MCQ 11 Mark
The least value of the function $f(x)=2 \cos x+x$ in the closed interval $\left[0, \frac{\pi}{2}\right]$ is
- A
- B
$\frac{\pi}{6}+\sqrt{3}$
- C
$\frac{\pi}{2}$
- D
The least value does not exist.
AnswerWe have, $f(x)=2 \cos x+x$
$
\begin{array}{l}
\Rightarrow f^{\prime}(x)=-2 \sin x+1 \\
\Rightarrow f^{\prime \prime}(x)=-2 \cos x
\end{array}
$
For critical points, $f^{\prime}(x)=0$
$
\begin{array}{l}
\Rightarrow \quad-2 \sin x+1=0 \\
\Rightarrow \sin x=\frac{1}{2}=\sin \left(\frac{\pi}{6}\right) \quad\left(\because x \in\left[0, \frac{\pi}{2}\right]\right) \\
\Rightarrow x=\frac{\pi}{6} \\
f^{\prime \prime}(x)\left(\text { at } x=\frac{\pi}{6}\right)=-2 \cos \frac{\pi}{6}=-\sqrt{3}<0
\end{array}
$
So, $x=\frac{\pi}{6}$ is the point of maxima .
Now, $f(0)=2$ and $f\left(\frac{\pi}{2}\right)=\frac{\pi}{2}=1.57$
$\Rightarrow$ Least value of $f(x)=1.57$ i.e., $\frac{\pi}{2}$
View full question & answer→MCQ 21 Mark
The value of $b$ for which the function $f(x)=x+\cos x+b$ is strictly decreasing over $R$ is
- A
$b<1$
- B
No value of $b$ exists
- C
$b \leq 1$
- D
$b \geq 1$
AnswerWe have, $f(x)=x+\cos x+b$
$\Rightarrow f^{\prime}(x)=1-\sin x \Rightarrow f^{\prime}(x) \geq 0 \forall x \in R$
$\Rightarrow \quad$ No such value of $b$ exists
View full question & answer→MCQ 31 Mark
The interval, in which function $y=x^3+6 x^2+6$ is increasing, is
- A
$(-\infty,-4) \cup(0, \infty)$
- B
$(-\infty,-4)$
- C
$(-4,0)$
- D
$(-\infty, 0) \cup(4, \infty)$
View full question & answer→MCQ 41 Mark
The total surface area (S) of the casted half cylinder will be
Answer$\begin{array}{l}\text {Total surface area, } S=\frac{2 \pi r(r+h)}{2}+2 r h \\ =\pi r^2+\pi r h+2 r h \\\end{array}$
View full question & answer→MCQ 51 Mark
The value of $x$ for which $\left(x-x^2\right)$ is maximum, is
- A
$3 / 4$
- B
$1 / 2$
- C
$1 / 3$
- D
$1 / 4$
AnswerLet $f(x)=x-x^2 \therefore f^{\prime}(x)=1-2 x$
For critical point, $f^{\prime}(x)=0 \Rightarrow 1-2 x=0 \Rightarrow x=1 / 2$
Now, at $x=1 / 2, f^{\prime \prime}(x)=-2<0$
So, $f(x)$ has maximum value at $x=1 / 2$.
View full question & answer→MCQ 61 Mark
The total cost of the train to travel $500 km$ at the most economical speed is
AnswerTotal cost for running the train for $500 km$
$=\frac{375}{4} v+\frac{600000}{v}$
$=\frac{375 \times 80}{4}+\frac{600000}{80}$=₹ 15000
View full question & answer→MCQ 71 Mark
The maximum value of $[x(x-1)+1]^{1 / 3}, 0 \leq x \leq 1$ is
- A
$0$
- B
$\frac{1}{2}$
- C
- D
$\sqrt[3]{\frac{1}{3}}$
AnswerLet $f(x)=[x(x-1)+1]^{1 / 3}, 0 \leq x \leq 1$
$\Rightarrow f^{\prime}(x)=\frac{2 x-1}{3\left(x^2-x+1\right)^{2 / 3}}$
For critical points, put $f^{\prime}(x)=0$
$\Rightarrow x=\frac{1}{2} \in[0,1]$
Now, $f(0)=1, f\left(\frac{1}{2}\right)=\left(\frac{3}{4}\right)^{1 / 3}$ and $f(1)=1$
$\therefore \quad$ Maximum value of $f(x)$ is 1 .
