datavorous/entrance-exam-dataset huggingface dataset:
If $C, R, L$ and $I$ denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time are :
(1) $C R$
(2) $\frac{L}{R}$
(3) $\sqrt{L C}$
(4) $L I^2$

Options:
A (1) and (2) only
B (1) and (3) only
C (1) and (4) only
D (1), (2) and (3) only

The correct answer is:
(1), (2) and (3) only

Solution:
$[C]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right],[R]=\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$[L]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]$ and $[I]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}\right]$
(1) $[C R]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(2) $\left[\frac{L}{R}\right]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(3) $[\sqrt{L C}]=\left(\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right] \times\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]\right)^{1 / 2}$
$=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(4) $\left[L I^2\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}^0\right]^2$
$=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^0\right]$
Hence, option (d) is correct.

Topic: Physics, Units and Dimensions, JEE Main
If $C_r={ }^n C_r$, then $C_0+C_4+C_8+C_{12}+\ldots=$

Options:
A $\frac{2^{\frac{n}{2}}\left[\sin \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]}{2}$
B $2^{\frac{n}{2}} \sin \frac{n \pi}{4}$
C $2^{n-1} \cos \frac{n \pi}{4}$
D $\frac{2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]}{2}$

The correct answer is:
$\frac{2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]}{2}$

Solution:
In given series difference of suffies is 4
$$
0-4=8-4=12-4=\ldots=4
$$
Now, $(\mathrm{l})^{\frac{1}{4}}=(\cos 0+i \sin 0)^{\frac{1}{4}}$
$$
\begin{aligned}
& =(\cos 2 r \pi+i \sin 2 r \pi)^{\frac{1}{4}} \\
& =\cos \frac{r \pi}{2}+i \sin \frac{r \pi}{2}
\end{aligned}
$$
where $r=0,1,2,3$
Four roots of unity are $1, i,-1,-i$
$$
\begin{aligned}
& =1, \alpha, \alpha, \alpha^2 \quad \text{say}\\
\text { and }(1+x)^n & =\sum_{r=0}^n{ }^n C_r x^r
\end{aligned}
$$
For $r=0,4,8,12 \ldots$ R.H.S of Eq. (v)
$$
\begin{array}{r}
{ }^n C_0(1+1+1+1)+{ }^n C_4\left(1+\alpha^4+\alpha^8+\alpha^{12}\right) \\
+{ }^n C_8\left(1+\alpha^8+\alpha^{16}+\alpha^{24}\right)+\ldots+ \\
=4\left({ }^n C_0+{ }^n C_4+{ }^n C_8 \ldots\right) \quad\left(\because \alpha^4=1\right) .
\end{array}
$$
L.H.S of Eq. (v)
$$
\begin{aligned}
& =2^n+(1+i)^n+\left(1+i^2\right)^n+\left(1+i^3\right)^n \\
& =2^n+(1+i)^n+0+(1-i)^n
\end{aligned}
$$
Since, $(1+i)^n=2^{\frac{n}{2}}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^n$
$$
=2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right]
$$
$$
\begin{aligned}
& \quad(1-i)^n=2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}-i \sin \frac{n \pi}{4}\right] \\
& \Rightarrow 2^n+(1+i)^n+(1-i)^n \\
& =2^n+2 \cdot 2^{\frac{n}{2}} \cos \frac{n \pi}{4} \\
& \Rightarrow 4 \cdot\left({ }^n C_0+{ }^n C_4+{ }^n C_8+{ }^n C_{12} \cdots\right) \\
& \quad=2 \cdot 2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]=\frac{2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]}{2}
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
If C (velocity of light), h (Planck's constant) and G (Universal gravitational constant) are taken as fundamental quantities, then the dimensional formula of mass is

