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Applied Maths MCQ

CBSE Model Paper 1 · CBSE STD 12 Science (CBSE - English Medium). 18 questions.

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18 questions · timed · auto-graded

MCQ 11 Mark
The order and degree of a differential equation $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^4+x^{\frac{1}{5}}=0$; respectively, are
  • A
    2 and 4
  • ✓
    2 and 1
  • C
    2 and 3
  • D
    3 and 3
Answer
Correct option: B.
2 and 1
(B) 2 and 1
order is 2 and degree is 1.
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MCQ 31 Mark
For a $3 \times 3$ matrix if $\operatorname{adj} A=2 A^{-1}$, find $\left|3 A A^T\right|$
  • ✓
    108
  • B
    12
  • C
    54
  • D
    8
Answer
Correct option: A.
108
(A) 108
$\operatorname{adj} A =2 A^{-1} \Rightarrow A^{-1}=\frac{1}{2}(\operatorname{adj} A)$
$
\therefore|A|=2
$
Now, $\left|3 AA ^{ T }\right|=3^3 \times| A |^2=108$
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MCQ 41 Mark
Shown below is a curve.


Image
$L_1$ is the tangent to any point $(x, y)$ on the curve.
$L_2$ is the line that connects the point $(x, y)$ to the origin.
The slope of $L_1$ is one third of the slope of $L_2$.
Then the differential equation, using the given conditions is:
  • ✓
    $\frac{d y}{d x}=\frac{y}{3 x}$
  • B
    $\frac{d y}{d x}=\frac{y}{x}$
  • C
    $\frac{d y}{d x}=\frac{x}{3 y}$
  • D
    $\frac{d y}{d x}=\frac{3 y}{x}$
Answer
Correct option: A.
$\frac{d y}{d x}=\frac{y}{3 x}$
(A) $\frac{d y}{d x}=\frac{y}{3 x}$
The slope of $L_1$ at any arbitrary point $(x, y)$ is $\frac{d y}{d x}$.
The slope of $L_2$ that connects the point $(x, y)$ to the origin is $\frac{y- 0 }{ x - 0 }=\frac{ y }{ x }$
Now,
$
\begin{aligned}
& \frac{d y}{d x}=\frac{1}{3} \times \frac{y}{x} \\
\therefore & \frac{d y}{d x}=\frac{y}{3 x} .
\end{aligned}
$
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MCQ 51 Mark
The present value of a sequence of payments of ₹ 800 made at the end of every 6 month and continuing forever. If money is worth 4%per annum compounded semi-annually, then the present value of the sequence is:
  • A
    ₹ 20000
  • ✓
    ₹ 40000
  • C
    ₹ 60000
  • D
    ₹ 80000
Answer
Correct option: B.
₹ 40000
B
R=₹ 800. $i=\frac{4}{200}=0.02$
$P=\frac{R}{i}=\frac{800}{0.02}$ = ₹40000.
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MCQ 61 Mark
The component of a time series attached to long term variations is termed as
  • A
    Seasonal variations
  • B
    Irregular variations
  • ✓
    Secular trend variations
  • D
    Cyclic variations
Answer
Correct option: C.
Secular trend variations
(C) Secular trend variations
Secular trend variations are considered as long-term variation, attributable to factor such as population change, technological progress and large –scale shifts in consumer tastes.
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MCQ 71 Mark
The values of $\frac{1}{x}$ for the given values of $x \in(-1,3)-\{0\}$ is
  • A
    $\left(-1, \frac{1}{3}\right) \cup(3, \infty)$
  • ✓
    $(-\infty,-1) \cup\left(\frac{1}{3}, \infty\right)$
  • C
    $\left(-\frac{1}{3}, 1\right)$
  • D
    $\left(-\frac{1}{3},-1\right)$
Answer
Correct option: B.
$(-\infty,-1) \cup\left(\frac{1}{3}, \infty\right)$
B
$
x \in(-1,3)-\{0\} \Rightarrow x \in(-1,0) \cup(0,3)
$
When $x \in(- 1 , 0 )$ then $\frac{ 1 }{ x } \in(-\infty,- 1 ) \ldots \ldots( i )$
When $x \in( 0 , 3 )$ then $\frac{ 1 }{ x } \in\left(\frac{ 1 }{ 3 }, \infty\right) \ldots .$. (ii)
From $(i) \&(i i)$, we have $\frac{1}{x} \in(-\infty,-1) \cup\left(\frac{1}{3}, \infty\right)$
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MCQ 81 Mark
If it is currently 6:00 pm in 12 hours clock then what will be the time after 375 hours?