View full question & answer→MCQ 81 Mark
The real function $f(x)=2 x^3-3 x^2-36 x+7$ is
- A
Strictly increasing in $(-\infty,-2)$ and strictly decreasing in $(-2, \infty)$
- B
Strictly decreasing in $(-2,3)$
- C
Strictly decreasing in $(-\infty, 3)$ and strictly increasing in $(3, \infty)$
- D
Strictly decreasing in $(-\infty,-2) \cup(3, \infty)$
View full question & answer→MCQ 91 Mark
Find the intervals in which the function $f$ given by $f(x)=x^2-4 x+6$ is strictly increasing.
- A
$(-\infty, 2) \cup(2, \infty)$
- B
$(2, \infty)$
- C
$(-\infty, 2)$
- D
$(-\infty, 2] \cup(2, \infty)$
AnswerWe have, $f(x)=x^2-4 x+6$
$
\Rightarrow f^{\prime}(x)=2 x-4
$
$\because f(x)$ is strictly increasing.
$
\begin{array}{ll}
\therefore & f^{\prime}(x)>0 \\
\Rightarrow & 2 x-4>0 \Rightarrow x>2 \\
\Rightarrow & x \in(2, \infty)
\end{array}
$
View full question & answer→MCQ 101 Mark
The volume ( $V$ ) of the casted half cylinder will be
- A
$\pi r^2 h$
- B
$\frac{1}{3} \pi r^2 h$
- C
$\frac{1}{2} \pi r^2 h$
- D
$\pi r^2(r+h)$
Answer$\because$ Volume of cylinder $=\pi r^2 h$
$\therefore \quad V=$ Volume of casted half cylinder $=(1 / 2) \pi r^2 h$
View full question & answer→MCQ 111 Mark
The function $(x-\sin x)$ decreases for
- A
all $x$
- B
$x<\frac{\pi}{2}$
- C
0 < X < $\frac{\pi}{4}$
- D
no value of $x$
AnswerLet $f(x)=x-\sin x$
Differentiating w.r.t. $x$, we get $f^{\prime}(x)=1-\cos x$
For function to be decreasing, $f^{\prime}(x)<0$
$\Rightarrow 1-\cos x<0 \Rightarrow \cos x>1$,
which is not possible, because maximum value of $\cos x$ is 1 .
$\therefore \quad f(x)=(x-\sin x)$ doesn't decrease at any value of $x$.
View full question & answer→MCQ 121 Mark
The ratio $h$ : $2 r$ for which $S$ to be minimum will be equal to
- A
$2 \pi: \pi+2$
- B
$2 \pi: \pi+1$
- C
$\pi: \pi+1$
- D
$\pi: \pi+2$
Answer$\because V=\frac{1}{2} \pi r^2 h$
and $S$ will be minimum, when $(\pi+2) V =\pi^2 r^3$
$\Rightarrow V=\frac{\pi^2 r^3}{\pi+2}$
From (i) and (ii), we get
$\Rightarrow \frac{1}{2} \pi r^2 h=\frac{\pi^2 r^3}{\pi+2} \Rightarrow \pi r^2 h(\pi+2)=2 \pi^2 r^3$
$\Rightarrow h(\pi+2)=2 \pi r \Rightarrow \frac{h}{2 r}=\frac{\pi}{\pi+2}$
Thus, required ratio i.e., $h: 2 r$ is $\pi: \pi+2$.