Options:
A h-12Â G12Â Co
B h12 C12 G-12
C h-12Â C12Â G-12
D h-12Â C-12Â G-12

The correct answer is:
h12 C12 G-12

Solution:
Let, M=CahbGc
ML0T0=LT-1a ML2T-1b M-1L3T-2c ......(i)
Where, h=EnergyFreqyency
=ML2 T-2T-1=ML2T-1
C=MetreSecond=LT-1
G=ForceÃdistance2mass2
=MLT-2 L2M2=M-1L3T-2
Comparing the coefficients M, L, T, of both sides we get
b-c=1 ......(ii)
a+2b+3c=0 ......(iii)
-a+b+2c=0 .....(iv)
Solve the equations (ii), (iii) and (iv), we get
a=12, b=12, c= -12
So, M=h12 C12 G-12

Topic: Physics, Units and Dimensions, JEE Main
If cartesian equation of the line is $x-1=2 y+3=3-z$, then its vector equation
is

Options:
A $\bar{r}=(\hat{\imath}-3 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+\hat{\jmath}-2)$
B $\bar{r}=(-\hat{\imath}-3 \hat{\jmath}+3 \hat{k})+\lambda\left(\hat{\imath}+\frac{1}{2} \hat{\jmath}-\hat{k}\right)$
C $\bar{r}=\left(-\hat{\imath}+\frac{3}{2} \hat{\jmath}-3 \hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})$
D $\bar{r}=\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+3 \hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})$

The correct answer is:
$\bar{r}=\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+3 \hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})$

Solution:
Given Cartesian equation is
$\begin{aligned}
x-1 &=2 y+3=3-z \\
\therefore \frac{x-1}{1} &=\frac{2\left(y+\frac{3}{2}\right)}{1}=\frac{-(z-3)}{1} \Rightarrow \frac{x-1}{1}=\frac{y+\frac{3}{2}}{\left(\frac{1}{2}\right)}=\frac{z-3}{-1}
\end{aligned}$
$\begin{array}{l}
\therefore 1, \frac{1}{2},-1 \quad \text { i.e. } 2,1,-2 \text { are d.r. of given line. } \\
\therefore \quad \mathrm{A}\left(1,-\frac{3}{2}, 3\right) \text { lies on given line. } \\
\therefore \quad \text { vector equation is } \\
\qquad \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})
\end{array}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If common tangents of x2+y2=r2 and x216+y29=1 forms a square, then the length of diagonal of the square is

Options:

The correct answer is:
10

Solution:
Director circle of x2+y2=r2 and x216+y29=1 is same
â x2+y2=2r2 & x2+y2=16+9 represents same circle
â 2r2=25âr=52
side of square = 2r=52
diagonal of square = 2(side of square)=252=10

Topic: Mathematics, Ellipse, JEE Main
If compound A reacts with B following first order kinetics with rate constant 2.011×10-3 s-1. The time taken by A (in seconds) to reduce from 7 g to 2 g will be _____ . (Nearest Integer) log5=0.698,log7=0.845,log2=0.301

Options:

The correct answer is:
623

Solution:
Given,
A+B→Pt=0 7gt=t 2g

at constant volume

From formula of first order kinetics:
⇒t=2.303 KlogA0At

=2·3032·011×10-3log72

=2·303×0.5442·011×10-3

=622.989

≈623

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If cos-123x+cos-134x=π2 x>34, then x is equal to :

Options:
A 14510
B 14511
C 14612
D 14512

The correct answer is:
14512

Solution:
We have, cos-123x+cos-134x=π2, x>34 ...i

∴ cos-134x=sin-116x2-94x

Put in equation i we get,
⇒cos-123x+sin-116x2-94x=π2 ...ii

We know sin-1x+cos-1x=π2

From equation ii we get,
⇒23x=16x2-94x
⇒832=16x2-9 ∵ x≠0
⇒16x2=649+9=1459
⇒x2=145144
⇒x=14512

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If $\cos ^{-1}\left(\frac{x}{2}\right)+\cos ^{-1}\left(\frac{y}{2}\right)=\theta$, then $9 x^2-12 x y \cos \theta+4 y^2=$

Options:
A $36 \sin ^2 \theta$
B $37 \sin ^2 \theta$
C $39 \sin ^2 \theta$
D $36 \cos ^2 \theta$

