MCQ 11 Mark
In which of the following intervals is $y=x^2 e^{-x}$ increasing ?
- A$(1,0)$
- B$(2,0)$
- C$(2,-\infty)$
- D$(0,2)$
Answer
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11 questions · timed · auto-graded
MCQ 21 Mark
The minimum value of function $f(x)=2 \cos x+x$ in interval $\left[0, \frac{\pi}{2}\right]$ :
- A2
- B$\frac{\pi}{6}+\sqrt{3}$
- ✓$\frac{\pi}{2}$
- DNone of these
Answer
View full question & answer→Correct option: C.
$\frac{\pi}{2}$
(C)$f(x)=2 \cos x+x \Rightarrow f^{\prime}(x)=-2 \sin x+1$
now, $f^{\prime}(x)=0 \Rightarrow-2 \sin x+1=0$
$\Rightarrow \quad \sin x=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow x=\frac{\pi}{6}$
$\Rightarrow \quad x=\frac{\pi}{6} \in\left[0, \frac{\pi}{2}\right]$
$\therefore \quad f(0)=2+0=2, f\left(\frac{\pi}{6}\right)=2 \cos \frac{\pi}{6}+\frac{\pi}{6}$
$=2\left(\frac{\sqrt{3}}{2}\right)+\frac{\pi}{6}=\sqrt{3}+\frac{\pi}{6}$
and $\quad f\left(\frac{\pi}{2}\right)=0+\frac{\pi}{2}=\frac{\pi}{2}$
The minimum value of $f(x)$ is $\frac{\pi}{2}$ at $x=\frac{\pi}{2}$.
now, $f^{\prime}(x)=0 \Rightarrow-2 \sin x+1=0$
$\Rightarrow \quad \sin x=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow x=\frac{\pi}{6}$
$\Rightarrow \quad x=\frac{\pi}{6} \in\left[0, \frac{\pi}{2}\right]$
$\therefore \quad f(0)=2+0=2, f\left(\frac{\pi}{6}\right)=2 \cos \frac{\pi}{6}+\frac{\pi}{6}$
$=2\left(\frac{\sqrt{3}}{2}\right)+\frac{\pi}{6}=\sqrt{3}+\frac{\pi}{6}$
and $\quad f\left(\frac{\pi}{2}\right)=0+\frac{\pi}{2}=\frac{\pi}{2}$
The minimum value of $f(x)$ is $\frac{\pi}{2}$ at $x=\frac{\pi}{2}$.
MCQ 31 Mark
In which of the following, function $f(x)=x^2-4 x$ +6 is increasing :
- A$(-\infty, 2) \cup(2, \infty)$
- B$(2, \infty)$
- C$(-\infty, 2)$
- D$(-\infty, 2] \cup(2, \infty)$
MCQ 41 Mark
Find the absolute maximum value of $f(x)=4 x-\frac{1}{2} x^2$ in interval $\left[-2, \frac{9}{2}\right]$.
- ✓8
- B9
- C6
- D10
Answer
View full question & answer→Correct option: A.
8
(A)Here $f(x)=4 x-\frac{1}{2} x^2$
$
\therefore \quad f^{\prime}(x)=4-x
$
now $f^{\prime}(x)=0 \Rightarrow 4-x=0 \Rightarrow x=4$
Thus, $\quad f(-2)=4(-2)-\frac{1}{2}(-2)^2=-8-2=-10$
$\begin{aligned}f(4) & =4(4)-\frac{1}{2}(4)^2=16-8=8 \\\text { and } \quad f\left(\frac{9}{2}\right) & =4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^2=18-\frac{81}{8}=\frac{63}{8}\end{aligned}$
Hence absolute maximum value is 8 at $x=4$.
$
\therefore \quad f^{\prime}(x)=4-x
$
now $f^{\prime}(x)=0 \Rightarrow 4-x=0 \Rightarrow x=4$
Thus, $\quad f(-2)=4(-2)-\frac{1}{2}(-2)^2=-8-2=-10$
$\begin{aligned}f(4) & =4(4)-\frac{1}{2}(4)^2=16-8=8 \\\text { and } \quad f\left(\frac{9}{2}\right) & =4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^2=18-\frac{81}{8}=\frac{63}{8}\end{aligned}$
Hence absolute maximum value is 8 at $x=4$.