View full question & answer→MCQ 131 Mark
For the given half-cylinder of volume $V$, the total surface area $S$ is minimum, when
- A
$(\pi+2) V=\pi^2 r^3$
- B
$(\pi+2) V=\pi^2 r^2$
- C
$2(\pi+2) V=\pi^2 r^3$
- D
$(\pi+2) \vee=\pi^2 r$
Answer$\because S=\pi r^2+\frac{2 V(\pi+2)}{\pi r} \Rightarrow \frac{d S}{d r}=2 \pi r-\frac{2 V(\pi+2)}{\pi} \times \frac{1}{r^2}$ For $S$ to be minimum, $\frac{d S}{d r}=0$ $\Rightarrow 2 \pi r=\frac{2 V(\pi+2)}{\pi r^2} \Rightarrow \pi^2 r^3=V(\pi+2)$
View full question & answer→MCQ 141 Mark
The total surface area $S$ can be expressed in terms of $V$ and $r$ as
- A
$2 \pi r+\frac{2 V(\pi+2)}{\pi r}$
- B
$\pi r+\frac{2 V}{\pi r}$
- C
$\pi r^2+\frac{2 V(\pi+2)}{\pi r}$
- D
$2 \pi r^2+\frac{2 V(\pi+2)}{\pi r}$
AnswerHere, $S=\pi r^2+\frac{2 V(\pi+2)}{\pi r}\left[\because V=\frac{1}{2} \pi r^2 h \Rightarrow \frac{2 V}{\pi r}=r h\right]$
View full question & answer→MCQ 151 Mark
The area of a trapezium is defined by function $f$ and given by $f(x)=(10+x) \sqrt{100-x^2}$, then the area when it is maximised is
- A
$75 cm ^2$
- B
$7 \sqrt{3} cm ^2$
- C
$75 \sqrt{3} cm ^2$
- D
$5 cm ^2$
AnswerWe have, $f(x)=(10+x) \sqrt{100-x^2}$
Which will give some real area if $-10<x<10$$
\begin{array}{l}
\Rightarrow f^{\prime}(x)=\frac{(10+x) \times(-2 x)}{2 \sqrt{100-x^2}}+\sqrt{100-x^2} \times 1 \\
\Rightarrow f^{\prime}(x)=\frac{-2 x^2-10 x+100}{\sqrt{100-x^2}}
\end{array}
$For critical points, put $f^{\prime}(x)=0$
$\begin{array}{l}
\Rightarrow x^2+5 x-50=0 \\
\Rightarrow(x+10)(x-5)=0 \\
\Rightarrow x=-10 \text { or } 5 \Rightarrow x=5 \quad[\because-10<x<10]
\end{array}
$Now, $f^{\prime \prime}(x)$
$
\begin{array}{l}
=\frac{\left(\sqrt{100-x^2}\right)(-4 x-10)+\left(2 x^2+10 x-100\right) \times \frac{1}{2} \frac{(-2 x)}{\sqrt{100-x^2}}}{\left(100-x^2\right)} \\
=\frac{2 x^3-300 x-1000}{\left(100-x^2\right)^{3 / 2}} \Rightarrow f^{\prime \prime}(5)=\frac{-30}{\sqrt{75}}<0
\end{array}
$$\therefore \quad$ Maximum area of trapezium
$=(10+5)(\sqrt{75})=75 \sqrt{3} cm ^2
$
View full question & answer→MCQ 161 Mark
A wire of length $20 cm$ is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is
AnswerLet $r$ be the radius, $\theta$ be the central angle and $/$ be the length of the circular sector.
Given, $1+2 r=20$
$\Rightarrow r \theta+2 r=20(\because l=r \theta) \Rightarrow \theta=\frac{20-2 r}{r}
$Let $A$ be the area of the circular sector.
$\therefore \quad A=\frac{\pi r^2 \theta}{2 \pi}=\frac{r^2}{2} \cdot\left(\frac{20-2 r}{r}\right)=r(10-r) \Rightarrow \frac{d A}{d r}=10-2 r
$
For maximum or minimum value of $A$, we have
$\frac{d A}{d r}=0 \Rightarrow r=5 \text { and } \frac{d^2 A}{d r^2}=-2<0
$$\therefore \quad$ Area is maximum at $r=5$
$\therefore \quad$ Maximum area, $A=5(10-5)=25 cm ^2$
View full question & answer→MCQ 171 Mark
The function $f(x)=\frac{x}{2}+\frac{2}{x}$ has a local minima at $x$ equal to
AnswerGiven, $f(x)=\frac{x}{2}+\frac{2}{x} \Rightarrow f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^2}$
For extremes $f^{\prime}(x)=0 \Rightarrow \frac{1}{2}-\frac{2}{x^2}=0 \Rightarrow x= \pm 2$
$f^{\prime \prime}(x)=\frac{4}{x^3}>0$ for $x=2$
$\therefore \quad x=2$ is the point of local minima.
