datavorous/entrance-exam-dataset huggingface dataset:
If $\alpha=\underset{x \rightarrow 0}{\operatorname{Lt}} \frac{x \cdot 2^x-x}{1-\cos x}$ and $\beta=\operatorname{Lt}_{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$, then

Options:
A $\alpha=\beta$
B $\alpha=2 \beta$
C $\alpha=\frac{\beta}{2}$
D $\alpha=3 \beta$

The correct answer is:
$\alpha=2 \beta$

Solution:
We have,
$$
\begin{aligned}
& \alpha=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
& =\lim _{x \rightarrow 0} \frac{1 \cdot 2^x+x \cdot 2^x \log 2-1}{\sin x} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
& =\lim _{x \rightarrow 0} \frac{2^x \log 2+2^x \log 2+x\left(\log 2^2 2^x\right.}{\cos x} \\
& =\lim _{x \rightarrow 0} \frac{2 \cdot 2^x \log 2+2^x \cdot x\left(\log 2^2\right.}{\cos x} \\
& \Rightarrow \alpha=2 \cdot \log 2 \\
& \text { Again, } \beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
& =\lim _{x \rightarrow 0} \frac{x 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}} \times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(x 2^x-x\right)\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)}{1+x^2-1+x^2} \quad\left[\frac{0}{0} \text { form }\right] \\
& =\lim _{x \rightarrow 0} \frac{\left(z^x-1\right)\left[\sqrt{1+x^2}+\sqrt{1-x^2}\right]}{2 x} \quad\left[\frac{0}{0} \text { form }\right] \\
& =\lim _{x \rightarrow 0} \frac{2^x \log 2\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)+\left(z^x-1\right)}{2} \quad\left[\frac{2 x}{\sqrt{1+x^2}}-\frac{2 x}{\sqrt{1-x^2}}\right] \\
& \Rightarrow \beta=\frac{\log 2(1+1)+0}{2}=\log 2 \\
& \therefore \quad \alpha=2 \beta \quad
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Main
If x0,y0,z0 is any solution of the system of equations 2x-y-z=1, -x-y+2z=1 and x-2y+z=2, then the value of x02-y02+1z0 where, z0≠0 is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Using Cramer’s rule, we get,
Δ=2-1-1-1-121-21
Using C1=C1+C2+C3
Δ=0-1-10-120-21=0
Δ1=1-1-11-122-21=0 (as C1 and C2 are identical)
Δ2=21-1-112121
Using C1=C1-C2+C3
=01-1012021=0
Δ3=2-11-1-111-22=0 (as C2 and C3 are identical)
Therefore, infinite solutions
Now, put z=λ, we get,
2x-y=1+λ and
-x-y=1-2λ
Solving these 2 equations, we get,
x=λ, y=1-λ
x02-y02+1z0=λ2-λ-12+1λ=2

Topic: Mathematics, Determinants, JEE Main
If $x>0, y>0, z>0, x y+y z+z x < 1$ and if $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi$, then $x+y+z$ equals to

Options:
A $0$
B $x y z$
C $3 x y z$
D $\sqrt{x y z}$

The correct answer is:
$x y z$

Solution:
$\begin{aligned} & \text { Given, } \tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{x+y+z-x y z}{1-x y-y z-z x}\right)=\pi \\ & \Rightarrow \quad \frac{x+y+z-x y z}{1-x y-y z-z x}=\tan \pi=0 \\ & \Rightarrow \quad x+y+z=x y z\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $x=\int_0^y \frac{d t}{\sqrt{1+t^2}}$, then $\frac{d^2 y}{d x^2}$ is equal to :

Options:
A
$y$

B
$\sqrt{1+y^2}$

C
$\frac{x}{\sqrt{1+y^2}}$

D
$y^2$

The correct answer is:
$y$

Solution:
$x=\int_0^y \frac{d t}{\sqrt{1+t^2}}$
$\Rightarrow 1=\frac{1}{\sqrt{1+y^2}} \cdot \frac{d y}{d x}$
$\left[\because\right.$ If $\mathrm{I}(x)=\int_{\phi(x)}^{\psi(x)} f(t) d t$, then $\frac{d \mathrm{I}(x)}{d x}=f\{\psi(x)\}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{1-\mathrm{y}^2}$
$\frac{d^2 y}{d x^2}=\frac{1}{2 \sqrt{1+y^2}} \cdot 2 y \cdot \frac{d y}{d x}=\frac{y}{\sqrt{1+y^2}} \cdot \sqrt{1+y^2}=y$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2013 (09 Apr Online)
If $\left|\begin{array}{ccc}x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4\end{array}\right|=0$, then $x=$

