MCQ 511 Mark
The function $\text{f(x)}=|\cos\text{x}|$ is:
- A
Differentiable at$\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
- ✓
Continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
- C
Neither differentiable nor continuous at $\text{x}=\text{n}\in\text{Z}$
- D
AnswerCorrect option: B. Continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
$\text{f(x)}=|\cos\text{x}|$
Given function is trigonometric function.
⇒ Hence, it is continuous.
Function is not differentiable at odd multiples of $\frac{\pi}{2}$
⇒ f(x) is not differentiable at $\text{x}=(2+\text{n}+1)\frac{\pi}{2}.$
View full question & answer→MCQ 521 Mark
Let $\text{f(x)=}\begin{cases}\frac{\text{x}^4-5\text{x}^2+4}{|(\text{x}-1)(\text{x-2})|},\text{x}\neq1,2\\6, \text{x}=1\\12,\text{x}=2\end{cases}$ Then f(x) is continuous on the set:
AnswerGiven:
$\text{f(x)=}\begin{cases}\frac{\text{x}^4-5\text{x}^2+4}{|(\text{x}-1)(\text{x-2})|},\text{x}\neq1,2\\6, \text{x}=1\\12,\text{x}=2\end{cases}$
Now,
$\Rightarrow\text{x}^4-5\text{x}^2+4=\text{x}^4-\text{x}^2-4\text{x}^2+4\\=\text{x}^2(\text{x}^2-1)-4(\text{x}^2-1)$
$=(\text{x}^2-1)(\text{x}^2-4)=(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)}{|(\text{x}-2)(\text{x}-1)|},&\text{x}\neq1,2\\6,& \text{x}=1\\12,&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}(\text{x}+1)(\text{x}+2),&\text{x}<1\\-(\text{x+1})(\text{x}+2),&1<\text{x}<2\$\text{x+1})(\text{x}+2),&\text{x}>2\\6,&\text{x=1}\\12,&\text{x}=2\end{cases}$
So,
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(1-\text{h}+1)(1-\text{h}+2)\\=2\times3=6$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(1+\text{h}+1)(1+\text{h}+2)\\=-2\times3=-6$
Also,
$\Rightarrow\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(2-\text{h}+1)(2-\text{h}+2)=-12$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2+\text{h}+1)(2+\text{h}+2)=12$
Thus,
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}$
$\therefore$ The only point of discontinuities of the function f(x) are x = 1 and x = 2. Hence, the given function is continuous on the set R - {1, 2}.
View full question & answer→MCQ 531 Mark
Choose the correct answers from the given four options:
The set of points where the function f given by $\text{f(x)}=|2\text{x}-1|\sin\text{x}$ is differentiable is:
AnswerCorrect option: B. $\text{R}-\Big\{\frac{1}{2}\Big\}$
We have, $\text{f(x)}=|2\text{x}-1|\sin\text{x}$
At $\text{x}=\frac{1}{2},\text{ f(x)}$ is not differentiable
Hence, f(x) is differentiable in $\text{R}-\Big\{\frac{1}{2}\Big\}$
$\because\ \text{Rf}'\Big(\frac{1}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{1}{2}+\text{h}\Big)-\text{f}\Big(\frac{1}{2}\Big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg|2\Big(\frac{1}{2}+\text{h}\Big)-1\bigg|\sin\Big(\frac{1}{2}+\text{h}\Big)-0}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{|2\text{h}|.\sin\Big(\frac{1+2\text{h}}{2}\Big)}{\text{h}}=2\sin\frac{1}{2}$
and $\text{Lf}'\Big(\frac{1}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{1}{2}-\text{h}\Big)-\text{f}\Big(\frac{1}{2}\Big)}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg|2\Big(\frac{1}{2}-\text{h}\Big)^{-1}\bigg|\sin\Big(\frac{1}{2}-\text{h}\Big)-0}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{|0-2\text{h}|-\sin\Big(\frac{1}{2}-\text{h}\Big)}{-\text{h}}=2\sin\Big(\frac{1}{2}\Big)$
$\therefore\ \text{Rf}'\Big(\frac{1}{2}\Big)\neq\text{Lf}'\Big(\frac{1}{2}\Big)$
So, f(x) is not differentiable at $\text{x}=\frac{1}{2}.$
View full question & answer→MCQ 541 Mark
Let $\text{f(x)}=\text{x}+\text{b}|\text{x}|+\text{c}|\text{x}|^4,$ where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if:
AnswerWe have,
$\text{f(x)}=\text{x}+\text{b}|\text{x}|+\text{c}|\text{x}|^4$
$\text{f(x)}=\begin{cases}\text{x}+\text{b}\text{x}+\text{c}\text{x}^4&\text{x}\geq0\\\text{x}+\text{b}\text{x}+\text{c}\text{x}^4&\text{x}<0\end{cases}$
Here, f(x) is differentiable at x = 0
$\therefore$ (LHL at x = 0) = (RHL at x = 0)
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{x}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{x}(0)}{\text{x}-0}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{a}-\text{bx}+\text{cx}^4-\text{a}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{a}+\text{bx}^4-\text{a}}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}-\text{b}(0-\text{h})+\text{c}(0-\text{h})^4-\text{a}}{0-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}(0+\text{h})+\text{c}(0+\text{h})^4-\text{a}}{0+\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}\text{h}+\text{c}\text{h}^4-\text{a}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}\text{h}+\text{c}\text{h}^4-\text{a}}{\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}\text{h}+\text{c}\text{h}^4}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}\text{h}+\text{c}\text{h}^4}{\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}(-\text{b}-\text{bh}^3)=\lim\limits_{\text{h}\rightarrow0}(\text{b}+\text{ch}^3)$
