JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 10 of 275
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The sum of the absolute maximum and minimum values of the function $f(x)=\lvert x^{2}-5x+6\rvert-3x+2$ in the interval $[-1,3]$ is equal to:
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Solution
Let $f(x)=3\sqrt{x-2}+\sqrt{4-x}$ be a real-valued function.
If $\alpha$ and $\beta$ are respectively the minimum and maximum values of $f$,
then $\alpha^2+2\beta^2$ is equal to:
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Solution
The number of complex numbers $z$ satisfying $|z|=1$ and $\left|\dfrac{z}{\overline{z}}+\dfrac{\overline{z}}{z}\right|=1$ is:
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Solution
Given $|z|=1 \Rightarrow z=e^{i\theta}$
$\frac{z}{\overline{z}}=e^{2i\theta},\quad \frac{\overline{z}}{z}=e^{-2i\theta}$
So,
$\left|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right| =|e^{2i\theta}+e^{-2i\theta}|=|2\cos 2\theta|$
Given $|2\cos 2\theta|=1 \Rightarrow |\cos 2\theta|=\frac{1}{2}$
Now in $[0,2\pi)$:
$2\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
⇒ $\theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6}$
But $z=e^{i\theta}$ is periodic with $2\pi$, so each valid $\theta$ gives **2 distinct $z$** over full cycle of $2\theta$
Hence total solutions $=4\times 2=8$
$\frac{z}{\overline{z}}=e^{2i\theta},\quad \frac{\overline{z}}{z}=e^{-2i\theta}$
So,
$\left|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right| =|e^{2i\theta}+e^{-2i\theta}|=|2\cos 2\theta|$
Given $|2\cos 2\theta|=1 \Rightarrow |\cos 2\theta|=\frac{1}{2}$
Now in $[0,2\pi)$:
$2\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
⇒ $\theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6}$
But $z=e^{i\theta}$ is periodic with $2\pi$, so each valid $\theta$ gives **2 distinct $z$** over full cycle of $2\theta$
Hence total solutions $=4\times 2=8$
Let $a>0$. If the function $f(x)=6x^3-45ax^2+108a^2x+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1x_2=54$, then $a+x_1+x_2$ is equal to
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Solution
$f(x)=6x^3-45ax^2+108a^2x+1$
$f'(x)=18x^2-90ax+108a^2$
$=18(x^2-5ax+6a^2)$
So critical points:
$x_1,x_2=2a,3a$
Given $x_1x_2=54$
⇒ $6a^2=54 \Rightarrow a^2=9$
$ \Rightarrow a=3$ (since $a>0$)
$x_1+x_2=5a=15$
$a+x_1+x_2=3+15=18$
$f'(x)=18x^2-90ax+108a^2$
$=18(x^2-5ax+6a^2)$
So critical points:
$x_1,x_2=2a,3a$
Given $x_1x_2=54$
⇒ $6a^2=54 \Rightarrow a^2=9$
$ \Rightarrow a=3$ (since $a>0$)
$x_1+x_2=5a=15$
$a+x_1+x_2=3+15=18$
Let $S = \{(x, y) \in \mathbb{R}^2 : \dfrac{y^2}{1 + r} - \dfrac{x^2}{1 - r} = 1 ; \, r \neq \pm 1 \}$.
Then $S$ represents :
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Solution
If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$,
then the sum of the first 15 terms of this A.P. is:
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Solution
If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha\log|x|+\beta x^{2}+x$ then
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Solution
Let $z$ be the complex number satisfying $|z - 5| \le 3$ and having maximum positive principal argument. Then $34\left|\frac{5z - 12}{5iz + 16}\right|^2$ is equal to:
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Solution
$|z - 5| \le 3$
For $\arg(z)$ to be maximum, $z$ lies at point $P$
$z = (4, 4)$
$= \left(4\left(\frac{4}{5}\right),; 4\left(\frac{3}{5}\right)\right)$
$= \left(\frac{16}{5},; \frac{12}{5}\right)$
Now
$34\left|\frac{5z - 12}{5iz + 16}\right|^2$
$= 34\left|\frac{(16 + 12i) - 12}{(16i + 12i^2) - 12i^2}\right|^2$
$= 34\left|\frac{4 + 12i}{16i + 4}\right|^2$
$= 34\left(\frac{16 + 144}{256 + 16}\right)$
$= 34\left(\frac{160}{272}\right) = 20$
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: $\boxed{\dfrac{5}{33}}$
Let C be the set of all complex numbers. Let ${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $ ${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $ and ${S_3} = \{ z \in C||z - \overline z | \ge 8\} $. Then the number of elements in ${S_1} \cap {S_2} \cap {S_3}$ is equal to :
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Solution
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