JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 4 of 275
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The integral $\displaystyle \int_{0}^{\pi}\sqrt{1+4\sin^{2}\frac{x}{2}-4\sin\frac{x}{2}}\,dx$ equals:
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Solution
Let $A = {x : |x^2 - 10| \le 6}$ and $B = {x : |x - 2| > 1}$. Then
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Solution
$|x^2 - 10| \le 6$
$-6 \le x^2 - 10 \le 6$
$4 \le x^2 \le 16$
$\Rightarrow A = [-4, -2] \cup [2, 4]$
$|x - 2| > 1$
$\Rightarrow B = (-\infty, 1) \cup (3, \infty)$
$A \cup B = (-\infty, 1) \cup (2, \infty)$
$A \cap B = [-4, -2] \cup [3, 4]$
$A - B = [2, 3]$
$B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$
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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Let X be a random variable such that the probability function of a distribution is given by $P(X = 0) = {1 \over 2},P(X = j) = {1 \over {{3^j}}}(j = 1,2,3,...,\infty )$. Then the mean of the distribution and P(X is positive and even) respectively are :
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Solution
The minimum value of the sum of the squares of the roots of
$x^{2}+(3-a)x+1=2a$ is:
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Solution
The area enclosed by the closed curve $\mathcal{C}$ given by the differential equation
$\dfrac{dy}{dx}+\dfrac{x+a}{\,y-2\,}=0,\quad y(1)=0$
is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $\mathcal{C}$ with the $y$-axis. If the normals at $P$ and $Q$ on $\mathcal{C}$ intersect the $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is:
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Solution
Consider a hyperbola $\text{H}$ having centre at the origin and foci on the x-axis.
Let $C_1$ be the circle touching the hyperbola $\text{H}$ and having the centre at the origin.
Let $C_2$ be the circle touching the hyperbola $\text{H}$ at its vertex and having the centre at one of its foci.
If areas (in sq units) of $C_1$ and $C_2$ are $36\pi$ and $4\pi$, respectively,
then the length (in units) of latus rectum of $\text{H}$ is:
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Solution
Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC divides the arc AC such that $\dfrac{\text{length of arc }AB}{\text{length of arc }BC}=\dfrac{1}{5}$, and $\overrightarrow{OC}=\alpha\,\overrightarrow{OA}+\beta\,\overrightarrow{OB}$, then $\alpha+\sqrt{2}\,(\sqrt{3}-1)\,\beta$ is equal to:
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Solution
Let radius $=1$ and take $O$ as origin
Arc $AC$ subtends $\frac{\pi}{2}$ ⇒ take $A(1,0),\ C(0,1)$
Given $\frac{AB}{BC}= \frac{1}{5}$ ⇒ $\angle AOB=\frac{\pi}{12}$
So $B=\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
$\overrightarrow{OC}=\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}$
$(0,1)=\alpha(1,0)+\beta\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
Comparing components:
$\alpha+\beta\cos\frac{\pi}{12}=0$
$\beta\sin\frac{\pi}{12}=1$
$\beta=\frac{1}{\sin\frac{\pi}{12}}=\frac{1}{\frac{\sqrt6-\sqrt2}{4}}=\sqrt6+\sqrt2$
$\alpha=-\beta\cos\frac{\pi}{12}=-(\sqrt6+\sqrt2)\cdot\frac{\sqrt6+\sqrt2}{4}=-2-\sqrt3$
Now,
$\alpha+\sqrt2(\sqrt3-1)\beta$
$=(-2-\sqrt3)+\sqrt2(\sqrt3-1)(\sqrt6+\sqrt2)$
$\sqrt2(\sqrt6+\sqrt2)=2\sqrt3+2$
So expression:
$=(-2-\sqrt3)+(2\sqrt3+2)(\sqrt3-1)$
$=(-2-\sqrt3)+(6-2\sqrt3+2\sqrt3-2)$
$=(-2-\sqrt3)+4=2-\sqrt3$
$\boxed{2-\sqrt3}$
Arc $AC$ subtends $\frac{\pi}{2}$ ⇒ take $A(1,0),\ C(0,1)$
Given $\frac{AB}{BC}= \frac{1}{5}$ ⇒ $\angle AOB=\frac{\pi}{12}$
So $B=\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
$\overrightarrow{OC}=\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}$
$(0,1)=\alpha(1,0)+\beta\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
Comparing components:
$\alpha+\beta\cos\frac{\pi}{12}=0$
$\beta\sin\frac{\pi}{12}=1$
$\beta=\frac{1}{\sin\frac{\pi}{12}}=\frac{1}{\frac{\sqrt6-\sqrt2}{4}}=\sqrt6+\sqrt2$
$\alpha=-\beta\cos\frac{\pi}{12}=-(\sqrt6+\sqrt2)\cdot\frac{\sqrt6+\sqrt2}{4}=-2-\sqrt3$
Now,
$\alpha+\sqrt2(\sqrt3-1)\beta$
$=(-2-\sqrt3)+\sqrt2(\sqrt3-1)(\sqrt6+\sqrt2)$
$\sqrt2(\sqrt6+\sqrt2)=2\sqrt3+2$
So expression:
$=(-2-\sqrt3)+(2\sqrt3+2)(\sqrt3-1)$
$=(-2-\sqrt3)+(6-2\sqrt3+2\sqrt3-2)$
$=(-2-\sqrt3)+4=2-\sqrt3$
$\boxed{2-\sqrt3}$
Let A={-3,-2,-1,0,1,2,3} and R be a relation on A defined by xRy iff 2x-y $\in\{0,1\}$. Let $l$ be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l+m+n is equal to
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Solution
Given $A=\{-3,-2,-1,0,1,2,3\}$ and $xRy \iff 2x-y\in\{0,1\}$
So $y=2x$ or $y=2x-1$
All valid pairs in $A$:
$(-1,-2),\ (-1,-3),\ (0,0),\ (0,-1),\ (1,2),\ (1,1),\ (2,3)$
So $l=7$
Reflexive pairs needed: $(x,x)$ for all 7 elements
Present: $(0,0),(1,1)$ ⇒ missing $5$ ⇒ $m=5$
For symmetry, add reverse of non-symmetric pairs:
Missing: $(-2,-1),\ (-3,-1),\ (-1,0),\ (2,1),\ (3,2)$ ⇒ $n=5$
$l+m+n=7+5+5=17$
$\boxed{17}$
So $y=2x$ or $y=2x-1$
All valid pairs in $A$:
$(-1,-2),\ (-1,-3),\ (0,0),\ (0,-1),\ (1,2),\ (1,1),\ (2,3)$
So $l=7$
Reflexive pairs needed: $(x,x)$ for all 7 elements
Present: $(0,0),(1,1)$ ⇒ missing $5$ ⇒ $m=5$
For symmetry, add reverse of non-symmetric pairs:
Missing: $(-2,-1),\ (-3,-1),\ (-1,0),\ (2,1),\ (3,2)$ ⇒ $n=5$
$l+m+n=7+5+5=17$
$\boxed{17}$
Let $z = \left(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}\right)^5 + \left(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}\right)^5.$
If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$, then :
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Solution
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