datavorous/entrance-exam-dataset huggingface dataset:
If ${ }^{2 n} C_3:{ }^n C_2=44: 3$, then for which of the following values of $r$, the value of ${ }^n C_r$ will be 15

Options:
A $r=3$
B $r=4$
C $r=6$
D $r=5$

The correct answer is:
$r=5$

Solution:
${ }^{15} C_{3 r}={ }^{15} C_{r+3} \Rightarrow{ }^{15} C_{15-3 r}={ }^{15} C_{r+3}$
$\Rightarrow 15-3 r=r+3 \Rightarrow r=3$.

Topic: Mathematics, Permutation Combination, JEE Main
If 2sinα1+cos2α=17 and 1-cos2β2=110, α, β∈0,π2, then tanα+2β, is equal to

Options:

The correct answer is:
1

Solution:
Given,

2sinα1+cos2α=17 and 1-cos2β2=110, α, β∈0,π2

Now using cos2θ=2cos2θ-1=1-2sin2θ, we can write

2sin⁡α2cos⁡α=17 and 2sin⁡β2=110

tan⁡α=17 and sin⁡β=110 or tan⁡β=13
tan2β=2tan⁡β1-tan2⁡β=2.131-19=34
tan⁡α+2β=tanα+tan2β1-tanαtan⁡2β=17+341-17·34=4+21282528=1

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2020 (08 Jan Shift 2)
If 2sinâ¡Î±1+cosâ¡Î±+sinâ¡Î±=34, then the value of 1-cosâ¡Î±+sinâ¡Î±1+sinâ¡Î± is equal to

Options:
A 43
B 34
C 14
D 74

The correct answer is:
34

Solution:
Let, x=2sinâ¡Î±1+cosâ¡Î±+sinâ¡Î±
Let, y=1-cosâ¡Î±+sinâ¡Î±1+sinâ¡Î±.

Now, xy=2sinâ¡Î±1+sinâ¡Î±1+sinâ¡Î±2-cos2â¡Î±
âxy=2sinâ¡Î±1+sinâ¡Î±1+sin2â¡Î±+2sinâ¡Î±-1+sin2â¡Î±=1
âx=y=34

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If 2sin3x+sin2xcosx+4sinx-4=0 has exactly 3 solutions in the interval 0,nπ2,n∈N, then the roots of the equation x2+nx+(n-3)=0 belong to :

Options:
A (0,∞)
B (-∞,0)
C -172,172
D Z

The correct answer is:
(-∞,0)

Solution:
Given: 2sin3x+sin2xcosx+4sinx-4=0

⇒2sin3x+2sinx·cos2x+4sinx-4=0

⇒2sin3x+2sinx·1-sin2x+4sinx-4=0

⇒2sin3x+2sinx-2sin3x+4sinx-4=0

⇒6sinx-4=0

⇒sinx=23

Now, for exactly three solution we get,

⇒n=5 (in the given interval)

So, x2+nx+n-3=0

⇒x2+5x+2=0

⇒x=-5±172

So, the required interval is (-∞,0).

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If 2+sinxdydx+y+1cosx=0 and y0=1, then yπ2 is equal to

Options:
A 13
B -23
C -13
D 43

The correct answer is:
13

Solution:
We have,

2+sinxdydx=-y+1cosx

⇒∫dyy+1=-∫cosxsinx+2dx

Let, sinx=t⇒cosxdx=dt

⇒logy+1=-∫dtt+2+C

where, C is the constant of integration.

⇒logy+1+logsinx+2-C=0

⇒logy+1sinx+2=C

Given, y0=1

⇒C=log4.

