Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — 3D — Equation of planes
Difficulty: Moderate

Question: Equation of a plane which passes through the point of
intersection of lines [IMAGE] and [IMAGE] and at greatest distance from the point (0, 0, 0) is


Correct option: 2

Solution:
Let a point [IMAGE] of the first line also lies
on the second line
Then [IMAGE] Hence the point of intersection P of the two lines is (4, 3, 5)
Equation of plane perpendicular to OP where O is (0, 0, 0) and passing
through P is
Subject: Mathematics — Circles — Equation of circle
Difficulty: Tough

Question: Equation of the circle which circumscribes the square formed by
the lines xy - 9x - 2y + 18 = 0 and xy - 5x - 6y + 30 = 0 is

1) x^2 + y^2 - 4x - 10y + 21 = 0
2) x^2 + y^2 - 8x - 14y + 57 = 0
3) x^2 + y^2 - 12x - 10y + 53 = 0
4) x^2 + y^2 - 7x - 15y + 64 = 0

Correct option: 2

Solution:
Given equations represent the lines x = 2, y = 9, x = 6, y = 5
which forms a square of each side equal to 4, vertex A(2, 5), C(6, 9).
Centre of the circle is the midpoint of AC and the radius is (1/2) AC.
Hence the equation of the circle is
$\Rightarrow$ x^2 + y^2 - 8x - 14y + 16 + 49 - 8 = 0
$\Rightarrow$ x^2 + y^2 - 8x - 14y + 57 =
Subject: Mathematics — Sequences and Series — Geometric progression
Difficulty: Moderate

Question: A geometric progression (G.P.) consists of even number of terms.
If the sum of all terms is times the sum of terms occupying odd
places. Then common ratio is


Answer: 4

Solution:
$a + ar + ar^{2} + \_\_\_\ ar^{2n - 1} = 5\lbrack a + ar^{2} + ar^{4} + \_\_\_\_ ar^{2n - 2})$
$\frac{r^{2n} - 1}{r - 1} = \frac{5(r^{2n} - 1)}{(r - 1)(r + 1)} \Rightarrow r = 4$
Subject: Mathematics — Permutations and Combinations — Divisibility
Difficulty: Tough

Question: n is selected from the set and the number
2^n + 3^n + 5^n is formed. Total number of ways of selecting n so
that the formed number is divisible by 4 is equal to

1) 50
2) 49
3) 48
4) 51

Correct option: 2

Solution:
If n is odd.
$3^{n} = 4\lambda_{1} - 1,5^{n} = 4\lambda_{2} + 1$
$\Rightarrow$ 2^n + 3^n + 5^n  is divisible by 4 if n > 2
Thus, n = 3, 5, 7, 9.....
Subject: Mathematics — Functions — Range of functions
Difficulty: Moderate

Question: If range of the function f(x) = sin^-1 x + 2tan^-1 x +
x^2 + 4x + 1 is then (p + q) equals


Answer: 4

Solution:
f (x) = sin^-1 x + 2tan^-1 x + x^2 + 4x + 1
domain of f (x) is 
f (x) 
f (x) is increasing function in its domain so range of f (x) is
$\lbrack - \pi - 2,\pi + 6\rbrack$
[IMAGE] p + q = 4
Subject: Mathematics — Differential equation — Formation of Differential equation
Difficulty: Moderate

Question: The differential equation representing the family of ellipses
having foci either on the x-axis or on the y-axis, centre at the origin
and passing through the point (0, 3) is:
ij fcUnq (0, 3)

1) xy y'- y^2 + 9 = 0
2) xy y" + x = 0
3) xy y' + y^2 - 9 = 0
4) x + y y" = 0

Correct option: 1

Solution:
Equation of ellipse
nh?kZ o`Ùk dk lehdj.k
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Passes (0, 3) ls xqtjrk 
$\Rightarrow$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$......(1)
$\frac{x}{a^{2}} = - \frac{y}{9}\frac{dy}{dx}$
$\frac{2x}{a^{2}} + \frac{2y}{9}\frac{dy}{dx} = 0$......(2)
$\frac{1}{a^{2}} = - \frac{y}{9x}y^{'}$
By lehdj.k (1) &o (2) ls equation is
$- \frac{xy}{9}y^{'} + \frac{y^{2}}{9} = 1$
$\Rightarrow$ xy y' - y^2 + 9 = 0
Subject: Mathematics — Functions — Greatest integer function
Difficulty: Tough