Options:
A $1,9$
B $-1,9$
C $-1,-9$
D $1,-9$

The correct answer is:
$1,-9$

Solution:
By $C_1 \rightarrow C_1+C_2+C_3$,
we have $(9+x)$ $\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4\end{array}\right|=0$
$\Rightarrow (x+9)$ $\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4\end{array}\right|=0$
\(\begin{aligned} & \Rightarrow(x+9)\left|\begin{array}{ccc}0 & 1-x & 0 \\ 0 & x-1 & 1-x \\ 1 & 3 & x+4\end{array}\right|=0 \\ & \Rightarrow(x+9) \cdot(1-x)^2=0 \\ & \Rightarrow \text { So, } x=1,1,-9 .\end{aligned}\)

Topic: Mathematics, Matrices, JEE Main
If $x=-1$ and $x=2$ are extreme points of $\mathrm{f}(x)=\alpha \log x+\beta x^2+x, \alpha$ and $\beta$ are constants, then the value of $\alpha^2+2 \beta$ is

Options:
A $-3$
B $3$
C $\frac{3}{2}$
D $5$

The correct answer is:
$3$

Solution:
According to the given condition, $\mathrm{f}^{\prime}(1)=0$ and $\mathrm{f}^{\prime}(2)=0$
$$
\mathrm{f}(x)=\alpha \log x+\beta x^2+x
$$
$$
\begin{array}{ll}
\therefore & \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \\
\therefore & \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 \\
& \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2
\end{array}
$$
$\therefore \quad$ From (i) and (ii), we get
$$
\begin{aligned}
& \beta=\frac{-1}{2} \text { and } \alpha=2 \\
\therefore \quad & \alpha^2+2 \beta=4-1=3
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
If x=-1 and x=2 are extreme points of fx=αlogx+βx2+x, then

Options:
A Î± = 2 , Î² = - 1 2
B α = 2 , β = 1 2
C α = - 6 , β = 1 2
D α = - 6 , β =- 1 2

The correct answer is:
Î± = 2 , Î² = - 1 2

Solution:
fx=αlogx+βx2+x
If x < 0
f⁡x=αlog-x+βx2+x
∴ f⁡'x=-α-x+2βx+1
If x > 0
f⁡x=αlogx+βx2+x
∴ f⁡'x=αx+2βx+1
f⁡ ′ - 1 = - α - 2 β + 1 = 0
f⁡ ′ 2 = α 2 + 4 β + 1 = 0
2 f⁡ ′ - 1 = - 2 α - 4 β + 2 = 0
and f'2=α2+4β+1=0
Adding,
- 3 α 2 + 3 = 0
∴ α = 2
∴ β = - 1 2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (06 Apr)
If x=-1 and x=2 are points of extrema of f x = Î± log x + Î² x 2 + x , then

Options:
A Î± = 2 , Î² = - 1 2
B α = 2 , β = 1 2
C α = - 6 , β = 1 2
D α = - 6 , β =- 1 2

The correct answer is:
Î± = 2 , Î² = - 1 2

Solution:
f x = Î± log x + Î²x 2 + x
If x<0
fâ¡ x = Î± log - x + Î² x 2 + x
â´ fâ¡ â² x = - Î± - x + 2 Î² x + 1
If x>0
fâ¡ x = Î± log x + Î² x 2 + x
â´ fâ¡ â² x = Î± x + 2 Î² x + 1
fâ¡ â² - 1 = - Î± - 2 Î² + 1 = 0
fâ¡ â² 2 = Î± 2 + 4 Î² + 1 = 0
2 fâ¡ â² - 1 = - 2 Î± - 4 Î² + 2 = 0
and fâ¡ â² 2 = Î± 2 + 4 Î² + 1 = 0
Adding,
- 3 Î± 2 + 3 = 0
â´ Î± = 2
â´ Î² = - 1 2

