datavorous/entrance-exam-dataset huggingface dataset:
If area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is 20 square units, then the area of the parallelogram having $3 \bar{a}$ $+\overline{\mathrm{b}}$ and $2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}$ as two adjacent sides in square units is

Options:
A 105
B 120
C 75
D 140

The correct answer is:
140

Solution:
We have $|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=20$ and we have to find value of
$$
\begin{aligned}
& |(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}})| \\
& (3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}) \\
& =6(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{b}} \times \overline{\mathrm{b}}) \\
& =0+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0=7(\overline{\mathrm{a}} \times \overline{\mathrm{b}})
\end{aligned}
$$
Hence area of the required parallelogram $=7 \times 20=140$

Topic: Mathematics, Vector Algebra, JEE Main
If $\alpha, \beta$ arethe roots of $a x^{2}+2 b x+c=0$ and $\alpha+\delta, \beta+\delta$ are the roots of $\mathrm{Ax}^{2}+2 \mathrm{Bx}+\mathrm{C}=0$, then what is $\left(\mathrm{b}^{2}-\mathrm{ac}\right) / \mathrm{B}^{2}-\mathrm{AC}$
equal to?

Options:
A $(\mathrm{b} / \mathrm{B})^{2}$
B $(a / \mathrm{A})^{2}$
C $\left(a^{2} b^{2}\right)\left(A^{2} B^{2}\right)$
D $(\mathrm{ab}) /(\mathrm{AB})$

The correct answer is:
$(a / \mathrm{A})^{2}$

Solution:
Since, $\alpha$ and $\beta$ are the roots of $a x^{2}+2 b x+c=0$
so, $k+\beta=-\frac{2 b}{a}$ and $\alpha \beta=\frac{c}{a}$
$\delta$ and $\beta+\delta$ are theroots of $A x^{2}+2 B x+C=0$
so, sum of the roots $=\alpha+\beta+2 \delta=-\frac{2 \mathrm{~B}}{\mathrm{~A}}$ and product
ofthe roots $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow-\frac{2 b}{a}+2 \delta=-\frac{2 B}{A}$
$\Rightarrow \delta=\frac{b}{a}-\frac{B}{A}$.. (i)
and $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \alpha \beta+(\alpha+\beta) \delta+\delta^{2}=\frac{\mathrm{C}}{\mathrm{A}}$...(ii)
Putting value of $\delta$ from equation (i) in equation (ii),
$\frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}}{\mathrm{a}}\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)+\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}^{2}}{\mathrm{a}^{2}}+\frac{2 \mathrm{bB}}{\mathrm{aA}}+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}-\frac{2 \mathrm{bB}}{\mathrm{aA}}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{B}^{2}}{\mathrm{~A}^{2}}-\frac{\mathrm{C}}{\mathrm{A}}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{\mathrm{B}^{2}-\mathrm{AC}}{\mathrm{A}^{2}}=\frac{\mathrm{b}^{2}-\mathrm{ac}}{\mathrm{a}^{2}}$
$\Rightarrow \frac{b^{2}-a c}{B^{2}-A C}=\left(\frac{a}{A}\right)^{2}$

Topic: Mathematics, Quadratic Equation, JEE Main
If arg z-10+6iz-4+2i=Ï4 (where z is a complex number), then the perimeter of the locus of z is

Options:
A 13Ï4 units
B 313Ï4 units
C 313Ï units
D 3Ï226 units

The correct answer is:
3Ï226 units

Solution:
Let A=4,2, B=10,6 and C is the center of locus of z (which is a circle)
⇒CA=CB=radius and ∠ACB=2π4=π2
⇒ΔACB is right-angled isosceles triangle
⇒r2+r2=10-42+6-22=36+16=52
⇒r2=26⇒r=26
⇒ perimeter=342πr
=3π226 units