$\Rightarrow-\text{b}=\text{b}$
$\Rightarrow2\text{b}=0$
$\Rightarrow\text{b}=0$
View full question & answer→MCQ 551 Mark
Let $f(x) = |x|$ and $g(x) = |x^3|,$ then:
- ✓
$f(x)$ and $g(x)$ both are continuous $\text{at x = 0}$
- B
$f(x)$ and $g(x)$ both are differentiable $\text{at x = 0}$
- C
$f(x)$ is differentiable but $g(x)$ is not differentiable $\text{at x = 0}$
- D
$f(x)$ and $g(x)$ both are not differentiable $\text{at x = 0}$
AnswerCorrect option: A. $f(x)$ and $g(x)$ both are continuous $\text{at x = 0}$
Absolute value function is continuous on $R.$
View full question & answer→MCQ 561 Mark
The points of discontinuity of the function $\text{f(x)}=\begin{cases}\frac{1}{5}(2\text{x}^2+3),&\text{x}\leq1\\6-5\text{x},&1<\text{x}<3\\\text{x}-3,&\text{x}\geq3\end{cases}$ is (are):
Answer$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}\frac{1}{5}(2\text{x}^2+3)=1$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}6-5\text{x}=1$
Function is continuou at x = 1
$\lim\limits_{\text{x}\rightarrow3^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}6-5\text{x}=-9$
$\lim\limits_{\text{x}\rightarrow3^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}\text{x}-3=0$
Function is discontinuous at x = 3
View full question & answer→MCQ 571 Mark
If $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}=$
AnswerWe have, $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y})$
$\Rightarrow\cos(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)=\frac{1}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\cos(\text{x}+\text{y})+\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}+\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}-\cos(\text{x}+\text{y})$
$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→MCQ 581 Mark
If $\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}},$ then $(2\text{xy}_1+\text{y})\text{y}_3=$
- ✓
$3(xy_2 + y_1) y_2$
- B
$3(xy_1 + y_2) y_2$
- C
$3(xy_1 + y_2) y_1$
- D
AnswerCorrect option: A. $3(xy_2 + y_1) y_2$
$\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}}$
$\Rightarrow(\text{x}^2+\text{c})\text{y}=\text{ax}+\text{b}$
Differentiating $\text{w.r.t.x},$ we get
$2\text{xy}+(\text{x}^2+\text{c})\frac{\text{dy}}{\text{dx}}=\text{a}$
Differentiating $\text{w.r.t.x},$ we get
$2\text{y}+2\text{xy}_1+2\text{xy}+(\text{x}^2+\text{c})\text{y}_2=0$
$\Rightarrow2\text{y}+4\text{xy}_1+\text{x}^2+\text{cy}_2=0$
Differentiating $\text{w.r.t.x},$ we get
$2\text{y}_1+4\text{y}_1+4\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3+2\text{xy}_2=0$
$\Rightarrow6\text{y}_1+6\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3=0$
$\Rightarrow6\text{y}_1+6\text{xy}_2+\Big(\frac{-2\text{y}-4\text{xy}_1}{\text{y}_2}\Big)\text{y}_3=0$
$[\because2\text{y}+4\text{xy}_1+(\text{x}^2+\text{c})\text{y}_2=0]$
$\Rightarrow6\text{y}_1\text{y}_2+6\text{x}(\text{y}_2)^2-2\text{y}-4\text{xy}_1\text{y}_3=0$
$\Rightarrow3\text{y}_1\text{y}_2+3\text{x}(\text{y}_2)^2-\text{y}-2\text{xy}_1\text{y}_3=0$
$\Rightarrow(\text{y}_1+\text{xy}_2)3\text{y}_2=(2\text{xy}_1+\text{y})\text{y}_3$
View full question & answer→MCQ 591 Mark
$\frac{\text{d}^{20}}{\text{dx}^{20}}(2\cos\text{x}\cos3\text{x})=$
- A
$2^{20}(\cos2\text{x}-2^{20}\cos4\text{x})$
- ✓
$2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$
- C
$2^{20}(\sin2\text{x}+2^{20}\sin4^\text{x})$
- D
$2^{20}(\sin2\text{x}-2^{20}\sin4^\text{x})$
AnswerCorrect option: B. $2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$
$\text{y}=2\cos\text{x}\cos3\text{x}=\cos(3\text{x}-\text{x})+\cos(3\text{x}+\text{x})=\cos2\text{x}+\cos4\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\sin2\text{x}-4\sin4\text{x}=-2(\sin2\text{x}+2\sin4\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\cos2\text{x}-16\cos4\text{x}=-2^2(\cos2\text{x}+2^2\cos4\text{x})$
$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}=2^3(\sin2\text{x}+2^3\sin4\text{x})$
$\Rightarrow\frac{\text{d}^4\text{y}}{\text{dx}^4}=2^3(2\cos2\text{x}+4\times2^3\cos4\text{x})=2^4(\cos2\text{x}+2^4\cos4\text{x})$
$\therefore\frac{\text{d}^{20}(\cos2\text{x}+\cos4\text{x})}{\text{dx}^{20}}=2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$
View full question & answer→MCQ 601 Mark
The value of $c$ in Rolle's theorem when $f(x) = 2x^3 - 5x^2 - 4x + 3, \text{x}\in\Big[\frac{1}{3},3\Big]$ is:
- ✓
$2$
- B
$-\frac{1}{3}$
- C
$-2$
- D
$\frac{2}{3}$
Answer$f(x) = 2x^3 - 5x^2 - 4x + 3$
Differentiating the given function with respect to $x,$ we get
$f'(x) = 6x^2 - 10x - 4$
$\Rightarrow f'(c) = 6c^2 - 10c - 4$
$\therefore f'(c) =0$
$\Rightarrow 3c^2 - 5c - 2 = 0$
$\Rightarrow 3c^2 - 6c + c - 2 = 0$
$\Rightarrow 3c(c - 2) + c - 2 = 0$
$\Rightarrow (3c + 1)(c - 2) = 0$
$\Rightarrow\text{c}=2, \frac{-1}{3}$
$\therefore\ \text{c}=2\in\Big(\frac{1}{3},3\Big)$
Thus, $\text{c}=2\in\Big(\frac{1}{3},3\Big)$ for which Rolle's theorem holds.