⇒logy+1sinx+2=log4 .....i

Now, put x=π2 in the equation i, we get, logy+1+log3-log4=0

⇒ logy+1=log43

⇒y+1=43

⇒y=43-1=13.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2017 (02 Apr)
If $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$, then $x=$

Options:
A $\frac{3 \pi}{4}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D None of these

The correct answer is:
$\frac{\pi}{4}$

Solution:
$\begin{aligned} & \text { We have } 2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cose} \alpha) \\ & \Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}(2 \operatorname{cosec} x) \\ & \frac{2 \cos x}{\sin ^2 x}=2 \operatorname{cosec} x \Rightarrow 2 \cos x=2 \sin x \\ & \text { or } \sin x=\cos x \\ & \Rightarrow x=\frac{\pi}{4} \\ & \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$, then $x=$

Options:
A $\frac{3 \pi}{4}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D None of these

The correct answer is:
$\frac{\pi}{4}$

Solution:
We have $2^{\tan ^{-1}(\cos x)}=\tan ^{-1}(2 \operatorname{cose} \alpha)$
$\begin{aligned}
& \Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}(2 \operatorname{cosec} x) \\
& \frac{2 \cos x}{\sin ^2 x}=2 \operatorname{cosec} x \Rightarrow 2 \cos x=2 \sin x \\
& \text { or } \sin x=\cos x \\
& \Rightarrow x=\frac{\pi}{4} .
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\alpha+\beta+\gamma=2 \theta$, then $\cos \theta+\cos (\theta-\alpha)$ $+\cos (\theta-\beta)+\cos (\theta-\gamma)$ is equal to

Options:
A $4 \sin \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}$
B $4 \cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \cos \frac{\gamma}{2}$
C $4 \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}$
D $4 \sin \alpha \cdot \sin \beta \cdot \sin \gamma$

The correct answer is:
$4 \cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \cos \frac{\gamma}{2}$

Solution:
Now,
$$
\begin{aligned}
& \cos (\theta-\alpha)+\cos (\theta-\beta)+\cos \theta+\cos (\theta-\gamma) \\
&= 2 \cos \left(\theta-\left(\frac{\alpha+\beta}{2}\right)\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\
&+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\theta-\frac{\gamma}{2}\right) \\
&=2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\beta-\alpha}{2}\right) \\
&+2 \cos \left(\frac{\gamma}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right) \\
&=2 \cos \left(\frac{\gamma}{2}\right)\left[\cos \left(\frac{\beta-\alpha}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)\right] \\
&=2 \cos \left(\frac{\gamma}{2}\right) 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \\
&= 4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\alpha+\beta+\gamma=2 \pi$, then

Options:
A $\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$
B $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}+\tan \frac{\beta}{2} \tan \frac{\gamma}{2}+\tan \frac{\gamma}{2} \tan \frac{\alpha}{2}=1$
C $\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=-\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$
D None of these

The correct answer is:
$\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$

Solution:
$\begin{aligned} & \text { We have } \alpha+\beta+\gamma=2 \pi \Rightarrow \frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=\pi \\ & \Rightarrow \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}\right)=\tan \pi=0\end{aligned}$
$\begin{aligned} & \Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}-\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}=0 \\ & \Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=f(x) \sqrt{2 x-1}+\mathrm{C},$ where $\mathrm{C}$ is a constant

of integration, then $f(x)$ is equal to:

Options:
A $\frac{1}{3}(x+1)$
B $\frac{2}{3}(x+2)$
C $\frac{2}{3}(x-4)$
D $\frac{1}{3}(x+4)$

The correct answer is:
$\frac{1}{3}(x+4)$

Solution:
Let $I=\int \frac{x+1}{\sqrt{2 x-1}} d x$

Put $\sqrt{2 x-1}=t$

$\begin{array}{l}

\therefore \quad 2 x-1=t^{2} \Rightarrow d x=t d t \\

I=\int \frac{\left(t^{2}+3\right)}{2} d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C \\

=\frac{(2 x-1)^{\frac{3}{2}}}{6}+\frac{3}{2}(2 x-1)^{\frac{1}{2}}+C \\

=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+C \\

=f(x) \cdot \sqrt{2 x-1}+C

\end{array}$

Hence, $f(x)=\frac{x+4}{3}$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$, then $\sin ^{-1}(\sin x)$ is equal to