Topic: Mathematics, Complex Number, JEE Main
If \( a=\sum_{n=0}^{\infty} \frac{x^{3 n}}{(3 n) !}, b=\sum_{n=1}^{\infty} \frac{x^{3 n-2}}{(3 n-2) !} \) and \( c=\sum_{n=1}^{\infty} \frac{x^{3 n-1}}{(3 n-1) !} \) then the value of \( a^{3}+b^{3}+c^{3}-3 a b c= \)

Options:
A \( 1 \)
B \( 0 \)
C \( -1 \)
D \( -2 \)

The correct answer is:
\( 1 \)

Solution:
a+b+c=∑n=0∞x3n3n!+∑n=1∞x3n-23n-2!+∑n=1∞x3n-13n-1!
=1+x+x22!+x33!+…=ex
a+bω+cω2=1+ωx+ω2x22!+ω2x33!+…=eωx ( ω is an imaginary cube root of unity)
And a+bω2+cω=eω2x
Now a3+b3+c3-3abc=a+b+c a+bω+cω2
a+bω2+cω
=exeωxeω2x=ex1+ω+ω2=e0.x=e0=1

Topic: Mathematics, Sequences and Series, JEE Main
If at $298 \mathrm{~K}$ the bond energies of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}$ and $\mathrm{H}-\mathrm{H}$ bonds are respectively $414,347,615$ and $435 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the value of enthalpy change for the reaction $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_3(\mathrm{~g})$ at $298 \mathrm{~K}$ will be

Options:
A
$-250 \mathrm{~kJ}$

B
$+125 \mathrm{~kJ}$

C
$-125 \mathrm{~kJ}$

D
$+250 \mathrm{~kJ}$

The correct answer is:
$-125 \mathrm{~kJ}$

Solution:
$\mathrm{CH}_2=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3-\mathrm{CH}_3$
$\Delta H=1(\mathrm{C}=\mathrm{C})+4(\mathrm{C}-\mathrm{H})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-6(\mathrm{C}-\mathrm{H})=1(\mathrm{C}=\mathrm{C})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-2(\mathrm{C}-\mathrm{H})$
$=615+435-347-2 \times 414=1050-1175=-125 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2003
If at a distance of 40 cm at an axial position of a dipole, the "magnetic potential" (analogous to electric potential) is 2.4Ã10-5 J A m-1, then the magnetic moment of the dipole is

Options:
A 28.6 A m2
B 32.2 A m2
C 38.4 A m2
D None of these

The correct answer is:
38.4 A m2

Solution:
Here, r=40 cm=0.4 m
θ=0° (an axial line)
M=?

As V=μ04πM cosθr2
⇒2.4×10-5=10-7×M×10.42
⇒M=38.4 A m2

Topic: Physics, Magnetic Properties of Matter, JEE Main
If at same temperature and pressure, the densities for two diatomic gases are respectively $d_1$ and $d_2$, then the ratio of velocities of sound in these gases will be

Options:
A $\sqrt{\frac{d_2}{d_1}}$
B $\sqrt{\frac{d_1}{d_2}}$
C $d_1 d_2$
D $\sqrt{d_1 d_2}$

The correct answer is:
$\sqrt{\frac{d_2}{d_1}}$

Solution:
Speed of sound
$v=\sqrt{\frac{\gamma P}{d}} \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{d_2}{d_1}} \quad(\ P-$ constant)

Topic: Physics, Waves and Sound, JEE Main
If \( A(\theta)=\left[\begin{array}{lr}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right] \) and \( A B=I \), then \( \left(\sec ^{2} \theta\right) B \) is equal to

Options:
A \( A(\theta) \)
B \( A\left(\frac{\theta}{2}\right) \)
C \( A(-\theta) \)
D \( A\left(-\frac{\theta}{2}\right) \)

The correct answer is:
\( A(-\theta) \)

Solution:
AB=I⇒ B=A-1
=11+tan2⁡θ 1 -tan⁡θtan⁡θ 1
=1sec2⁡θ 1 -tan⁡θtan⁡θ 1
⇒ (sec2⁡θ) B=1 -tan⁡θtan⁡θ 1=A(-θ)

