Question 11 Mark
If $y=x \log x$ then value of $\frac{d^2 y}{d x^2}$ :
Answer(B)
$\begin{aligned} & & y & =x \log x \\ \Rightarrow & & \frac{d y}{d x} & =x \cdot \frac{1}{x}+\log x \cdot 1 \\ \Rightarrow & & \frac{d y}{d x} & =1+\log x \\ \Rightarrow & & \frac{d^2 y}{d x^2} & =0+\frac{1}{x}=\frac{1}{x}\end{aligned}$
View full question & answer→Question 21 Mark
If $3 x+2 y=\sin x$ then $\frac{d y}{d x}$ :
Answer$(C) \ \frac{d}{d x}(3 x+2 y)$
$=\frac{d}{d x}(\sin x)$
$\Rightarrow 3+2 \frac{d y}{d x}=\cos x$
$\Rightarrow \frac{d y}{d x}=\frac{\cos x-3}{2}$
View full question & answer→Question 31 Mark
If function $f$ is defined such that
$
f(x)=\left\{\begin{array}{cc}
\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\
3, & \text { if } x=\frac{\pi}{2}
\end{array}\right.
$is continuous at $x=\frac{\pi}{2}$, then value of $k$ :
Answer(C)
$
\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{2}} f(x) & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x} \\
& =\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}=\lim _{h \rightarrow 0} \frac{-k \sin h}{-2 h} \\
& =\frac{k}{2} \lim _{k \rightarrow 0} \frac{\sin h}{h} \\
& =\frac{k}{2}(1)=\frac{k}{2}
\end{aligned}
$
for continuity at $x=\frac{\pi}{2}$
$
\begin{aligned}
& & \lim _{x \rightarrow \frac{\pi}{2}} f(x) & =f\left(\frac{\pi}{2}\right)\\
\Rightarrow & & \frac{k}{2} & =3 \\
\therefore & & k & =6
\end{aligned}
$
View full question & answer→Question 41 Mark
If $\left(x^2+y^2\right)^2=x y$, then $\frac{d y}{d x}$ :
Answer$(C)$ Here $\left(x^2+y^2\right)^2=x y$
Differentiating $w.r.t. x^ 2\left(x^2+y^2\right)\left(2 x+2 y \frac{d y}{d x}\right)=x \frac{d y}{d x}+y$
$\Rightarrow 4\left(x^2+y^2\right) x+4\left(x^2+y^2\right) y \frac{d y}{d x}=x \frac{d y}{d x}+y$
$\Rightarrow \quad\left[4 y\left(x^2+y^2\right)-x\right] \frac{d y}{d x}=y-4 x\left(x^2+y^2\right)$
Hence, $\frac{d y}{d x}=\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}$
View full question & answer→Question 51 Mark
If $\sin y=x \cos (a+y)$, then $\frac{d x}{d y}$ :
Answer(A)Here $\quad x=\frac{\sin y}{\cos (a+y)}$$
\begin{aligned}
\therefore \quad \frac{d x}{d y} & =\frac{\cos (a+y) \cos y-\sin y[-\sin (a+y)}{\cos ^2(x \times y)} \\
& =\frac{\cos (a+y) \cos y+\sin (a+y) \sin y}{\cos ^2(a+y)} \\
& =\frac{\cos (a+y-y)}{\cos ^2(a+y)}=\frac{\cos y}{\cos ^2(a+y)}
\end{aligned}
$
View full question & answer→Question 61 Mark
If $y=\sin \left(m \sin ^{-1} x\right)$ in which of the option is correct :
Answer$(B)$ Given that : $y=\sin \left(m \sin ^{-1} x\right)$
$\therefore \frac{d y}{d x} =\cos \left(m \sin ^{-1}\right) \cdot \frac{d}{d x}\left(m \sin ^{-1} x\right)$
$ =\cos \left(m \sin ^{-1}\right) \cdot \frac{m}{\sqrt{1-x^2}}$
$ =\frac{m \cos \left(m \sin ^{-1}\right) x}{\sqrt{1-x^2}}$
$\Rightarrow \sqrt{1-x^2} \frac{d y}{d x}=m \cos \left(m \sin ^{-1} x\right)$
On squaring $\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=m^2\left(1-y^2\right)$
again differentiating $\text{w.r.t.}\ x$
$\left(1-x^2\right) 2\left(\frac{d y}{d x}\right) \frac{d^2 y}{d x^2}+(-2 x)$
$\left(\frac{d y}{d x}\right)^2=m^2(-2 y) \frac{d y}{d x}$
$\Rightarrow \quad\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=-m^2 y \quad\left[\right.$ divide by $\left.\frac{d y}{d x}\right]$
hence $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+m^2 y=0$
View full question & answer→Question 71 Mark
If $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta$ then $\frac{d y}{d x}$ :
