Applied Maths MCQ

(c) $y_c=a+b x$
Explanation: $y_c=a+b x$

(b) $- xe ^{-x}+C$
$
\begin{array}{l}
\text { Explanation: } I=\int(x-1) e^{-x} \\
=\int xe^{-x} dx-\int e^{-x} dx \\
=-xe^{-x}-\int 1 \cdot(-) e^{-x} dx-\int e^{-x} dx+c \\
=-xe^{-x}+\int e^{-x} dx-\int e^{x} dx+c \\
=-xe^{-x}+C
\end{array}
$

(a) none of these
Explanation: $Z=4 x+2 y$
Subject to constraints
$
\begin{array}{l}
2 x+3 y \leq 18 \\
x+y \geq \text { and } \\
x, y \geq 0
\end{array}
$

(d) $17 km / hr$
Explanation: 12 km upstream in $48 min \Rightarrow$ it will cover 15 km in 1 hr Speed of stream $=2 km / hr$
$\therefore$ Speed of boat in still water $=15+2=17 km / hr$

(d) $[-7,3]$
$
\begin{array}{l}
\text { Explanation: }|x+2| \leq 5 \\
\Rightarrow-5 \leq x+2 \leq 5 \\
\Rightarrow-7 \leq x \leq 3 \\
\Rightarrow x \in[-7,-3]
\end{array}
$

(c) 5
Explanation: $(18 \times 10)(\bmod 7)=18(\bmod 7) \times 10(\bmod 7)$
$
\begin{array}{l}
=4(\bmod 7) \times 3(\bmod 7) \\
=12(\bmod 7)=5
\end{array}
$

(d) I
Explanation: Given that $A^2=A$
Calculating value of $(I+A)^3-7 A$ :
$
\begin{array}{l}
(I+A)^3-7 A=I^3+A^3+3 I^2 A+3 I A^2-7 A \\
=I+A^2 \cdot A+3 A+3 A+3 A^2-7 A\left(I^n=I \text { and } I \cdot A=A\right) \\
=I+A \cdot A+3 A+3 A-7 A\left(A^2=A\right) \\
=I+A+3 A+3 A-7 A
\end{array}
$
Hence, $(I+A)^3-7 A=I$

(c) 20 m
Explanation: To reach the winning post A will have to cover a distance of $(500-140) m =360 m$ While A covers 3 m , B covers 4 m .
While A covers $360 m, B$ covers $=\frac{4 \times 360}{3}=480 m$
$\therefore$ A wins by 20 m .

(d) $x y=C$
Explanation: $xy = C$

(d) $9: 4$
Explanation: Let events A: a heart is drawn
Event B: a King card is drawn.
The probability of winning the bet $= P ( A$ or B $)$
$
\begin{array}{l}
P(A \text { or } B)=P(A)+P(B)-P(A \cap B) \\
=\frac{13}{52}+\frac{4}{52}-\frac{1}{52} \text { (There is one king of heart) } \\
=\frac{13+4-1}{52} \frac{16}{52}=\frac{4}{13} \\
\therefore \text { Probability of losing the bet }=1-\frac{4}{13}=\frac{9}{13}
\end{array}
$
The odds against an event are the ratio of the number of ways the event cannot happen to the number of ways it can happen.
$\therefore$ the odds against drawing a heart or a king are $\frac{9}{13}: \frac{4}{13}=9: 4$.

(b) $\frac{15}{16}$
Explanation: $n =4, p = q =\frac{1}{2}$
$
\begin{array}{l}
P(X \geq 1)=1-P(X=0) \\
P(X \geq 1)=1-\left(\frac{1}{2}\right)^4 \\
P(X \geq 1)=\frac{15}{16}
\end{array}
$

(b) ₹ 1400
$\begin{array}{l}\text { Explanation: As PV }=\frac{500}{0.04}\left[1-(1.04)^{-3}\right] \\ =12500[1-0.888]\end{array}$
= 12500 × 0.112 = ₹ 1400

(d) $n_1+n_2-2$
Explanation: $n _1+ n _2-2$