Topic: Mathematics, Matrices, JEE Main
If B0=-4-3-3 1 0 1 4 4 3, Bn=adjBn-1, ∀n∈N and I is an identity matrix of order 3, then B1+B3+B5+B7+B9 is equal to

Options:
A B0
B 5B0
C 25B0
D 5I

The correct answer is:
5B0

Solution:
B02=B0B0=-4-3-3101443-4-3-3101443

⇒B0B0=100010001=I=B0-1B0
Also, B0=1
B1=adjB0=B0B0-1=B0
B2=adjB1=B0
B3=B5=B7=B9=B0
Hence, B1+B3+B5+B7+B9=5B0

Topic: Mathematics, Matrices, JEE Main
If B=∫1ex+1dx=-f(x)+C, where, C is the constant of integration and ef0=2, then the value of ef-1 is

Options:
A 4
B e+1
C 2e
D 0

The correct answer is:
e+1

Solution:
The given integral is B=â«e-x1+e-xdx
Let, e-x+1=t
â-e-xdx=dt
â´B=â«-dtt=-lnâ¡t+C
âB=-lnâ¡e-x+1+C
â´fx=lnâ¡e-x+1
Thus, ef-1=elnâ¡e+1=e+1

Topic: Mathematics, Indefinite Integration, JEE Main
If $b^2, a^2, c^2$ are in A.P., then $a+b, b+c, c+a$ will be in

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

The correct answer is:
H.P.

Solution:
Given that $b^2, a^2, c^2$ are in A.P.
Therefore $a^2-b^2=c^2-a^2$
$\Rightarrow(a-b)(a+b)=(c-a)(c+a)$
$\Rightarrow \frac{a-b}{c+a}=\frac{c-a}{a+b} \Rightarrow$ $\frac{b-a+c-c}{(c+a)(b+c)}=\frac{a+b-b-c}{(b+c)(a+b)}$
$\Rightarrow \frac{1}{b+c}-\frac{1}{a+b}=\frac{1}{c+a}-\frac{1}{b+c}$
$\Rightarrow \frac{1}{a+b}, \frac{1}{b+c}, \frac{1}{c+a}$ are in A.P.
Hence $(a+b),(b+c),(c+a)$ are in H.P.

Topic: Mathematics, Sequences and Series, JEE Main
If B=52α1021α3-1 is the inverse of a 3×3 matix A, then the sum of all values of α for which detA+1=0, is:

Options:
A 2
B 1
C 0
D -1

The correct answer is:
1

Solution:
Since, A=-1
Hence, B=1|A|=-1
⇒B=5-5-2α-α-2α
=2α2-2α-25

1+A=0
α2-α-12=0
Sum =1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If $\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$, then $a, b, c$ are in

Options:
A A.P.
B G.P.
C H.P.
D In G.P. and H.P. both

The correct answer is:
H.P.

Solution:
Since the reciprocals of $a$ and ${ }^C$ occur on RHS, let us first assume that ${ }^{a, b, c}$ are in H.P.
So that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
$\Rightarrow \quad \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=d$, say
$\Rightarrow \frac{a-b}{a b}=d=\frac{b-c}{b c} \Rightarrow a-b=a b d$ and $b-c=b c d$
Now LHS $=-\frac{1}{a-b}+\frac{1}{b-c}=-\frac{1}{a b d}+\frac{1}{b c d}$
$=\frac{1}{b d}\left(\frac{1}{c}-\frac{1}{a}\right)=\frac{1}{b d}(2 d) \Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}={ }_{\text {RHS }}$
$\therefore a, b, c$ are in H.P. is verified.