View full question & answer→MCQ 181 Mark
Given a curve $y=7 x-x^3$ and $x$ increases at the rate of 2 units per second. The rate at which the slope of the curve is changing, when $x=5$ is
AnswerGiven curve is $y=7 x-x^3$
Differentiating both sides w.r.t. $x$, we get
$
\frac{d y}{d x}=7-3 x^2
$
Let $m$ be the slope, then $m=7-3 x^2$
Differentiating both sides w.r.t. t, we get
$
\Rightarrow \frac{d m}{d t}=-\frac{6 x d x}{d t}
$
It is given that $\frac{d x}{d t}=2$ units/sec and $x=5$
$
\Rightarrow \frac{d m}{d t}=-6(5)(2)=-60
$
View full question & answer→MCQ 191 Mark
The function $f(x)=k x$ - $\sin x$ is strictly increasing for
AnswerGiven, $f(x)=k x-\sin x$
$
\begin{array}{l}
\Rightarrow f^{\prime}(x)=k-\cos x>0 \quad(\because f(x) \text { is strictly increasing } \therefore f(x)>0) \\
\Rightarrow k>\cos x \\
\therefore k>1 \quad(\because \cos x \in[-1,1])
\end{array}
$
View full question & answer→MCQ 201 Mark
The function $f(x)=x^3-3 x^2+12 x-18$ is :
- A
strictly decreasing on $R$
- B
strictly increasing on $R$
- C
neither strictly increasing nor strictly decreasing on $R$
- D
strictly decreasing on $(-\infty, 0)$
Answer$\begin{array}{l}\text {} f(x)=x^3-3 x^2+12 x-18 \\ \Rightarrow \quad f^{\prime}(x)=3 x^2-6 x+12=3\left(x^2-2 x+1^2\right)+9 \\ =3(x-1)^2+3^2>0 \forall x \in R \\ \Rightarrow \quad f(x) \text { is strictly increasing on } R\end{array}$
View full question & answer→MCQ 211 Mark
In which of these intervals is the function $f(x)=3 x^2-4 x$ strictly decreasing?
AnswerCorrect option: A. $(-\infty, 0)$
$(-\infty, 0)$
View full question & answer→MCQ 221 Mark
The interval in which the function $f(x)=2 x^3+9 x^2+$ $12 x-1$ is decreasing, is
- A
$(-1, \infty)$
- B
$(-2,-1)$
- C
$(-\infty,-2)$
- D
$[-1,1]$
AnswerWe have, $f(x)=2 x^3+9 x^2+12 x-1$$
\Rightarrow f^{\prime}(x)=6 x^2+18 x+12
$
For decreasing, $f^{\prime}(x)<0$
$
\begin{array}{ll}
\therefore & 6 x^2+18 x+12<0 \\
\Rightarrow & x^2+3 x+2<0 \Rightarrow(x+1)(x+2)<0 \Rightarrow-2<x<-1
\end{array}
$So, $f(x)$ is decreasing, if $x \in(-2,-1)$.
View full question & answer→MCQ 231 Mark
If $f(x)=a(x-\cos x)$ is strictly decreasing in $R$, then ' $a$ ' belongs to
- A
$\{0\}$
- B
$(0, \infty)$
- C
$(-\infty, 0)$
- D
$(-\infty, \infty)$
AnswerWe have, $f(x)=a(x-\cos x)$
$
\therefore \quad f^{\prime}(x)=a(1+\sin x)
$
It is given that $f(x)$ is a strictly decreasing function.
$
\begin{array}{l}
\therefore \quad f^{\prime}(x)<0 \Rightarrow a(1+\sin x)<0 \\
\text { But }(1+\sin x) \geq 0 \text { as }-1 \leq \sin x \leq 1 \\
\therefore \quad 1+\sin x<0 \text { is not possible. } \\
\Rightarrow \quad a<0 \therefore a \in(-\infty, 0)
\end{array}
$
View full question & answer→MCQ 241 Mark
The function $f(x)=x^3+3 x$ is increasing in interval
- A
$(-\infty, 0)$
- B
$(0, \infty)$
- C
$R$
- D
$(0,1)$
Answerf(x)=$x^3+3 x$
For increasing, we must have $f^{\prime}(x)>0$
$
\therefore f^{\prime}(x)=3 x^2+3>0 \Rightarrow 3\left(x^2+1\right)>0
$
$\Rightarrow x^2+1>0$, which is true $\forall x \in R$.
View full question & answer→MCQ 251 Mark
The function $f(x)=x^5-5 x^4+5 x^3-1$ has
- A
one minima and two maxima
- B
two minima and one maxima
- C
two minima and two maxima
- ✓
one minima and one maxima
AnswerCorrect option: D. one minima and one maxima
(d) : Given, $f(x)=x^5-5 x^4+5 x^3-1$
Then, $f^{\prime}(x)=5 x^4-20 x^3+15 x^2$
$
\Rightarrow f^{\prime \prime}(x)=20 x^3-60 x^2+30 x
$
Now, $f^{\prime}(x)=0 \Rightarrow 5 x^2\left(x^2-4 x+3\right)=0$
$
\begin{array}{l}
\Rightarrow \quad x=0,1,3 \\
f^{\prime \prime}(1)=-10<0 \text { and } f^{\prime \prime}(3)=90>0 \\
f^{\prime \prime}(0)=0 \text { and } f^{\prime \prime \prime}(0) \neq 0
\end{array}
$
So, $x=0$ is a point of inflexion.