Hence, the required value of $c$ is $2.$
View full question & answer→MCQ 611 Mark
Let $\text{y}=\sqrt{\sin\text{x}+\text{y}},$ then $\frac{\text{dy}}{\text{dx}}=$
- A
$\frac{\sin\text{x}}{2\text{y}-1}$
- B
$\frac{\sin\text{x}}{1-2\text{y}}$
- C
$\frac{\cos\text{x}}{1-2\text{y}}$
- ✓
$\frac{\cos\text{x}}{2\text{y}-1}$
AnswerCorrect option: D. $\frac{\cos\text{x}}{2\text{y}-1}$
$\text{y}=\sqrt{\sin\text{x}+\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\times\Big(\cos\text{x}+\frac{\text{dy}}{\text{dx}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}+\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\Big(\frac{2\text{y}-1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}-1}$
View full question & answer→MCQ 621 Mark
If $\text{y}=(\sin^{-1}\text{x})^2,$ then $(1-\text{x}^2)\text{y}_2$ is equal to:
- ✓
$xy_1 + 2$
- B
$xy_1 - 2$
- C
$-xy_1 + 2$
- D
AnswerCorrect option: A. $xy_1 + 2$
Here,
$\text{y}=(\sin^{-1}\text{x})\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)^\frac{3}{2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{\text{xy}_1}{(1-\text{x}^2)}$
$\Rightarrow\text{y}_2(1-\text{x}^2)=2+\text{xy}_1$
View full question & answer→MCQ 631 Mark
Write the number of points where $f(x) = |x + 2| + |x - 3|$ is not differentiable:
View full question & answer→MCQ 641 Mark
The function $\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}$
- A
Discontinuous at only one point.
- B
Discontinuous exactly at two points.
- ✓
Discontinuous exactly at three points.
- D
AnswerCorrect option: C. Discontinuous exactly at three points.
Given,
$\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}$
$\Rightarrow\text{f(x)}=\frac{4-\text{x}^2}{\text{x}(4-\text{x}^2)}$
$\Rightarrow\text{f(x)}=\frac{1}{\text{x}},\text{x}\neq0 $ and $4-\text{x}^2\neq0 $ or $ \text{x}\neq0,\pm2$
Clearly, f(x) is defined and continuous for all real numbers except $\left\{0,\pm2\right\}$
Therefore, f(x) is discontinuous exactly at three points.
View full question & answer→MCQ 651 Mark
If $\text{f}(\text{x})=\sqrt{\text{x}^2+6\text{x}+9},$ then f'(x) is equal to:
AnswerWe have, $\text{f}(\text{x})=\sqrt{\text{x}^2+6\text{x}+9}$
$=\sqrt{(\text{x}+3)^2}$
$=|\text{x}+3|$
$\text{f}(\text{x})=\begin{cases}\text{x}+ 3,\text{x}\geq-3\\ -\text{x}-3,\text{x}<-3\end{cases}$
$\Rightarrow\text{f}'(\text{x})=\begin{cases} 1,\text{x}\geq-3\\ -1,\text{x}<-3\end{cases}$
$\therefore$ f'(x) = -1 for x < -3
View full question & answer→MCQ 661 Mark
If $\text{x}=\text{t}^2,\text{y}=\text{t}^3$ Then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$
- A
$\frac{3}{2}$
- B
$\frac{3}{4\text{t}}$
- C
$\frac{3}{2\text{t}}$
- ✓
$\frac{3\text{t}}{2}$
AnswerCorrect option: D. $\frac{3\text{t}}{2}$
$\text{x}=\text{t}^2\Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}$
$\text{y}=\text{t}^3\Rightarrow\frac{\text{dy}}{\text{dt}}=3\text{t}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}^2}{2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}}{2}$
View full question & answer→MCQ 671 Mark
If $\text{f(x)}=\begin{cases}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$ then f(x) is:
- A
Continuous as well as differentiable at x = 0
- B
Continuous but not differentiable at x = 0
- C
Differentiable but not continuous at x = 0
- ✓
Answer$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}}=\lim\limits _{\text{x}\rightarrow0}\frac{1}{1+\text{e}^{\frac{-1}{\text{x}}}}=1\ \Big(\because\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-1}{\text{x}}}=0\Big)$
$\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}\neq\text{f}(0)$
Function is not continuous,
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(-\text{h})-\text{f}(0)}{-\text{h}}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{e}^{\frac{1}{\text{h}}}}-0}{-\text{h}}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{-\Big(1+\text{e}^{\frac{1}{\text{h}}}\Big)\text{h}}=-\infty$
Similarly,
$\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\infty$
View full question & answer→MCQ 681 Mark
If the fucnction $\text{f(x)}=\begin{cases}(\cos\text{x})^{\frac{1}{\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuouse at x = 0, then the value of k is:
Answer$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\text{x}}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(1+\cos \text{x}-1)^{\frac{1}{\text{x}}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\Big(1-2\sin^2\frac{\text{x}}{2}\Big)^\frac{1}{\text{x}}$
$\text{f(0)}=\lim\limits\Big(1-2\sin^2\frac{\text{x}}{1}\Big)^{\frac{1}{-2\sin^2\frac{\text{x}}{2}}\times\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$
$\text{f(0)}= \lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$
$\text{f(0)}=\lim \limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text{x}}{2}}{\text{x}}\times\sin\frac{\text{x}}{2}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text {x}}{2}}{\frac{\text{x}}{2}}\times\frac{1}{2}\sin\frac{\text{x}}{2}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{1\times\sin\frac{1}{2}}=\text{e}^0=1$
$\Rightarrow\text{k}=1$
View full question & answer→MCQ 691 Mark
The function $f(x) = e|x|$ is:
AnswerCorrect option: A. Continuous everywhere but not differentiable at $x = 0$
View full question & answer→MCQ 701 Mark
If $f(x) = |x^2 - 9x + 20|$, then $f(x)$ is equal to:
AnswerCorrect option: C. $-2x + 9,$ if $4 < x < 5$
We have$, f(x) = |x^2 - 9x + 20|$
$\text{f}(\text{x})=\begin{Bmatrix} \text{x}^2-9\text{x}+20, & -\infty<\text{x}\leq4 \\ -\big(\text{x}^2-9\text{x}+20\big), & 4<\text{x}<5 \\ \text{x}^2-9\text{x}+20, & 5\leq\text{x}<\infty \end{Bmatrix}$
$\Rightarrow\text{f}\ '(\text{x})=\begin{Bmatrix} 2\text{x}-9\text{x}, & -\infty<\text{x}\leq4 \\ 2\text{x}-9, & 4<\text{x}<5 \\ 2\text{x}-9, & 5\leq\text{x}<\infty \end{Bmatrix}$
$\therefore f\ '(x) = -2x + 9$ for $4 < x < 5$
View full question & answer→MCQ 711 Mark
The value of b for which the function $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$ is continuous at every point of its domain, is:
AnswerGiven, f(x) is continuous at every point of its domain. So, it is continuous at x = 1.