The correct answer is:
$36 \sin ^2 \theta$

Solution:
We have,
$\begin{aligned}
& \cos ^{-1}\left(\frac{x}{2}\right)+\cos ^{-1}\left(\frac{y}{3}\right)=\theta \\
& \Rightarrow \cos ^{-1}\left[\frac{x}{2} \cdot \frac{y}{3}-\sqrt{\left(1-\left(\frac{x}{2}\right)^2\right)\left(1-\left(\frac{y}{3}\right)^2\right)}\right]=\theta \\
& \Rightarrow \frac{x}{2} \cdot \frac{y}{3}-\sqrt{\left(1-\left(\frac{x}{2}\right)^2\right)\left(1-\left(\frac{y}{3}\right)^2\right)}=\cos \theta \\
& \quad=\frac{x y}{6}-\cos \theta=\sqrt{\left(1-\left(\frac{x}{2}\right)^2\right)\left(1-\left(\frac{y}{3}\right)^2\right)}
\end{aligned}$
squaring both sides, we get
$\begin{aligned}
& \Rightarrow\left(\frac{x y}{6}-\cos \theta\right)^2=\left(1-\frac{x^2}{4}\right)\left(1-\frac{y^2}{9}\right) \\
& \begin{aligned}
\Rightarrow \frac{x^2 y^2-12 x y \cos \theta+36 \cos ^2 x}{36}=\frac{\left(4-x^2\right)\left(9-y^2\right)}{36} & =36-9 x^2-4 y^2+x^2 y^2 \\
\Rightarrow x^2 y^2-12 x y \cos \theta+36 \cos ^2 x & =36\left(1-\cos ^2 \theta\right)
\end{aligned} \\
& \Rightarrow 9 x^2-12 x y \cos \theta+4 y^2=36 \sin ^2 \theta
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If \( \cos ^{-1} \lambda+\cos ^{-1} \mu+\cos ^{-1} \nu=3 \pi \) then \( \lambda \mu+\mu \nu+\nu \lambda \) is equal to.

Options:
A \( -3 \)
B \( 0 \)
C \( 3 \)
D \( -1 \)

The correct answer is:
\( 3 \)

Solution:
Given equation:

cos-1λ+cos-1μ+cos-1γ=3π

As. range of cos-1x is 0,π

⇒ cos-1λ=π; cos-1μ=π ; cos-1γ=π

Only possible solution of given equation

⇒ λ=cosπ=-1

⇒ μ=cosπ=-1

⇒ γ=cosπ=-1

Value of equation

λμ+μγ+γλ=-1×-1+-1×-1+-1×-1

= 1+1+1=3

⇒λμ+μγ+γλ=3

Hence solved

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$, then $4 x^2-4 x y \cos \alpha+y^2$ is equal to

Options:
A
$2 \sin 2 \alpha$

B
4

C
$4 \sin ^2 \alpha$

D
$-4 \sin ^2 \alpha$

The correct answer is:
$4 \sin ^2 \alpha$

Solution:
$$
\begin{aligned}
& \cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha \\
& \cos ^{-1}\left(\frac{x y}{2}+\sqrt{\left(-x^2\right)\left(1-\frac{y^2}{4}\right)}\right)=\alpha \\
& \cos ^{-1}\left(\frac{x y+\sqrt{4-y^2-4 x^2+x^2 y^2}}{2}\right)=\alpha \\
& \Rightarrow 4-y^2-4 x^2+x^2 y^2=4 \cos ^2 \alpha+x^2 y^2-4 x y \cos \alpha \\
& \Rightarrow 4 x^2+y^2-4 x y \cos \alpha=4 \sin ^2 \alpha
\end{aligned}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2005
If $\cos ^{-1} x+\cos ^{-1} y=2 \pi$, then $\sin ^{-1} x+\sin ^{-1} y$ is equal to

Options:
A $\pi$
B $-\pi$
C $\frac{\pi}{2}$
D None of these

The correct answer is:
$-\pi$

Solution:
$\begin{aligned} & \cos ^{-1} x+\cos ^{-1} y=2 \pi \\ & \Rightarrow \frac{\pi}{2}-\sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y=2 \pi \\ & \Rightarrow \pi-\left(\sin ^{-1} x+\sin ^{-1} y\right)=2 \pi \\ & \Rightarrow \sin ^{-1} x+\sin ^{-1} y=-\pi\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then the value of $x^{2025}+x^{2026}+x^{2027}$ is