Topic: Mathematics, Sequences and Series, JEE Main
If $b \gt a$, then the equation $(x-a)(x-b)=1$ has

Options:
A Both roots in $[a, b]$
B Both roots in $(-\infty, a)$
C Both roots in $(b,+\infty)$
D One root in $(-\infty, a)$ and the other in $(b,+\infty)$

The correct answer is:
One root in $(-\infty, a)$ and the other in $(b,+\infty)$

Solution:
The equation is $x^2-(a+b) x+a b-1=0$
$\therefore$ discriminant $=(a+b)^2-4(a b-1)=(b-a)^2+4 \gt 0$
$\therefore$ both roots are real. Let them be $\alpha, \beta$ where
$\alpha=\frac{(a+b)-\sqrt{(b-a)^2+4}}{2}, \beta=\frac{(a+b)+\sqrt{(b-a)^2+4}}{2}$
Clearly, $\alpha \lt \frac{(a+b)-\sqrt{(b-a)^2}}{2}=\frac{(a+b)-(b-a)}{2}=a$
$(\because b\gta)$
and $\beta \gt \frac{(a+b)+\sqrt{(b-a)^2}}{2}=\frac{a+b+b-a}{2}=b$
Hence, one root $\alpha$ is less than $a$ and the other root $\beta$ is greater than $b$.

Topic: Mathematics, Quadratic Equation, JEE Main
If $b$ and $c$ are the lengths of the segments of any focal chord of a parabola $y^2=4 a x$, then the length of the semi-latus rectum is :

Options:
A $\frac{b c}{b+c}$
B $\sqrt{b c}$
C $\frac{b+c}{2}$
D $\frac{2 b c}{b+c}$

The correct answer is:
$\frac{2 b c}{b+c}$

Solution:
Since the semi latus rectum of a parabola is the harmonic mean between the segments of any focal chord of the parabola. $\therefore l$ is the harmonic mean between $b$ and $c$. Hence,
$l=\frac{2 b c}{b+c}$

Topic: Mathematics, Parabola, JEE Main
If B,C are square matrices of same order such that C2=BC-CB and B2=-I, where I is an identity matrix, then the inverse of matrix C-B is

Options:
A C
B C+B
C C-B
D I

The correct answer is:
C+B

Solution:
Given, C2-BC+CB=0

⇒C2-BC+CB-B2+B2=0
⇒C-BC+B+B2=0

⇒C-BC+B=I (as B2=-I )
Hence, C-B-1=C+B

Topic: Mathematics, Matrices, JEE Main
If B is a 3×3 matrix such that B2=0, then det. I+B50-50B is equal to :

Options:
A 1
B 2
C 3
D 50

The correct answer is:
1

Solution:
B2=0⇒B4=B6=B8=....=B50=0 and B3=B2B=0B=0
⇒ I+B50=I+C150B (higher powers are zero)
∴ det.I+B50-50B
=det. I+50C1B-50B
=det.I
=1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2014 (09 Apr Online)
If $b$ is the first term of an infinite G. P whose sum is five, then $b$ lies in the interval.

Options:
A
$(-\infty,-10)$

B
$(10, \infty)$

C
$(0,10)$

D
$(-10,0)$

The correct answer is:
$(0,10)$

Solution:
First term $=b$ and common ratio $=\mathrm{r}$
For infinite series, Sum $=\frac{b}{1-r}=5$ $\Rightarrow b=5(1-\mathrm{r})$
So, interval of $b=(0,10)$ as, $-1 < \mathrm{r} < 1$ for infinite GP.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If b is the first term of an infinite geometric progression whose sum is five, then b lies in the interval

Options:
A 10,Â â
B -â,-10
C -10,Â 0
D 0, 10

The correct answer is:
0, 10

Solution:
If b is the first term and r is the common ratio of an infinite G.P. and sum is 5, then

5=b1-r

⇒1-r=b5

⇒r=1-b5

⇒r=5-b5

∴-1<r<1

∴-1<5-b5<1

⇒-5<5-b<5

⇒-10<-b<0

⇒0<b<10

⇒b∈0, 10

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (15 Apr)
If B is the magnetic field at the centre of a circular loop of area A and its magnetic dipole moment is found to be nBAmÎ¼0Ï, then what is the value of n+m?

Options:

The correct answer is:
3.5

Solution:
Let r be the radius of the circular loop.