$\therefore f(x)$ has maximum at $x=1$ and minimum at $x=3$.
View full question & answer→MCQ 261 Mark
Find all the points of local maxima and local minima of the function $f(x)=(x-1)^3(x+1)^2$.
- A
$1,-1,-1 / 5$
- B
$1,-1$
- C
$1,-1 / 5$
- D
$-1,-1 / 5$
Answer$
\begin{array}{l}
\text { (a) : Let } y=f(x)=(x-1)^3(x+1)^2 \\
\Rightarrow \quad \frac{d y}{d x}=3(x-1)^2(x+1)^2+2(x+1)(x-1)^3 \\
=(x-1)^2(x+1)\{3(x+1)+2(x-1)\} \\
=(x-1)^2(x+1)(5 x+1)
\end{array}
$
For local maximum or local minimum, we have
$
\begin{aligned}
& \frac{d y}{d x}=0 \Rightarrow(x-1)^2(x+1)(5 x+1)=0 \\
\Rightarrow & x=1 \text { or } x=-1 \text { or } x=-\frac{1}{5}
\end{aligned}
$
View full question & answer→MCQ 271 Mark
For what value of $a, f(x)=-x^3+4 a x^2+2 x-5$ is decreasing $\forall x$ ?
Answer(d) : $f(x)=-x^3+4 a x^2+2 x-5$
$
\therefore f^{\prime}(x)=-3 x^2+8 a x+2
$
Since, $f(x)$ is decreasing, $\forall x$, therefore
$
\begin{array}{l}
f^{\prime}(x)<0 \\
\Rightarrow \quad-3 x^2+8 a x+2<0 \\
\end{array}
$
From above, it is clear that decreasingness of $f(x)$ will be depend on the value of $a$ and $x$.
View full question & answer→MCQ 281 Mark
Find the minimum value of $f(x)=2 x^3-24 x+107$ in the interval $[1,3]$.
Answer(b) : We have, $f(x)=2 x^3-24 x+107$ in $[1,3]$
$
\Rightarrow f^{\prime}(x)=6 x^2-24
$
Now, $f^{\prime}(x)=0 \Rightarrow 6 x^2-24=0 \Rightarrow x= \pm 2$
But, $x=-2 \notin[1,3]$
So, $x=2$ is the only critical point.
Now, $f(1)=2-24+107=85$
$
f(2)=2(2)^3-24 \times 2+107=75
$
and $f(3)=2(3)^3-24(3)+107=89$
Hence, the maximum value of $f(x)$ is 89 at $x=3$ and the minimum value is 75 at $x=2$.
View full question & answer→MCQ 291 Mark
The function $f$ defined by $f(x)=4 x^4-2 x+1$ is increasing for
- A
$x<1$
- B
$x>0$
- C
$x<\frac{1}{2}$
- ✓
$x>\frac{1}{2}$
AnswerCorrect option: D. $x>\frac{1}{2}$
(d) : We have, $f(x)=4 x^4-2 x+1 \Rightarrow f^{\prime}(x)=16 x^3-2$
The function is increasing if $f^{\prime}(x)>0$
$
\Rightarrow 16 x^3-2>0 \Rightarrow x>\frac{1}{2}
$
View full question & answer→MCQ 301 Mark
$3 x^2-6 x+5$ is an increasing function, if
- A
- B
- ✓
$x>1$
- D
-1 < x < $-\frac{1}{2}$
Answer(c) : Let $f(x)=3 x^2-6 x+5$
Differentiating w.r.t. $x$, we get $f^{\prime}(x)=6 x-6$
Since, it is increasing function.
$
\Rightarrow 6 x-6>0 \Rightarrow(x-1)>0 \Rightarrow x>1
$
View full question & answer→MCQ 311 Mark
If $x$ is real, the minimum value of $x^2-8 x+$ 17 is
Answer(c) : Let $f(x)=x^2-8 x+17 \Rightarrow f^{\prime}(x)=2 x-8$
For minimum or maximum, we have $f^{\prime}(x)=0$
$
\Rightarrow 0=2 x-8 \Rightarrow x=4
$
Now, $f^{\prime \prime}(x)=2$
At $x=4, f^{\prime \prime}(x)$ is + ve
$\therefore \quad x=4$ is local minima point.