Options:
A $-1$
B $0$
C $1$
D $3$

The correct answer is:
$-1$

Solution:
$\begin{aligned} & \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi \\ & \text { Since } 0 \leq \cos ^{-1} x \leq \pi \\ & \\ & 0 \leq \cos ^{-1} y \leq \pi \text { and } 0 \leq \cos ^{-1} z \leq \pi \\ & \\ & \text { Here, } \cos ^{-1} x=\cos ^{-1} y=\cos ^{-1} z=\pi \\ & \Rightarrow x=y=z=\cos \pi=-1 \\ & \therefore \quad x^{2005}+x^{2026}+x^{2027} \\ & =(-1)^{2025}+(-1)^{2026}+(-1)^{2027} \\ & =-1+1-1 \\ & =-1\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then $x y+y z+z x=$

Options:
A $0$
B $1$
C $3$
D $-3$

The correct answer is:
$3$

Solution:
Given $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$
$0 \leq \cos ^{-1} x \leq \pi$
$\therefore 0 \leq \cos ^{-1} y \leq \pi$ and $0 \leq \cos ^{-1} z \leq \pi$
Here $\cos ^{-1} x=\cos ^{-1} y=\cos ^{-1} z=\pi$
$\Rightarrow x=y=z=\cos \pi=-1$
$\begin{aligned}
\therefore x y+y z+z x & =(-1)(-1)+(-1)(-1)+(-1)(-1) \\
& =1+1+1=3 .
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi$, then

Options:
A $x^2+y^2+z^2+x y z=0$
B $x^2+y^2+z^2+2 x y z=0$
C $x^2+y^2+z^2+x y z=1$
D $x^2+y^2+z^2+2 x y z=1$

The correct answer is:
$x^2+y^2+z^2+2 x y z=1$

Solution:
$\begin{aligned}
& \text { Given that } \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi \\
& \Rightarrow \cos ^{-1}(x)+\cos ^{-1}(y)+\cos ^{-1}(z)=\cos ^{-1}(-1) \\
& \Rightarrow \cos ^{-1}(x)+\cos ^{-1}(y)=\cos ^{-1}(-1)-\cos ^{-1}(z) \\
& \Rightarrow \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{\left.1-y^2\right)}=\cos ^{-1}\{(-1)(z)\}\right. \\
& \Rightarrow x y-\sqrt{\left(1-x^2\right)\left(1-y^2\right)}=-z \\
& \Rightarrow(x y+z)=\sqrt{\left(1-x^2\right)\left(1-y^2\right)}
\end{aligned}$
Squaring both sides we get $x^2+y^2+z^2+2 x y z=1$
Trick : Put
$x=y=z=\frac{1}{2}$
so that
$\cos ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}=\pi$
Obviously (4) holds for these values of $x, y, z$.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If cos-1â¡x-x22+x34-â¦+sin-1â¡x2-x42+x64-â¦=Ï2 for 0<x<2, then x is equal to

Options:

The correct answer is:
1

Solution:
Since, sin-1â¡x+cos-1â¡x=Ï2 for xâ¤1.
â´x-x22+x34-â¦=x2-x42+x64-â¦
âx1+x2=x21+x220<x<2
âx2+x=x22+x2â2x+x3=2x2+x3
âx2=xâx=0,1
But x2=xâ 0,â´x=1.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If \( \cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha \), then \( 4 x^{2}-4 x y \cos \alpha+y^{2} \) is equal to

Options:
A \( 4 \)
B \( 2 \sin 2 \alpha \)
C \( -4 \sin ^{2} \alpha \)
D \( 4 \sin ^{2} \alpha \)

The correct answer is:
\( 4 \sin ^{2} \alpha \)

