datavorous/entrance-exam-dataset huggingface dataset:
If (1+x+x2 )25 = a0+a1 x+a2 x2+ .............a50 x50, then the value of a0+a2+a4+ .........+a50 is

Options:
A odd and of the form 3n
B odd and of the form ( 3n−1 )
C odd and of the form ( 3n+1 )
D even

The correct answer is:
even

Solution:
Putting x=1 , we get,

3 25 = a 0 + a 1 + a 2 +.....+ a 50 ....(1)
Putting x=−1 , we get,

1= a 0 − a 1 + a 2 +........+ a 50 ...(2)
adding equation (1) & (2), we get,
⇒ a 0 + a 2 +……+ a 50 = 3 25 +1 2 = ( 1+2 ) 25 +1 2
⇒ a 0 + a 2 + a 4 +.....+ a 50 = ​( 25 C 0 + ​ 25 C 1 ·2+ ​ 25 C 2 · 2 2 +…+ ​ 25 C 25 · 2 25 )+1 2 = 2[ 1+ ​ 25 C 1 + ​ 25 C 2 ·2+…+ ​ 25 C 25 · 2 24 ] 2
⇒ a 0 + a 2 + a 4 +......+ a 50 =2[ 13+   25 C 2 +......+   25 C 25 .2 23 ]
a 0 + a 2 + a 4 +......+ a 50 =2K ⇒ even

Topic: Mathematics, Binomial Theorem, JEE Main
If $1, z_1, z_2, \ldots, z_{n-1}$ are the $n$th roots of unity, $\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_{n-1}\right)$ is equal to

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

The correct answer is:
$n$

Solution:
We have given that, $1, z_1, z_2, z_3, \ldots . z_{n-1}$ are the $n$th root of unity.
$\begin{aligned}
& \therefore \quad z^n-1=(z-1)\left(z-z_1\right)\left(z-z_2\right), \ldots .,\left(z-z_{n-1}\right) \\
& (z-1)\left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1\right) \\
& =(z-1)\left(z-z_1\right)+\ldots . .\left(z-z_{n-1}\right) \\
& \left(z^{n-1}+z^{n-2}+\ldots . .+z^2+z+1\right)=\left(z-z_1\right) \\
& \left(z-z_1\right)=\left(z-z_{n-1}\right)
\end{aligned}$
Put $z=1$, we get
$\begin{aligned}
& (1+1+\ldots .+n \text { times }) \\
& =\left(1-z_1\right)\left(1-z_2\right)+\ldots .\left(1-z_{n-1}\right) \\
& n=\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_{n-1}\right)
\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\left|\begin{array}{ccc}2 & 4 & 0 \\ 0 & 5 & 16 \\ 0 & 0 & 1+p\end{array}\right|=20$, then what is the value of $p$?

Options:
A 0
B 1
C 2
D 5

The correct answer is:
1

Solution:
$\left|\begin{array}{ccc}2 & 4 & 0 \\ 0 & 5 & 16 \\ 0 & 0 & 1+p\end{array}\right|=20$
On expanding along $\mathrm{C}_{1}$, $2\{5(1+p)-0\}=20$
$\Rightarrow 1+p=2$
$\Rightarrow p=1$

Topic: Mathematics, Determinants, JEE Main
If 2∫01tan-1xdx=∫01cot-11-x+x2dx, then ∫01tan-11-x+x2dx is equal to

Options:
A π2+ln2
B ln2
C π2-ln4
D ln4

The correct answer is:
ln2

Solution:
From the given equation
2∫01tan-1⁡xdx= ∫01π2-tan-1⁡1-x+x2dx
⇒2∫01tan-1⁡xdx= ∫01π2 dx- ∫01tan-1⁡1-x+x2dx
⇒∫01tan-1⁡1-x+x2dx=π2-2 ∫01tan-1⁡xdx …i
Let I1= ∫01tan-1⁡xdx
=tan-1⁡xx01-∫0111+x2xdx
=π4- ∫01x1+x2dx

=π4-12 ∫012x1+x2dx

=π4-12 ln1+x201
=π4-12ln⁡2
By equation i,
∫01tan-1⁡1-x+x2dx= π2-2π4-12ln2=ln⁡2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2016 (09 Apr Online)
If $(2,0)$ is the vertex and $y$-axis the directrix of a parabola, then its focus is