Topic: Mathematics, Application of Derivatives, JEE Main
If x1 and x2 represent the mole fractions of a component A in the vapour phase and liquid mixture respectively and pAo and pBo represent vapour pressure of pure A and B, then total vapour pressure of liquid mixture is

Options:
A pAox1x2
B pAox2x1
C pBox1x2
D pBox2x1

The correct answer is:
pAox2x1

Solution:
According to Dalton's law of partial pressures of gases,
pApTotal=x1
pA=x1ptotal ...(i)
and according to Raoult's law
pA=pAox2 ...(ii)
Now by equating (i) and (ii), we get
pTotalx1=pAox2
pTotal=pAox2x1

Topic: Chemistry, Solutions, JEE Main
If x=1 is a critical point of the function f(x)=3x2+ax-2-aex, then

Options:
A x=1 and x=-23 are local minima of f
B x=1 andx=-23 is a local maxima of f
C x=1 is a local maxima andx=-22 is a local minima of f
D x=1 is a local minima andx=-23 are local maxima of f

The correct answer is:
x=1 is a local minima andx=-23 are local maxima of f

Solution:
f(x)=3x2+ax-2-aex
f'(x)=3x2+ax-2-aex+ex(6x+a)=ex3x2+(a+6)x-2

∵ x=1 is a critical point ∴ f'(1)=0

∴3+a+6-2=0
a=-7

∴f'(x)=ex3x2-x-2=ex3x2-3x+2x-2=ex(3x+2)(x-1)

∴ maxima at x=-2 3∴ minima at x=1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If $x=\frac{1-t}{1+t}, y=\frac{2 t}{1+t}$, then $\frac{d^{2} y}{d x^{2}}=$

Options:
A $\frac{2 t}{(1+t)^{2}}$
B $\frac{1}{(1+t)^{4}}$
C $\frac{2 t}{(1+t)^{2}}$
D 0

The correct answer is:
0

Solution:
We have, $x=\frac{1-t}{1+t}$
$$
\frac{d x}{d t}=\frac{-(1+t)-(1-t)}{(1+t)^{2}}=\frac{-2}{1+t^{2}} \text { and } y=\frac{2 t}{1+t}
$$
$$
\frac{d y}{d t}=\frac{(1+t) 2-2 t}{(1+t)^{2}}=\frac{2}{(1+t)^{2}}
$$
Now, $\quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=-1$ and $\quad \frac{d^{2} y}{d x^{2}}=0$

Topic: Mathematics, Differentiation, JEE Main
If $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$, then $\frac{d y}{d x}$ is equal to :

Options:
A $-\frac{y}{x}$
B $\frac{y}{x}$
C $-\frac{x}{y}$
D $\frac{x}{y}$

The correct answer is:
$-\frac{x}{y}$

Solution:
Let $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$

Put $t=\tan \theta$, we get

$x=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ and $y=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$\Rightarrow x=\cos 2 \theta$ and $y=\sin 2 \theta$

$\therefore \frac{\mathrm{dx}}{\mathrm{d} \theta}=-2 \sin 2 \theta$ and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=2 \cos 2 \theta$

Now, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{dx}}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$

Topic: Mathematics, Differentiation, JEE Main
If x=1-t21+t2 and y=2t1+t2,then dydx is equal to

Options:
A -yx
B yx
C -xy
D xy

The correct answer is:
-xy

Solution:
∵x=1-t21+t2 and y=2t1+t2
Put t=tan⁡θ in both the equations, we get
x=1-tan2⁡θ1+tan2⁡θ=cos⁡2θ ...(i)
and y=2tan⁡θ1+tan2⁡θ=sin⁡2θ ...(ii)
On differentiating both the eqs.(i) and (ii), we get
dxdθ=-2sin⁡2θ and dydθ=2cos⁡2θ
Therefore, dydx=dydθdxdθ=-cos⁡2θsin⁡2θ=-xy

Topic: Mathematics, Differentiation, JEE Main
If $|x| \lt 1$, then the sum of the series $1+2 x+3 x^2+4 x^3+\ldots \ldots \ldots . . \infty$ will be