Answer(B)Given : $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta$$\begin{aligned}\therefore \quad \frac{d x}{d \theta} & =-2 \sin \theta+2 \sin 2 \theta \text { and } \\\frac{d y}{d \theta} & =2 \cos \theta-2 \cos 2 \theta \\\therefore \quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{2 \cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-2 \sin \theta} \\& =\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}\end{aligned}$
View full question & answer→Question 81 Mark
If function $f(x)=\left\{\begin{array}{cl}\frac{e^{3 x}-e^{-5 x}}{x} & , \text { if } x \neq 0 \\ k & , \text { if } x=0\end{array}\right.$ is continuous then value of $k$ :
Answer(D)
$
\begin{aligned}
\lim _{x \rightarrow 0} f(x) & =\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-5 x}}{x} \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right)+\left(1-e^{-5 x}\right)}{x} \\
& =\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-5 x}-1}{x} \\
& =\lim _{3 x \rightarrow 0} \frac{e^{3 x}-1}{3 x} \cdot 3-\lim _{-5 x \rightarrow 0} \frac{e^{-5 x}-1}{-5 x} \cdot(-5) \\
& =(1)(3)-(1)(-5) \\
& =3+5=8
\end{aligned}
$
for continuity at $x=0$,
$
\begin{aligned}
& & \lim _{x \rightarrow 0} f(x) & =f(0) \\
\Rightarrow & & 8 & =k \\
\therefore & & k & =8
\end{aligned}
$
View full question & answer→Question 91 Mark
Differentiation of $\log \left[\log \left(\log x^5\right)\right] \text{w.r.t.} \ x$ is :
Answer$(D)$ Suppose $y=\log \left[\log \left(\log x^5\right)\right]$
$\therefore \frac{d y}{d x} =\frac{1}{\log \left(\log x^5\right)} \cdot \frac{d}{d x}\left[\log \left(\log x^5\right)\right]$
$ =\frac{1}{\log \left(\log x^5\right)} \cdot \frac{1}{\log x^5} \frac{d}{d x}\left(\log x^5\right)$
$=\frac{1}{\log \left(\log x^5\right)} \cdot \frac{1}{\log x^5} \cdot \frac{1}{x^5} \cdot \frac{d}{d x}\left(x^5\right)$
$=\frac{1}{\log \left(\log x^5\right)} \cdot \frac{1}{\log x^5} \cdot \frac{1}{x^5} \cdot 5 x^4$
$=\frac{5}{x\left(\log x^5\right) \log \left(\log x^5\right)}$
View full question & answer→Question 101 Mark
The derivative of function $\cos (\sin x)$ is :
Answer(D)$-\cos x \sin (\sin x)$
View full question & answer→Question 111 Mark
If $y=x \log _e x$ then value of $\frac{d^2 y}{d x^2}$ :
Answer(D)Given that $y=x \log _e x$
$\therefore \quad \frac{d y}{d x}=1 \cdot \log x+x \cdot \frac{1}{x}=1+\log _e x$
again differentiating
$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(1+\log _c x\right)=0+\frac{1}{x}=\frac{1}{x}$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{1}{x}$
Hence correct option is (D).
View full question & answer→Question 121 Mark
If $f(x)=\frac{1-\cos x}{x^2}$, is continuous at $x=0$, then $f(0)$ equals to:
Answer(A)given function is continuous at $x=0$ so,$
f(0)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(0-h)
$
$
\begin{aligned}
\lim _{h \rightarrow 0} f(0+h) & =\lim _{h \rightarrow 0} \frac{1-\cos (0+h)}{(0+h)^2}=\lim _{h \rightarrow 0} \frac{1-\cos h}{h^2} \\
& =\lim _{h \rightarrow 0} \frac{2 \sin ^2 h / 2}{h^2}=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{\sin h / 2}{h / 2}\right)^2 \\
& =\frac{1}{2} \times 1=\frac{1}{2}
\end{aligned}
$
hence $\quad f(0)=\frac{1}{2}$
hence correct option is (A).
View full question & answer→Question 131 Mark
For which value of $k$, this funciton is continuous at $x=1$.$
f(x)=\left\{\begin{array}{cc}
\frac{x^2-3 x+2}{x-1}, & x \neq 1 \\
k, & x=1
\end{array}\right.