Options:
A $(2,0)$
B $(-2,0)$
C $(4,0)$
D $(-4,0)$

The correct answer is:
$(4,0)$

Solution:
Vertex $=(2,0) \Rightarrow$ focus is $(2+2,0)=(4,0)$

Topic: Mathematics, Parabola, JEE Main
If $\frac{1}{2} \leq x \leq 1$, then $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ is equal to

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{3}$
C $\pi$
D $0$

The correct answer is:
$\frac{\pi}{3}$

Solution:
$\begin{aligned} & \therefore \cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{\sqrt{3}}{2} \sqrt{1-x^2}\right) \\ & =\cos ^{-1} x+\cos ^{-1}\left(x \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \sqrt{1-x^2}\right) \\ & =\cos ^{-1} x+\cos ^{-1}\left(\frac{1}{2}\right)-\cos ^{-1} x \\ & =\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If α,-π2<α<π2 is the solution of 4cosθ+5sinθ=1, then the value of tanα is

Options:
A 10-106
B 10-1012
C 10-1012
D 10-106

The correct answer is:
10-1012

Solution:
Given: 4cosθ+5sinθ=1

⇒41-tan2θ21+tan2θ2+52tanθ21+tan2θ2=1

⇒5tan2θ2-10tanθ2-3=0

⇒tanθ2=10±100-45-32×5

⇒tanθ2=10±16010

⇒tanθ2=5±405

It is given that, α∈-π2,π2

⇒α2∈-π4,π4

⇒tanα2∈-1,1

⇒tanα2=5-405

We know that, tanθ=2tanθ21-tan2θ2

⇒tanα=25-4051-5-4052

⇒tanα=25-4052405-85

⇒tanα=5-4040-4

⇒tanα=40-2024

⇒tanα=10-1012

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If $\quad 2^3+4^3+6^3+\ldots+2(n)^3=h n^2(n+1)^2$, then $h$ is equal to

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

The correct answer is:
$2$

Solution:
$\begin{aligned} \text { Let } S_n=2^3+4^3 & +6^3+\ldots+(2 n)^3 \\ & =\Sigma(2 n)^3=8 \Sigma n^3=8 \quad\left[\frac{n(n+1)}{2}\right]^2 \\ \Rightarrow 2 n^2(n+1)^2 & =h n^2(n+1)^2 \\ \Rightarrow \quad h & =2\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
If π2<α<3π4, then 2tan⁡α+1cos2α is equal to

Options:
A -1+tan⁡α
B -1-tan⁡α
C 1+tan⁡α
D 1-tan⁡α

The correct answer is:
-1-tan⁡α

Solution:
2tanâ¡Î±+1cos2Î±
=2tanâ¡Î±+sec2Î±=2tanâ¡Î±+1+tan2Î±
=1+tanâ¡Î±2=1+tanâ¡Î±
Since, tanâ¡Î±<-1 when Ï2<Î±<3Ï4
â1+tanâ¡Î±<0
â1+tanâ¡Î±=-1-tanâ¡Î±

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\left[\begin{array}{cc}-2 & 5 \\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{c}3 \\ -1\end{array}\right]$, then $(x, y)$ is

Options:
A $(1,2)$
B $(-1,2$
C $(1,-2$
D $(2,1)$

The correct answer is:
$(2,1)$

Solution:
$\left[\begin{array}{cc}-2 & 5 \\ 3 & -1\end{array}\right]_{2 x \times 2}\left[\begin{array}{l}x \\ y\end{array}\right]_{2 \times 1}=\left[\begin{array}{cc}1 & 2 \\ 3 & 4\end{array}\right]_{2 \times 2}\left[\begin{array}{c}3 \\ -1\end{array}\right]_{2 \times 1}$ $\Rightarrow\left[\begin{array}{c}-2 x+5 y \\ 3 x-y\end{array}\right]=\left[\begin{array}{l}1 \\ 5\end{array}\right]$
$\Rightarrow-2 x+5 y=1,3 x-y=5$
or $y=3 x-5 \quad \text{...(i)}$
$\Rightarrow \quad-2 x+5(3 x-5)=1$
$\Rightarrow \quad-2 x+15 x-25=1$
$\Rightarrow \quad 13 x=26 \Rightarrow x=2$
Substituting $x=2$ in Eq. (i), we get
$$
y=6-5=1
$$
Hence, $(x, y)=(2,1)$

