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Question 22 Marks
Differentiate the given function w.r.t. $\mathrm{x}: \frac{\sqrt{(x-3)\left(x^{2}+4\right)}}{3 x^{2}+4 x+5}$
Answer
Let $\mathrm{y}=\sqrt{\frac{(x-3)\left(x^{2}+4\right)}{3 x^{2}+4 x+5}}$, taking logarithm of both sides and applying the properties of logarithm, we get
$\log y=\frac{1}{2}\left[\log (x-3)+\log \left(x^{2}+4\right)-\log \left(3 x^{2}+4 x+5\right)\right]$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2}\left[\frac{1}{x-3} \cdot 1+\frac{1}{x^{2}+4} \cdot 2 x-\frac{1}{3 x^{2}+4 x+5} \cdot(3 \cdot 2 \mathrm{x}+4)\right]$
$\Rightarrow \frac{d y}{d x}=\frac{y}{2}\left(\frac{1}{x-3}+\frac{2 x}{x^{2}+4}-\frac{6 x+4}{3 x^{2}+4 x+5}\right)$
$\therefore \frac{d y}{d x}=\frac{1}{2} \sqrt{\frac{(x-3)\left(x^{2}+4\right)}{3 x^{2}+4 x+5}}\left[\frac{1}{x-3}+\frac{2 x}{x^{2}+4}-\frac{6 x+4}{3 x^{2}+4 x+5}\right]$
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Question 32 Marks
If $\mathrm{y}=\mathrm{x}+\frac{1}{x}$, prove that $x^{2} \frac{d y}{d x}-\mathrm{xy}+2=0$.
Answer
Let assume: $y=x+\frac{1}{x}...(1)$
Diff. (1), w.r.t. 'x', we get
$\frac{d y}{d x}=\frac{d}{d x}(x)+\frac{d}{d x}\left(\frac{1}{x}\right)$
$\frac{d y}{d x}=1+\left(\frac{-1}{x^{2}}\right)$
$ \frac{d y}{d x}=1-\frac{1}{x^2} \ldots \ldots (2)$
We have to prove
$x^{2} \frac{d y}{d x}-x y+2=0$
L.H.S. $x^{2} \frac{d y}{d x}-x y+2$
$=x^{2}\left(1-\frac{1}{x^{2}}\right)-x\left(\frac{x+1}{x}\right)+2$
Using (1) and (2)
$=x^{2}-1-x^{2}-1+2$
$=0=$ R.H.S
Hence proved.
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Question 42 Marks
A clock gains 5 seconds in 2 minutes and was set right at 9:00 a.m. If it shows 2:30 in the afternoon on the same day. What is the correct time?
Answer
Given that, the clock gains 5 seconds in 2 minutes
$\Rightarrow$ it gains $12 \times 5=60$ seconds i.e. 1 minute in $12 \times 2=24$ minutes
$\Rightarrow$ when the incorrect clock moves 25 minutes, the correct clock moves 24 minutes.
Now, from 9:00 a.m. to 2:30 p.m. on the same day the time passed by incorrect clock
$=5$ hours 30 minutes $=330$ minutes.
When an incorrect clock moves 330 minutes, the correct clock moves
$=\frac{24}{25} \times 330$ minutes $=316$ minutes 48 seconds
$=5$ hours 16 minutes 48 seconds
Hence, the correct time is 2:6:48 p.m.
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Question 62 Marks
If RAHUL is coded as 22-5-12-25-16, then how will you code VIRAT?
Answer
Here R is coded as 22 which is its actual positions $18+4$.
Similarly, A is coded as $1+4$ i.e. 5 .
H is coded as $8+4=12$, U is coded as $21+4=25$ and L is coded as $12+4=16$.
V is equivalent to $22+4=26$
I is equivalent to $9+4=13$
$R$ is equivalent to $18+4=22$
A is equivalent to $1+4=5$
and T is equivalent to $20+4=24$
$\therefore$ 'VIRAT' will be coded as 26-13-22-5-24
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Question 72 Marks
A and $B$ together can build a wall in 30 days. If $A$ is twice as good a workman as $B$, in how many days will $A$ alone finish the work?
Answer
Since A is twice as good a workman as B,
A's one day work = B's 2 days work
$\Rightarrow$ B's one day work = A's $\frac{1}{2}$ day work ..(i)
Since A and B together can build a wall in 30 days,
$\therefore$ A's one day work + B's one day work $=\frac{1}{30}$
$\Rightarrow$ A's one day work + A's $\frac{1}{2}$ day work $=\frac{1}{30}$ [using (i)]
$\Rightarrow$ A's $1+\frac{1}{2}$ i.e. $\frac{3}{2}$ days work $=\frac{1}{30}$
$\Rightarrow$ A's one day work $=\frac{2}{3} \times \frac{1}{30}=\frac{1}{45}$
$\therefore$ An alone can complete the work in 45 days.

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2 Marks Questions - Applied Maths STD 11 Science Questions - Vidyadip