$\therefore \quad$ Minimum value $=(4)^2-8(4)+17=16-32+17=1$
View full question & answer→MCQ 321 Mark
$f(x)=x-[x]$ in the interval $[0,1]$ is
- ✓
- B
- C
neither increasing nor decreasing
- D
Answer(a) : Given $f(x)=x-[x], x \in[0,1]$
But for $0 \leq x \leq 1,[x]=0, \therefore f(x)=x-0=x$ in $[0,1]$.
Let $x_1, x_2 \in[0,1]$ be such that $x_1$ < $x_2$
=> $f\left(x_1\right)$ < $f\left(x_2\right)$ => $f$ is increasing in $[0,1]$.
View full question & answer→MCQ 331 Mark
The interval in which the function $y=x^3+5 x^2-1$ is decreasing, is
- A
$\left(0, \frac{10}{3}\right)$
- B
$(0,10)$
- ✓
$\left(\frac{-10}{3}, 0\right)$
- D
AnswerCorrect option: C. $\left(\frac{-10}{3}, 0\right)$
(c) : Given, $y=x^3+5 x^2-1$
$
\Rightarrow \frac{d y}{d x}=3 x^2+10 x=x(3 x+10)
$
For function to be decreasing, $\frac{d y}{d x}<0$
x(3x + 10) < 0 => $\frac{-10}{3}$< x < 0
View full question & answer→MCQ 341 Mark
The function $f(x)=x+\frac{4}{x}$ has
- A
a local maxima at $x=2$ and local minima at $x=-2$
- ✓
local minima at $x=2$, and local maxima at $x=-2$
- C
absolute maxima at $x=2$ and absolute minima at $x=-2$
- D
absolute minima at $x=2$ and absolute maxima at $x=-2$
AnswerCorrect option: B. local minima at $x=2$, and local maxima at $x=-2$
(b) : Given, $f(x)=x+\frac{4}{x}$
$
f^{\prime}(x)=1-\frac{4}{x^2} \quad \therefore f^{\prime}(x)=0 \Rightarrow x= \pm 2
$
and $f^{\prime \prime}(x)=\frac{8}{x^3}$ which is $>0$ for $x=2$ and $<0$ for $x=-2$
$\therefore f(x)$ has local minima at $x=2$ and local maxima at $x=-2$.
View full question & answer→MCQ 351 Mark
Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are
Answer(c) : Let two parts be $x$ and $20-x$
$
\begin{array}{l}
\therefore \quad P=x^3(20-x)=20 x^3-x^4 \\
\Rightarrow \frac{d P}{d x}=60 x^2-4 x^3 \Rightarrow \frac{d^2 P}{d x^2}=120 x-12 x^2 \\
\therefore \quad \frac{d P}{d x}=0 \Rightarrow 60 x^2-4 x^3=0 \\
\Rightarrow \quad x=0,15 \\
{\left[\frac{d^2 P}{d x^2}\right]_{x=15}=-900<0} \\
\end{array}
$
$\therefore \quad P$ is maximum at $x=15$
$\therefore \quad$ Two parts are $(15,5)$.
View full question & answer→MCQ 361 Mark
The value of $b$ for which the function $f(x)=\sin x-b x+c$ is strictly decreasing for $x \in R$ is given by
- A
$b<1$
- B
$b \geq 1$
- ✓
$b>1$
- D
$b \leq 1$
Answer(c) : $f(x)=\sin x-b x+c$
$\Rightarrow f^{\prime}(x)=\cos x-b<0$ for all $x \in R$
Hence cosx < b => b > 1.
View full question & answer→MCQ 371 Mark
If the train has travelled a distance of $500 km$, then the total cost of running the train is given by function
- A
$\frac{15}{16} v+\frac{600000}{v}$
- B
$\frac{375}{4} v+\frac{600000}{v}$
- C
$\frac{5}{16} v^2+\frac{150000}{v}$
- D
$\frac{3}{16} v+\frac{6000}{v}$
AnswerLet total cost of running the train be $C$.