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f}\text{(x)}=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Big(4(1+\text{h})^2+3\text{b}(1+\text{h})\Big)=5(1)-4$
$\Rightarrow4+3\text{b}=1$
$\Rightarrow-3=3\text{b}$
$\Rightarrow \text{b} = -1$
View full question & answer→MCQ 721 Mark
If $\text{f(x)}=\begin{cases}\frac{36^\text{x}-9^\text{x}-4\text{x}+1}{\sqrt{2}-\sqrt{1+\cos\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at $x = 0,$ these $k$ equals.
AnswerCorrect option: C. $16\sqrt{2}$ in $6$ in $3$
$\text{k}=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{ \text{x}}+1}{\sqrt2-\sqrt{1+\cos\text{x}}}$
consider,
$=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{\sqrt{2}-\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(4\times9)^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(4^{\text{x}}\times9^{\text{x}})-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\cos\text{x})}{1-\cos^2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\cos\text{x})}{\sin^2\text{x}}$
dividing by $x^2$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{9^{\text{x}}-1}{\text{x}}{\times\frac{4^{\text{x}}-1}{\text{x}}\times\big(\sqrt{2}+\sqrt{1+\cos\text{x}}\big)(1+\cos\text{x})}}{\frac{\sin^2\text{x}}{\text{x}^2}}$
$=(\log9)(\log4)\big(\sqrt{2}+\sqrt{1+1}\big)(1+1)$
$=4\sqrt{2}(\log9)(\log4)$
$=4\sqrt{2}(2\log3)(2\log2)$
$=16\sqrt{2}(\log3)(\log2)$
View full question & answer→MCQ 731 Mark
If f(x) = |x - 3| and g(x) = fof(x), then for x > 10, g'(x) is equal to:
AnswerFor, x > 10
f(x) = |x - 3| = x - 3
g(x) = fof (x) = |x - 3| - 3
= x - 3 - 3
= x - 6
$\therefore$ g'(x) = 1
View full question & answer→MCQ 741 Mark
If $\text{f(x)}=\begin{cases}\frac{1-\cos10\text{x}}{\text{x}^2},&\text{x}<0\\\text{a},&\text{x}=0\\\frac{\sqrt{\text{x}}}{\sqrt{625+\sqrt{\text{x}}}-25},&\text{x}>0\end{cases}$ then the value of so that f(x) may be continuous at x = 0 is:
AnswerIf f(x) is continuous at x = 0, then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(-10\text{h}))}{(-\text{h})^2}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(10\text{h}))}{\text{h}^2}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(2\sin^2(5\text{h}))}{\text{h}^2}=\text{a}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{2\times25(\sin^2(5\text{h}))}{25\text{h}^2}=\text{a}$
$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\frac{(\sin^2(5\text{h}))}{(5\text{h})^2}=\text{a}$
$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(5\text{h})}{5\text{h}}\Big)^2=\text{a}$
$\Rightarrow\text{a}=50$
View full question & answer→MCQ 751 Mark
If the function f(x) defined by $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, then k =
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f(0)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}}\Big)=\text{k}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3\log(1+3\text{x})}{3\text{x}}-\frac{2\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)-2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)+2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{-2\text{x}}\Big)=\text{k}$
$\Rightarrow3\times1+2\times1=\text{k}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$
$\Rightarrow\text{k}=3+2$
$\Rightarrow\text{k}=5$
View full question & answer→MCQ 761 Mark
If $\text{y}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x},$ then $\frac{\text{dy}}{\text{dx}}=$
- ✓
$\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}$
- B
$\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\log\Big(1+\frac{1}{\text{x}}\Big)$
- C
$\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big\{\log(\text{x}+1)-\frac{\text{x}}{\text{x}+1}\Big\}$
- D
$\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big\{\log\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}\Big\}$
AnswerCorrect option: A. $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}$
Let $\text{y}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x}$
Taking log on both sides,
$\log\text{y}=\text{x}\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\bigg(\frac{1}{1+\frac{1}{\text{x}}}\bigg)\frac{\text{d}}{\text{dx}}\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\times\frac{\text{x}}{\text{x}+1}\Big(-\frac{1}{\text{x}^2}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{\text{x}+1}\times-\frac{1}{\text{x}^2}+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}+1}+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{-1}{\text{x}+1}+\log\Big(1+\frac{1}{\text{x}}\Big)\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)-\frac{-1}{\text{x}+1}\Big]$
View full question & answer→MCQ 771 Mark
$\frac{\text{d}}{\text{dx}}\bigg[\log\bigg\{\text{e}^\text{x}\Big(\frac{\text{x}-2}{\text{x}+2}\Big)^\frac{3}{4}\bigg\}\bigg]$ equals:
- ✓
$\frac{\text{x}^2-1}{\text{x}^2-4}$
- B
$1$
- C
$\frac{\text{x}^2+1}{\text{x}^2-4}$
- D
$\text{e}^\text{x}\frac{\text{x}^2-1}{\text{x}^2-4}$
AnswerCorrect option: A. $\frac{\text{x}^2-1}{\text{x}^2-4}$
Let, $\text{y}=\frac{\text{d}}{\text{dx}}\bigg[\log\bigg\{\text{e}^\text{x}\Big(\frac{\text{x}-2}{\text{x}+2}\Big)^\frac{3}{4}\bigg\}\bigg]$
$\Rightarrow\text{y}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\log\text{e}+\frac{3}{4}\log\Big(\frac{\text{x}-2}{\text{x}+2}\Big)\Big]$
$\Rightarrow\text{y}=\frac{\text{d}}{\text{dx}}\Big[\text{x}+\frac{3}{4}\log\Big(\frac{\text{x}-2}{\text{x}+2}\Big)\Big]$