Topic: Mathematics, Matrices, JEE Main
If $(2,3,5)$ is one end of a diameter of the sphere $x^2+y^2+z^2-6 x-12 y-2 z+20=0$, then the coordinates of the other end of the diameter are

Options:
A
$(4,9,-3)$

B
$(4,-3,3)$

C
$(4,3,5)$

D
$(4,3,-3)$

The correct answer is:
$(4,9,-3)$

Solution:
Coordinates of centre $(3,6,1)$
Let the coordinates of the other end of diameter are $(\alpha, \beta, \gamma)$
then $\frac{\alpha+2}{2}=3, \frac{\beta+3}{2}=6, \frac{\gamma+5}{2}=1$
Hence $\alpha=4, \beta=9$ and $\gamma=-3$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2007
If 2,3,9,5,2,1,1,λ,8 and λ,2,3 are coplanar, then the product of all possible values of λ is

Options:
A 212
B 598
C 578
D 958

The correct answer is:
958

Solution:
Given,

Points A2,3,9, B5,2,1, C1,λ,8 & Dλ,2,3 are coplanar,

Now we know that condition of coplanarity is AB→AC→AD→=0

⇒3-1-8-1λ-3-1λ-2-1-6=0

⇒3-6λ-3-1+16+λ-2-81-λ-3λ-2=0

⇒3-6λ+17+4+λ-8+8λ-3λ-2=0

⇒-18λ+51+4+λ-8+8λ2-40λ+48=0

⇒8λ2-57λ+95=0

So product of roots will be 958

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (29 Jul Shift 2)
If 2+3i is one of the roots of the equation 2x3-9x2+kx-13=0, k∈R, then the real root of this equation (where i2=-1) :

Options:
A Exists and is equal to 12
B Does not exist
C Exists and is equal to 1
D Exists and is equal to -12

The correct answer is:
Exists and is equal to 12

Solution:
If 2+3i in one of the roots, then 2-3i would be other.
Since coefficients of the equation are real.

Let γ be the third root, then product of roots → α β γ =132
2+3i 2-3i ⋅γ=132
4+9⋅γ=132
⇒γ=12
The value of k will come if we put x=12 in the equation
2⋅18-94+k⋅12-13=0
⇒k2=15
⇒k=30
∴ Equation will become 2x3-9x2+30x-13=0
⇒αβ+βγ+γα=302=15
⇒2+3i12+2-3i12+2+3i 2-3i=15
⇒1+i2+1-i2+13=15
⇒15=15
Hence roots exists and is equal to 12 '

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2015 (10 Apr Online)
If $\left(2+\frac{x}{3}\right)^{55}$ is expanded in the ascending powers of $x$ and the coefficients of powers of $x$ in two consecutive terms of the expansion are equal, then these terms are:

Options:
A
$7^{\text {th }}$ and $8^{\text {th }}$

B
$8^{\text {th }}$ and $9^{\text {th }}$

C
$28^{\text {th }}$ and $29^{\text {th }}$

D
$27^{\text {th }}$ and $28^{\text {th }}$

The correct answer is:
$7^{\text {th }}$ and $8^{\text {th }}$

Solution:
Let $r^{\text {th }}$ and $(r+1)^{\text {th }}$ term has equal coefficient
$$
\begin{aligned}
&\left(2+\frac{x}{3}\right)^{55}=2^{55}\left(1+\frac{x}{6}\right)^{55} \\
&r^{\text {th }} \text { term }=2^{5555} C_r\left(\frac{x}{6}\right)^r
\end{aligned}
$$
Coefficient of $x^r$ is $2^{55}{ }^{55} C_r \frac{1}{6^r}$
$$
(r+1)^{\text {th }} \text { term }=2^{5555} C_{r+1}\left(\frac{x}{6}\right)^{r+1}
$$
Coefficient of $x^{r+1}$ is $2^{55}{ }^{55} C_{r+1} \cdot \frac{1}{6^{r+1}}$
Both coefficients are equal
$$
2^{5555} C_r \frac{1}{6^r}=2^{5555} C_{r+1} \frac{1}{6^{r+1}}
$$
$$
\begin{aligned}
&\frac{1}{|r| 55-r}=\frac{1}{\lfloor r+1 \mid 54-r} \cdot \frac{1}{6} \\
&6(r+1)=55-r \\
&6 r+6=55-r \\
&7 r=49 \\
&r=7 \\
&(r+1)=8
\end{aligned}
$$
Coefficient of $7^{\text {th }}$ and $8^{\text {th }}$ terms are equal.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2014 (12 Apr Online)
If $\left[\begin{array}{cc}2 & -3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]=\left[\begin{array}{cc}1 & 4 \\ 2 & -5\end{array}\right]$, then $(a, b, c, d)=$