Options:
A $x$
B $-x$
C $\pi+x$
D $\pi-x$

The correct answer is:
$\pi-x$

Solution:
We have $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$
$\begin{aligned}
& \Rightarrow \frac{-\pi}{2} \leq x-\pi \leq \frac{\pi}{2} \Rightarrow \frac{-\pi}{2} \leq \pi-x \leq \frac{\pi}{2} \\
& \Rightarrow \sin ^{-1}\{\sin (\pi-x)\}=\pi-x
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If ∫2x2+5x2+adx=fx, where fx is a polynomial or ratio of polynomials, then the number of possible value(s) of a is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
fx=â«2x2+a+5-2ax2+adx
=2x+5-2aâ«dxx2+a
If fx is polynomial, a=52
If fx is ratio of polynomials, a=0
Therefore, the possible values of a is 2.

Topic: Mathematics, Indefinite Integration, JEE Main
$\text { If } 2^{x}+2^{y}=2^{x+y} \text {, then } \frac{d y}{d x}=$

Options:
A $2^{x-y} \frac{2^{y}-1}{2^{x}-1}$
B $2^{x-y} \frac{2^{y}-1}{1-2^{x}}$
C $\frac{2^{x}+2^{y}}{2^{x}-2^{y}}$
D None of these

The correct answer is:
None of these

Solution:
On differentiating

$\begin{array}{l}

2^{x} \log 2+2^{y} \log 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\

=2^{x} \cdot 2^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \log 2+2^{y} \cdot 2^{x} \log 2 \\

\Rightarrow 2^{x}+2^{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2^{x+y} \frac{\mathrm{dy}}{\mathrm{dx}}+2^{x+y} \\

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x+y}-2^{x}}{2^{y}-2^{x+y}} \\

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x}+2^{y}-2^{x}}{2^{y}-2^{x}-2^{y}}=-2^{y-x}

\end{array}$

Topic: Mathematics, Differentiation, JEE Main
If $2 x+3 y=5$ is the perpendicular bisector of the line segment joining the points $A\left(1, \frac{1}{3}\right)$ and $B$, then $B$ is equal to

Options:
A $\left(\frac{21}{13}, \frac{49}{39}\right)$
B $\left(\frac{17}{13}, \frac{31}{39}\right)$
C $\left(\frac{7}{13}, \frac{49}{39}\right)$
D $\left(\frac{21}{13}, \frac{31}{39}\right)$

The correct answer is:
$\left(\frac{21}{13}, \frac{31}{39}\right)$

Solution:
Let $\quad l_1 \equiv 2 x+3 y=5$
Since, line $A B \perp I_1$
$\therefore$ Slope of $l_1$ is
$$
\begin{array}{r}
m_1 \text { say }=\frac{-2}{3} \\
\therefore \text { Slope of } A B=\frac{-1}{(-2 / 3)}=\frac{3}{2}
\end{array}
$$
Equation of line $A B$ is
$$
\begin{array}{cc}
& \left(y-\frac{1}{3}\right)=\frac{3}{2}(x-1) \\
\Rightarrow \quad & 3 x-2 y=\frac{7}{3}
\end{array}
$$
Equation of line $I_1$ is
$$
2 x+3 y=5
$$
From Eqs. (i) and (ii), we get
$$
\Rightarrow \quad \begin{array}{rl}
13 x & =17 \\
x & x=\frac{17}{13}
\end{array}
$$
From Eq. (i), we get
$$
3 y=5-\frac{34}{13} \Rightarrow y=\frac{65-34}{13 \times 3}=\frac{31}{39}
$$
So, mid-point $P \rightarrow\left(\frac{17}{13}, \frac{31}{39}\right)$
Coordinate of point $B$
$$
\begin{aligned}
& =\left(\frac{17}{13} \times 2-1, \frac{31}{39} \times 2-\frac{1}{3}\right) \\
& =\left(\frac{34-13}{13}, \frac{62-13}{39}\right)=\left(\frac{21}{13}, \frac{49}{39}\right)
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
If $2 x+3 y-5 z=7, x+y+z=6,3 x-4 y+2 z=1$, then $x=$