$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3}{4\Big(\frac{\text{x}-2}{\text{x}+2}\Big)}\times\frac{(\text{x}+2)\times1-(\text{x}-2)\times1}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3(\text{x}+2)}{4(\text{x}-2)}\times\frac{\text{x}+2-\text{x}+2}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3(\text{x}+2)}{4(\text{x}-2)}\times\frac{4}{(\text{x}+2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\frac{3}{(\text{x}^2-4)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-4+3}{\text{x}^2-4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-1}{\text{x}^2-4}$
View full question & answer→MCQ 781 Mark
If $\text{f(x)}=|\log_{10}\text{x}|\text{fx}=\log_{10}\text{x},$ then at $x = 1:$
- $f(x)$ is continuous and $\text{f}\ '(1^+)=\log_{10}\text{e}$
- $f(x)$ is continuous and $\text{f}\ '(1^+)=\log_{10}\text{e}$
- $f(x)$ is continuous and $\text{f}\ '(1^-)=-\log_{10}\text{e}$
- $f(x)$ is continuous and $\text{f}\ '(1^-)=-\log_{10}\text{e}$
- A
$a$ and $b$
- B
$a$ and $c$
- C
$b$ and $c$
- ✓
$a$ and $d$
AnswerCorrect option: D. $a$ and $d$
Given,
$\text{f(x)}=|\log_{10}\text{x}|=\bigg|\frac{\log_{\text{e}}\text{x}}{\log_{\text{e}}10}\bigg|$
$=|(\log_{\text{e}}\text{x})\times(\log_{10}\text{e})|$
$=(\log_{10}\text{e})|\log_{10}\text{x}|$
$\text{f}\ '(1^+)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{{10}}\text{e})|\log_{\text{e}}(1+\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1|}{\text{h}}$
$=(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$
$=(\log_{10}\text{e})\times1$
$=(\log_{10}\text{e})$
Also,
$f\ '(1^-)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{10}\text{e})|\log_{\text{e}}(1-\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1}{\text{h}}$
$=-(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$
$=(\log_{10}\text{e})\times1=-(\log_{10}\text{e})$
View full question & answer→MCQ 791 Mark
If $\text{f(x)}=\text{x}\sin\frac{1}{\text{x}},\text{ x}\neq0,$ then the value of the function at x = 0, so that the function is continuous at x = 0, is:
Answer$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\times\sin\frac{1}{\text{x}}$
$-1\leq\sin\frac{1}{\text{x}}\leq1$
$\text{x}\times(-1)\leq\text{x}\sin\frac{1}{\text{x}}\leq\text{x}$
$\lim\limits_{\text{x}\rightarrow0}-\text{x}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\frac{1}{\text{x}}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}$
$\text{f}(0)=0$
View full question & answer→MCQ 801 Mark
The values of the constants a, b and for which the function $\text{f(x)}=\begin{cases}(1+\text{ax})^{\frac{1}{\text{x}}},&\text{x}>0\\\text{b},&\text{x}=0\\\frac{(\text{x}+\text{c})^{\frac{1}{2}}-1}{(\text{x}+1)^{\frac{1}{2}}-1},&\text{x}>0\end{cases}$ may be continuous at x = 0, are:
- A
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=-\frac{2}{3},\text{ c}=1$
- B
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\frac{2}{3},\text{ c}=-1$
- ✓
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$
- D
AnswerCorrect option: C. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}(1+\text{ax})^{\frac{1}{\text{x}}}$
$\text{b}=\lim\limits_{\text{x}\rightarrow{\text{a}}}(1+\text{ax})^{\frac{1}{\text{ax}}\times\text{a}}$
$\text{b}=\text{e}^{\text{a}}$
$\text{a}=\log_{\text{e}}\text{b}$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow\text{a}^+}\frac{(\text{x}+\text{c})^{\frac{1}{3}}-1}{(\text{x}+1)^{\frac{1}{2}}-1}$
Here, $\text{c}=1$
$\text{x}+1=\text{y}$
$\text{x}\rightarrow0\Rightarrow\text{y}\rightarrow1$
$\text{f}(0)=\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}^{\frac{1}{2}}-1}$
$\text{b}=\lim\limits_{\text{y}\rightarrow1}\frac{\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}-1}}{\frac{\text{y}^{\frac{1}{2}}-1}{\text{y}-1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$
$\text{a}=\log\text{b}=\log\frac{2}{3}$
View full question & answer→MCQ 811 Mark
If $\text{y}^\frac{1}{\text{n}}+\text{y}-^\frac{1}{\text{n}}=2\text{x}$ then find $(\text{x}^2-1)\text{y}_2+\text{xy}_1=$
AnswerCorrect option: A. $-n^2y$
$\text{y}^\frac{1}{\text{n}}+\text{y}-^\frac{1}{\text{n}}=2\text{x}$
Differentiating both sides we get
$\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}-1}-\text{y}^{\frac{1}{\text{n}}-1}\Big)=2$
$\Rightarrow\text{y}_1\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)=2\text{ny}$
Again differentiating both sides we get
$\text{y}_2\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)+\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}-1}\Big)=2\text{ny}_1$
$\Rightarrow\text{ny}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+\frac{\text{y}^2_1}{\text{y}}\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)=2\text{n}^2\text{y}_1$
$\Rightarrow\text{nyy}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$
$\Rightarrow\text{nyy}_2\frac{2\text{ny}}{\text{y}_1}+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$
$\Rightarrow\frac{\text{n}^2\text{y}^2\text{y}_2}{\text{y}_1^2}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{4\text{x}^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow(\text{x}^2-1)\text{y}_2+\text{xy}_1=\text{n}^2\text{y}$
View full question & answer→MCQ 821 Mark
If $\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$ then the value of a, 0 < r < is equals to:
- A
$\frac{\text{n}!}{\text{r}!}$
- B
$\frac{(\text{n}-\text{r})!}{\text{r}!}$
- ✓
$\frac{\text{n}!}{(\text{n}-\text{r})!}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\frac{\text{n}!}{(\text{n}-\text{r})!}$
$\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$