Solution:
Given,

cos-1x-cos-1y2=α ...i
let A=cos-1x, B= cos-1y2
⇒ cosA=x, cosB=y2 ...ii
Using equation i ⇒ A - B = α
⇒ cosA-B=cos α
⇒ cos A cos B + sin A sin B = cos α
⇒ xy 2 + 1 - x 2 1 - y 2 4 = cos α Using equation ii
⇒ cos α - xy 2 2 = 1 - x 2 1 - y 2 4
⇒ cos2α+x2y24-xy cos α=1-y24-x2+x2y24

⇒ y24+x2-xy cos α=1-cos2α
⇒ x 2 + y 2 4 - xy cos α = sin 2 α
⇒ 4 x 2 - 4 xy cos α + y 2 = 4 sin 2 α .

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If cos-1y2=logex55,y<2, then

Options:
A x2y''+xy'-25y=0
B x2y''-xy'-25y=0
C x2y''-xy'+25y=0
D x2y''+xy'+25y=0

The correct answer is:
x2y''+xy'+25y=0

Solution:
Given,

cos-1y2=logex55

Using property logxm=mlogx we get,

cos-1y2=5logex5

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

-11-y24·y'2=5·1x5·15

⇒-y'4-y2=5x

⇒-xy'=54-y2

Again differentiating w.r.t x we get,

⇒-xy''-y'=5·124-y2-2yy'

⇒xy''+y'=5yy'4-y2

⇒xy''+y'=5y·-5x

⇒x2y''+xy'=-25y

⇒x2y''+xy'+25y=0

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2022 (27 Jun Shift 1)
If $\cos 2 B=\frac{\cos (A+C)}{\cos (A-C)}$, then $\tan A, \tan B, \tan C$ are in

Options:
A A.P.
B G.P.
C H.P.
D None of these

The correct answer is:
G.P.

Solution:
$\begin{aligned}
& \cos 2 B=\frac{\cos (A+C)}{\cos (A-C)}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \\
& \Rightarrow \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{1-\tan A \tan C}{1+\tan A \tan C} \\
& \Rightarrow 1+\tan ^2 B-\tan A \tan C-\tan A \tan C \tan ^2 B \\
&= 1-\tan ^2 B+\tan A \tan C-\tan A \tan C \tan ^2 B \\
& \Rightarrow 2 \tan ^2 B=2 \tan A \tan C \Rightarrow \tan ^2 B=\tan A \tan C
\end{aligned}$
Hence, $\tan A, \tan B$ and $\tan C$ will be in G.P.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If cos⁡2x+2cos⁡x=1, then sin2⁡x2-cos2⁡x is equal to

Options:

The correct answer is:
1

Solution:
cos⁡2x+2cos⁡x=1
2cos2⁡x-1+2cos⁡x=1
cos2⁡x+cos⁡x-1=0
cos⁡x=-1±52
cos⁡x=5-12⇒cos2⁡x=5+1-254=3-52
sin2⁡x2-cos2⁡x
=1-cos2⁡x2-cos2⁡x
=1-3-522-3-52
=2-3+524-3+52
=5-125+12=44=1

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\cos 3 \theta=\alpha \cos \theta+\beta \cos ^3 \theta$, then $(\alpha, \beta)=$

Options:
A $(3,4)$
B $(4,3)$
C $(-3,4)$
D $(3,-4)$

The correct answer is:
$(-3,4)$

Solution:
Given that $\cos 3 \theta=\alpha \cos \theta+\beta \cos ^3 \theta$ But $\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta \Rightarrow(\alpha, \beta)=(-3,4)$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\cos \theta=\frac{3}{5}$ and $\cos \phi=\frac{4}{5}$, where $\theta$ and $\phi$ are positive acute angles, then $\cos \frac{\theta-\phi}{2}=$

Options:
A $\frac{7}{\sqrt{2}}$
B $\frac{7}{5 \sqrt{2}}$
C $\frac{7}{\sqrt{5}}$
D $\frac{7}{2 \sqrt{5}}$