MCQ 51 Mark
Function $f(x)=2 x^3-15 x^2+36 x+6$ is increasing in which of interval :
- ✓$(-\infty, 2) \cup(3, \infty)$
- B$(-\infty, 2)$
- C$(-\infty, 2] \cup[3, \infty)$
- D$[3, \infty)$
Answer
View full question & answer→Correct option: A.
$(-\infty, 2) \cup(3, \infty)$
(A)
$
\begin{aligned}
f(x) & =2 x^3-15 x^2+36 x+6 \\
f^{\prime}(x) & =6 x^2-30 x+36 \\
& =6\left(x^2-5 x+6\right)
\end{aligned}
$
$
=6(x-2)(x-3)
$
$f(x)$ is increasing if $f^{\prime}(x)>0$
$
\begin{array}{lc}
\Rightarrow & 6(x-2)(x-3)>0 \\
\Rightarrow & x \in(-\infty 2) \cup(3, \infty)
\end{array}
$
$
\begin{aligned}
f(x) & =2 x^3-15 x^2+36 x+6 \\
f^{\prime}(x) & =6 x^2-30 x+36 \\
& =6\left(x^2-5 x+6\right)
\end{aligned}
$
$
=6(x-2)(x-3)
$
$f(x)$ is increasing if $f^{\prime}(x)>0$
$
\begin{array}{lc}
\Rightarrow & 6(x-2)(x-3)>0 \\
\Rightarrow & x \in(-\infty 2) \cup(3, \infty)
\end{array}
$
MCQ 61 Mark
Function $y=x^2 e^{-x}$ is decreasing in which of interval :
- A$(0,2)$
- B$(2, \infty)$
- C$(-\infty, 0)$
- ✓$(-\infty, 0) \cup(2, \infty)$
Answer
View full question & answer→Correct option: D.
$(-\infty, 0) \cup(2, \infty)$
(D) Given $\quad y=f(x)=x^2 e^{-x}$$
\begin{aligned}
\therefore \quad f^{\prime}(x) & =x^2 e^{-x}(-1)+2 x e^{-x} \\
& =x e^{-x}(2-x)
\end{aligned}
$
now $e^{-x}>0 \quad \forall x \in R$
Hence $f(x)$ will be decreasing if$
\begin{aligned}
f^{\prime}(x)<0 \\
\Rightarrow x(2-x)<0 \\
\Rightarrow x(x-2)>0 \\
\Rightarrow &x \in(-\infty, 0) \cup(2, \infty)
\end{aligned}
$
\begin{aligned}
\therefore \quad f^{\prime}(x) & =x^2 e^{-x}(-1)+2 x e^{-x} \\
& =x e^{-x}(2-x)
\end{aligned}
$
now $e^{-x}>0 \quad \forall x \in R$
Hence $f(x)$ will be decreasing if$
\begin{aligned}
f^{\prime}(x)<0 \\
\Rightarrow x(2-x)<0 \\
\Rightarrow x(x-2)>0 \\
\Rightarrow &x \in(-\infty, 0) \cup(2, \infty)
\end{aligned}
$
MCQ 71 Mark
A function $f: R \rightarrow R$ is defined such that $f(x)=x^3$ +1 , then function :
- Ais not any maximum value
- Bis not any minimum value
- Cis maximum and minimum value
- ✓is neither maximum nor minimum value
Answer
View full question & answer→Correct option: D.
is neither maximum nor minimum value
(D)
It is clear from graph, function is neither maximum nor minimum value.
It is clear from graph, function is neither maximum nor minimum value.
MCQ 81 Mark
Maximum value of $=\left(\frac{1}{x}\right)^x$:
- ✓$e^{1 / e}$
- B$e$
- C$\left(\frac{1}{e}\right)^{1 / e}$
- D$e^{\varepsilon}$
Answer
View full question & answer→Correct option: A.