Then, $C=\frac{3}{16} v^2 t+1200 t$
Now, distance covered $=500 km$
$
\Rightarrow \text { Time }=\frac{500}{v} hrs
$
$\therefore$ Total cost of running the train for $500 km$
$
\begin{array}{l}
=\frac{3}{16} v^2\left(\frac{500}{v}\right)+1200\left(\frac{500}{v}\right) \\
\Rightarrow C=\frac{375}{4} v+\frac{600000}{v}
\end{array}
$
View full question & answer→MCQ 381 Mark
The function $f(x)=x-\frac{1}{x}, x \in R, x \neq 0$ is
- ✓
increasing for all $x \in R$
- B
decreasing for all $x \in R$
- C
increasing for all $x \in(0, \infty)$
- D
neither increasing nor decreasing
AnswerCorrect option: A. increasing for all $x \in R$
(a) : $f(x)=x-\frac{1}{x}$
$\therefore f^{\prime}(x)=1+\frac{1}{x^2}>0$ for all $x \in R, x \neq 0$
$\therefore \quad f(x)$ is increasing for all $x \in R$, where $x \neq 0$.
View full question & answer→MCQ 391 Mark
If $y=x^3+x^2+x+1$, then $y$
- A
- B
- ✓
neither have a local minimum nor local maximum
- D
AnswerCorrect option: C. neither have a local minimum nor local maximum
(c) : Let $f(x)=y=x^3+x^2+x+1$
Then, $f^{\prime}(x)=3 x^2+2 x+1$.
For a maximum or minimum, we have
$
f^{\prime}(x)=0 \Rightarrow 3 x^2+2 x+1=0
$
But, this equation gives imaginary values of $x$.
So, $f^{\prime}(x) \neq 0$ for any real value of $x$.
Hence, $f(x)$ does not have a maximum or minimum.
View full question & answer→MCQ 401 Mark
Find the maximum value of $f(x)=\sin (\sin x)$ for all $x \in R$.
- A
$-\sin 1$
- B
$\sin 6$
- ✓
$\sin 1$
- D
$-\sin 3$
AnswerCorrect option: C. $\sin 1$
(c) : We have, $f(x)=\sin (\sin x), x \in R$
Now, $-1 \leq \sin x \leq 1$ for all $x \in R$
$\Rightarrow \sin (-1) \leq \sin (\sin x) \leq \sin 1$ for all $x \in R$
$[\because \sin x$ is an increasing function on $[-1,1]]$
$\Rightarrow \quad-\sin 1 \leq f(x) \leq \sin 1$ for all $x \in R$
This shows that the maximum value of $f(x)$ is $\sin 1$.
View full question & answer→MCQ 411 Mark
The function $f(x)=x+\sin x$ is
Answer(a) : $\because f(x)=x+\sin x$
Differentiating w.r.t. $x$, we get $f^{\prime}(x)=1+\cos x$
$f^{\prime}(x) \geq 0$ for all values of $x$
$(\because \cos x$ is lying between -1 to 1 )
$\therefore f(x)$ is always increasing.
View full question & answer→MCQ 421 Mark
$y=x(x-3)^2$ decreases for the values of $x$ given by
- ✓
- B
- C
$x>0$
- D
0 < x < $\frac{3}{2}$
Answer(a) : $y=x(x-3)^2$
$
\begin{aligned}
\Rightarrow \quad & \frac{d y}{d x}=x \cdot 2(x-3)+(x-3)^2 \\
& =(x-3)(2 x+x-3)=(x-3)(3 x-3)=3(x-3)(x-1)
\end{aligned}
$
For decreasing function, $\frac{d y}{d x}<0$
=> (x - 3) (x - 1 )< 0 = > 1 < x < 3
View full question & answer→MCQ 431 Mark
Let $f: R \rightarrow R$ be defined by $f(x)=2 x+\cos x$, then $f$
- A
has a minimum at $x=\pi$
- B
has a maximum, at $x=0$
- C
- ✓
is an increasing function
AnswerCorrect option: D. is an increasing function
(d) $: f(x)=2 x+\cos x$
$
f^{\prime}(x)=2-\sin x
$
We know that, $-1 \leq \sin x \leq 1$
$
-1 \leq-\sin x \leq 1 \Rightarrow 1 \leq 2-\sin x \leq 3
$
$\Rightarrow f^{\prime}(x)>0 \therefore f(x)$ is always increasing.
View full question & answer→MCQ 441 Mark
The smallest value of the polynomial $x^3-18 x^2+96 x$ in $[0,9]$ is
Answer(b) : $f(x)=x^3-18 x^2+96 x$
$
\begin{array}{l}
\Rightarrow f^{\prime}(x)=3 x^2-36 x+96 \\
\therefore \quad f^{\prime}(x)=0 \Rightarrow x^2-12 x+32=0 \Rightarrow x=8,4
\end{array}
$
Now, $f(0)=0, f(4)=160, f(8)=128, f(9)=135$
So, smallest value of $f(x)$ is 0 at $x=0$.
