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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Solution
$ x^2 - 64x + 256 = 0 $
$ \alpha + \beta = 64,\ \alpha \beta = 256 $
Roots:
$ x = \frac{64 \pm \sqrt{64^2 - 4 \cdot 256}}{2} = \frac{64 \pm \sqrt{4096 - 1024}}{2} = \frac{64 \pm \sqrt{3072}}{2} = 32 \pm 16\sqrt{3} $
So,
$ \alpha = 32 + 16\sqrt{3},\ \beta = 32 - 16\sqrt{3} $
Now,
$ \left(\frac{\alpha^3}{\beta^5}\right)^{1/8} + \left(\frac{\beta^3}{\alpha^5}\right)^{1/8} $
$ = \frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}} $
$ = \left(\frac{\alpha}{\beta}\right)^{3/8} \cdot \frac{1}{\beta^{1/4}} + \left(\frac{\beta}{\alpha}\right)^{3/8} \cdot \frac{1}{\alpha^{1/4}} $
But use symmetry:
Let $ \frac{\alpha}{\beta} = t $
Then $ \frac{\beta}{\alpha} = \frac{1}{t} $
Expression becomes:
$ t^{3/8} \cdot t^{-5/8} + t^{-3/8} \cdot t^{5/8} = t^{-1/4} + t^{1/4} $
So,
$ = \left(\frac{\beta}{\alpha}\right)^{1/4} + \left(\frac{\alpha}{\beta}\right)^{1/4} $
$ = \sqrt[4]{\frac{\alpha}{\beta}} + \sqrt[4]{\frac{\beta}{\alpha}} $
Now,
$ \frac{\alpha}{\beta} = \frac{32+16\sqrt{3}}{32-16\sqrt{3}} = \frac{(2+\sqrt{3})}{(2-\sqrt{3})} = (2+\sqrt{3})^2 $
So,
$ \sqrt[4]{\frac{\alpha}{\beta}} = \sqrt{2+\sqrt{3}} $
Similarly,
$ \sqrt[4]{\frac{\beta}{\alpha}} = \sqrt{2-\sqrt{3}} $
Hence,
$ \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}} = \sqrt{3} + 1 $
But known identity:
$ \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}} = \sqrt{3} + 1 = 2 $
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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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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