JEE Main 2021 Mathematics Question Paper with Answer and Solution

(A) Given $f(x) = \int_{a}^{x} g(t) \, dt$. By the Fundamental Theorem of Calculus,$f'(x) = g(x)$.
Since $f(x) = 0$ has $5$ distinct roots in $(a, b)$,by Rolle's Theorem,$f'(x) = 0$ must have at least $5 - 1 = 4$ distinct roots in $(a, b)$.
Thus,$g(x) = 0$ has at least $4$ distinct roots in $(a, b)$.
Now,by Rolle's Theorem applied to $g(x)$,since $g(x) = 0$ has at least $4$ distinct roots,$g'(x) = 0$ must have at least $4 - 1 = 3$ distinct roots in $(a, b)$.
The equation $g(x) g'(x) = 0$ is satisfied if $g(x) = 0$ or $g'(x) = 0$.
Since $g(x) = 0$ has at least $4$ roots and $g'(x) = 0$ has at least $3$ roots,the total number of distinct roots for $g(x) g'(x) = 0$ is at least $4 + 3 = 7$ roots in $(a, b)$.

(A) For $0 \leq x \leq \pi$,$f(x) = \max \{\sin t : 0 \leq t \leq x\}$. Since $\sin t$ increases from $0$ to $1$ on $[0, \pi/2]$ and decreases from $1$ to $0$ on $[\pi/2, \pi]$,we have:
$f(x) = \sin x$ for $0 \leq x \leq \pi/2$,and $f(x) = 1$ for $\pi/2 < x \leq \pi$.
For $x > \pi$,$f(x) = 2 + \cos x$.
Now,check continuity at $x = \pi$:
Left-hand limit: $\lim_{x \to \pi^-} f(x) = 1$.
Right-hand limit: $\lim_{x \to \pi^+} f(x) = 2 + \cos(\pi) = 2 - 1 = 1$.
Since $f(\pi) = 1$,the function is continuous at $x = \pi$.
Check differentiability:
At $x = \pi/2$: Left derivative is $\cos(\pi/2) = 0$,right derivative is $0$. So,$f$ is differentiable at $x = \pi/2$.
At $x = \pi$: Left derivative is $0$ (since $f(x)=1$ for $x \in [\pi/2, \pi]$). Right derivative is $-\sin(\pi) = 0$. So,$f$ is differentiable at $x = \pi$.
For $x > \pi$,$f(x) = 2 + \cos x$,which is differentiable everywhere. Thus,$f$ is differentiable everywhere in $(0, \infty)$.

(B) Given curves are $y = x+2$ and $y = x^2$.
To find the points of intersection,set $x^2 = x+2$.
$x^2 - x - 2 = 0$
$(x-2)(x+1) = 0$
So,$x = 2$ and $x = -1$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$:
$A = \int_{-1}^{2} (x+2 - x^2) \, dx$
$A = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$
$A = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 - \frac{-1}{3} \right)$
$A = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$
$A = \left( 6 - \frac{8}{3} \right) - \left( \frac{3 - 12 + 2}{6} \right)$
$A = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$.

(B) Given differential equation: $(x-x^{3}) dy = (y+yx^{2}-3x^{4}) dx$
Rearranging the terms:
$(x-x^{3}) dy - y(1+x^{2}) dx = -3x^{4} dx$
Divide by $x(1-x^{2})$ or rearrange to identify the structure:
$x dy - x^{3} dy = y dx + yx^{2} dx - 3x^{4} dx$
$x dy - y dx = x^{3} dy + yx^{2} dx - 3x^{4} dx$
Divide by $x^{2}$:
$\frac{x dy - y dx}{x^{2}} = \frac{x^{3} dy + yx^{2} dx}{x^{2}} - 3x^{2} dx$
$d(\frac{y}{x}) = d(xy) - d(x^{3})$
Integrating both sides:
$\frac{y}{x} = xy - x^{3} + C$
Given $y(3)=3$,substitute $x=3, y=3$:
$\frac{3}{3} = 3(3) - 3^{3} + C$
$1 = 9 - 27 + C$
$1 = -18 + C \Rightarrow C = 19$
So,the equation is $\frac{y}{x} = xy - x^{3} + 19$
For $x=4$:
$\frac{y}{4} = 4y - 64 + 19$
$\frac{y}{4} = 4y - 45$
$y = 16y - 180$
$15y = 180$
$y = 12$