  • A
    6 am
  • B
    6 pm
  • ✓
    9 am
  • D
    9 pm
Answer
Correct option: C.
9 am
(C) 9 am
$
\begin{array}{l}
375 \text { hours }=(24 \times 15+15) \text { hours } \\
\therefore 375(\bmod 24)=15
\end{array}
$
Therefore, it will be 9 am after 375 hours.
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MCQ 91 Mark
In a large consignment of electric bulbs 5% of a batch of batteries are defective. A random sample of 80 is taken for inspection with replacement. Then the Variance of the number of defectives in the sample, is
Answer
Correct option: B.
$\frac{19}{5}$
(B) $\frac{19}{5}$
This is a binomial distribution with $n= 8 0 , p=5 \%=\frac{ 1 }{ 2 0 }$. If $X$ is the binomial random variable for the number of defectives then $X$ is $B\left( 8 0 , \frac{ 1 }{ 2 0 }\right)$.
So, $\sigma^2=n p q=80 \times \frac{1}{20} \times \frac{19}{20}=\frac{19}{5}$.
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MCQ 101 Mark
A fish jumps out of the water surface and follows the parabolic path $y=6 x-x^2-8 ; 2 \leq x \leq 4$ . The fish reaches the highest height in its path at ( 3,1)  . The slope of the path of the fish at ( 3,1 ) is
  • ✓
    0
  • B
    1
  • C
    2
  • D
    3
Answer
Correct option: A.
0
(A) 0
The equation of the parabolic path $y=6 x-x^2-8 ; 2 \leq x \leq 4$
$\begin{array}{l}\frac{d y}{d x}=6-2 x \\ \Rightarrow{\frac{d y}{d x_{x=3}}}_x=6-2 \times 3=0\end{array}$
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MCQ 111 Mark
The original cost of a machine is ₹1200000 and the scarp value of the machine after a useful life of 3 years is ₹300000, then the book value of the machine at the end 2 years is
  • A
    ₹100000
  • B
    ₹250000
  • ✓
    ₹ 600000
  • D
    ₹800000
Answer
Correct option: C.
₹ 600000
(C) ₹ 600000
Annual depreciation $=\frac{1200000-300000}{3}$ =₹ 300000
$\therefore$ Book value of the asset at the end of 2 years =₹ $(1200000-2 \times 300000)$ =₹ 600000.
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MCQ 121 Mark
A player rolls one fair die. If the die shows an odd number, the player wins the value that appears on the die, else loses half the value that appears on it. The expected gain of the player is
  • A
    $-\frac{1}{2}$
  • B
    𝟎
  • ✓
    $\frac{1}{2}$
  • D
    1
Answer
Correct option: C.
$\frac{1}{2}$
(C) $\frac{1}{2}$
Number on the die$x_i$$p_i$$p_i x_i$
1 1$\frac{1}{6}$$\frac{1}{6}$
2-1$\frac{1}{6}$$-\frac{1}{6}$
3 3$\frac{1}{6}$$\frac{3}{6}$
4-2$\frac{1}{6}$$-\frac{2}{6}$
5 5$\frac{1}{6}$$\frac{5}{6}$
6 3$\frac{1}{6}$$-\frac{3}{6}$
Expected gain $=E(X)=\sum p_i x_i=\frac{3}{6}=\frac{1}{2}$
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MCQ 141 Mark
For the purpose of $t$ - test of significance, a random sample of size $(n) 2025$ is drawn from a normal population, then the degree of freedom $(v)$ is
  • A
    $2025^{2025}$
  • B
    $2024^{2025}$
  • C
    2025
  • ✓
    2024
Answer
Correct option: D.
2024
(D) 2024
Here $n = 2 0 2 5$
$\therefore$ Degree of freedom $= 2 0 2 5 - 1 = 2 0 2 4$.
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MCQ 151 Mark
The absolute minimum value of the function $f ( x )= 4 x -\frac{ 1 }{ 2 } x ^2$ in the interval Image is:
Answer
(C)  -10
$ f(x)=4 x-\frac{1}{2} x^2 $
Being a polynomial function f(x)  is differentiable $\forall x \in\left(-2, \frac{9}{2}\right)$
$f^{\prime}(x)=4-x$
$f^{\prime}(x)=4-x=0 \Rightarrow x=4$
For the function $f(x)=4 x-\frac{1}{2} x^2$ in the interval Image , the end points are $x=-2 \& x=\frac{9}{2}$
$\therefore$ The absolute minimum value of the function $f(x)=4 x-\frac{1}{2} x^2$ in the interval Image  is 
$\operatorname{Min}\left\{f(-2), f(4), f\left(\frac{9}{2}\right)\right\}=\operatorname{Min}\left\{-10,8, \frac{63}{8}\right\}=-10$