The correct answer is:
$\frac{7}{5 \sqrt{2}}$

Solution:
We have $\cos \theta=\frac{3}{5}$ and $\cos \phi=\frac{4}{5}$
$\begin{aligned}
& \begin{aligned}
\text {Therefore } \cos (\theta-\phi) & =\cos \theta \cos \phi+\sin \theta \sin \phi \\
& =\frac{3}{5} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}
\end{aligned} \\
& \text { But } 2 \cos ^2\left(\frac{\theta-\phi}{2}\right)=1+\cos (\theta-\phi)=1+\frac{24}{25}=\frac{49}{50} \\
& \therefore \cos ^2\left(\frac{\theta-\phi}{2}\right)=\frac{49}{50} \text {. Hence, } \cos \left(\frac{\theta-\phi}{2}\right)=\frac{7}{5 \sqrt{2}}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If cos⁡5θ=5cosθ⁡-20cos3⁡θ+acos5⁡θ+b, then the value of a+b is equal to

Options:
A 20
B 16
C -16
D 15

The correct answer is:
16

Solution:
cos⁡5θ=cos⁡4θ+θ
=cos⁡4θcos⁡θ-sin⁡4θsin⁡θ
=22cos2θ-12-1cos⁡θ-2.2sin⁡θcos⁡θ2cos2θ-1sin⁡θ
=cos⁡θ24cos4⁡θ+1-4cos2⁡θ-1-4sin2⁡θcos⁡θ2cos2⁡θ-1
=8cos5⁡θ-8cos3⁡θ+2cos⁡θ-cos⁡θ-41-cos2θcos⁡θ2cos2⁡θ-1
=8cos5⁡θ-8cos3⁡θ+cos⁡θ-4cos⁡θ2cos2⁡θ-1-2cos4⁡θ+cos2⁡θ
=16cos5⁡θ-20cos3⁡θ+5cos⁡θ
Clearly, a=16, b=0

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If∫cosθ5+7sinθ-2cos2θdθ=Aloge|B(θ)|+C, where C is a constant of integration, then B(θ)A can be:

Options:
A 2sinθ+1sinθ+3
B 2sinθ+15(sinθ+3)
C 5(sinθ+3)2sinθ+1
D 5(2sinθ+1)sinθ+3

The correct answer is:
5(2sinθ+1)sinθ+3

Solution:
I=∫cosθ2sin2θ+7sinθ+3dθ

sinθ=t ⇒ cosθdθ=dt

=12∫1t2+72t+32dt=12∫1t+742-542dt=15ln2t+1t+3+c=15ln2sinθ+1sinθ+3+c

so A=15

B(θ)=5(2sinθ+1)sinθ+3

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If $\cos A=\frac{\sqrt{3}}{2}$, then $\tan 3 A=$

Options:
A 0
B $1 / 2$
C 1
D $\infty$

The correct answer is:
$\infty$

Solution:
We have
$\cos A=\frac{\sqrt{3}}{2} \Rightarrow A=30^{\circ}$
$\Rightarrow \tan 3 A=\tan 90^{\circ}=\infty$.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\cos A=\frac{3}{4}$, then $32 \sin \frac{A}{2} \cos \frac{5}{2} A=$

Options:
A $\sqrt{7}$
B $-\sqrt{7}$
C 7
D -7

The correct answer is:
$-\sqrt{7}$

Solution:
$\begin{aligned}
\cos A & =\frac{3}{4} \Rightarrow \sin A=\frac{\sqrt{7}}{4} \\
\text { L.H.S } & =16(\sin 3 A-\sin 2 A) \\
& =16 \sin A\left(3-4 \sin ^2 A-2 \cos A\right) \\
& =16 \cdot \frac{\sqrt{7}}{4}\left(3-4 \cdot \frac{7}{16}-2 \cdot \frac{3}{4}\right)=-\sqrt{7}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\cos (A+B)=\alpha \cos A \cos B+\beta \sin A \sin B$, then $(\alpha, \beta)=$

Options:
A $(-1,-1)$
B $(-1,1)$
C $(1,-1)$
D $(1,1)$

The correct answer is:
$(1,-1)$

Solution:
Given $\cos (A+B)=\alpha \cos A \cos B+\beta \sin A \sin B$
$\begin{aligned}
& \text { But } \cos (A+B)=\cos A \cos B-\sin A \sin B \\
& \Rightarrow \alpha=1, \beta=-1 .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If cosα+cosβ=32 and sinα+sinβ=12 and θ is the arithmetic mean of α & β, then sin2θ+cos2θ is equal to:

