JEE MAIN 2020 Previous Year Questions (PYQs) – Page 1 of 15
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The region represented by {z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1} is also given by the inequality :{z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1}
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Let a , b, c , d and p be any non zero distinct real numbers such that(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
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$\lim_{x \to 1} \left( \dfrac{\int_{0}^{(x-1)^{2}} t \cos(t^{2}) \, dt}{(x-1)\sin(x-1)} \right)$
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If I1 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$ andI2 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$ such that I2 = $\alpha $I1then $\alpha $ equals to :
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The position of a moving car at time t is given by f(t) = at2 + bt + c, t > 0, where a, b and c are realnumbers greater than 1. Then the average speed of the car over the time interval [t1, t2] isattained at the point :
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If $\alpha $ and $\beta $ be two roots of the equation x2 – 64x + 256 = 0. Then the value of${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$ is :
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If $\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$ and $\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$(n, a > 1) then the standard deviation of n observations x1, x2, ..., xn is :
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If {p} denotes the fractional part of the number p, then $\left\{ {{{{3^{200}}} \over 8}} \right\}$, is equal to :
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If f(x + y) = f(x)f(y) and $\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$ , x, y $ \in $ N, where N is the set of all natural number, then thevalue of${{f\left( 4 \right)} \over {f\left( 2 \right)}}$ is :
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Solution
Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :
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Solution
Let the 11 consecutive natural numbers be
$1, 2, 3, \dots, 11.$
Total ways to choose any 3 numbers = $\displaystyle \binom{11}{3} = 165.$
Now, we need to count the number of 3-number selections that can form an arithmetic progression (A.P.) with positive common difference.
For an A.P., let the middle term be $a$ and common difference be $d>0$. Then the three terms are: $(a-d,\ a,\ a+d)$ These must all lie between $1$ and $11$.
That means $1 \le a-d$ and $a+d \le 11$ ⟹ $d \le \min(a-1,\ 11-a)$
Now we count possible values of $d$ for each $a$:
Total = $1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25.$
Hence, number of favorable triplets = $25.$
Therefore, $\displaystyle P = \frac{25}{165} = \frac{5}{33}.$
Final Answer: $\boxed{\dfrac{5}{33}}$
Now, we need to count the number of 3-number selections that can form an arithmetic progression (A.P.) with positive common difference.
For an A.P., let the middle term be $a$ and common difference be $d>0$. Then the three terms are: $(a-d,\ a,\ a+d)$ These must all lie between $1$ and $11$.
That means $1 \le a-d$ and $a+d \le 11$ ⟹ $d \le \min(a-1,\ 11-a)$
Now we count possible values of $d$ for each $a$:
| $a$ | $\min(a-1,\ 11-a)$ | Possible $d$ values |
|---|---|---|
| 1 | 0 | – |
| 2 | 1 | 1 |
| 3 | 2 | 1,2 |
| 4 | 3 | 1,2,3 |
| 5 | 4 | 1,2,3,4 |
| 6 | 5 | 1,2,3,4,5 |
| 7 | 4 | 1,2,3,4 |
| 8 | 3 | 1,2,3 |
| 9 | 2 | 1,2 |
| 10 | 1 | 1 |
| 11 | 0 | – |
Hence, number of favorable triplets = $25.$
Therefore, $\displaystyle P = \frac{25}{165} = \frac{5}{33}.$
Final Answer: $\boxed{\dfrac{5}{33}}$
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