Question: If [x] denotes the greatest integer [IMAGE] x,
then the system of linear equations

1) Have infinitely many solutions if
$\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right)$and has a unique
solutions if $\theta \in \left( \pi,\frac{7\pi}{6} \right)$.
2) Has a unique solution if
$\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right) \cup \left( \pi,\frac{7\pi}{6} \right)$
3) Has a unique solution if
$\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right)$ and have
infinitely many solutions if
$\theta \in \left( \pi,\frac{7\pi}{6} \right)$.
4) Have infinitely many solutions if
$\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right) \cup \left( \pi,\frac{7\pi}{6} \right)$

Correct option: 1

Solution:
if $\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right)$
⇒ $cos\theta \in \left( - \frac{1}{2},0 \right) \Rightarrow - cos\theta \in (0,1/2)$
$sin\theta \in \left( \frac{\sqrt{3}}{2},1 \right)\& cot\theta \in \left( - \frac{1}{\sqrt{3}},0 \right)$
If
$\theta \in \left( \pi,\frac{7\pi}{6} \right) \Rightarrow cos\theta \in \left( - 1,\frac{- \sqrt{3}}{2} \right) \Rightarrow - cos\theta \in \left( \frac{\sqrt{3}}{2},1 \right)$
$sin\theta \in \left( - \frac{1}{2},0 \right)\& cot\theta \in (\sqrt{3},\infty)$
Hence in
$\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right) \Rightarrow \lbrack sin\theta\rbrack = 0;\lbrack - cos\theta\rbrack = 0;\lbrack cot\theta\rbrack = - 1$
$Hence\ 0 = 0\& - x + y = 0Have\ infinitely\ solutions$
$\begin{matrix}
\ \\
\ \ 
\end{matrix}in\ \theta \in \left( \pi,\frac{7\pi}{6} \right) \Rightarrow \lbrack sin\theta\rbrack = - 1;\lbrack - cos\theta\rbrack = 0;\lbrack cot\theta\rbrack = 1,2,3,\ldots\ldots\ldots\ldots\ldots$
Hence $- x - 0.y = 0 \Rightarrow x = 0$
$\lbrack cot\theta\rbrack x + y = 0 \Rightarrow 1 \cdot x + y = 0\ or\ 2 \cdot x + y = 0\ or\ $.........
each line well cut x=0 at exactly one point. Hence unique solution.
Subject: Mathematics — Continuity and derivability — Continuity and derivability
Difficulty: Moderate

Question: Let $f(x) = \left\{ \begin{matrix}
\int_{0}^{x}\mspace{2mu}\{ 1 + |1 - t|\} dt,\ \ \ \ x > 2 \\
5x - 7,\ \ \ \ x \leq 2
\end{matrix} \right.\ $
Then f(2) is

1) Continuous but not differentiable
2) is
3) Non continuous and non differentiable
4) Differentiable but not continuous

Correct option: 1

Solution:
$f(x) = \left\{ \begin{matrix}
\int_{0}^{x}\mspace{2mu}\{ 1 + |1 - t|\} dt, & x > 2 \\
5x - 7, & x \leq 2
\end{matrix} \right.\ $,
$f(x) = \int_{0}^{1}\mspace{2mu}(1 + (1 - t))dt + \int_{1}^{x}\mspace{2mu}$(1 +
(1 - t )) dt
$$= \left( 2 - - \right)$$
$$\therefore f(x) = 1 + 
> 5x - 7\ \ \ \ x \leq 2 \\
> 1 + \ \ \ x > 2
> \right.\ $$
L = R = V = 2 at x = 2 [IMAGE] continuous
$$= - h) - 7 - = 5$$
 \rightarrow = 2$$
$L^{'} \pm R^{'}$ so not differentiable at x = 2