Options:
A 35
B 75
C 45
D 85

The correct answer is:
75

Solution:
Given
cos α + cos β = 32
sin α + sin β = 12
⇒2cos(α+β2)cos(α-β2)=32 ......(i) and
2sin⁡(α+β2)cos⁡(α-β2)=12 ....(ii)
Dividing (ii) and (i)
tan α + β2
∴ tan⁡θ =13
sin2θ+cos⁡2θ = 2tan⁡θ1+tan2⁡θ+ 1-tan2⁡θ1+tan2⁡θ=2131+19+1-191+19
= 23+89109=1410=75

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2015 (11 Apr Online)
If cos Î± + cos Î² = a , sin Î± + sin Î² = b and Î± - Î² = 2 Î¸ , then cos 3 Î¸ cos Î¸ is equal to

Options:
A a2+b2-2
B a2+b2-3
C 3-a2-b2
D a2+b24

The correct answer is:
a2+b2-3

Solution:
cos 3 Î¸ cos Î¸ = 4 cos 2 Î¸ - 3 = 2 1 + cos 2 Î¸ - 3
= 2 cos 2 Î¸ - 1 = 2 cos Î± - Î² - 1
Now, cos 2 Î± + sin 2 Î± + cos 2 Î² + sin 2 Î² + 2 cos Î± - Î² = a 2 + b 2
â 2 cos Î± - Î² = a 2 + b 2 - 2
â cos 3 Î¸ cos Î¸ = a 2 + b 2 - 3

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If \(\cos (f(x))=\frac{1-x^2}{1+x^2}\) and \(\tan (g(x))=\frac{3 x-x^3}{1-3 x^2}\), then \(\frac{d f}{d g}=\)

Options:
A \(\frac{3}{2}\)
B \(\frac{1+x^2+2 x^3}{\left(1-3 x^2\right)^2}\)
C \(\frac{2}{3}\)
D \(\frac{x^2+x^3}{\left(1+x^2\right)\left(1-3 x^2\right)}\)

The correct answer is:
\(\frac{2}{3}\)

Solution:
Given, \(\cos (f(x))=\frac{1-x^2}{1+x^2}\)
Put \(\quad x=\tan \theta\)
\(\begin{aligned}
\cos (f(x)) & =\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} \\
\cos (f(x)) & =\cos 2 \theta \\
f(x) & =2 \theta=2 \tan ^{-1} x \\
\tan (g(x)) & =\frac{3 x-x^3}{1-3 x^2}
\end{aligned}\)
Put \(\quad x=\tan \theta\)
\(\begin{aligned}
& \tan (g(x))=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta} \\
& \tan (g(x))=\tan 3 \theta \\
& \therefore \quad g(x)=3 \theta=3 \tan ^{-1} x \\
& \therefore \quad \frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{2}{3} \\
& \therefore \quad \frac{d f}{d g}=\frac{2}{3}
\end{aligned}\)
Hence, answer is (c).

Topic: Mathematics, Differentiation, JEE Main
If \( \cos h(2 x)=199 \), then \( \cot h x= \)

Options:
A \( \frac{5}{3 \sqrt{11}} \)
B \( \frac{5}{6 \sqrt{11}} \)
C \( \frac{7}{3 \sqrt{11}} \)
D \( \frac{10}{3 \sqrt{11}} \)

The correct answer is:
\( \frac{10}{3 \sqrt{11}} \)