(B) Using the vector triple product formula,$\vec{a} = \vec{b} \times (\vec{b} \times \vec{c}) = (\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}$.
Given $|\vec{a}| = \sqrt{2}$,$|\vec{b}| = 1$,$|\vec{c}| = 2$,and $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta = 1 \cdot 2 \cdot \cos \theta = 2 \cos \theta$.
Substituting these values,$\vec{a} = (2 \cos \theta) \vec{b} - (1)^2 \vec{c} = 2 \cos \theta \vec{b} - \vec{c}$.
Now,calculate $|\vec{a}|^2 = |2 \cos \theta \vec{b} - \vec{c}|^2 = (2 \cos \theta)^2 |\vec{b}|^2 + |\vec{c}|^2 - 2(2 \cos \theta) (\vec{b} \cdot \vec{c})$.
$|\vec{a}|^2 = 4 \cos^2 \theta (1) + 4 - 4 \cos \theta (2 \cos \theta) = 4 \cos^2 \theta + 4 - 8 \cos^2 \theta = 4 - 4 \cos^2 \theta$.
Since $|\vec{a}|^2 = (\sqrt{2})^2 = 2$,we have $2 = 4 - 4 \cos^2 \theta$.
$4 \cos^2 \theta = 2 \Rightarrow \cos^2 \theta = \frac{1}{2}$.
Since $0 < \theta < \frac{\pi}{2}$,$\cos \theta = \frac{1}{\sqrt{2}}$,which implies $\theta = \frac{\pi}{4}$.
Therefore,$1 + \tan \theta = 1 + \tan(\frac{\pi}{4}) = 1 + 1 = 2$.

(A) Let $X$ be the number of correct answers. Since the student guesses,$X$ follows a binomial distribution $B(n=8, p=1/2)$.
We want to find the smallest $n$ such that $P(X \geq n) < \frac{1}{2}$.
$P(X \geq n) = \sum_{r=n}^{8} {}^{8}C_{r} (\frac{1}{2})^{r} (\frac{1}{2})^{8-r} = \frac{1}{2^{8}} \sum_{r=n}^{8} {}^{8}C_{r} < \frac{1}{2}$.
$\sum_{r=n}^{8} {}^{8}C_{r} < 2^{7} = 128$.
We know that $\sum_{r=0}^{8} {}^{8}C_{r} = 2^{8} = 256$.
So,$\sum_{r=n}^{8} {}^{8}C_{r} = 256 - \sum_{r=0}^{n-1} {}^{8}C_{r} < 128$.
$\sum_{r=0}^{n-1} {}^{8}C_{r} > 128$.
For $n=5$,$\sum_{r=0}^{4} {}^{8}C_{r} = {}^{8}C_{0} + {}^{8}C_{1} + {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} = 1 + 8 + 28 + 56 + 70 = 163$.
Since $163 > 128$,the condition holds for $n=5$.
For $n=4$,$\sum_{r=0}^{3} {}^{8}C_{r} = 1 + 8 + 28 + 56 = 93$,which is not greater than $128$.
Thus,the smallest value of $n$ is $5$.

(B) The given equation is $x^{3}-3x^{2}y-xy^{2}+3y^{3}=0$.
Factoring the expression:
$x^{2}(x-3y) - y^{2}(x-3y) = 0$
$(x^{2}-y^{2})(x-3y) = 0$
$(x-y)(x+y)(x-3y) = 0$
This implies $x=y$ or $x=-y$ or $x=3y$.
Since $x, y \in N$,$x=-y$ is impossible as $x, y > 0$.
Thus,the relation is $R = \{(x, y) \in N \times N : x=y \text{ or } x=3y\}$.
$1$. Reflexivity: For any $x \in N$,$(x, x) \in R$ because $x=x$ is true. Thus,$R$ is reflexive.
$2$. Symmetry: Consider $(3, 1) \in R$ because $3=3(1)$. However,$(1, 3) \notin R$ because $1 \neq 3$ and $1 \neq 3(3)=9$. Thus,$R$ is not symmetric.
$3$. Transitivity: Consider $(9, 3) \in R$ (since $9=3(3)$) and $(3, 1) \in R$ (since $3=3(1)$). For $R$ to be transitive,$(9, 1)$ must be in $R$. But $9 \neq 1$ and $9 \neq 3(1)=3$. Thus,$R$ is not transitive.
Therefore,$R$ is reflexive but neither symmetric nor transitive.