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

The correct answer is:
$\frac{1}{(1-x)^2}$

Solution:
This is an A.G.P.
Let $S=1+2 x+3 x^2+\ldots \ldots \infty$
$\Rightarrow x . S=x+2 x^2+\ldots \ldots \infty$
Subtracting $(1-x) S=1+x+$ $x^2+\ldots \ldots \ldots \infty=\frac{1}{1-x}$
$\therefore S=\frac{1}{(1-x)^2}$
Aliter: use $S=\left[1+\frac{r}{1-r} \times\right.$ diff. of A.P. $] \frac{1}{1-r}$

Topic: Mathematics, Sequences and Series, JEE Main
If x+1x=1, λ=x4000+1x4000 and μ be the digit at the unit place of the number 22n+1, where n∈N and n>1, then the value of λ + μ is equal to

Options:

The correct answer is:
6

Solution:
∵ x+1x=1⇒x2-x+1=0
⇒ x + ω x + ω 2 = 0
∴ x = - ω ,⁡ - ω 2 , where ω and ω2 are complex cube roots of unity
if x=-ω, then λ = - ω 4 0 0 0 + 1 - ω 4 0 0 0 = ω 4 0 0 0 + 1 ω 4 0 0 0
⇒ λ = ω + 1 ω = ω 2 + 1 ω = - ω ω = - 1 ∵ ω3=1and 1+ω+1ω2=0
and also for x = - ω 2 , ⁡ λ = - 1
Now, for n>1, ⁡2n=4k, ⁡k∈N⁡
∴ 22n=24k=16k=a number with last digit 6
∴ μ = the digit at the unit place of the number 22n+1
i.e. μ=6+1=7
Hence, λ+μ=-1+7=6

Topic: Mathematics, Complex Number, JEE Main
If $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3$ and $\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3$ are both in G.P. with the same common ratio, then the points $\left(\mathrm{x}_1, \mathrm{y}_1\right),\left(\mathrm{x}_2, \mathrm{y}_2\right)$ and $\left(\mathrm{x}_3\right.$, $\mathrm{y}_3$ )

Options:
A
are vertices of a triangle

B
lie on a straight line

C
lie on an ellipse

D
lie on a circle

The correct answer is:
lie on a straight line

Solution:
Taking co-ordinates as $\left(\frac{\mathrm{x}}{\mathrm{r}}, \frac{\mathrm{y}}{\mathrm{r}}\right) ;(\mathrm{x}, \mathrm{y}) \&(\mathrm{xr}, \mathrm{yr})$. Above coordinates satisfy the relation $\mathrm{y}=\mathrm{mx}$ Therefore lies on the straight line.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2003
If x1, x2, x3, x4, x5.......xn are n observations such that âi=1nxi2=400 and âi=1nxi=100, then the possible value of n among the following is

Options:
A 18
B 20
C 24
D 27

The correct answer is:
27

Solution:
We know that,
σ2≥0
⇒Σxi2n-Σxin2≥0
⇒400n-10000n2≥0
⇒n≥25

Topic: Mathematics, Statistics, JEE Main
If x1, x2, x3...x34 are numbers such that xi=xi+1=150, ∀i∈1,2,3,4,....9 and xi+1-xi=-2, ∀i∈10,11,....33, then median of x1, x2, x3......x34 is

Options:
A 134
B 135
C 148
D 150

The correct answer is:
135

Solution:
Total no. of term =34
So mean of 17th and 18th term is median
x10+n=136,x18=134
Hence, median =135

Topic: Mathematics, Statistics, JEE Main
If $\left|\begin{array}{lll}x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c\end{array}\right|=0$, then $a, b, c$ are in

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

The correct answer is:
A.P.