$
Answer$(B)$
value of function at $x=1$
$f(1)=k$
Value of $\text{R.H.L.}$
$\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} \frac{(1+h)^2-3(1+h)+2}{1+h-1}$
$=\lim _{h \rightarrow 0} \frac{1+2 h+h^2-3-3 h+2}{h}$
$=\lim _{h \rightarrow 0} \frac{h^2-h}{h}=\lim _{h \rightarrow 0}[h-1]=-1$
because function is continuous at $x=1$, so
$f(1)=\text{ R . H . L .}$
$\therefore k=-1$
Hence correct option is $(B).$
View full question & answer→Question 141 Mark
Derivative of $\cos ^{-1}\left(4 x^3-3 x\right)$ :
Answer$(B)$ Put $x=\cos \theta$$\cos ^{-1}\left(4 x^3-3 x\right)=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right)$
$\Rightarrow \cos ^{-1}(\cos 3 \theta)=3 \theta=3 \cos ^{-1} x[\because x=\cos \theta]$
$\therefore \frac{d}{d x}\left(3 \cos ^{-1} x\right)=3 \frac{d}{d x}\left(\cos ^{-1} x\right)=3\left\{\frac{-1}{\sqrt{1-x^2}}\right\}$
$=\frac{-3}{\sqrt{1-x^2}}$
Hence correct option is $(B).$
View full question & answer→Question 151 Mark
If function $f(x)=\left\{\begin{array}{c}\frac{x^2-16}{x-4}, x \neq 4 \\ k, x=4\end{array}\right.$, is continuous at $x=4$, then value of $k$ is :
Answer(C)value of function of $x=4$
$
f(4)=k
$
value of R.H.L. at $x=4$$
\begin{aligned}
\lim _{x \rightarrow 4^{+}} f(x) & =\lim _{h \rightarrow 0} f(4+h)=\lim _{h \rightarrow 0} \frac{(4+h)^2-16}{4+h-4} \\
& =\lim _{h \rightarrow 0} \frac{16+8 h+h^2-16}{h} \\
& =\lim _{h \rightarrow 0} \frac{h(8+h)}{h}=8
\end{aligned}
$
$\because$ function is continuous at $x=4$.
$\therefore \quad f(4)=\lim _{h \rightarrow 0} f(4+h)$
$\Rightarrow \quad k=8$
View full question & answer→Question 161 Mark
Function $f(x)=\left\{\begin{array}{c}x+\lambda, \text { if } x<3 \\ 4, \text { if } x=3 \\ 3 x-5, \text { if } x>3\end{array}\right.$, is continuous at $x=3$, then value of $\lambda$ is :
Answer(A)value of function at $x=3$
$
f(3)=4
$
value of L.H.L.$
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}[3-h+\lambda]=3+\lambda
$because function is continuous.$
\therefore f(3)=\lim _{h \rightarrow 0} f(3-h) \quad \therefore 4=3+\lambda
$
$
\lambda=1
$
Hence correct option is (A).
View full question & answer→Question 171 Mark
What will be the value of $k$ for which $f(x)=$
$\left\{\begin{array}{c}\frac{\sin 3 x}{x}, x \neq 0 \\ k \quad, x=0\end{array}\right.$ is continuous at $x=0$.
Answer(A) At $x=0, f(0)=k$$
\begin{aligned}
\lim _{h \rightarrow 0} f(0+h) & =\lim _{h \rightarrow 0} \frac{\sin 3(0+h)}{0+h}=\lim _{h \rightarrow 0} \frac{\sin 3 h}{h} \\
& =3 \lim _{h \rightarrow 0} \frac{\sin 3 h}{3 h}=3 \times 1=3 \\
\lim _{h \rightarrow 0} f(0-h) & =\lim _{h \rightarrow 0} \frac{\sin 3(0-h)}{0-h}=\lim _{h \rightarrow 0} \frac{\sin (-3 h)}{-h}
\end{aligned}
$$
\begin{array}{r}
=\lim _{h \rightarrow 0}\left\{\frac{-\sin (3 h)}{-3 h} \times 3\right\} \\
=3 \lim _{h \rightarrow 0} \frac{\sin 3 h}{3 h}=3
\end{array}
$
function is continuous at $x=0$$
\therefore \quad f(0)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(0-h) \Rightarrow k=3
$
View full question & answer→Question 181 Mark
If function $f(x)=\left\{1+x, x \leq 2k-\frac{x}{2}, x>2\right.$, is continuous at $x=2$, then value of $k$ will be :
Answer$(D)$ value of function at $x=2$
$f(2) =1+2=3$
$f(2+h) =k-\left(\frac{2+h}{2}\right)$
$f(2-h) =1+(2-h)=(3-h)$
Value of $\text{R.H.L.}$
$f(2+0)=\lim _{h \rightarrow 0} f(2+h) $
$=\lim _{h \rightarrow 0}\left[k-\left(\frac{2+h}{2}\right)\right]$
$ =k-1$
Value of $\text{L.H.L.}$
$f(2-0)=\lim _{h \rightarrow 0} f(2-h)$
$=\lim _{h \rightarrow 0}[3-h]=3$
Function is continuous at $x=2$, so.
$f(2) =\text { R.H.L. }=\text { L.H.L. }$
$\Rightarrow 3 =k-1=3 $
$\therefore k=4$
Hence correct option is $(D).$
View full question & answer→