Options:
A $(1,6,2,5)$
B $(1,2,7,5)$
C $(1,2,-7,5)$
D $(-1,-2,7,-5)$

The correct answer is:
$(1,2,-7,5)$

Solution:
$\left[\begin{array}{cc}2 & -3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]$$=\left[\begin{array}{cc}1 & 4 \\ 2 & -5\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]$$=\left[\begin{array}{cc}2 & -3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{cc}1 & 4 \\ 2 & -5\end{array}\right]=\left[\begin{array}{cc}1 & -7 \\ 2 & 5\end{array}\right]$
$\Rightarrow(a, b, c, d)=(1,2,-7,5)$.

Topic: Mathematics, Matrices, JEE Main
If 2.5×10-6 N average force is exerted by a light wave on a non-reflecting surface of 30 cm2 area during 40 min of time span, the energy flux of light just before it falls on the surface is _______ W cm-2.(Round off to the Nearest Integer), (Assume complete absorption and normal incidence conditions are there)

Options:

The correct answer is:
25

Solution:
F=IAc

⇒I=FcA=2.5×10-6×3×10830=25 W cm-2

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If $(-2, 6)$ is the image of the point $(4, 2)$ with respect to the line $L = 0$, then $L$ is equal to

Options:
A $6x - 4y - 7 = 0$
B $2x + 3y - 5 = 0$
C $3x - 2y + 5 = 0$
D $3x - 2y + 10 = 0$

The correct answer is:
$3x - 2y + 5 = 0$

Solution:
Slope of the line $C D$
$$
=\frac{6-2}{-2-4}=\frac{4}{-6}=-\frac{2}{3}
$$
Slope of the line $A B=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}$
Mid-point of $C D$ is $O\left(\frac{4-2}{2}, \frac{2+6}{2}\right)$
$$
=O(1,4)
$$
Equation of line $A B$ is,
$$
\begin{aligned}
y-4 & =\frac{3}{2}(x-1) \\
\Rightarrow \quad 2 y-8 & =3 x-3 \\
\Rightarrow \quad 3 x-2 y+5 & =0
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
If 2, 7, 9 and 5 are subtracted respectively from four numbers in geometric progression, then the resulting numbers are in arithmetic progression. The smallest of the four numbers is

Options:
A –24
B –12
C 6
D 3

The correct answer is:
–24

Solution:
Let the numbers be a,ar,ar2,ar3 ,
⇒a-2,ar-7,ar2-9 and ar3-5 are in A.P.
⇒a-2+ar2-9=2ar-7 and ar-7+ar3-5=2ar2-9
⇒a+ar2=2ar-3 and ar+ar3=2ar2-6
⇒1+r2=2r-3a and 1+r2=2r-6ar
⇒3a=6ar⇒r=2
⇒3a=2r-1-r2=4-1-4=-1⇒a=-3
So, the numbers are –3,–6,–12,–24

Topic: Mathematics, Sequences and Series, JEE Main
If 210+29·31+28·32+……+2·39+310=S-211, then S is equal to

Options:
A 311-212
B 311
C 3112+210
D 2.311

The correct answer is:
311

Solution:
S'=210+29·3+28·32+…….+2·39+310

S'=2101+32+322+…….+329+3210

G.P. →a=1,r=32,n=11

S'=210·3211-132-1=211311211-1

=311-211

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If 22020+2021 is divided by 9, then the remainder obtained is

Options:
A 0
B 1
C 3
D 7

The correct answer is:
3

Solution:
22020=2.22019=2.23673=29-1673
=29673-673C1â9672+673C2â9671+673C6729-1
=2 (a multiple of 9-1 )
22020+2021=9K+2019
9K+2019 when divided by 9 gives remainder 3

Topic: Mathematics, Binomial Theorem, JEE Main
If 2a1, 2a2, 2a3..... 2ar are in geometric progression, then a1a2a3an+1an+2an+3a2n+1a2n+2a2n+3 is equal to