$\Rightarrow\text{e}^\text{x}(\text{nx}^{\text{n-1}}-\text{a}_1(\text{n}-1)\text{x}^{\text{n-2}}+\text{a}_2(\text{n}-2)\text{x}^{\text{n}-3}+...+(-1)^{\text{n}-1}\text{a}_{\text{n}-1}+\text{x}^\text{a}-\text{a}_1\text{x}^{\text{n-2}}+...+(-1)^\text{n}\text{a}_\text{n})=\text{x}^\text{n}\text{e}^\text{x}$
$\Rightarrow\text{e}^\text{x}(\text{x}^\text{n}+(\text{n}-\text{a}_1)\text{x}^{\text{n}-1}-(\text{a}_1(\text{n-1})-\text{a}_2)\text{x}^{\text{n}-2}\\+(\text{a}_2(\text{n}-2)-\text{a}_3)\text{x}^{\text{n}-3}-...)=\text{x}^\text{n}\text{e}^\text{x}$
on comparing both sides we get
$\text{n}-\text{a}_1=0$
$\Rightarrow\text{a}_1=\text{n}$
$\text{a}_1(\text{n}-1)-\text{a}_2=0$
$\Rightarrow\text{a}_2=\text{a}_1(\text{n}-1)=\text{n}(\text{n}-1)$
$\text{a}_2(\text{n}-2)-\text{a}_3=0$
$\Rightarrow\text{a}_3=\text{a}_2(\text{n}-2)=\text{n}(\text{n}-1)(\text{n}-2)$
So,
$\text{a}_\text{r}=\text{n}(\text{n-1})(\text{n}-2)...(\text{n}-(\text{r}-1)=\frac{\text{n}!}{(\text{n}-\text{r})!}$
View full question & answer→MCQ 831 Mark
If $\text{f}(\text{x})=\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}}^2},$ then $(1-\text{x})^2\text{f}''(\text{x})-\text{xf}(\text{x})=$
AnswerHere,
$\text{f}(\text{x})=\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}}^2}$
$\Rightarrow\sqrt{1-\text{x}}^2\text{f}(\text{x})=\sin^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\sqrt{1-\text{x}^2}\text{f}'(\text{x})-\frac{\text{x f}{(\text{x})}}{\sqrt{1-\text{x}}^2}=\frac{1}{\sqrt{1-\text{x}}^2}$
$\Rightarrow(1-\text{x}^2)\text{f}'(\text{x})-\text{xf}(\text{x})=1$
View full question & answer→MCQ 841 Mark
The function $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
- A
- ✓
is not continuous at x = 0
- C
is not continuous at x = 0, but can be made continuous at x = 0
- D
AnswerCorrect option: B. is not continuous at x = 0
Given, $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},\text{x}\neq0\\0,\text{x}=0\end{cases}$
We have
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1}\Bigg)$
if $\text{e}^\frac{1}{\text{x}}=\text{t},$ then
$\text{x}\rightarrow0, \text{t}\rightarrow\infty$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{t}\rightarrow\infty}\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$
$=\lim\limits_{\text{t}\rightarrow\infty}\Bigg(\frac{1-\frac{1}{\text{t}}}{1+\frac{1}{\text{t}}}\Bigg)=\frac{1-0}{1+0}=1$
Also, f(0) = 0
$\because\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}\neq\text{f}(0)$
Hence, f(x) is discontinuons at x = 0.
View full question & answer→MCQ 851 Mark
Let $\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$ then, f is:
- A
- ✓
- C
- D
Everywhere diffrentiable.
Answer$\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$
$\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\text{x}+1}{\text{x}+1}=0$
Similarly,
$\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{x}+1}{\text{x}+1}=0$
Function is diffrentiable at x = -1.
View full question & answer→MCQ 861 Mark
If $\sin\text{y}=\text{x}\cos(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
- ✓
$\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
- B
$\frac{\cos\text{a}}{\cos^2(\text{a}+\text{y})}$
- C
$\frac{\sin^2\text{y}}{\cos\text{a}}$
- D
$\text{None of these.}$
AnswerCorrect option: A. $\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
We have, $\sin\text{y}=\text{x}\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})\big]$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=1\times\cos(\text{a}+\text{y})-\text{x}\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})-\text{x}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\big[\cos\text{y}+\text{x}\sin(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\Big[\cos\text{y}+\frac{\sin\text{y}}{\cos(\text{a}+\text{y})}\times\sin(\text{a}+\text{y})\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\begin{bmatrix}\because\sin\text{y}=\text{x}\cos(\text{a}+\text{y}) \\ \because\text{x}=\frac{\sin\text{y}}{\cos(\text{a}+\text{y})} \end{bmatrix}$
$\Rightarrow\Big[\frac{\cos(\text{a}+\text{y})\cos\text{y}+\sin\text{y}\sin(\text{a}+\text{y})}{\cos(\text{a}+\text{y})}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\cos(\text{a}+\text{y}-\text{y})}{\cos(\text{a}+\text{y})}\times\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
View full question & answer→MCQ 871 Mark
If $\text{y}=\log\sqrt{\tan\text{x}},$ then the value of $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=\frac{\pi}{4}$ is givne by:
- A
$\infty$
- ✓
$1$
- C
$0$
- D
$\frac{1}{2}$
AnswerWe have, $\text{y}=\log\sqrt{\tan\text{x}}$$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\tan\text{x}}}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\tan\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\tan\text{x}}}\times\frac{1}{2\sqrt{\tan\text{x}}}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\tan\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2\text{x}}{2\tan\text{x}}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\Big[\sec\big(\frac{\pi}{4}\big)\Big]^2}{2\tan\big(\frac{\pi}{4}\big)}=\frac{2}{2\times1}=1$
View full question & answer→MCQ 881 Mark
Choose the correct answers from the given four options:The function $\text{f(x)}=\text{e}^{|\text{x}|}$ is:
AnswerCorrect option: A. Continuous everywhere but not differentiable at $x = 0.$
Let $\text{u(x)}=|\text{x}|$ and $\text{v(x)}=\text{e}^\text{x}$
$\therefore\ \text{f(x)}=\text{vou(x)}=\text{v}[\text{u(x)]}$
$=\text{v}|\text{x}|=\text{e}^{|\text{x}|}$
Since$, u(x)$ and $v(x)$ are both continuous functions.