∴ A=πr2

⇒r=Aπ

Magnetic field at the centre of the loop is

B=μ0I2r

⇒B

=μ0I2Aπ⇒I=2Bμ0Aπ

Now, magnetic moment of the loop is

M=IA

=2Bμ0Aπ·A

=2Bμ0A32π

⇒n=2 and m=1.5

∴ n+m=3.5

Topic: Physics, Magnetic Effects of Current, JEE Main
If B is the magnetic field at the centre of a circular loop of area A and its magnetic dipole moment is found to be nBAmμ0π, then what is the value of n+m?

Options:

The correct answer is:
3.5

Solution:
Let r be the radius of the circular loop.

∴ A=πr2

⇒r=Aπ

Magnetic field at the centre of the loop is

B=μ0I2r

⇒B

=μ0I2Aπ⇒I=2Bμ0Aπ

Now, magnetic moment of the loop is

M=IA

=2Bμ0Aπ·A

=2Bμ0A32π

⇒n=2 and m=1.5

∴ n+m=3.5

Topic: Physics, Magnetic Effects of Current, JEE Main
If bn=∫0π2cos2nxsinxdx,n∈ℕ, then

Options:
A b3-b2,b4-b3,b5-b4 are in an A.P. with common difference-2
B 1b3-b2,1b4-b3,1b5-b4 are in an A.P. with common difference 2
C b3-b2,b4-b3,b5-b4 are in a G.P.
D 1b3-b2,1b4-b3,1b5-b4 are in an A.P. with common difference -2

The correct answer is:
1b3-b2,1b4-b3,1b5-b4 are in an A.P. with common difference -2

Solution:
Given,

bn=∫0π21+cos2nxsinxdx

bn+1-bn=∫0π2cos2n+1x-cos2nxsinxdx

=∫0π2-sin2n+1xsinxsinxdx

=cos(2n+1)x2n+10π2=-12n+1

So,1b3-b2=-5

1b4-b3=-7

1b5-b4=-9

So,1b3-b2,1b4-b3,1b5-b4 are in A.P. with c.d=-2

Topic: Mathematics, Definite Integration, JEE Main
If bn=∫0π2cos2nxsinxdx,n∈ℕ, then

Options:
A b3-b2,b4-b3,b5-b4 are in an A.P. with common difference-2
B 1b3-b2,1b4-b3,1b5-b4 are in an A.P. with common difference 2
C b3-b2,b4-b3,b5-b4 are in a G.P.
D 1b3-b2,1b4-b3,1b5-b4 are in an A.P. with common difference -2

The correct answer is:
1b3-b2,1b4-b3,1b5-b4 are in an A.P. with common difference -2

Solution:
Given,

bn=∫0π21+cos2nxsinxdx

bn+1-bn=∫0π2cos2n+1x-cos2nxsinxdx

=∫0π2-sin2n+1xsinxsinxdx

=cos(2n+1)x2n+10π2=-12n+1

So,1b3-b2=-5

1b4-b3=-7

1b5-b4=-9

So,1b3-b2,1b4-b3,1b5-b4 are in A.P. with c.d=-2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (25 Jun Shift 2)
If $\alpha, \beta, \gamma$ be the angles which a line makes with the positive direction of co-ordinate axes, then $\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=$

Options:
A 2
B 1
C 3
D 0

The correct answer is:
2

Solution:
Since $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \Rightarrow \Sigma \sin ^2 \alpha=3-1=2$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If $\alpha, \beta, \gamma$ be the direction angles of a vector and $\cos \alpha=\frac{14}{15}, \quad \cos \beta=\frac{1}{3}$ then $\cos \gamma=$

Options:
A $\pm \frac{2}{15}$
B $\frac{1}{5}$
C $\pm \frac{1}{15}$
D None of these

The correct answer is:
$\pm \frac{2}{15}$

Solution:
$\begin{aligned}
\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma & =1 \\
& \Rightarrow \cos \gamma=\sqrt{1-\left(\frac{14}{15}\right)^2-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}-\left(\frac{196}{225}\right)}= \pm \frac{2}{15}
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If λ be the ratio of the roots of the quadratic equation in x, 3m2x2+mm-4x+2=0, then the least value of m for which λ+1λ=1, is :

Options:
A 2-3
B -2+2
C 4-23
D 4-32

The correct answer is:
4-32

Solution:
Let α and β are the roots of given equation 3m2x2+mm-4x+2=0.