Options:
A $\left|\begin{array}{ccc}2 & -5 & 7 \\ 1 & 1 & 6 \\ 3 & 2 & 1\end{array}\right| \div\left|\begin{array}{ccc}7 & 3 & -5 \\ 6 & 1 & 1 \\ 1 & -4 & 2\end{array}\right|$
B $\left|\begin{array}{ccc}-7 & 3 & -5 \\ -6 & 1 & 1 \\ -1 & -4 & 2\end{array}\right| \div\left|\begin{array}{ccc}2 & 3 & -5 \\ 1 & 1 & 1 \\ 3 & -4 & 2\end{array}\right|$
C $\left|\begin{array}{ccc}7 & 3 & -5 \\ 6 & 1 & 1 \\ 1 & -4 & 2\end{array}\right| \div\left|\begin{array}{ccc}2 & 3 & -5 \\ 1 & 1 & 1 \\ 3 & -4 & 2\end{array}\right|$
D None of these

The correct answer is:
$\left|\begin{array}{ccc}7 & 3 & -5 \\ 6 & 1 & 1 \\ 1 & -4 & 2\end{array}\right| \div\left|\begin{array}{ccc}2 & 3 & -5 \\ 1 & 1 & 1 \\ 3 & -4 & 2\end{array}\right|$

Solution:
For the given set of equation, by Cramer's Rule
$x=\frac{D_x}{D}=\left|\begin{array}{ccc}7 & 3 & -5 \\ 6 & 1 & 1 \\ 1 & -4 & 2\end{array}\right| \div \left|\begin{array}{ccc}2 & 3 & -5 \\ 1 & 1 & 1 \\ 3 & -4 & 2\end{array}\right|$.

Topic: Mathematics, Determinants, JEE Main
If $2 x+3 y+12=0$ and $x-y+4 \lambda=0$ are conjugate with respect to the parabola $y^2=8 x$, then $\lambda$ is equal to

Options:
A $2$
B $-2$
C $3$
D $-3$

The correct answer is:
$-3$

Solution:
Using the condition that if two lines $l_1 x+m_1 y+n_1=0$ and $l_2 x+m_2 y+n_2=0$ are conjugate w.r.t. parabola $y^2=4 a x$, then
$$
l_1 n_2+l_2 n_1=2 a m_1 m_2
$$
Given conjugate lines are $2 x+3 y+12=0$ and $x-y+4 \lambda=0$ and equation of parabola is $y^2=8 x$.
Here, $l_1=2, m_1=3, n_1=12 ; l_2=1, m_2=-1$, $n_2=4 \lambda$ and $a=2$
$\therefore$ From Eq. (i),
$$
\begin{gathered}
2 \times 4 \lambda+1 \times 12=2 \times 2 \times 3 \times(-1) \\
8 \lambda=-12-12 \Rightarrow \lambda=-3
\end{gathered}
$$

Topic: Mathematics, Parabola, JEE Main
If 2+xxx2x2+xx2x2x2+x=16x-ax-bx-cx-d be an identity in x where a,b,c,d are independent of x, then the value of 1325abcd is

Options:

The correct answer is:
24.96

Solution:
Since, the given equation is an identity in x, putting x=0, we get,
200020002=16abcd
⇒abcd=48⇒1325abcd=13×4825=24.96

Topic: Mathematics, Determinants, JEE Main
If 2x=y15+y-15 and x2-1d2ydx2+λxdydx+ky=0, then λ+k is equal to

Options:
A 26
B -24
C -23
D -26

The correct answer is:
-24

Solution:
Given y15+y-15=2x
⇒15y-45-15y-65dydx=2
⇒y′y15-y-15=10y
⇒y′ 2x2-1=10y

⇒y′ x2-1=5y
Differentiating again
y′′2x2-1+y′22x2x2-1=10y'
⇒y′′x2-1+xy′=5x2-1 y′
⇒y′′ x2-1+xy′-25y=0
⇒λ=1, k=-25

