim\limits_{\text{x} \rightarrow 3^{+}} \dfrac{|\text{x}-3|}{\text{x}-3}$ equals:
Answerfor $ \text{x}=30^+,$
$ \text{ x}-3 > 0$
Let $\text{L}=\displaystyle \lim_{\text{x}-3^+}\dfrac {|\text{x}-3|}{\text{x}-3}$
$=\displaystyle \lim_{\text{x}\rightarrow 3^+}$
$=\dfrac{(\text{x}-3)}{(\text{x}-3)}$
$=\lim\limits_{\text{x}\rightarrow 3^+}(1)$
$=1$
View full question & answer→MCQ 1291 Mark
$\lim_\limits{\text{x} \rightarrow 1}{1−\text{x}+[\text{x}+1]+[1−\text{x}]}$, where $[x]$ denotes greatest integer function, is
AnswerSubstitute $x = 1 + t$
$ \text{L.H.S} \lim_\limits{\text{t}\rightarrow o^{-}} (-\text{t}+[2+\text{t}]+[-\text{t}])$
$= 0 + 1 + 0 = 1$
$ \text{R.H.S} \lim_\limits{\text{t}\rightarrow o^{-}} (-\text{t}+[2+\text{t}]+[-\text{t}])$
$= 0 + 2 - 1 = 1$
$\text{L.H.S = R.H.S}$
View full question & answer→MCQ 1301 Mark
$\lim\limits_{\text{n}\rightarrow \infty } \frac{{\text{np}\ \text{sin}^2(\text{n}!)}}{\text{n}+1}, 0 < p < 1$ is equal to:
View full question & answer→MCQ 1311 Mark
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin7\text{x}}{\sin3\text{x}}$ equals:
- ✓
$\frac{7}{3}$
- B
$\frac{10}{3}$
- C
$\frac{14}{3}$
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{7}{3}$
View full question & answer→MCQ 1321 Mark
What is the value of $\text{ddx} (\sin x \tan x)?$
- ✓
$\sin x + \tan x \sec x$
- B
$\cos x + \tan x \sec x$
- C
$\sin x + \tan x$
- D
$\sin x + \tan x \sec^2x$
AnswerCorrect option: A. $\sin x + \tan x \sec x$
We follow product rule $\frac{\text{d}}{\text{dx}}(\text{f}.\text{g})=\text{g.}\frac{\text{d}}{\text{dx}}(\text{f})+(\text{f})\frac{\text{d}}{\text{dx}}(\text{f}.\text{g})$
Here, $f = \sin x$ and $g = \tan x$
$\frac{\text{d}}{\text{dx}} (\sin x \tan x) = \cos x \tan x + \sec^2 x \sin x$
$\frac{\text{d}}{\text{dx}} (\sin x \tan x) = \sin x + \tan x \sec x$
View full question & answer→MCQ 1331 Mark
What is the derivative of $f(x) = x|x|?$
- A
$|x| + x$
- B
$2x$
- ✓
$2|x|$
- D
$-2|x|$
AnswerCorrect option: C. $2|x|$
View full question & answer→MCQ 1341 Mark
If $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points on the curve $y = ax^2+ bx + c$, then as perLagrange’s mean value theorem whichof the following is correct?
- A
At least one point $C(x_3, y_3)$ where the tangent will be intersecting the chord $AB$
- B
At least one point $C(x_3, y_3)$ where the tangent will be overlapping to the chord $AB$
- C
At least two points where the tangent will be parallel to the chord $AB$
- ✓
At least one point $C(x_3, y_3)$ where the tangent will be parallel to the chord $AB$
AnswerCorrect option: D. At least one point $C(x_3, y_3)$ where the tangent will be parallel to the chord $AB$
Here, $y = f(x) = ax^2+ bx + c$
As $f(x)$ is a polynomial function, it is continuous and differentiable for all $x.$
So, according to geometrical interpretation of mean value theorem there,
will be at least one point $C(x_3, y_3)$ between $A(x_1, y_1)$ and $B(x_2, y_2)$ where,
tangent will be parallel chord $AB.$
View full question & answer→MCQ 1351 Mark
if $f(x) = x, x < 0: f(x) = 0, x = 0; f(x) = x, x > 0,$ then $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})$ is equal to:
AnswerGiven: $\text{f}(\text{x})= \left\{ \begin{matrix} \text{x} \\ 0 \\ \text{x}^ 2 \end{matrix}\begin{matrix}\quad \text{x}<0 \\\quad \text{x}=0 \\ \quad \text{x}>0 \end{matrix} \right\}$
$ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})=?$
Sol: left hand $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x}) x = 0$ right hand limit $\rightarrow \lim_\limits{\text{x} \rightarrow 0}\text{f}(\text{x})=\lim_\limits{\text{x} \rightarrow0},\text{x}=0 $
right hand limit → $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})=\lim_\limits{\text{x}\rightarrow 0}\text{x}^2=0$
$\text{LHL = RHL f(0) = 0}$
$\text{LHL = RHL = f(0) = 0}$
View full question & answer→MCQ 1361 Mark
If $n$ is a positive integer, then ($\sqrt{3} + 1)^{2n+1} + (\sqrt{3} - 1)^{2n+1}$ is:
AnswerSince $n$ is a positive integer, assume $n = 1$
$(\sqrt{3}+1)^3 + (\sqrt{3}−1)^3$
$ = {3\sqrt{3} + 1 + 3\sqrt{3}(\sqrt{3} + 1)} + {3\sqrt{3} - 1 - 3\sqrt{3}(\sqrt{3} - 1)}$
$ = 3\sqrt{3} + 1 + 9 + 3\sqrt{3} + 3\sqrt{3}- 1 - 9 + 3\sqrt{3}$
$= 12\sqrt{3}$, which is an irrational number.