∴α+β=-mm-43m2 & αβ=23m2 .........i

Given αβ=λ & λ+1λ=1

⇒αβ+βα=1

⇒α+β2=3αβ

Using equation i, we get

-mm-43m22=3×23m2

⇒m=4±32

Hence, the least value of m is 4-32.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (12 Jan Shift 1)
If $\alpha, \beta$ be the roots of the quadratic equation $a x^2+b x+c=0$ and $k$ be a real number, then the condition so that $\alpha \lt k \lt \beta$ is given by

Options:
A $a c\gt0$
B $a k^2+b k+c=0$
C $a c \lt 0$
D $a^2 k^2+a b k+a c \lt 0$

The correct answer is:
$a^2 k^2+a b k+a c \lt 0$

Solution:
Here $a x^2+b x+c=a(x-\alpha)(x-\beta)$
Since $\alpha, \beta$ be the roots of $a x^2+b x+c=0$.
Also $\alpha \lt k \lt \beta$, so $a(k-\alpha)(k-\beta) \lt 0$
Also $a^2 k^2+a b k+a c=a\left(a k^2+b k+c\right)$
$=a^2(k-\alpha)(k-\beta) \lt 0 \Rightarrow a^2 k^2+a b k+a c \lt 0$

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha, \beta$ be the roots of the quadratic equation $\mathrm{x}^2+\mathrm{x}+1=0$ then the equation whose roots are $\alpha^{19}, \beta^7$ is

Options:
A $\mathrm{x}^2-\mathrm{x}+1=0$
B $\mathrm{x}^2-\mathrm{x}-1=0$
C $\mathrm{x}^2+\mathrm{x}-1=0$
D $x^2+x+1=0$

The correct answer is:
$x^2+x+1=0$

Solution:
Hints : Roots are $\omega, \omega^2$
Let $\alpha=\omega, \beta=\omega^2$ $\alpha^{19}=\omega, \beta^7=\omega^2$
$\therefore$ Equation remains same i.e. $\mathrm{x}^2+\mathrm{x}+1=0$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2010
If Bond energies of C-C, C=C, C-H bonds are 83 Kcal, 140 Kcal & 99 Kcal resp. The calculate the heat of formation of benzene.

Heat of atomistion of C=170.9 KcalHeat of atomisation of H=52.1 Kcal

Options:
A -65 Kcal
B -70 Kcal
C -75 Kcal
D -80 Kcal

The correct answer is:
-75 Kcal

Solution:
Aim: 6C s+3H2gâC6H6g
ÎH = ?
For reactant
Heat of atomization of 6 moles of C=6Ã170.9
Hear of atomization of 6 moles of H=6Ã52.1
For product
Heat of formation of 6 moles of C-H bonds =-6 Ã99
Heat of formation of 3 moles of C-C bonds =-3Ã83
Heat of formation of 3 moles of C=C bonds =-3Ã140
On adding :
ÎH = 6Ã170.9 + 6Ã52.1+ 6Ã â99 +3Ã â83 + 3Ã â140 = â75Kcal .