(A) Given that $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$.
Subtracting the second equation from the first,we get:
$A^{5}-A^{3} B^{2} = B^{5}-A^{2} B^{3}$
$A^{3}(A^{2}-B^{2}) = B^{5}-A^{2} B^{3}$
Rearranging the terms:
$A^{3}(A^{2}-B^{2}) + B^{3}(A^{2}-B^{2}) = 0$
$(A^{3}+B^{3})(A^{2}-B^{2}) = 0$
Since $(A^{2}-B^{2})$ is an invertible matrix,we can post-multiply by its inverse $(A^{2}-B^{2})^{-1}$:
$(A^{3}+B^{3})(A^{2}-B^{2})(A^{2}-B^{2})^{-1} = 0 \cdot (A^{2}-B^{2})^{-1}$
$A^{3}+B^{3} = 0$
Therefore,the determinant of the zero matrix is $|A^{3}+B^{3}| = |0| = 0$.

(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.
We can observe that $A = I + N$,where $I$ is the identity matrix and $N = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
Note that $N^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $N^3 = 0$.
Using the binomial expansion,$A^n = (I+N)^n = I + nN + \frac{n(n-1)}{2}N^2$.
Thus,$A^n = \begin{bmatrix} 1 & n & n + \frac{n(n-1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & n & \frac{n^2+n}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix}$.
The sum of elements of $A^n$ is $S_n = 1 + n + \frac{n^2+n}{2} + 0 + 1 + n + 0 + 0 + 1 = 3 + 2n + \frac{n^2+n}{2} = 3 + \frac{5n+n^2}{2}$.
Sum of elements of $M = \sum_{n=1}^{20} S_n = \sum_{n=1}^{20} (3 + \frac{5}{2}n + \frac{1}{2}n^2) = 3(20) + \frac{5}{2} \frac{20(21)}{2} + \frac{1}{2} \frac{20(21)(41)}{6}$.
$= 60 + 525 + 1435 = 2020$.

(D) The equation of the line passing through $Q(3,-4,-5)$ and $R(2,-3,1)$ is given by $\frac{x-3}{2-3} = \frac{y-(-4)}{-3-(-4)} = \frac{z-(-5)}{1-(-5)} = r$.
This simplifies to $\frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = r$.
Any point on this line is given by $(x, y, z) = (-r+3, r-4, 6r-5)$.
Since this point lies on the plane $2x+y+z=7$,we substitute the coordinates into the plane equation:
$2(-r+3) + (r-4) + (6r-5) = 7$.
$-2r + 6 + r - 4 + 6r - 5 = 7$.
$5r - 3 = 7 \Rightarrow 5r = 10 \Rightarrow r = 2$.
Substituting $r=2$ back into the point coordinates,we get the point of intersection $T = (-2+3, 2-4, 6(2)-5) = (1, -2, 7)$.
The distance between $P(3,4,4)$ and $T(1,-2,7)$ is given by the distance formula:
$PT = \sqrt{(1-3)^2 + (-2-4)^2 + (7-4)^2}$.
$PT = \sqrt{(-2)^2 + (-6)^2 + (3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.

(B) Let $I = \int_{0}^{\pi} \sin^{3} x e^{-\sin^{2} x} dx$. Since $\sin^{3}(\pi - x) = \sin^{3} x$ and $\sin^{2}(\pi - x) = \sin^{2} x$,we have $I = 2 \int_{0}^{\pi/2} \sin^{3} x e^{-\sin^{2} x} dx$.
Using $\sin^{3} x = \sin x (1 - \cos^{2} x)$,$I = 2 \int_{0}^{\pi/2} \sin x e^{-\sin^{2} x} dx - 2 \int_{0}^{\pi/2} \sin x \cos^{2} x e^{-\sin^{2} x} dx$.
For the second integral,let $u = \cos x$,$du = -\sin x dx$. However,it is easier to use $v = \sin x$,$dv = \cos x dx$.
Alternatively,rewrite as $I = 2 \int_{0}^{\pi/2} \sin x (1 - \sin^{2} x) e^{-\sin^{2} x} dx$. Let $t = \sin x$,$dt = \cos x dx$.
Using integration by parts on $2 \int_{0}^{\pi/2} \sin x \cos^{2} x e^{-\sin^{2} x} dx$,let $u = \cos x$ and $dv = \sin x \cos x e^{-\sin^{2} x} dx$. Then $du = -\sin x dx$ and $v = \frac{1}{2} e^{-\sin^{2} x}$.
Evaluating the integral leads to $I = 2 - \frac{3}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$.
Comparing with $\alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$,we get $\alpha = 2$ and $\beta = 3$.
Thus,$\alpha + \beta = 2 + 3 = 5$.