Solution:
As given $\left|\begin{array}{lll}x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c\end{array}\right|=0$
$=\left|\begin{array}{ccc}-1 & -1 & x+3 \\ -1 & -1 & x+4 \\ a-b & b-c & x+c\end{array}\right|=0 \quad \begin{aligned} & C_1 \rightarrow C_1-C_2 \\ & C_2 \rightarrow C_2-C_3\end{aligned}$
$\Rightarrow\left|\begin{array}{ccc}0 & 0 & -1 \\ -1 & -1 & x+4 \\ a-b & b-c & x+c\end{array}\right|=0$, by $R_1 \rightarrow R_1-R_2$
$\Rightarrow(-1)(-b+c+a-b)=0$
$\Rightarrow \quad 2 b-a-c=0 \Rightarrow a+c=2 b$ i.e., $a, b, c$ are in A.P.
Trick : In such type of problem, put any suitable value of $x$ i.e. 0 , so that the determinant.
$\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 4 \\ a & b & c\end{array}\right|=0$
$\begin{aligned} & \Rightarrow 1(3 c-4 b)-2(2 c-4 a)+3(2 b-3 a)=0 \\ & \Rightarrow-c+2 b-a=0 \Rightarrow 2 b=a+c \text {. Hence the result. }\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
If x1, x2, ….., xn and 1h1, 1h2, ….., 1hn are two A.P.s such that x3=h2=8 & x8=h7=20, then x5·h10 is equal to

Options:
A 3200
B 1600
C 2650
D 2560

The correct answer is:
2560

Solution:
Let d1 be the common difference of the A.P. x1,x2,…, then x8-x3=5 d1=12⇒

⇒x5=x3+2 d1=8+2×125=12.8

Let d2 be the common difference of the other sequence then

5 d2=120-18=-340⇒d2

⇒1 h10=1 h7+3 d2=1200⇒h10=200

⇒x5·h10=12.8×200=2560

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (15 Apr)
If $x_1, x_2, \ldots ., x_n$ and $\frac{1}{h_1}, \frac{1}{h^2}, \ldots . . \frac{1}{h_n}$ are two A.P's such that $x_3=h_2=8$ and $x_8=h_7=20$, then $x_5 . h_{10}$ equals.

Options:
A
2560

B
2650

C
3200

D
1600

The correct answer is:
2560

Solution:
Suppose $d_1$ is the common difference of the A.P.
$x_1, x_2, \ldots . x_n$ then
$$
\begin{aligned}
&\because x_8-x_3=5 d_1=12 \Rightarrow d_1=\frac{12}{5}=2.4 \\
&\Rightarrow x_5=x_3+2 d_1=8+2 \times \frac{12}{5}=12.8
\end{aligned}
$$
Suppose $d_2$ is the common difference of the
A.P $\frac{1}{h_1}, \frac{1}{h_2}, \ldots . \cdot \frac{1}{h_n}$ then
$$
\begin{aligned}
&5 d_2=\frac{1}{20}-\frac{1}{8}=\frac{-3}{40} \Rightarrow d_2=\frac{-3}{200} \\
&\because \frac{1}{h_{10}}=\frac{1}{h_7}+3 d_2=\frac{1}{200} \Rightarrow h_{10}=200 \\
&\Rightarrow x_5 . h_{10}=12.8 \times 200=2560
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If ϕ(x)=1x∫π4x42sint-3ϕ'(t)dt,x>0 then ϕ'π4 is equal to

Options:
A 46+π
B 86+π
C 8π
D 46-π

The correct answer is:
86+π

Solution:
Given,

ϕ(x)=1x∫π/4x42sint-3ϕ'(t)dt

Differentiate both sides w.r.t. x we get,

ϕ'(x)=-12x3/2∫π/4x42sint-3ϕ'(t)dt+1x42sinx-3ϕ'(x)

Put x=π4

ϕ'(π4)=-12π43/2×0+4π42×12-3ϕ'(π4)

⇒π4ϕ'(π4)+3ϕ'(π4)=4

⇒ϕ'π43+π4=4

⇒ϕ'π4=86+π

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (31 Jan Shift 2)
If $\phi(x)=\int_{1 / x}^{\sqrt{x}} \sin \left(t^2\right) d t$, then $\phi^{\prime}(1)=$

Options:
A $\sin 1$
B $2 \sin 1$
C $\frac{3}{2} \sin 1$
D None of these

The correct answer is:
$\frac{3}{2} \sin 1$

Solution:
$\begin{aligned} \phi^{\prime}(x) & =\sin x \frac{d}{d x} \sqrt{x}-\sin \frac{1}{x^2} \frac{d}{d x}\left(\frac{1}{x}\right) \\ & =\sin x \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{x^2} \sin \frac{1}{x^2} \\ \Rightarrow \phi^{\prime}(1) & =\frac{1}{2} \sin 1+\sin 1=\frac{3}{2} \sin 1\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
If x+1xxxx+λxxxx+λ2=98103x+81, then λ, λ3 are the roots of the equation