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2017 (09 Apr Online)
If $(2 x-y+1)+i(x-2 y-1)=2-3 i$, then the multiplicative inverse of $(x-i y)$ is

Options:
A $\frac{15}{41}+\frac{12}{41} i$
B $\frac{6}{29}+\frac{15}{29} i$
C $\frac{15}{29}+\frac{6}{29} i$
D $\frac{12}{41}+\frac{15}{41} i$

The correct answer is:
$\frac{12}{41}+\frac{15}{41} i$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Complex Number, JEE Main
If 2xy+3yx=20, then dydx at 2,2 is equal to:

Options:
A -2+loge83+loge4
B -3+loge164+loge8
C -3+loge82+loge4
D -3+loge42+loge8

The correct answer is:
-2+loge83+loge4

Solution:
Given,

2xy+3yx=20

Now differentiating both sides w.r.t x, we get

2xyy'lnx+yx+3yxlny+xy·y'=0 here ln=loge

Putting x=2 & y=2, we get

⇒8ln2·y'+1+12lny+y'=0

⇒8ln2·y'+1+12lny+y'=0

⇒y'=-3ln2+22ln2+3

⇒y'=-2+ln83+ln4

Topic: Mathematics, Differentiation, JEE Main
If 2xy+3yx=20, then dydx at 2,2 is equal to:

Options:
A -2+loge83+loge4
B -3+loge164+loge8
C -3+loge82+loge4
D -3+loge42+loge8

The correct answer is:
-2+loge83+loge4

Solution:
Given,

2xy+3yx=20

Now differentiating both sides w.r.t x, we get

2xyy'lnx+yx+3yxlny+xy·y'=0 here ln=loge

Putting x=2 & y=2, we get

⇒8ln2·y'+1+12lny+y'=0

⇒8ln2·y'+1+12lny+y'=0

⇒y'=-3ln2+22ln2+3

⇒y'=-2+ln83+ln4

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2023 (06 Apr Shift 1)
If \(2 x-k y+3=0,3 x-y+1=0\) are conjugate lines with respect to \(5 x^2-6 y^2=15\), then \(k=\)

Options:
A 6
B 4
C 3
D 2

The correct answer is:
6

Solution:
We know that,
If two lines \(l_1 x+m_1 y+n_1=0\) and
\(l_2 x+m_2 y+n_2=0\) are conjugate lines with respect to the hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text {, then } a^2 l_1 l_2-b^2 m_1 m_2=n_1 n_2\)
For hyperbola,
\(5 x^2-6 y^2=15 \Rightarrow \frac{x^2}{3}-\frac{y^2}{5 / 2}=1\)
Here, \(a^2=3, b^2=\frac{5}{2}, l_1=2, l_2=3\),
and
\(m_1=-k, m_2=-1\)
\(\begin{array}{lc}
\text {and } & n_1=3, n_2=1 \\
\therefore & 3 \times 2 \times 3-\frac{5}{2}(-k)(-1)=3 \times 1 \\
\Rightarrow & 18-\frac{5}{2} k=3 \Rightarrow 15=\frac{5}{2} k \Rightarrow k=6
\end{array}\)

Topic: Mathematics, Hyperbola, JEE Main
If 2y-a is the harmonic mean between yâx and yâz , then x-a, y-a and z-a are in

Options:
A arithmetic progression
B geometric progression
C harmonic progression
D None of these

The correct answer is:
geometric progression

Solution:
It is given that, y-x, 2y-a and y-z are in harmonic progression.
â 1y-x,12y-a,1y-z are in arithmetic progression.
â 12y-a-1y-x=1y-z-12y-a
â 2a-y-xy-x=y+z-2ay-z
â x-a+y-ax-a-y-a=y-a+z-ay-a-z-a
â x-ay-a=y-az-a
Hence, x-a,y-a and z-a are in geometric progression

Topic: Mathematics, Sequences and Series, JEE Main
If \(2 y \cos \theta=x \sin \theta\) and \(2 x \sec \theta-y \operatorname{cosec} \theta=3\), then \(x^2+4 y^2=\)