Options:
A 25
B 23
C 0
D None of these

The correct answer is:
0

Solution:
2a1, 2a2,2a3..... are in G.P
â a2- a1=a3- a2=a4- a3=.....
â a1, a2, a3 ..... are in A.P .
â a2n+1 + a1=2an+1, a2 + a2n+2=2an+2,
a3 + a2n+3=2an+3
So, by operation R2 âR2-12 (R1+R3)
âÎ=a1a2a3000a2n+1a2n+2a2n+3=0

Topic: Mathematics, Determinants, JEE Main
If $2 a+3 b+6 c=0(a, b, c \in R)$ then the quadratic equation $a x^2+b x+c=0$ has

Options:
A
at least one root in [0, 1]

B
at least one root in [2, 3]

C
at least one root in [4, 5]

D
none of these

The correct answer is:
at least one root in [0, 1]

Solution:
Let $f(x)=\frac{a x^3}{3}+\frac{b x^2}{2}+c x \Rightarrow f(0)=0$ and $f(1)=\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6}=0$
Also $f(x)$ is continuous and differentiable in $[0,1]$ and $\left[0,1]\right.$. So by Rolle's theorem, $f^{\prime}(x)=0$. i.e. $a x^2+b x+c=0$ has at least one root in $[0,1]$

Topic: Mathematics, Quadratic Equation, JEE Main
If $2 a+3 b+6 c=0(a, b, c \in R)$ then the quadratic equation $a x^2+b x+c=0$ has

Options:
A
at least one root in [0, 1]

B
at least one root in [2, 3]

C
at least one root in [4, 5]

D
none of these

The correct answer is:
at least one root in [0, 1]

Solution:
Let $f(x)=\frac{a x^3}{3}+\frac{b x^2}{2}+c x \Rightarrow f(0)=0$ and $f(1)=\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6}=0$
Also $f(x)$ is continuous and differentiable in $[0,1]$ and $\left[0,1]\right.$. So by Rolle's theorem, $f^{\prime}(x)=0$. i.e. $a x^2+b x+c=0$ has at least one root in $[0,1]$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2002
If $2 a+3 b+6 c=0$, then at least one root of the equation $a x^2+b x+c=0$ lies in the interval

Options:
A
$(0,1)$

B
$(1,2)$

C
$(2,3)$

D
$(1,3)$

The correct answer is:
$(0,1)$

Solution:
Let $f^{\prime}(x)=a x^2+b x+c \Rightarrow f(x)=\frac{a x^3}{3}+\frac{b x^2}{2}+c x+d$ $\Rightarrow f(x)=\frac{1}{6}\left(2 a x^3+3 b x^2+6 c x+6 d\right)$, Now $f(1)=f(0)=d$, then according to Rolle's theorem $\Rightarrow f^{\prime}(x)=a x^2+b x+c=0$ has at least one root in $(0,1)$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2004
If $\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$
$$
=\alpha(a+b() b+c() c+a) \neq 0
$$
then $\alpha$ is equal to

Options:
A
$a+b+c$

B
$a b c$

C
4

D
1

The correct answer is:
4

Solution:
Let $\Delta=\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$
Applying $C_1+C_3$ and $C_2+C_3$
$$
\Delta=\left|\begin{array}{ccc}
-a+c & 2 a+b+c & a+c \\
2 b+a+c & -b+c & b+c \\
a-c & b-c & -2 c
\end{array}\right|
$$
Now, applying $R_1+R_3$ and $R_2+R_3$
$$
\Delta=\left|\begin{array}{ccc}
0 & 2(a+b & ) a-c \\
2(a+b & 0 & b-c \\
a-c & b-c & -2 c
\end{array}\right|
$$
On expanding, we get
$$
\begin{aligned}
\Delta= & -2(a+b)\{-2 c[2(a+b)] \\
& -(a-c)(b-c)\} \\
& +(a-c)[2(a+b)(b-c)] \\
\Delta= & 8 \mathrm{c}(a+b)(a+b) \\
& +4(a+b)(a-c)(b-c)
\end{aligned}
$$
$$
\begin{aligned}
& =4(a+b)\left[2 a c+2 b c+a b-b c-a c+c^2\right] \\
& =4(a+b)\left[a c+b c+a b+c^2\right] \\
& =4(a+b)[c(a+c)+b(a+c)] \\
& =4(a+b)(b+c)(c+a) \\
& =\alpha(a+b)(b+c)(c+a)
\end{aligned}
$$
Hence, $\alpha=4$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2012 (12 May Online)
If 2 and 6 are the roots of the equation $a x^2+b x+1=0$, then the quadratic equation, whose roots are $\frac{1}{2 a+b}$ and $\frac{1}{6 a+b}$, is :

