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Prakhar Gupta 3 years, 7 months ago
Rishab Jain 3 years, 7 months ago
Kapil Adhikari 3 years, 7 months ago
Yes, Word problem questions are deleted for session 2020-21
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Vivek Chaurasia 3 years, 7 months ago
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Kapil Adhikari 3 years, 7 months ago
Y=sin(logx)
dy/dx=cos(logx) .1/x)
Now,by quotient rule
d²y/dx²=[x(-sin(logx).1/x)- cos(logx).1]/x²
d²y/dx²=[-sin(logx)-cos(logx)] /x²
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Queen Manisha 3 years, 7 months ago
Queen Manisha 3 years, 7 months ago
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Sia ? 3 years, 7 months ago
Let there be a natural number n,
We know that n divides n, which implies nRn.
So, Every natural number is related to itself in relation R.
Thus relation R is reflexive .
Let there be three natural numbers a,b,c and let aRb,bRc
aRb implies a divides b and bRc implies b divides c, which combinedly implies that a divides c i.e. aRc.
So, Relation R is also transitive .
Let there be two natural numbers a,b and let aRb,
aRb implies a divides b but it can't be assured that b necessarily divides a.
For ex, 2R4 as 2 divides 4 but 4 does not divide 2 .
Thus Relation R is not symmetric .
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Sia ? 3 years, 7 months ago
{tex}\text{Let 2 white ball be denoted by}~ w_1, w_{2}~ \& ~red~ ball~ by~ \mathrm{r}{/tex}
Then the sample space is
{tex}\mathrm{S}=\left\{\left(\mathrm{w}_{1}, \mathrm{w}_{1}\right),\left(\mathrm{w}_{2}, \mathrm{w}_{2}\right),\left(\mathrm{w}_{1}, \mathrm{w}_{2}\right)\right. \left(w_{2}, w_{1}\right),\left(r, w_{1}\right),\left(w_{1}, r\right) \left(r, w_{2}\right),\left(w_{2}, r\right) (r, r)\}{/tex}
Let X: Number of red balls in two draws
We can see there are 0 red, 1 red, and 2 red balls
{tex}So, X=0~ or ~X=1 ~or~ X=2{/tex}
Posted by ਸਹਿਜ ਸਿੰਘ 3 years, 8 months ago
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Prakhar Gupta 3 years, 7 months ago
0Thank You