Options:
A 4x2+24x-27=0
B 4x2-24x-27=0
C 4x2+24x+27=0
D 4x2-24x+27=0

The correct answer is:
4x2-24x+27=0

Solution:
Given that, x+1xxxx+λxxxx+λ2=98103x+81

Put x=0 as x∈R

⇒1000λ000λ2=981030+81

On expanding the determinant we get,

⇒λ3=98×81

⇒λ3=9323

⇒λ=92 and λ3=96

Now the required quadratic equation is x2-92+96x+92×96=0

⇒x2-6x+274=0

⇒4x2-24x+27=0

Hence, the required quadratic equation is 4x2-24x+27=0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If $\int \frac{x^2-x+1}{x^2+1} e^{\cot ^{-1} x} d x=A(x) e^{\cot ^{-1} x}+C$, then $A(x)$ is equal to :

Options:
A
$-x$

B
$x$

C
$\sqrt{1-x}$

D
$\sqrt{1+x}$

The correct answer is:
$x$

Solution:
Let $\mathrm{I}=\int \frac{x^2-x+1}{x^2+1} \cdot e^{\cot ^{-1} x} d x$
Put $x=\cot t \Rightarrow-\operatorname{cosec}^2 t d t=d x$
Now, $1+\cot ^2 t=\operatorname{cosec}^2 t$
$\therefore$
$\mathrm{I}=\int \frac{e^t\left(\cot ^2 t-\cot t+1\right)}{\left(1+\cot ^2 t\right)}\left(-\operatorname{cosec}^2 t\right) d t$
$$
\begin{aligned}
& =-\int e^t\left(\operatorname{cosec}{ }^2 t-\cot t\right) d t \\
& =\int e^t\left(\cot t-\operatorname{cosec}^2 t\right) d t \\
& =e^t \cot t+\mathrm{C} \\
& =e^{\cot ^{-1} x}(x)+\mathrm{C} \equiv \mathrm{A}(x) \cdot e^{\cot ^{-1} x}+\mathrm{C} \\
\Rightarrow \mathrm{A}(x) & =x
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2013 (22 Apr Online)
If $\int \frac{d x}{x^2+2 x+2}=f(x)+c$, then $f(x)$ is equal to :

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

The correct answer is:
$\tan ^{-1}(x+1)$

Solution:
Let $I=\int \frac{d x}{x^2+2 x+2}$
$=\int \frac{d x}{x^2+2 x+1+1}=\int \frac{d x}{1+(x+1)^2}$
$=\tan ^{-1}(x+1)+c$
But $I=f(x)+c$
$\therefore \quad f(x)=\tan ^{-1}(x+1)$

Topic: Mathematics, Indefinite Integration, JEE Main
If $x^2+2 x+2 x y+m y-3$ has two rational factors, then the value of $m$ will be

Options:
A $-6,-2$
B $-6,2$
C $6,-2$
D $6,2$

The correct answer is:
$6,-2$

Solution:
Given expression $\quad x^2+2 x+2 x y+m y-3$ can be written as $x^2+2 x(1+y)+(m y-3)$
But factors are rational, so $B^2-4 A C$ is a perfect square.
Now $\left.4(1+y)^2-(m y-3)\right\} \geq 0$
$\Rightarrow 4\left\{y^2+1+2 y-m y+3\right\} \geq 0$
$\Rightarrow y^2+2 y-m y+4 \geq 0$
Hence $2 y-m y= \pm 4 y$ {as it is perfect square }
Now taking (-) sign, we get $m=6$

Topic: Mathematics, Quadratic Equation, JEE Main
If $x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$, then $x^2(x-4)^2$ is equal to :

Options:
A 7
B 4
C 2
D 1

The correct answer is:
1

Solution:
$\because \quad x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}$
$=\frac{2+\sqrt{3}}{\sqrt{4-3}}=2+\sqrt{3}$
$\therefore x^2(x-4)^2=(2+\sqrt{3})^2(2+\sqrt{3}-4)^2$
$=(\sqrt{3}+2)^2(\sqrt{3}-2)^2$
$=\left[(\sqrt{3})^2-(2)^2\right]^2$
$=(3-4)^2=(-1)^2$
$=1$