So$, f(x)$ is also continuous function but $u(x) = |x|$ is not differentiable at $x = 0,$
where as $v(x) = e^x$ is differentiable at everywhere.
Hence$, f(x)$ is continuous everywhere but not differentiable at $x = 0.$
View full question & answer→MCQ 891 Mark
If $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}^3(\text{x}^3-\text{y})^3,$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
- ✓
$\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
- B
$\frac{\text{y}^2}{\text{x}^2}\sqrt{\frac{1-\text{y}^6}{1+\text{x}^6}}$
- C
$\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{x}^6}{1-\text{y}^6}}$
- D
$\text{None of these.}$
AnswerCorrect option: A. $\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
We have, $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}(\text{x}^3-\text{y}^3)$Putting $\text{x}^3=\sin\text{A}\text{ and y}^3=\sin\text{B}$
$\Rightarrow\sqrt{1-\sin^2\text{A}}+\sqrt{1-\sin2\text{B}}=\text{a}(\sin\text{A}-\sin\text{B})$
$\Rightarrow\cos\text{A}+\cos\text{B}=\text{a}(\sin\text{A}-\sin\text{B})$
$\Rightarrow2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{b}}{2}\Big) \\ =2\text{a}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)$
$\Rightarrow\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{a}$
$\Rightarrow\frac{\text{A}+\text{B}}{2}=\cot^{-1}(\text{a})$
$\Rightarrow\text{A}+\text{B}=2\cot^{-1}(\text{a})$
$\Rightarrow\sin^{-1}\text{x}^{3}-\sin^{-1}\text{y}^3=2\cot^{-1}(\text{a})$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}\times\frac{\text{d}}{\text{dx}}(\text{x}^3)-\frac{1}{\sqrt{1-\text{y}^6}}\times\frac{\text{d}}{\text{dx}}(\text{y}^3)=0$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}3\text{x}^2-\frac{1}{\sqrt{1-\text{y}^6}}\times3\text{y}^2\times\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
View full question & answer→MCQ 901 Mark
The function $\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$
AnswerCorrect option: C. Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0
Given function is
$\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$
Consider,
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^-}\text{f(x)}=\frac{1}{\text{n}^2}$
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}}1=1$
Hence, fnuction is discontinuous at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0
View full question & answer→MCQ 911 Mark
The function $\text{f(x)}=\sin^{-1}(\cos\text{x})$ is:
Answer$\text{f(x)}=\sin^{-1}(\cos\text{x})$
$\text{f(x)}=\sin^{-1}\Big[\sin\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$
$\text{f(x)}=\frac{\pi}{2}-\text{x}$
Function is continuous at x = 0.
View full question & answer→MCQ 921 Mark
If $\text{x}=2\text{ at},\text{y}=\text{at}^2,$ where a is a constant, then $\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{ at}\ \text{x}=\frac{1}{2}$ is:
- ✓
$\frac{1}{2}\text{a}$
- B
- C
- D
AnswerCorrect option: A. $\frac{1}{2}\text{a}$
Here,
$\text{x}=2\text{ at},\text{y}=\text{at}^2,$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{dt}}=2\text{a}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=2\text{at}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2\text{at}}{2\text{a}}=\text{t}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=1\times\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{a}}$
Now, $\Big[\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big]_{\text{x}=\frac{1}{2}}=\frac{1}{2\text{a}}$
View full question & answer→MCQ 931 Mark
If $\sin^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\log\text{a}$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
- A
$\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}$
- ✓
$\frac{\text{y}}{\text{x}}$
- C
$\frac{\text{x}}{\text{y}}$
- D
$\text{None of these.}$
AnswerCorrect option: B. $\frac{\text{y}}{\text{x}}$
$\sin^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\log\text{a}$$\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\sin(\log\text{a})=\text{k}$
$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$
$\text{x}^2-\text{y}^2=\text{kx}^2+\text{ky}^2$
$\text{x}^2-\text{kx}^2=\text{ky}^2+\text{y}^2$
$(1-\text{k})\text{x}^2=(\text{x}+1)\text{y}^2$
$\frac{1-\text{x}}{\text{k}+1}=\frac{\text{y}^2}{\text{x}^2}\ .....(\text{i})$
Consider,
$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$
Differentaiting with resepect to x,
$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{kx}+2\text{ky}\frac{\text{dy}}{\text{dx}}$
$2\text{x}-2\text{kx}=2\text{ky}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}$
$2\text{x}(1-\text{k})=2\text{y}(\text{k}+1)\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(1-\text{k})}{\text{y}(\text{k}+1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\times\frac{\text{y}^2}{\text{x}^2}\ .....(\because\text{from(i)})$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
View full question & answer→MCQ 941 Mark
If $x = t^2, y = t^3$, then $\frac{\text{d}^2\text{y}}{\text{dx}^2} =$
- A
$\frac{3}{2}$
- ✓
$\frac{3}{4\text{t}}$
- C
$\frac{3}{2\text{t}}$
- D
$\frac{3\text{t}}{2}$
AnswerCorrect option: B. $\frac{3}{4\text{t}}$
$\text{x}=\text{t}^2\Rightarrow\frac{\text{dx}}{\text{dt}} = 2\text{t}$
$\text{y}=\text{t}^3\Rightarrow\frac{\text{dy}}{\text{dt}} = 3\text{t}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{3\text{t}^2}{2\text{t}} = \frac{3\text{t}}{2}$
Hence, $\frac{\text{d}^2\text{y}}{\text{dx}^2} = \frac{3}{4\text{t}}$
View full question & answer→MCQ 951 Mark