Options:
A 2
B 4
C 1
D none

The correct answer is:
4

Solution:
We have \(2 \mathrm{y} \cos \theta=\mathrm{x} \sin \theta\)
or \(\frac{\cos \theta}{\mathrm{x}}=\frac{\sin \theta}{2 \mathrm{y}}=\mathrm{k}\) (say)
Then \(\cos \theta=\mathrm{kx}\) and \(\sin \theta=2 \mathrm{ky}\) ...(1)
Again \(2 \mathrm{x} \sec \theta-\mathrm{y} \operatorname{cosec} \theta=3\)
or \(\frac{2 \mathrm{x}}{\cos \theta}-\frac{\mathrm{y}}{\sin \theta}=3\)
or \(\frac{2 \mathrm{x}}{\mathrm{kx}}-\frac{\mathrm{y}}{2 \mathrm{ky}}=3 \quad\) [from(1)]
or \(\frac{2}{\mathrm{k}}-\frac{1}{2 \mathrm{k}}=3\), giving \(\mathrm{k}=\frac{1}{2}\)
We now get \(\cos \theta=\frac{\mathrm{x}}{2}\) and \(\sin \theta=\mathrm{y}\)
Squaring and adding we get
\(\begin{aligned}
& \cos ^2 \theta+\sin ^2 \theta=\frac{x^2}{4+y^2} \\
& \Rightarrow \frac{x^2}{4}+y^2=1, \text { or } x^2+4 y^2=4
\end{aligned}\)

Topic: Mathematics, Trigonometric Equations, JEE Main
If 2y=cot-13cosx+sinxcosx-3sinx2 ∀x∈0,π2, then dydx is equal to

Options:
A π6-x
B 2x-Ï3
C x-Ï6
D None of these

The correct answer is:
None of these

Solution:
2y=cot-13cosx+sinxcosx-3sinx2

=cot-13+tanx1-3tanx2

=cot-1tanπ3+x2

=cot-1cotπ2-π3+x 2

⇒2y=π6-x2, x∈0,π6π+π6-x2, x∈π6,π2

⇒ 2dydx=2π6-x.-1⇒dydx=x-π6, x∈0,π6

And dydx=x-7π6, x∈π6,π2
Left Hand and Right Hand Derivatives are not same so function is non derivable at x=π6.

Hence, dydx does not exist for all values in the given interval.

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If 2y=cot-13cosx+sinxcosx-3sinx2 ∀x∈0,π2, then dydx is equal to

Options:
A π6-x
B 2x-Ï3
C x-Ï6
D None of these

The correct answer is:
None of these

Solution:
2y=cot-13cosx+sinxcosx-3sinx2

=cot-13+tanx1-3tanx2

=cot-1tanπ3+x2

=cot-1cotπ2-π3+x 2

⇒2y=π6-x2, x∈0,π6π+π6-x2, x∈π6,π2

⇒ 2dydx=2π6-x.-1⇒dydx=x-π6, x∈0,π6

And dydx=x-7π6, x∈π6,π2
Left Hand and Right Hand Derivatives are not same so function is non derivable at x=π6.

Hence, dydx does not exist for all values in the given interval.

Topic: Mathematics, Differentiation, JEE Main
If 2y=cot-1⁡3cos⁡x+sin⁡xcos⁡x-3sin⁡x2 ∀x∈0,π2, then dydx is equal to

Options:
A π6-x
B 2x-π3
C x-π6
D π3-x

The correct answer is:
x-π6

Solution:
2y=cot-1⁡3cosx+sinxcos⁡x-3sin⁡x2
=cot-1⁡3+tanx1-3tanx2
=cot-1⁡tan⁡π3+x2
=cot-1⁡cot⁡ π2-π3+x 2
2y=π6-x2, x∈0,π6
=π+π6-x2, x∈π6,π2
⇒ 2dydx=2π6-x.-1⇒dydx=x-π6, x∈0,π6
dydx=x-7π6, x∈π6,π2
x-π6 is matching with one of the given options.