Topic: Mathematics, Quadratic Equation, JEE Main
If x2-3x+2 is a factor of x4-ax2+b, then the equation whose roots are a and b is

Options:
A x2-9x-20=0
B x2-9x+20=0
C x2+9x-20=0
D x2+9x+20=0

The correct answer is:
x2-9x+20=0

Solution:
If x2-3x+2 is a factor of x4-ax2+b, then the equation whose roots are a and b is

The remainder is zero of equation x2-3x+2

x2-3x+2=0

x2-2x-x+2=0

xx-2-1x-2=0

x-2x-1=0

If x-1=0

x=1

Putting value of x in equation x4-ax2+b=0 remainder be zero

14-a12+b=0

-a+b=-1...............i

If, x-2=0

x=2

Putting value of x=2

24-a22+b=0

-4a+b=-16.......................ii

After solving there equations i & ii, the value of a=5, b=4

Here, α=5 & β=4

Now, the new quadratic equation

x2-α+β+α.β=0

x2-5+4x+5×4=0

Hence, the equation whose roots are a & b

x2-9x+20=0

Topic: Mathematics, Quadratic Equation, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 2)
If $\frac{x-4}{x^2-5 x+6}$ can be expanded in the ascending powers of $x$, then the coefficient of $x^3$ is

Options:
A $\frac{-73}{648}$
B $\frac{73}{648}$
C $\frac{71}{648}$
D $\frac{-71}{648}$

The correct answer is:
$\frac{-73}{648}$

Solution:
$\begin{aligned} & \frac{x-4}{x^2-5 x+6}=\frac{x-4}{(x-2)(x-3)} \\ & =\frac{2}{(x-2)}-\frac{1}{(x-3)}=2(x-2)^{-1}-(x-3)^{-1} \\ & =2(-2)^{-1}\left(1-\frac{x}{2}\right)^{-1}-(-3)^{-1}\left(1-\frac{x}{3}\right)^{-1} \\ & =-\left[1+\left(\frac{x}{2}\right)+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3+\ldots\right] \\ & \quad+\frac{1}{3}\left[1+\left(\frac{x}{3}\right)+\left(\frac{x}{3}\right)^2+\left(\frac{x}{3}\right)^3+\ldots\right] \\ & \therefore \text { Coefficient of } x^3 \text { in } \frac{x-4}{x^2-5 x+6} .\end{aligned}$
$\begin{aligned} & =-\left(\frac{1}{2}\right)^3+\frac{1}{3}\left(\frac{1}{3}\right)^3 \\ & =-\frac{1}{8}+\frac{1}{81}=\frac{-81+8}{648} \\ & =-\frac{73}{648}\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
If x2+9y2-4x+3=0, x,y∈R, then x and y respectively lie in the intervals

Options:
A -13,13 and-13,13
B 1, 3 and -13,13
C -13,13 and 1, 3
D 1, 3 and 1, 3

The correct answer is:
1, 3 and -13,13

Solution:
The Given equation is x2+9y2-4x+3=0

⇒9y2+0y+x2-4x+3=0

Make quadratic of y, we have D≥0 As it gives real values

⇒0-4×9×x2-4x+3≥0

⇒x2-3x-x+3≤0

⇒x-3x-1≤0

x∈1,3

Now making quadratic in x equation is x2-4x+3+9y2=0

D≥0

16-4×3+9y2≥0

⇒4-3-9y2≥0

⇒9y2≤1

⇒y∈-13,13

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If $x^2-10 a x-11 b=0$ have roots $c$ and $d$ and $x^2-10 c x-11 d=0$ have roots $a$ and $b$, then, find $a+b+c+d$.