If $\text{f}(\text{x})=\Big(\frac{\text{x}^\text{l}}{\text{x}^\text{m}}\Big)^{\text{l}+\text{m}}\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\text{m}+\text{n}}\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\text{n}+1},$ the $f\ '(x)$ is equal to:
AnswerWe have $\text{f}(\text{x})=\Big(\frac{\text{x}^\text{l}}{\text{x}^\text{m}}\Big)^{\text{l}+\text{m}}\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\text{m}+\text{n}}\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\text{n}+1}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}-\text{m})(\text{l}+\text{m})}\times\text{x}^{(\text{m}-\text{n})(\text{m}+\text{n})}\times\text{x}^{(\text{n}-\text{l})(\text{n}-\text{l})}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{\text{l}^2-\text{m}^2}\times\text{x}^{\text{m}^2-\text{n}^2}\times\text{x}^{\text{n}^2-\text{l}^2}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}^2-\text{m}^2+\text{m}^2-\text{n}^2+\text{n}^2-\text{l}^2)}$
$\Rightarrow\text{f}(\text{x})=\text{x}^0$
$\Rightarrow\text{f}(\text{x})=1$
$\Rightarrow\text{f}\ '(\text{x})=0$
View full question & answer→MCQ 961 Mark
If $\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})(\cos3\text{x}+\text{i}\sin3\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})\ \text{and}\ \text{f}(1)=1, $ then f1 is equals to:
- A
$$$\frac{\text{n}(\text{n}+1)}{2}$
- B
$\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
- ✓
$-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
- D
$\text{none of these}$
AnswerCorrect option: C. $-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
$\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos\text{x}+\text{i}\sin\text{x})^2...(\cos\text{x}+\text{i}\sin\text{x})^\text{n}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{1+2+3........\text{n}}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{\frac{\text{n}(\text{n}+1)}{2}}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^\text{a}$
$\Rightarrow\text{f}(\text{ax})=(\cos\text{ax}+\text{i}\sin\text{ax})...1$
$\Rightarrow\text{f}(1)=(\cos\text{a}+\text{i}\sin\text{a})$
$\Rightarrow1=(\cos\text{a}+\text{i}\sin\text{a})...2\ [\because\text{f}(1)=1]$
Differentiating eqn.1, we get
$\text{f}'(\text{x})=\text{a}(-\sin\text{ax}+\text{i}\cos\text{ax})$
$\Rightarrow\text{f}''(\text{x})=\text-{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$
$\Rightarrow\text{f}''\text{x}=\text{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$
$\Rightarrow\text{f}''(\text{x})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{ax}+\text{i}\sin\text{ax})$
$\Rightarrow\text{f}''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{a}+\text{i}\sin\text{a})$
$\Rightarrow\text{f}''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}\ [\text{using }2]$
View full question & answer→MCQ 971 Mark
If $\text{x}=\text{a}\cos\ \text{nt}-\text{b}\sin\ \text{nt}$ then $\frac{\text{d}^2\text{x}}{\text{dt}^2}$ is:
- A
$n^2x$
- ✓
$-n^2x$
- C
$-nx$
- D
$nx$
AnswerCorrect option: B. $-n^2x$
Here
$\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
Differentiating $\text{w.r.t.t},$ we get
$\frac{\text{dx}}{\text{dt}}=-\text{an}\sin\text{nt}-\text{bn}\cos\text{nt}$
Differentiating $\text{w.r.t.t},$ we get
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{an}^2\cos\text{nt}+\text{bn}^2\sin\text{nt}$
$=-\text{n}^2\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
$=-\text{n}^2\text{x}$
View full question & answer→MCQ 981 Mark
If from Lagrange's mean value theorem, we have
$\text{f}\ '(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then:
- A
$\text{a}<\text{x}_1\leq\text{b}$
- B
$\text{a}\leq\text{x}_1<\text{b}$
- ✓
$\text{a}<\text{x}_1<\text{b}$
- D
$\text{a}\leq\text{x}_1\leq\text{b}$
AnswerCorrect option: C. $\text{a}<\text{x}_1<\text{b}$
We have
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$
In the Lagrange's mean value theorem, $\text{c}\in(\text{a},\text{b})$ such that $\text{f}\ '(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
So, if there is $x_1$ such that $\text{f}\ '(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then $\text{x}_1\in(\text{a},\text{b})$
$\Rightarrow\text{a}<\text{x}_1<\text{b}$
View full question & answer→MCQ 991 Mark
Choose the correct answers from the given four options:
The function $\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}$ is:
- A
Discontinuous at only one point.
- B
Discontinuous at exactly two points.
- ✓
Discontinuous at exactly three points.
- D
AnswerCorrect option: C. Discontinuous at exactly three points.
We have, $\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}=\frac{(4-\text{x}^2)}{\text{x}(4-\text{x}^2)}$
$=\frac{(4-\text{x}^2)}{\text{x}(2^2-\text{x}^2)}=\frac{4-\text{x}^2}{\text{x}(2+\text{x})(2-\text{x})}$
Clearly, f(x) is discontinuous at exactly three points x = 0, x = -2 and x = 2.
View full question & answer→MCQ 1001 Mark
Let f(x) = |x| + |x - 1|, then:
- ✓
f(x) is continuous at x = 0, as well as at x = 1
- B
f(x) is continuous at x = 0, but not at x = 1
- C
f(x) is continuous at x = 0, but not at x = 0
- D
AnswerCorrect option: A. f(x) is continuous at x = 0, as well as at x = 1
Since modulus function is everywhere continuous |x| and |x - 1| are also everywhere continuous.
Also,
It is known that if f and g are continuous functions, then f + g will also be continuous.
Thus, |x| + |x - 1| is everywhere continuous.
Hence, f(x) is continuous at x = 0 and x = 1, x = 0 and x = 1.
View full question & answer→