JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 1 of 275
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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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Solution
The number of real solutions of the equation, x2 $-$ |x| $-$ 12 = 0 is :
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Solution
Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \alpha \hat{k}$, $\alpha > 0$.
If the projection of $\vec{a} \times \vec{b}$ on the vector $-\hat{i} + 2\hat{j} - 2\hat{k}$ is $30$, then $\alpha$ is equal to:
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Solution
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :
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Solution
If the system of equations
$x+\big(\sqrt{2}\sin\alpha\big)y+\big(\sqrt{2}\cos\alpha\big)z=0$
$x+(\cos\alpha)y+(\sin\alpha)z=0$
$x+(\sin\alpha)y-(\cos\alpha)z=0$
has a non-trivial solution, then $\alpha\in\left(0,\frac{\pi}{2}\right)$ is equal to:
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Solution
Let $\mathrm{I}(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$. If $\mathrm{I}(37)-\mathrm{I}(24)=\frac{1}{4}\left(\frac{1}{\mathrm{~b}^{\frac{1}{13}}}-\frac{1}{\mathrm{c}^{\frac{1}{13}}}\right), \mathrm{b}, \mathrm{c} \in \mathcal{N}$, then $3(\mathrm{~b}+\mathrm{c})$ is equal to
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Solution
Let $t=\frac{x-11}{x+15}$
Then, $dx=\frac{26}{(1-t)^2}dt$
Also, $(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}=t^{\frac{11}{13}}(x+15)^2$
and $x+15=\frac{26}{1-t}$
So, $I=\int \frac{\frac{26}{(1-t)^2}dt}{t^{\frac{11}{13}}\cdot \frac{26^2}{(1-t)^2}}$
$I=\frac{1}{26}\int t^{-\frac{11}{13}}dt$
$I=\frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}}$
$I=\frac14\left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$
Now, $I(37)-I(24)=\frac14\left[\left(\frac{26}{52}\right)^{\frac{2}{13}}-\left(\frac{13}{39}\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\left(\frac12\right)^{\frac{2}{13}}-\left(\frac13\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\frac{1}{4^{\frac{1}{13}}}-\frac{1}{9^{\frac{1}{13}}}\right]$
So, $b=4,\ c=9$
$3(b+c)=3(4+9)=39$
$\boxed{39}$
Then, $dx=\frac{26}{(1-t)^2}dt$
Also, $(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}=t^{\frac{11}{13}}(x+15)^2$
and $x+15=\frac{26}{1-t}$
So, $I=\int \frac{\frac{26}{(1-t)^2}dt}{t^{\frac{11}{13}}\cdot \frac{26^2}{(1-t)^2}}$
$I=\frac{1}{26}\int t^{-\frac{11}{13}}dt$
$I=\frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}}$
$I=\frac14\left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$
Now, $I(37)-I(24)=\frac14\left[\left(\frac{26}{52}\right)^{\frac{2}{13}}-\left(\frac{13}{39}\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\left(\frac12\right)^{\frac{2}{13}}-\left(\frac13\right)^{\frac{2}{13}}\right]$
$=\frac14\left[\frac{1}{4^{\frac{1}{13}}}-\frac{1}{9^{\frac{1}{13}}}\right]$
So, $b=4,\ c=9$
$3(b+c)=3(4+9)=39$
$\boxed{39}$
Let the three sides of a triangle be on the lines 4x-7y+10=0, x+y=5 and 7x+4y=15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x=0, y=0 and x+y=1 is
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Solution
Triangle vertices:
$A\left(-\frac{5}{3},\frac{20}{3}\right),\ B(1,2),\ C\left(\frac{25}{11},\frac{30}{11}\right)$
Slope $AC=-1$ ⇒ altitude from $B$ has slope $1$ ⇒ $y=x+1$
Solving with another altitude gives orthocentre $H_1=(1,2)$
Second triangle $(0,0),(1,0),(0,1)$ ⇒ right angled ⇒ $H_2=(0,0)$
Distance $=\sqrt{(1)^2+(2)^2}=\sqrt{5}$
$\boxed{\sqrt{5}}$
$A\left(-\frac{5}{3},\frac{20}{3}\right),\ B(1,2),\ C\left(\frac{25}{11},\frac{30}{11}\right)$
Slope $AC=-1$ ⇒ altitude from $B$ has slope $1$ ⇒ $y=x+1$
Solving with another altitude gives orthocentre $H_1=(1,2)$
Second triangle $(0,0),(1,0),(0,1)$ ⇒ right angled ⇒ $H_2=(0,0)$
Distance $=\sqrt{(1)^2+(2)^2}=\sqrt{5}$
$\boxed{\sqrt{5}}$
If the angle of intersection at a point where two circles with radii 5cm and 12cm intersect is $90^\circ$, then the length (in cm) of their common chord is:
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Solution
$AB = \dfrac{2 \cdot r_1 \cdot r_2}{\sqrt{r_1^2 + r_2^2}}$
$ = \dfrac{2 \times 5 \times 12}{\sqrt{25 + 144}} = \dfrac{120}{\sqrt{169}}$
$ = \dfrac{120}{13}$
$\boxed{AB = \dfrac{120}{13} \text{ cm}}$
$\displaystyle \lim_{x\to 0}\frac{\sin(\pi \cos^{2}x)}{x^{2}}$ is equal to :
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Solution
If the area of the region ${(x,y) : 1 - 2x \le y \le 4 - x^2,; x \ge 0,; y \ge 0}$ is $\frac{\alpha}{\beta}$, $\alpha, \beta \in \mathbb{N}$, $\gcd(\alpha,\beta) = 1$, then the value of $(\alpha + \beta)$ is:
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Solution
Required area
$= \frac{2}{3}\times 8 - \frac{1}{2}\times \frac{1}{2}\times 1$
$= \frac{16}{3} - \frac{1}{4} = \frac{61}{12} = \frac{\alpha}{\beta}$
$\Rightarrow \alpha + \beta = 73$
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