JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 15 of 275
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Let the maximum value of $(\sin^{-1}x)^2 + (\cos^{-1}x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m\pi^2}{n}$, where $\gcd(m,n)=1$. Then m+n is equal to ______.
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Solution
$(\sin^{-1}x)^2 + (\cos^{-1}x)^2$
$= (\sin^{-1}x + \cos^{-1}x)^2 - 2\sin^{-1}x\cos^{-1}x$
$= \frac{\pi^2}{4} - 2(\sin^{-1}x)\left(\frac{\pi}{2} - \sin^{-1}x\right)$
$= 2\left(\sin^{-1}x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$
Maximum occurs at $\sin^{-1}x = -\frac{\pi}{3}$
$\Rightarrow 2\left(\frac{\pi}{3} + \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$
$= 2\left(\frac{7\pi}{12}\right)^2 + \frac{\pi^2}{8}$
$= \frac{49\pi^2}{72} + \frac{9\pi^2}{72} = \frac{29\pi^2}{36}$
$\Rightarrow m = 29,; n = 36$
$\Rightarrow m+n = 65$
Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?
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Solution
Let there be two families with 3 members each (say Family A and Family B), and one family with 4 members (Family C).
Step 1: Since members of the same family must sit together, treat each family as a single block.
Thus, there are 3 blocks: A, B, and C.
They can be arranged in $3! = 6$ ways.
Step 2: Now arrange members within each family:
Family A (3 members): $3!$ ways
Family B (3 members): $3!$ ways
Family C (4 members): $4!$ ways
Step 3: Total number of arrangements = $3! \times 3! \times 3! \times 4!$
Step 4: Simplify:
$3! = 6$ and $4! = 24$
$\Rightarrow (3!)^3\times4!$
∴ Total number of ways = $\boxed{5184}$
Step 1: Since members of the same family must sit together, treat each family as a single block.
Thus, there are 3 blocks: A, B, and C.
They can be arranged in $3! = 6$ ways.
Step 2: Now arrange members within each family:
Family A (3 members): $3!$ ways
Family B (3 members): $3!$ ways
Family C (4 members): $4!$ ways
Step 3: Total number of arrangements = $3! \times 3! \times 3! \times 4!$
Step 4: Simplify:
$3! = 6$ and $4! = 24$
$\Rightarrow (3!)^3\times4!$
∴ Total number of ways = $\boxed{5184}$
If $\sin \theta + \cos \theta = {1 \over 2}$, then 16(sin(2$\theta$) + cos(4$\theta$) + sin(6$\theta$)) is equal to :
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If the line $x-1=0$ is a directrix of the hyperbola $kx^{2}-y^{2}=6$, then the hyperbola passes through the point:
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Solution
The area of the region given by $\{(x,y):\, xy\le 8,\ 1\le y\le x^{2}\}$ is:
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Solution
Let $y=y(x)$ be the solution of the differential equation
$(x^{2}+4)^{2}\,dy+(2x^{3}y+8xy-2)\,dx=0.$
If $y(0)=0$, then $y(2)$ is equal to:
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Solution
If the area of the region $\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq \mathrm{a}+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}>0\right\}$ is $\frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}$, then the value of $a$ is
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A line passing through the point $A(-2,0)$ touches the parabola $P: y^2=x-2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $\overline{AB}$, parabola $P$ and the $x$-axis is:
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Solution
The value of $\cos \dfrac{\pi}{22}\cdot \cos \dfrac{\pi}{23}\cdot \ldots \cdot \cos \dfrac{\pi}{210}\cdot \sin \dfrac{\pi}{210}$ is –
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Solution
A value of $\theta \in \left( {0,{\pi \over 3}} \right)$, for which
$\left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$, is :
$\left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$, is :
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Solution
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