Options:

The correct answer is:
1210

Solution:
Here, $a+b=10 c$
$$
c+d=10 a
$$
On subtracting Eq. (ii) from Eq. (i), we get, $\quad(a-c)+(b-d)=10(c-a)$ $\Rightarrow \quad(b-d)=11(c-a)$
Since, $c$ is the root of $x^2-10 a x+11 b=0$
$$
\Rightarrow \quad c^2-10 a c-11 b=0
$$
Similarly, $a$ is the root of
$$
\begin{aligned}
\Rightarrow & x^2-10 c x+11 d=0 \\
& a^2-10 c a-11 d=0
\end{aligned}
$$
On subtracting Eq. (v) from Eq. (iv), we get
$$
\begin{aligned}
& \left(c^2-a^2\right)=11(b-d) \\
& \therefore \quad(c+a)(c-a)=11 \times 11(c-a) \quad\{\text { from (i) \& (ii) } \\
& \Rightarrow \quad c+a=121 . \\
& \therefore \quad a+b+c+d=10 c+10 a \\
& =10(c+a)=1210 \\
&
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
If $x^2+a x+10=0$ and $x^2+b x-10=0$ have a common root, then $a^2-b^2$ is equal to

Options:
A $10$
B $20$
C $30$
D $40$

The correct answer is:
$40$

Solution:
Let $\alpha$ be a common root, then
$\alpha^2+a \alpha+10=0$ ...(i)
and $\alpha^2+b \alpha-10=0$ ...(ii)
form (i) - (ii),
$(a-b) \alpha+20=0 \Rightarrow \alpha=-\frac{20}{a-b}$
Substituting the value of $\alpha$ in (i), we get
$\left(-\frac{20}{a-b}\right)^2+a\left(-\frac{20}{a-b}\right)+10=0$
$\Rightarrow 400-20 a(a-b)+10(a-b)^2=0$
$\Rightarrow 40-2 a^2+2 a b+a^2+b^2-2 a b=0$
$\Rightarrow a^2-b^2=40$.

Topic: Mathematics, Quadratic Equation, JEE Main
If x2ay3b=e5m, x3cy4d=e2n, Î1=5m3b2n4d, Î2=2a5m3c2n and Î3=2a3b3c4d, then the values of x and y are

Options:
A Δ1Δ3,Δ2Δ3
B Δ2Δ1,Δ3Δ1
C logΔ1Δ3,logΔ2Δ3
D eΔ1Δ3,eΔ2Δ3

The correct answer is:
eΔ1Δ3,eΔ2Δ3

Solution:
Taking log in both the equations, we get,
2alog⁡x+3blog⁡y=5m
3clog⁡x+4dlog⁡y=2n
By Cramer’s Rule,
log⁡x=Δ1Δ3⇒x=eΔ1Δ3
log⁡y=Δ2Δ3⇒y=eΔ2Δ3

Topic: Mathematics, Determinants, JEE Main
If x=2cosec-1t and y=2sec-1t, t≥1, then dydx is equal to

Options:
A yx
B -yx
C -xy
D xy

The correct answer is:
-yx

Solution:
Given, x=2cosec-1t=2cosec-1t2 and y=2sec-1t=2sec-1t2, then we have

x·y=2sec−1t+cosec−1t2

And, we know that sec-1x+cosec-1x=π2,

⇒xy=2π4

On differentiating, both sides with respect to x, and applying product rule, we get

xdydx+y=0

⇒dydx=-yx.

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2018 (16 Apr Online)
If $(x-2)$ is a common factor of the expressions $x^2+a x+b$ and $x^2+c x+d$, then $\frac{b-d}{c-a}$ is equal to

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

The correct answer is:
$2$

Solution:
Since, $(x-2)$ is a common factor of the expressions $x^2+a x+b$ and $x^2+c x+d$
$\begin{array}{rlrl}\Rightarrow & 2 a+b =2 c+d \\ \Rightarrow & b-d =2(c-a) \\ \Rightarrow & \frac{b-d}{c-a} =2\end{array}$

Topic: Mathematics, Quadratic Equation, JEE Main
If x2+mx+1x2+x+1<3 for all real x, then

Options:
A m<-1
B -1<m<6
C -1<m<5
D m>6

The correct answer is:
-1<m<5

Solution:
We have x2+x+1=x+122+34>0
So, -3<x2+mx+1x2+x+1<3
⇒-3x2+x+1<x2+mx+1<3x2+x+1
⇒4x2+m+3x+4>0 and 2x2+3-mx+2>0 for real x
⇒m+32-4×4×4<0 and 3-m2-4×2×2<0
⇒m+3+8m+3-8<0 and m-3+4m-3-4<0
⇒m+11m-5<0 and m+1m-7<0
⇒-11<m<5 and -1<m<7
Hence, -1<m<5

Topic: Mathematics, Functions, JEE Main