Topic: Mathematics, Differentiation, JEE Main
If \( 2(y-a) \) is the harmonic mean between \( y-x \) and \( y-z \), then \( x-a, y-a \) and \( z-a \) are in

Options:
A arithmetic progression
B geometric progression
C harmonic progression
D none of these

The correct answer is:
geometric progression

Solution:
It is given that, y-x, 2y-a and y-z are in harmonic progression.
â 1y-x,12y-a,1y-z are in arithmetic progression.
â 12y-a-1y-x=1y-z-12y-a
â 2a-y-xy-x=y+z-2ay-z
â x-a+y-ax-a-y-a=y-a+z-ay-a-z-a
â x-ay-a=y-az-a
Hence, x-a,y-a and z-a are in geometric progression

Topic: Mathematics, Sequences and Series, JEE Main
If -3,-1 is the largest interval in which the function fx=x3+6x2+ax+2 is decreasing, then a is equal to (where, . denotes the greatest integer function)

Options:
A 8
B 9
C 10
D 11

The correct answer is:
9

Solution:
f ′ x =3 x 2 +12x+a
if fx is decreasing in −3,−1,f'x<0, ∀x∈−3,−1
∴ f ′ −3 = f ′ −1 =0
⇒a3=-3-1
⇒a=9 (using product of roots)

Topic: Mathematics, Application of Derivatives, JEE Main
If Ï<Î¸<3Ï2 and cosâ¡Î¸=-35, then tanÎ¸4 is equal to

Options:
A 5-12
B 5+12
C -5+14
D -5-14

The correct answer is:
5+12

Solution:
cosâ¡Î¸=-35 ⇒2cos2Î¸2-1=-35
2cos2Î¸2=1-35=25
cosÎ¸2=Â±15
⇒cosÎ¸2=-15 âµÏ2<Î¸2<3Ï4
Now, tan2Î¸4=1-cosÎ¸21+cosÎ¸2=1+151-15=5+15-1
tan2Î¸4=5+124=5+122
⇒tanÎ¸4=5+12 âµÏ4<Î¸4<3Ï8

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If 32sin2α-1,14 and 34-2sin2α are the first three terms of an A.P. for some α , then the sixth term of this A.P. is

Options:
A 66
B 81
C 65
D 78

The correct answer is:
66

Solution:
If a, b, c are in AP, then b is A.M of a & c.

∴ 2 b=a+c

⇒28=32sin2α-1+34-2sin2α

Putting, 32sin2α=x we get,

28=x3+81x⇒x2-84x+243=0

⇒(x-3)(x-81)=0

∴ 32sin2α=3 or 34

sin2α=12, ∵sin2α≠2

Terms are 1,14,27,…… then

T6=1+5(13)=66

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If 0,3+5 is a point on the ellipse whose foci are 2,3 and -2,3, then the length of the semi-major axis is

Options:

The correct answer is:
3

Solution:
If P is a point on the ellipse having S & S' as foci, then

PS+P S ′ =2a
Where 2a= length of the major-axis
⇒22+5+22+5=2a
⇒a=3= semi-major axis

Topic: Mathematics, Ellipse, JEE Main
If 3644k is the term, independent of x, in the binomial expansion of x4-12x212, then k is equal to

Options:

The correct answer is:
55

Solution:
Given that x4-12x212

rth term of an expression is Tr+1 =Crnxn-ryr Tr+1=Cr12x412-r-12x2r=Cr121412-r(-12)r·x12-r-2r
⇒ 12-3r=0
⇒ r=4

So term independent from x is T5=C412·148·34·44=C412·3444

We know that Crn=n!n-r!r!