View full question & answer→MCQ 451 Mark
The interval on which the function $f(x)=2 x^3+9 x^2+12 x-1$ is decreasing is
- A
$[-1, \infty)$
- ✓
$[-2,-1]$
- C
$(-\infty,-2]$
- D
$[-1,1]$
AnswerCorrect option: B. $[-2,-1]$
(b) : $f(x)=2 x^3+9 x^2+12 x-1$
$
f^{\prime}(x)=6 x^2+18 x+12=6\left(x^2+3 x+2\right)=6(x+2)(x+1)
$
For decreasing function, $f^{\prime}(x) \leq 0$
$
\begin{array}{l}
\therefore \quad 6(x+2)(x+1) \leq 0 \\
\Rightarrow \quad(x+2)(x+1) \leq 0 \Rightarrow-2 \leq x \leq-1
\end{array}
$
View full question & answer→MCQ 461 Mark
If $V=\frac{4}{3} \pi r^3$, at what rate in cubic units is $V$ increasing when $r=10$ and $\frac{d r}{d t}=0.01$ ?
- A
$\pi$
- B
$4 \pi$
- C
$40 \pi$
- D
$4 \pi / 3$
Answer$
\begin{array}{l}
\text { (b) : Given, } V=\frac{4}{3} \pi r^3 \\
\Rightarrow \quad \frac{d V}{d t}=\frac{4}{3} \times 3 \pi r^2 \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t} \\
\Rightarrow \quad \frac{d V}{d t}=4 \pi(10)^2 \times 0.01 \quad\left[\because r=10, \frac{d r}{d t}=0.01\right] \\
=4 \pi \times 100 \times 0.01=4 \pi \\
\end{array}
$
View full question & answer→MCQ 471 Mark
Find the interval in which $f(x)=\log (1+x)-\frac{x}{1+x}$ is increasing.
- ✓
$(0, \infty)$
- B
$(-\infty, 0)$
- C
$(-\infty, 3)$
- D
AnswerCorrect option: A. $(0, \infty)$
(a) : Here, $f^{\prime}(x)=\frac{x}{(1+x)^2}$
So, critical point is $x=0$ only.
and disjoint intervals are $(-\infty, 0)$ and $(0, \infty)$.
So, $f(x)$ is increasing in $(0, \infty)$ and decreasing in $(-\infty, 0)$.
View full question & answer→MCQ 481 Mark
The function $f(x)=\cot ^{-1} x+x$ increases in the interval
- A
$(1, \infty)$
- B
$(-1, \infty)$
- C
$(0, \infty)$
- ✓
$(-\infty, \infty)$
AnswerCorrect option: D. $(-\infty, \infty)$
(d) : $f(x)=\cot ^{-1} x+x$
$
\Rightarrow f^{\prime}(x)=\frac{-1}{1+x^2}+1 \Rightarrow f^{\prime}(x)=\frac{x^2}{1+x^2} \geq 0 \text {, for } x \in R
$
$\therefore f(x)$ is increasing on $(-\infty, \infty)$.
View full question & answer→MCQ 491 Mark
$2 x^3-6 x+5$ is an increasing function, if
- A
- B
- ✓
$x<-1$ or $x>1$
- D
-1 < x < $-\frac{1}{2}$
AnswerCorrect option: C. $x<-1$ or $x>1$
(c) : Let $f(x)=2 x^3-6 x+5$
On differentiating w.r.t. $x$, we get $f^{\prime}(x)=6 x^2-6$
Since, it is increasing function.
$
\begin{array}{l}
\Rightarrow \quad 6 x^2-6>0 \Rightarrow(x-1)(x+1)>0 \\
\Rightarrow \quad x>1 \text { or } x<-1
\end{array}
$
View full question & answer→MCQ 501 Mark
The function $f(x)=\frac{x}{\log x}$ increases in the interval
- A
$(0, \infty)$
- B
$(0, e)$
- ✓
$(e, \infty)$
- D
AnswerCorrect option: C. $(e, \infty)$
(c) : $f(x)=\frac{x}{\log x}$ is defined for $x>0$ and $x \neq 1$
Also, $f^{\prime}(x)=\frac{\log x \cdot 1-x \cdot \frac{1}{x}}{(\log x)^2}=\frac{\log x-1}{(\log x)^2}$
$
\therefore f^{\prime}(x)>0 \Rightarrow \log x>1 \Rightarrow x>e \therefore x \in(e, \infty) \text {.}
$
View full question & answer→