View full question & answer→MCQ 1371 Mark
$\lim_\limits{\text{x} \rightarrow \infty}\frac{2\sqrt{\text{x}}+3\sqrt[3]{\text{x}}+4\sqrt[4]{\text{x}}+...+\text{n}\sqrt[\text{n}]{\text{x}}}{\sqrt{(2\text{x}-3}+{\sqrt[3]{(2\text{x}-3)}+{\sqrt[4]{(2\text{x}-3)}+...+}{\sqrt[\text{n}]{(2\text{x}-3)}}}}$ is equal to:
AnswerCorrect option: C. $\sqrt{2}$
$ \lim_\limits{\text{x} \rightarrow \infty}\frac{2\sqrt{\text{x}}+3\sqrt[3]{\text{x}}+4\sqrt[4]{\text{x}}+...+\text{n}\sqrt[\text{n}]{\text{x}}}{\sqrt{(2\text{x}-3}+{\sqrt[3]{(2\text{x}-3)}+{\sqrt[4]{(2\text{x}-3)}+...+}{\sqrt[\text{n}]{(2\text{x}-3)}}}}$
Dividing numerator and denominator by$ \sqrt{\text{x}}$
$=\frac{2}{\sqrt{2}}$
$=\sqrt{2}$
View full question & answer→MCQ 1381 Mark
Find the value of $\lim\limits_{\text{x} \rightarrow 0} \frac{2\text{x}^2 + 3\text{x} + 4}{2}$
AnswerLet $ \lim\limits_{\text{x} \rightarrow 0} \frac{2\text{x}^2 + 3\text{x} + 4}{2}$ This is not an indeterminate form,
Therefore, $\text{L}=\lim\limits_{\text{x}\rightarrow 0}\dfrac {2(0)+3(0)+4}{2}=\dfrac {4}{2}\text{L}=2$
View full question & answer→MCQ 1391 Mark
$\lim\limits_{X\rightarrow 0} 2\sin\frac{23\text{x}}{x^2}$ is equal to:
View full question & answer→MCQ 1401 Mark
The coefficient of $x^n$ in the expansion of $(1 - 2x + 3x^2- 4x^3+ ........)^{-n}$ is:
AnswerCorrect option: B. $ \frac{(2\text{n})!}{(\text{n}!)^2}$
We have,
$(1 - 2x + 3x^2 - 4x^3 + ........)^{-n}$
$= \{(1 + x)^{-2}\}^{-n}$
$= (1 + x)^{2n}$
So, the coefficient of $x^nC_3= ^{2n}C_n$
$=\frac{(2\text{n})!}{(\text{n}!)^2}$
View full question & answer→MCQ 1411 Mark
Choose the correct answer.
If $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$ where $[.]$ denotes the greatest integer function. then $\lim\limits_{\text{x} \rightarrow 0}\text{f}(\text{x})$ is equal to :
AnswerGiven $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$
$\text{L}.\text{H}.\text{H}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0-\text{h}]}{[0-\text{h}]} $
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin[\text{-h}]}{[-\text{h}]}=-1$
$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0+\text{h}]}{[0+\text{h}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[\text{h}]}{[\text{h}]}=1$
$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$
So, the limit does not exist.
View full question & answer→MCQ 1421 Mark
Let $f(x) = (x - a) (x - b) (x - c), a < b < c.$ Then $f(x) = 0$ has two roots. At which interval does these roots belongs?