Options:
A $2 x^2+11 x+12=0$
B $x^2+8 x+12=0$
C $4 x^2+14 x+12=0$
D $x^2+10 x+16=0$

The correct answer is:
$x^2+8 x+12=0$

Solution:
$\begin{aligned} & \text { Sum }=8=-\frac{b}{a} \\ & \text { Product }=12=\frac{1}{a} \quad \Rightarrow a=\frac{1}{12} \\ & b=-\frac{2}{3} \\ & 2 a+b=\frac{2}{12}-\frac{2}{3}=-\frac{1}{2} \\ & 6 a+b=\frac{6}{12}-\frac{2}{3}=-\frac{1}{6} \\ & \text { sum }=-8 \\ & P=12 \\ & x^2+8 x+12=0\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (04 Apr Shift 1)
If $2 \cos \theta+\sin \theta=1\left(\theta \neq \frac{\pi}{2}\right)$, then $7 \cos \theta+6 \sin \theta$ is equal to:

Options:
A
$\frac{1}{2}$

B
2

C
$\frac{11}{2}$

D
$\frac{46}{5}$

The correct answer is:
$\frac{46}{5}$

Solution:
Given $2 \cos \theta+\sin \theta=1$
Squaring both sides, we get
$$
\begin{aligned}
&(2 \cos \theta+\sin \theta)^2=1^2 \\
&\Rightarrow 4 \cos ^2 \theta+\sin ^2 \theta+4 \sin \theta \cos \theta=1 \\
&\Rightarrow 3 \cos ^2 \theta+\left(\cos ^2 \theta+\sin ^2 \theta\right)+4 \sin \theta \cos \theta \\
&=1 \\
&\Rightarrow 3 \cos ^2 \theta+\not \lambda+4 \sin \theta \cos \theta=1 \\
&\Rightarrow 3 \cos ^2 \theta+4 \sin \theta \cos \theta=0 \\
&\Rightarrow \cos \theta(3 \cos \theta+4 \sin \theta)=0 \\
&\Rightarrow 3 \cos \theta+4 \sin \theta=0 \\
&\Rightarrow 3 \cos \theta=-4 \sin \theta \\
&\Rightarrow \frac{-3}{4}=\tan \theta=\sqrt{\sec ^2 \theta-1}=\frac{-3}{4} \\
&\left(\because \tan ^2 \theta=\sqrt{\sec ^2 \theta-1}\right) \\
&\Rightarrow \sec ^2 \theta-1=\left(\frac{-3}{4}\right)^2=\frac{9}{16} \\
&\Rightarrow \sec 2 \theta=\frac{9}{16}+1=\frac{25}{16} \Rightarrow \sec \theta=\frac{5}{4}
\end{aligned}
$$
or $\cos \theta=\frac{4}{5}$
Now, $\sin ^2 \theta+\cos ^2 \theta=1$
$$
\begin{aligned}
&\Rightarrow \sin ^2 \theta+\left(\frac{4}{5}\right)^2=1 \\
&\sin ^2 \theta+\frac{4}{5}=1 \Rightarrow \sin ^2 \theta=1-\frac{16}{25}=\frac{9}{25} \\
&\sin \theta=\pm \frac{3}{5}
\end{aligned}
$$
Taking $\quad\left(\sin \theta=+\frac{3}{5}\right) \quad$ because $\left(\sin \theta=-\frac{3}{5}\right)$ cannot satisfy the given equation.
Therefore; $7 \cos \theta+6 \sin \theta$
$$
=7 \times \frac{4}{5}+6 \times \frac{3}{5}=\frac{28}{5}+\frac{18}{5}=\frac{46}{5}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2014 (11 Apr Online)
If 2+cos⁡x3+ydydx+sin⁡x=0 and y0=1, then yπ3 is equal to

Options:
A 43
B 73
C 13
D 1

The correct answer is:
13

Solution:
dyy+3=-sin⁡x2+cos⁡xdx
⇒ln⁡y+3=ln⁡2+cos⁡x+ln⁡C
⇒y+3=C(2+cos⁡x)
Now, at x=0, y=1
⇒4=3C⇒C=43
⇒y=432+cos⁡x-3
Hence, yπ3=432+12−3=23⋅5−3=13