=55×3644

They given that it is =k 3644

So required value of k = 55.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If $(3, \lambda \lambda \lambda \lambda)$ and $(5,6)$ are the conjugate points to the curve $x^{2}+y^{2}=3$, then $\lambda \lambda \lambda \lambda$ is

Options:
A $-1$
B 1
C $-2$
D 2

The correct answer is:
$-2$

Solution:
If the $(3, \lambda \lambda \lambda \lambda)$ and $(5,6)$ are the conjugate points to the curve $x^{2}+y^{2}=3$, then
$\begin{aligned} x_{1} x_{2}+y_{1} y_{2} &=3 \\ \Rightarrow \quad(3)(5)+(\lambda \lambda \lambda \lambda) \cdot(6) &=3 \end{aligned}$
$\Rightarrow \quad 15+6(\lambda \lambda \lambda \lambda)=3$
$\Rightarrow 6(\lambda \lambda \lambda \lambda)=-12$
$\Rightarrow \lambda \lambda \lambda \lambda=-2$

Topic: Mathematics, Circle, JEE Main
If $\frac{\pi}{3}, \theta$ are the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$, then $\tan \theta=$

Options:
A $-\sqrt{3}$
B $\sqrt{3}$
C -1
D $\frac{1}{\sqrt{2}}$

The correct answer is:
$-\sqrt{3}$

Solution:
Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$
Let eccentricity of ellipse be ' $e$ ' Then
$b^2=a^2\left(1-e^2\right)$
Here, $\quad b^2=12, a^2=16$
$\therefore \quad 12=16\left(1-e^2\right)$
$1-e^2=\frac{3}{4}$
or $e^2=1-\frac{3}{4}=\frac{1}{4} \text { or } e=\frac{1}{2}$
If $\alpha, \beta$ are the eccentric angles of the ends of a focal chord of the ellipse, then eccentricity is given by
$e=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)}$
Here, $\alpha=\frac{\pi}{3}, \beta=\theta$, and $e=\frac{1}{2}$
$\frac{1}{2}=\frac{\cos \left(\frac{\pi / 3-\theta}{2}\right)}{\cos \left(\frac{\pi / 3+\theta}{2}\right)}$
Multiplying in $N^r$ and $D^r$ by $2 \sin \left(\frac{\pi / 3+\theta}{2}\right)$
$\frac{1}{2}=\frac{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3-\theta}{2}\right)}{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3+\theta}{2}\right)}$
$=\frac{\sin \left(\frac{\pi / 3+\theta+\pi / 3-\theta}{2}\right)+\sin \left(\frac{\pi / 3+\theta-\pi / 3+\theta}{2}\right)}{\sin (\pi / 3+\theta)}$
$\left\{\because 2 \sin A \cos B=\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)\right.$ and $2 \sin A \cos A=\sin 2 A\}$
$\frac{1}{2}=\frac{\sin \frac{\pi}{3}+\sin \theta}{\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta}$
$\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta=2\left(\sin \frac{\pi}{3}+\sin \theta\right)$
$\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=2\left(\frac{\sqrt{3}}{2}+\sin \theta\right)$
$\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\sqrt{3}+2 \sin \theta$
or $\frac{\sqrt{3}}{2} \cos \theta=\frac{3}{2} \sin \theta+\sqrt{3}$
$\cos \theta=\sqrt{3} \sin \theta+2$ or $\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta=1$
or $\quad \cos \frac{\pi}{3} \cos \theta-\sin \frac{\pi}{3} \sin \theta=1$
or $\cos \left(\frac{\pi}{3}+\theta\right)=1$ or $\cos \left(\frac{\pi}{3}+\theta\right)=\cos 0^{\circ}$
On comparing both sides, we get
$\frac{\pi}{3}+\theta=0^{\circ} \text { or } \theta=-\frac{\pi}{3} \quad \therefore \quad \tan \theta=\tan \left(\frac{-\pi}{3}\right)$
$\tan \theta=-\sqrt{3}$

Topic: Mathematics, Ellipse, JEE Main
If 3cos2x=3-1cosx+1, then number of solutions of the given equation when x∈0,π2 is ________.

Options:

The correct answer is:
1

Solution:
3cosx2-3cosx+cosx-1=0

⇒3cosx+1cosx-1=0

⇒cosx=1 or cosx=-13 (reject)

⇒x=0 only

Hence, only one solution.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2021 (26 Feb Shift 1)