Topic: Chemistry, Thermodynamics (C), JEE Main
If both mean and the standard deviation of 50 observations $x_1, x_2, \ldots ., x_{50}$ are equal to 16 , then mean of $\left(x_1-5\right)^2,\left(x_2-5\right)^2, \ldots \ldots .,\left(x_{50}-5\right)^2$ is

Options:
A 378
B 377
C 357
D 397

The correct answer is:
377

Solution:
$\vec{x}=16 \text { and S.D. }=16$
if we subtract 5 from each observations mean will become $16-5=11$ and S.D. remains unchanged i.e., 16
$\begin{aligned}
& \Rightarrow \sqrt{\frac{\sum\left(x_i-5\right)^2}{50}-11^2}=16 \\
& \Rightarrow \frac{\sum\left(x_i-5\right)^2}{50}=256+121=377
\end{aligned}$
hence, the required mean is 377

Topic: Mathematics, Statistics, JEE Main
If both mean and the standard deviation of 50 observations $x_1, x_2, \ldots ., x_{50}$ are equal to 16 , then mean of $\left(x_1-5\right)^2,\left(x_2-5\right)^2, \ldots \ldots .,\left(x_{50}-5\right)^2$ is

Options:
A 378
B 377
C 357
D 397

The correct answer is:
377

Solution:
$\vec{x}=16 \text { and S.D. }=16$
if we subtract 5 from each observations mean will become $16-5=11$ and S.D. remains unchanged i.e., 16
$\begin{aligned}
& \Rightarrow \sqrt{\frac{\sum\left(x_i-5\right)^2}{50}-11^2}=16 \\
& \Rightarrow \frac{\sum\left(x_i-5\right)^2}{50}=256+121=377
\end{aligned}$
hence, the required mean is 377

Topic: Mathematics, Statistics, MHT CET, MHT CET 2022 (08 Aug Shift 2)
If both roots of x2-2ax+a2+a-3=0 are less than 3, then

Options:
A a<2
B 2≤a≤3
C 3<a≤4
D a>4

The correct answer is:
a<2

Solution:
Both roots are less than 3
⇒D≥0,-B2A<3, f3>0
D=4a2-4a2+a-3=-4a-3≥0
⇒a≤3….i
-B2A=2a2=a<3….ii
f3=9-6a+a2+a-3>0
a2-5a+6>0
a-2a-3>0⇒a<2 or a>3…..iii
Taking the intersection of i,ii,iii, we get,
a<2

Topic: Mathematics, Quadratic Equation, JEE Main
If both the mean and the standard deviation of 50 observations x1, x2, …, x50 are equal to 16, then the mean of x1-42, x2-42, …,x50-42 is

Options:
A 525
B 480
C 400
D 380

The correct answer is:
400

Solution:
For observations x1, x2, ...........x50

We have, mean, x¯=∑xi 50=16 .....i

And, standard deviation, σ=∑xi250-x¯2=16

⇒∑xi250=162+x¯2

⇒∑xi250=162+162=512 .....ii

Now, the mean value of x1-42, x2-42, ..., x50-42 will be

=∑xi-4250=∑xi2-8 ∑xi+16×5050

=∑xi250-8∑xi50+16

Using i and ii,

=512-8×16+16=400.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If both the mean and the standard deviation of 50 observations x1, x2, â¦, x50 are equal to 16, then the mean of x1-42, x2-42, â¦,x50-42 is

Options:
A 525
B 480
C 400
D 380

The correct answer is:
400

Solution:
For observations x1, x2, ...........x50
Mean, x-=∑xi 50=16 . . . . . (i)
Variance, σ2=∑xi250-(x-)2=162
⇒∑xi250=162+(x-)2=162+162=512 . . . . . (ii)
So, the mean value of (x1-4)2, (x2-4)2, .......(x50-4)2 will be
=∑(xi-4)250=∑xi2-8∑xi+16×5050
=∑xi250-8∑xi50+16=512-8×16+16=400 (using (i) and (ii))

Topic: Mathematics, Statistics, JEE Main
If both the roots of $k\left(6 x^2+3\right)+r x+2 x^2-1=0$ and $6 k\left(2 x^2+1\right)+p x+4 x^2-2=0$ are common, then $2 r-p$ is equal to

Options:
A $-1$
B $0$
C $1$
D $2$

The correct answer is:
$0$

Solution:
Given equation can be written as
$(6 k+2) x^2+r x+3 k-1=0$ ...(i)
and $2(6 k+2) x^2+p x+2(3 k-1)=0$ ...(ii)
Condition for common roots is
$\frac{12 k+4}{6 k+2}=\frac{p}{r}=\frac{6 k-2}{3 k-1}=2$ or
$2 r-p=0$