Topic: Mathematics, Differential Equations, JEE Main
If 2 data sets having 10 and 20 observations have coefficients of variation 50 and 60 respectively and arithmetic means 30 and 25 respectively, then the combined variance of those 30 observations is

Options:
A 20753
B 20759
C 10009
D 10753

The correct answer is:
20759

Solution:
Given,

n1=10, x¯1=30, CV1=50
n2=20, x¯1=25, CV2=60
CV1=σ1x¯1×100⇒50=σ130×100⇒σ1=15
CV2=σ2x¯2×100⇒60=σ125×100⇒σ2=15
Combined mean =10×30+20×2530=803
d1=x¯1-x¯=103, d2=x¯-x¯2=53
Combined variance =n1σ12+n2σ22+n1d12+n2d22n1+n2
=2250+4500+10009+500930
=622527=20759

Topic: Mathematics, Statistics, JEE Main
If 2 distinct numbers are between 0 to 180 (both inclusive) and the probability that their average is 60 is k, then 1086k is equal to

Options:

The correct answer is:
4

Solution:
Let 2 distinct number are a,b
a+b2=60⇒a+b=120
if a=0 then b=120
a=1 then b=119
.
.
.
a=59 then b=61
favourable case = 60
Total case =181C2
Required probability =60181C2=2543=k
1086k=4

Topic: Mathematics, Probability, JEE Main
If $2 \mathrm{f}(\mathrm{x})=\mathrm{f}^{\prime}(x)$ and $\mathrm{f}(0)=3$, then the value of $\mathrm{f}(2)$ is

Options:
A $3 e^{2}$
B $2 e^{3}$
C $4 e^{3}$
D $3 e^{4}$

The correct answer is:
$3 e^{4}$

Solution:
We have $f^{\prime}(x)=2 f(x)$
$$
\begin{array}{l}
\therefore \int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{d} \mathrm{x}=\int 2 \mathrm{~d} \mathrm{x} \\
\therefore \log |\mathrm{f}(\mathrm{x})|=2 \mathrm{x}+\mathrm{c}
\end{array}
$$
Now $f(0)=3$
$$
\begin{array}{l}
\therefore|\log 3|=0+c \Rightarrow c=\log 3 \\
\therefore \log |f(x)|=2 x+\log 3
\end{array}
$$
When $x=2$,
$$
\begin{aligned}
& \log |\mathrm{f}(2)|=2(2)+\log 3=4+\log 3 \\
\therefore & \mathrm{f}(2)=\mathrm{e}^{4+\log 3}=\mathrm{e}^{4} \cdot \mathrm{e}^{\log 3}=3 \mathrm{e}^{4}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If 2fx+y=fxâfy for all real x,ây, where f â² 0 =3 and f4=25, then the value of f â² 4 is equal to

Options:

The correct answer is:
37.5

Solution:
f ′ 4 = lim h→0 f 4+h −f 4 h
=limh→02f4+h-2f4+02h
=limh→0f4⋅fh-f4⋅f02h
Using  2fx+y=fx⋅fy , ∀ x,y
f'4=f42limh→0fh-f0h
=f42×f'0=252×3=37.5

Topic: Mathematics, Application of Derivatives, JEE Main
If 2f xy = f x y + f y x for all x,âyâR and f 1 =3, then the value of Î£ r = 1 1 0 f r is equal to

Options:
A 3 2 3 1 0 - 1
B 3 2 3 9 - 1
C 3 1 0 - 1 2
D 12310-1

The correct answer is:
3 2 3 1 0 - 1

Solution:
From given functional equation, 2f xy = f x y + f y x ,ââx,âyâR
putting y=1 ,
2f x =f x + f 1 x
f x = 3 x
â´ Î£ r = 1 1 0 f r = Î£ r = 1 1 0 3 r = 3 3 1 0 - 1 3 - 1 = 3 2 3 1 0 - 1

Topic: Mathematics, Functions, JEE Main
If $2 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved to make $200 \mathrm{ml}$ solution at $25^{\circ} \mathrm{C}$, the molarity $(\mathrm{M})$ at $90^{\circ} \mathrm{C}$ is

Options:
A $\mathrm{M} < 0.25$
B $0.5>M>0.25$
C $\mathrm{M}=0.25$
D $0.5 < \mathrm{M} < 1.0$

The correct answer is:
$\mathrm{M} < 0.25$

Solution:
Concentration of the solution at $20^{\circ} \mathrm{C}$
$$
=\frac{\mathrm{n}}{\mathrm{V}}=\frac{2 / 40}{200 / 1000}=0.25(\mathrm{M}) \quad[\mathrm{Mw} \text { of } \mathrm{NaOH}=40]
$$
As temp. increases the volume of the solution increases but the no. of moles remains the same. Hence, concentration of the solution decreases.