(C) Given vectors are $\vec{a}=(1, -\alpha, \beta)$,$\vec{b}=(3, \beta, -\alpha)$,and $\vec{c}=(-\alpha, -2, 1)$,where $\alpha, \beta \in \mathbb{Z}$.
From $\vec{a} \cdot \vec{b} = -1$:
$(1)(3) + (-\alpha)(\beta) + (\beta)(-\alpha) = -1$
$3 - 2\alpha\beta = -1 \Rightarrow 2\alpha\beta = 4 \Rightarrow \alpha\beta = 2$.
From $\vec{b} \cdot \vec{c} = 10$:
$(3)(-\alpha) + (\beta)(-2) + (-\alpha)(1) = 10$
$-3\alpha - 2\beta - \alpha = 10 \Rightarrow -4\alpha - 2\beta = 10 \Rightarrow 2\alpha + \beta = -5$.
Since $\alpha\beta = 2$ and $\beta = -5 - 2\alpha$,we substitute $\beta$:
$\alpha(-5 - 2\alpha) = 2 \Rightarrow -5\alpha - 2\alpha^2 = 2 \Rightarrow 2\alpha^2 + 5\alpha + 2 = 0$.
$(2\alpha + 1)(\alpha + 2) = 0$. Since $\alpha$ is an integer,$\alpha = -2$.
Then $\beta = -5 - 2(-2) = -5 + 4 = -1$.
Now,calculate the scalar triple product $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & -(-2) & -1 \\ 3 & -1 & -(-2) \\ -(-2) & -2 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix}$.
$= 1(-1 + 4) - 2(3 - 4) - 1(-6 + 2)$
$= 1(3) - 2(-1) - 1(-4) = 3 + 2 + 4 = 9$.

(B) Given the differential equation: $dy = e^{\alpha x + y} dx$.
Rearranging the terms to separate the variables:
$e^{-y} dy = e^{\alpha x} dx$.
Integrating both sides:
$\int e^{-y} dy = \int e^{\alpha x} dx$
$-e^{-y} = \frac{e^{\alpha x}}{\alpha} + C \quad \dots(i)$
Using the condition $y(0) = \log_{e}(\frac{1}{2}) = -\log_{e} 2$:
$-e^{-(-\log_{e} 2)} = \frac{e^{\alpha(0)}}{\alpha} + C$
$-e^{\log_{e} 2} = \frac{1}{\alpha} + C$
$-2 = \frac{1}{\alpha} + C \quad \dots(ii)$
Using the condition $y(\log_{e} 2) = \log_{e} 2$:
$-e^{-\log_{e} 2} = \frac{e^{\alpha \log_{e} 2}}{\alpha} + C$
$-\frac{1}{2} = \frac{2^{\alpha}}{\alpha} + C \quad \dots(iii)$
Subtracting equation $(ii)$ from equation $(iii)$:
$(-\frac{1}{2}) - (-2) = \frac{2^{\alpha}}{\alpha} - \frac{1}{\alpha}$
$\frac{3}{2} = \frac{2^{\alpha} - 1}{\alpha}$
For $\alpha = 2$:
$\frac{2^{2} - 1}{2} = \frac{4 - 1}{2} = \frac{3}{2}$.
Thus,the value of $\alpha$ is $2$.

(A) Let $I = \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,but it is more convenient to write $8 - \sin 2x = 9 - (1 + \sin 2x) = 9 - (\sin x + \cos x)^2$.
So,$I = \int \frac{\cos x - \sin x}{\sqrt{9 - (\sin x + \cos x)^2}} dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{3^2 - t^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we have $I = \sin^{-1}(\frac{t}{3}) + c$.
Substituting $t = \sin x + \cos x$ back,we get $I = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c$.
Comparing this with the given expression $a \sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$,we find $a = 1$ and $b = 3$.
Thus,the ordered pair $(a, b)$ is $(1, 3)$.

(A) Let the point be $P(a, 6, 9)$ and its mirror image be $P'(20, b, -a-9)$.
The midpoint $M$ of $PP'$ is $\left(\frac{a+20}{2}, \frac{6+b}{2}, \frac{9-a-9}{2}\right) = \left(\frac{a+20}{2}, \frac{6+b}{2}, -\frac{a}{2}\right)$.
Since $M$ lies on the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} = k$,we have:
$\frac{\frac{a+20}{2}-3}{7} = \frac{\frac{6+b}{2}-2}{5} = \frac{-\frac{a}{2}-1}{-9} = k$.
From the first and third parts: $\frac{a+14}{14} = \frac{a+2}{18} \implies 18a + 252 = 14a + 28 \implies 4a = -224 \implies a = -56$.
Substituting $a = -56$ into the line equation: $\frac{-56+14}{14} = \frac{6+b-4}{10} \implies -3 = \frac{b+2}{10} \implies b+2 = -30 \implies b = -32$.
Thus,$|a+b| = |-56 - 32| = |-88| = 88$.