$e^{1 / e}$
(A)$y=\left(\frac{1}{x}\right)^x$
$\therefore \quad \log y=\log \left(\frac{1}{x}\right)^x=-x \log x$
$\therefore \quad \frac{1}{y} \frac{d y}{d x}=-x \frac{1}{x}-\log x=-(1+\log x)$
$\Rightarrow \quad \frac{d y}{d x}=-y(1+\log x)$
now $\frac{d y}{d x}=0 \Rightarrow 1+\log x=0$
$\Rightarrow \log x=-1 \Rightarrow x=e^{-1}=\frac{1}{e}$
hence maximum value $=\left(\frac{1}{1 / e}\right)^{1 / e}=e^{1 / e}$
$\therefore \quad \log y=\log \left(\frac{1}{x}\right)^x=-x \log x$
$\therefore \quad \frac{1}{y} \frac{d y}{d x}=-x \frac{1}{x}-\log x=-(1+\log x)$
$\Rightarrow \quad \frac{d y}{d x}=-y(1+\log x)$
now $\frac{d y}{d x}=0 \Rightarrow 1+\log x=0$
$\Rightarrow \log x=-1 \Rightarrow x=e^{-1}=\frac{1}{e}$
hence maximum value $=\left(\frac{1}{1 / e}\right)^{1 / e}=e^{1 / e}$
MCQ 91 Mark
If a particle is moving in a straight line. In $t$ time, it will cover $x$ distance according to $x=2 t^2-5 t+$ 1 then find the velocity after 5 sec .
- ✓15
- B20
- C25
- D26
Answer
View full question & answer→Correct option: A.
15
(A)
$
\begin{aligned}
x & =2 t^2-5 t+1 \\
V & =\frac{d x}{d t}=4 t-5 \quad \therefore\left(\frac{d x}{d t}\right)_{t=5}=4 \times 5-5=15
\end{aligned}
$
Hence correct answer is (A).
$
\begin{aligned}
x & =2 t^2-5 t+1 \\
V & =\frac{d x}{d t}=4 t-5 \quad \therefore\left(\frac{d x}{d t}\right)_{t=5}=4 \times 5-5=15
\end{aligned}
$
Hence correct answer is (A).
MCQ 101 Mark
A stone throw upward having speed equation $S =9.8 t-4.9 t^2$ where S in $m / s$. What will be the time to achieve maximum height ?
- A2 sec
- B1$1 / 2 sec$
- ✓1 sec
- D$1 / 2 sec$
Answer
View full question & answer→Correct option: C.
1 sec
(C)
$
\begin{aligned}
S & =9.8 t-4.9 t^2 \\
\frac{d S}{d t} & =9.8-9.8 t
\end{aligned}
$
To achieve maximum height, final velocity will be zero.$
\begin{aligned}
0 & =9.8-9.8 t \\
t & =\frac{9.8}{9.8}=1 sec
\end{aligned}
$
Hence correct option is (C).
$
\begin{aligned}
S & =9.8 t-4.9 t^2 \\
\frac{d S}{d t} & =9.8-9.8 t
\end{aligned}
$
To achieve maximum height, final velocity will be zero.$
\begin{aligned}
0 & =9.8-9.8 t \\
t & =\frac{9.8}{9.8}=1 sec
\end{aligned}
$
Hence correct option is (C).
MCQ 111 Mark
Speed equation of a particle is $S =t^2-4 t+5$, Find the time having zero speed.
- A1 sec
- ✓2 sec
- C3 sec
- D4 sec
Answer
View full question & answer→Correct option: B.
2 sec
(B)
$
\begin{aligned}
S & =t^2-4 t+5 \\
\frac{d S}{d t} & =2 t-4
\end{aligned}
$
ATQ$
\Rightarrow 2 t-4=0 \Rightarrow t=2 sec
$
Hence, correct option is (B).
$
\begin{aligned}
S & =t^2-4 t+5 \\
\frac{d S}{d t} & =2 t-4
\end{aligned}
$
ATQ$
\Rightarrow 2 t-4=0 \Rightarrow t=2 sec
$
Hence, correct option is (B).