Topic: Chemistry, Solutions, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
If 2, h1, h2,....., h20, 6 are in harmonic progression and 2, a1, a2,...., a20, 6 are in arithmetic progression, then the value of a3h18 is equal to

Options:
A 6
B 12
C 3
D 9

The correct answer is:
12

Solution:
12,1h1,1h2,....,1h20,16 are in A.P.
⇒16=12+21d⇒16-1221=d
⇒d=-163
⇒1h18=12+18d=12+182-1637=12-27
1h18=314⇒h18=143
Also, 2,a1,a2,.....,a20,6 are in A.P.
⇒6=2+21D⇒D=421
⇒a3=2+3D=2+3×4217=187
⇒a3h18=1867×1423=12

Topic: Mathematics, Sequences and Series, JEE Main
If $2+i \sqrt{3}$ is a root of the equation $x^2+p x+q=0$, where $p$ and $q$ are real, then $(p, q)=$

Options:
A $(-4,7)$
B $(4,-7)$
C $(4,7)$
D $(-4,-7)$

The correct answer is:
$(-4,7)$

Solution:
Since $2+i \sqrt{3}$ is a root, therefore $2-i \sqrt{3}$ will be other root. Now, sum of the roots $=4=-p$ and product of roots $=7=q$. Hence $(p, q)=(-4,7)$

Topic: Mathematics, Quadratic Equation, JEE Main
If $(2, k)$ is a point on the parabola passing through the points $(1,-3),(-1,5),(0,2)$ and having its axis parallel to the $Y$-axis, then $k$ is equal to

Options:
A -10
B 3
C -7
D 5

The correct answer is:
-10

Solution:
Let equation of parabola is
Eq. (i) passes through $(1,-3),(-1,5)$ and $(0,2)$.
From Eqs. (ii), (iii) and (iv),
From Eq. (i), $\quad \begin{aligned} & a=-1, b=-4, c= \\ & y=-x^2-4 x+2\end{aligned}$
$\because(2, k)$ lies on the above parabola.
$$
\therefore \quad k=-4-8+2=-10
$$

Topic: Mathematics, Parabola, JEE Main
If 2 moles of diatomic gas and 1 mole of monatomic gas are mixed, then the ratio of specific heats for the mixture is

Options:
A 7 3
B 5 4
C 19 13
D 15 19

The correct answer is:
19 13

Solution:
μ 1 = 1 , γ 1 =5 / 3 (for monatomic gas) and μ 2 = 2 , γ 2 = 7 / 5 (for diatomic gas).
Form formula
γ mixture = μ 1 γ 1 γ 1 - 1 + μ 2 γ 2 γ 2 - 1 μ 1 γ 1 - 1 + μ 2 γ 2 - 1
= 1 × 5 3 5 3 - 1 + 2 × 7 5 7 5 - 1 1 5 3 - 1 + 2 7 5 - 1 = 5 3 + 7 3 2 + 5 = 1 9 1 3

Topic: Physics, Thermodynamics, JEE Main
If α2nα2n+2α2n+4β2nβ2n+2β2n+4γ2nγ2n+2γ2n+4=1β2-1α21γ2-1β21α2-1γ2 {where α2, β2 and γ2 are all distinct}, then the value of n is equal to

Options:
A 4
B -4
C 3
D -2

The correct answer is:
-2

Solution:
Given, α2nβ2nγ2n 1α2α41β2β41γ2γ4=α2-β2β2-γ2(γ2-α2)α4β4γ4
⇒α2nβ2nγ2n[α2-β2β2-γ2γ2-α2]=α2-β2β2-γ2(γ2-α2)αβγ4
On comparison, we get,
n=-2

Topic: Mathematics, Determinants, JEE Main