Topic: Mathematics, Quadratic Equation, JEE Main
If both the roots of the equation 4x2-2x+m=0 lie in the interval -1,1, then

Options:
A -3<m<-2
B 0<m<2
C 2<m<52
D -2<m≤14

The correct answer is:
-2<m≤14

Solution:
Coefficient of x2 is positive
So, D≥0,f1>0,f-1>0
D≥0⇒4-4×4×m≥0⇒m≤14
f-1>0⇒4+2+m>0⇒m>-6
f1>0⇒4-2+m>0⇒m>-2
Taking intersection we get m∈-2,14

Topic: Mathematics, Quadratic Equation, JEE Main
If both the roots of the equation x2+a-1x+a=0 are positive, then the complete solution set of real values of a is

Options:
A 0,∞
B 0,1
C 0,3-22
D 3-22,1

The correct answer is:
0,3-22

Solution:
Both roots are positive ⇒D>0, S>0, P>0
Sum of the roots, S>0⇒-a-1>0⇒a-1<0⇒a<1
Product of roots, P>0⇒a>0
Discriminant, D>0⇒a-12-4a>0⇒a2+1-6a>0
a2-6a+1>0⇒a>3+22 or a<3-22
Taking intersection we get a∈0,3-22

Topic: Mathematics, Quadratic Equation, JEE Main
If both the roots of the equation x2-mx+1=0 are less than unity, then

Options:
A m≤-2
B m>2
C -1≤m,≤3
D 0≤m≤52

The correct answer is:
m≤-2

Solution:
fx=x2-mx+1 and α≤β are the roots of fx=0.
Now, α<β<1 implies that f1 and the coefficients of x2 have the same sign.

This gives f1>0
⇒1-m+1>0⇒m<2
Also, the discriminant is m2-4≥0.

So m≤-2 or m≥+2.
Taking intersection, we get,

m≤-2.
Also, note that if m=-2, the roots are -1,-1.

Topic: Mathematics, Quadratic Equation, JEE Main
If both the roots of the quadratic equation $x^2-2 k x+k^2+k-5=0$ are less than 5 , then $\mathrm{k}$ lies in the interval

Options:
A
$(5,6]$

B
$(6, \infty)$

C
$(-\infty, 4)$

D
$[4,5]$

The correct answer is:
$(-\infty, 4)$

Solution:
$$
\begin{aligned}
& \frac{-b}{2 a} < 5 \\
& f(5)>0 \\
& \Rightarrow k \in(-\infty, 4) .
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
If both the roots of the quadratic equation $x^2-2 k x+k^2+k-5=0$ are less than 5 , then $\mathrm{k}$ lies in the interval

Options:
A
$(5,6]$

B
$(6, \infty)$

C
$(-\infty, 4)$

D
$[4,5]$

The correct answer is:
$(-\infty, 4)$

Solution:
$$
\begin{aligned}
& \frac{-b}{2 a} < 5 \\
& f(5)>0 \\
& \Rightarrow k \in(-\infty, 4) .
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2005
If both the roots of the quadratic equation x2-mx+4=0 are real and distinct and they lie in the interval 1, 5, then m lies in the interval:
Note: In the actual JEE paper interval was 1, 5

Options:
A -5,-4
B 3, 4
C (5, 6)
D 4, 5

The correct answer is:
4, 5

Solution:
Given roots of x2-mx+4=0 are distinct and lies in 1, 5, Following conditions must be true
(i) D>0⇒m2-16>0

⇒m-4m+4>0

⇒m∈-∞, -4∪4, ∞
(ii) f1>0⇒m<5
(iii) f5>0⇒m<295
(iv) 1<-b2a<5⇒2<m<10
Taking intersection of all conditions, we get m∈4, 5

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (09 Jan Shift 2)