- A
Both the roots in $(a, b)$
- ✓
At least one root in $(a, b)$ and at least one root in $(b, c)$
- C
Both the roots in $(b, c)$
- D
Neither in $(a, b)$ nor in $(b, c)$
AnswerCorrect option: B. At least one root in $(a, b)$ and at least one root in $(b, c)$
$f(x)$ being a polynomial is continuous and differentiable for all real values of $x.$
We also have $f(a) = f(b) = f(c).$
If we apply Rolle’s theorem to $f(x)$ in $[a, b]$ and $[b, c]$ we will observe that $f(x) = 0$
will have at least one root in $(a, b)$ and at least one root in $(b, c).$
But $f(x)$ is a polynomial of degree two,
so that $f(x) = 0$
can’t have more than two roots.
It implies that exactly one root of $f(x) = 0$
will lie in $(a, b)$ and exactly one root of $f(x) = 0$ will lie in $(b, c).$
Let $y = f(x)$ be a polynomial function of degree $n.$
If $f(x) = 0$ has real roots only,
then $f(x) = 0, f(x) = 0, … , f^{n-1}(x) = 0$ will have real roots.
It is in fact the general version of above mentioned application,
because if $f(x) = 0$ have all real roots, then between two consecutive roots of $f(x) = 0,$
exactly one root of $f(x) = 0$ will lie.
View full question & answer→MCQ 1431 Mark
$ \lim_\limits{\text{x} \rightarrow 1}\frac{(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{n}})}{[(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{}})]^2},n \in N,$ equals:
AnswerCorrect option: B. $^{2\text{n}}{\text{C}}_\text{n}$
$\lim_\limits{\text{x} \rightarrow 1}\frac{(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{n}})}{[(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{}})]^2}$
$\lim_\limits{\text{x} \rightarrow 1}\frac{\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{2\text{n}})}{(1-\text{x})}}{\Big[\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{\text{n}})}{(1-\text{x})}\Big]^2}$
$=\frac{1\times2\times3\times\ldots(2\text{n})}{(1\times2\times3\ldots \text{n})^2} = \frac{(2n)!}{\text{n}!\text{n}!}={}^{2\text{n}}\text{C}_\text{n}$
Hence, option $B$ is correct.
View full question & answer→MCQ 1441 Mark
$f(x) = x − 1+ x − 3$ then $f (2) =:$
View full question & answer→MCQ 1451 Mark
Choose the correct answer. If $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is equal to:
AnswerGiven $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{-(\sin\text{x}+\cos\text{x})^{2}(\sin\text{x}+\cos\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{\sin^{2}\text{x}+\cos^{2}\text{x}+2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{-2}{(\sin\text{x}-\cos\text{x})^{2}}$
$\therefore \Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-2}{(\sin\text{0}-\cos0)^{2}}=\frac{-2}{(-1)^{2}}=2$
View full question & answer→MCQ 1461 Mark
$\lim x\rightarrow 1x−1xn−1$ is equal to
Answer$= {\lim _\limits{\text{x}\to 1}}\frac{{{\text{x}}^{\text{n}}}-1}{\text{x}-1}$
$ =\lim_\limits{\text{x}\to 1}\frac{\left( \text{x}-1 \right)\left( {{\text{x}}^{\text{n}-1}}+{{\text{x}}^{\text{n} -2}}+.....+\text{x}+1 \right)}{\text{x}-1}$
$ =\lim_\limits{\text{x}\to 1}\sum\limits_{\text{i}=0}^{\text{n}-1}{{{\text{x}}^{\text{i}}}}$
$ =\sum\limits_{\text{i}=0}^{\text{n}-1}{{{1}^{\text{i}}}}$
$ =\sum\limits_{\text{i}=0}^{\text{n}-1}{{{1}^{\text{}}}}$
$ = \text{n}$
Hence, this is the answer.
View full question & answer→MCQ 1471 Mark
$\lim\limits_{\text{x}\rightarrow1}(1+\cos\pi)\cot^2\pi\text{x}:$
- A
$1$
- B
$-1$
- ✓
$\frac{1}{2}$
- D
$0$
AnswerCorrect option: C. $\frac{1}{2}$
View full question & answer→MCQ 1481 Mark
$\lim\limits_{θ→0} \sin\text{m}^2\frac{θ}{θ}$ is equal to:
View full question & answer→MCQ 1491 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
AnswerGiven $\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{1+\tan^{2}\text{x}-2}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\tan\text{x}-1}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{(\tan\text{x}+1)(\tan\text{x}-1)}{(\tan\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}(\tan\text{x}+1)$
$=\tan\frac{\pi}{4}+1$
$=1+1$
$=2$
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Given that $f(x)$ is a differentiable function of $x$ and that $f(x) f(y) = f(x) + f(y) + f(xy) −2$ and that $f(2) = 5.$Then $f(3)$ is equal to?
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