Solution:
Given, cos h2x=199

Now,cos h2x=e2x+e-2x2

⇒e2x+e-2x2=199

⇒e2x+e-2x=398

Now, cot hx=ex+e-xex-e-x

a+b2=a2+b2+2ab

⇒ex+e-x2=e2x+e-2x+2exe-x

⇒ex+e-x2=398+2=400

⇒ex+e-x=400=20

a-b2=a2+b2-2ab

⇒ex-e-x2=e2x+e-2x-2exe-x

⇒ex-e-x2=398-2=396

⇒ex-e-x=396=611

Therefore, cot hx=ex+e-xex-e-x=20611=10311

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\cos P=\frac{1}{7}$ and $\cos Q=\frac{13}{14}$, where $P$ and $Q$ both are acute angles. Then the value of $P-Q$ is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $75^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
Given,
$\cos P=\frac{1}{7}, \cos Q=\frac{13}{14}$
$\begin{aligned}
& \therefore \cos (P-Q)=\cos P \cos Q+\sin P \sin Q \\
& =\frac{1}{7} \cdot \frac{13}{14}+\frac{\sqrt{48}}{7} \cdot \frac{\sqrt{27}}{14}=\frac{13+36}{98}=\frac{1}{2}=\cos 60^{\circ} \\
& \Rightarrow P-Q=60^{\circ}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\cos x+\cos y+\cos \alpha=0$ and $\sin x+\sin y+\sin \alpha=0$, then $\cot \left(\frac{x+y}{2}\right)=$

Options:
A $\sin \alpha$
B $\cos \alpha$
C $\cot \alpha$
D $\sin \left(\frac{x+y}{2}\right)$

The correct answer is:
$\cot \alpha$

Solution:
Given equation $\cos x+\cos y+\cos \alpha=0$ and $\sin x+\sin y+\sin \alpha=0$. The given equation may be written as $\cos x+\cos y=-\cos \alpha$ and $\sin x+\sin y=-\sin \alpha$. Therefore
$\begin{aligned}
& 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\cos \alpha \\
& 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\sin \alpha \\
& \frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} \\
& =\frac{\cos \alpha}{\sin \alpha} \Rightarrow \cot \left(\frac{x+y}{2}\right)=\cot \alpha \\
&
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If ∫cosxdxsin3x1+sin6x23=fx1+sin6x1λ+c, where c is a constant of integration, then λfπ3 is equal to

Options:
A -98
B 2
C 98
D -2

The correct answer is:
-2

Solution:
Let, sinx=t

cosxdx=dt
I=∫dtt31+t623=∫dtt71+1t623

Put 1+1t6=r3

⇒dtt7=-12r2dr

I=-12∫r2drr2=-12r+c

⇒I=-12sin6x+1sin6x13+c

⇒I=-12sin2x1+sin6x13+c

So, fx=-12cosec2x and λ=3

λfπ3=-2

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2020 (08 Jan Shift 1)
If ∫cosx-sinx8-sin2x dx=asin-1sinx+cosx b+c, where c is a constant of integration, then the ordered pair a,b is equal to:

Options:
A 1,-3
B 3,1
C -1,3
D 1,3

The correct answer is:
1,3

Solution:
∫cosx-sinx8-sin2xdx

=∫cosx-sinx9-sinx+cosx2dx

Let sinx+cosx=t

∫dt9-t2=sin-1t3+c

=sin-1sinx+cosx3+c

So a=1, b=3

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If cosâ¡x=tanâ¡y, cosâ¡y=tanâ¡z and cosâ¡z=tanâ¡x, then sinâ¡x=2sinâ¡Î¸ where Î¸ is (where, x,y,z,Î¸ are acute angles)

Options:
A 15o
B 18o
C 2212o
D 75o

The correct answer is:
18o

Solution:
cos2x=tan2y=sec2y-1=cot2z-1
So, 1+cos2x=cot2z=cos2zsin2z=cos2z1-cos2z
=tan2x1-tan2x=sin2xcos2x-sin2x
⇒2-sin2x=sin2x1-2sin2x
⇒2-sin2x1-2sin2x=sin2x
⇒2sin4x-6sin2x+2=0
⇒sin2x=3±9-42=3±52
but sin2x∈0,1 so rejecting sin2x=3+52
we have, sin2x=3-52=5-122
⇒sin⁡x=5-12⇒2sin⁡θ=5-12
⇒sin⁡θ=5-14⇒θ=18o

Topic: Mathematics, Trigonometric Equations, JEE Main