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MCQ 161 Mark
Let the cost function for a manufacturer is given by $C(x)=\frac{x^3}{3}-x^2+2 x$ (In rupees)
  • ✓
    The marginal cost decreases from 0 to 1 and then increases onwards.
  • B
    The marginal cost increases from 0 to 1 and then decreases onwards.
  • C
    Marginal cost decreases as production level increases from zero.
  • D
    Marginal cost increases as production level increases from zero.
Answer
Correct option: A.
The marginal cost decreases from 0 to 1 and then increases onwards.
(A) The marginal cost decreases from 0 to 1 and then increases onwards.
The cost function for a manufacturer is given by $C(x)=\frac{x^3}{3}-x^2+2 x$ (in rupees). The marginal cost function is given by $M C ( x )=\frac{d C}{d x }= x ^2- 2 x + 2$ $ M C^{\prime}(x)=2 x-2 $
So, the marginal cost decreases from 0 to 1 and then increases onwards
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MCQ 171 Mark
Sampling which provides for a known non-zero equal chance of selection is
  • ✓
    Systematic sampling
  • B
    Convenience sampling
  • C
    Quota sampling
  • D
    Purposive sampling
Answer
Correct option: A.
Systematic sampling
(A) Systematic sampling
Systematic Sampling as it is a type of probability sampling while others are types of non-probability sampling. (When selection of objects from the population is random, then objects of the population have an equal probability i.e., has a known non-zero equal chance of selection. In other words, in probability sampling, sample units are selected at random.)
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MCQ 181 Mark
The area (in sq units) bounded by the curve $y=\sqrt{x}$, the $x$-axis, $x=1$ and $x=4$ is
  • A
    $\frac{11}{3}$
  • B
    $\frac{1}{4}$
  • ✓
    $\frac{14}{3}$
  • D
    $\frac{13}{3}$
Answer
Correct option: C.
$\frac{14}{3}$
(C) $\frac{14}{3}$
The required area is given by $\left|\int_1^4(\sqrt{x}) d x\right|=\left[\frac{X^{\frac{3}{2}}}{\frac{3}{2}}\right]_1^4\left|=\left|\frac{2}{3}(8-1)\right|=\frac{ 1 4 }{3}\right.$ squnits.
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MCQ - Applied Maths STD 12 Science Questions - Vidyadip