datavorous/entrance-exam-dataset huggingface dataset:
The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe $100 \mathrm{Hz}$. The frequency of the third harmon of the closed pipe is

Options:
A $100 \mathrm{Hz}$
B $200 \mathrm{Hz}$
C $300 \mathrm{Hz}$
D $150 \mathrm{Hz}$

The correct answer is:
$300 \mathrm{Hz}$

Solution:
Let the length of closed organ pipe 1 be. Then,
$$
v_{n}=(2 n-1) \frac{v}{4 l}
$$
Its third harmonics (put $n=2$ ) in above question.
$$
v_{3}=\frac{3}{4} \frac{v}{1}
$$
Now, the length of open organ pipe is 21 , then
$$
v_{n}=n \frac{v}{2(2 l)}=\frac{n v}{4 l}
$$
Its fundamental frequency,
$$
v_{0}=\frac{v}{2 l}=100
$$
$$
v_{3}=\frac{3 v}{4 l}=3 \times 100=300 \mathrm{Hz}
$$

Topic: Physics, Waves and Sound, JEE Main
The length of chord of contact of the tangents drawn from the point $(2,5)$ to the parabola $y^2=8 x$, is

Options:
A $\frac{1}{2} \sqrt{41}$
B $\sqrt{41}$
C $\frac{3}{2} \sqrt{41}$
D $2 \sqrt{41}$

The correct answer is:
$\frac{3}{2} \sqrt{41}$

Solution:
Equation of chord of contact of tangent drawn from a point $\left(x_1, y_1\right)$ to parabola $y^2=4 a x$ is $y y_1=2 a\left(x+x_1\right)$ so that $5 y=2 \times 2(x+2) \Rightarrow 5 y=4 x+8$. Point of intersection of chord of contact with parabola
$y^2=8 x$ are $\left(\frac{1}{2}, 2\right),(8,8)$, so that length $=\frac{3}{2} \sqrt{41}$

Topic: Mathematics, Parabola, JEE Main
The length of common chord of the circles $(x-a)^2+y^2=a^2$ and $x^2+(y-b)^2=b^2$ is

Options:
A $2 \sqrt{a^2+b^2}$
B $\frac{a b}{\sqrt{a^2+b^2}}$
C $\frac{2 a b}{\sqrt{a^2+b^2}}$
D None of these

The correct answer is:
$\frac{2 a b}{\sqrt{a^2+b^2}}$

Solution:
Equation of common chord is $a x-b y=0$.
Now length of common chord $=2 \sqrt{r_1^2-p_1^2}=2 \sqrt{r_2^2-p_2^2}$
where $r_1$ and $r_2$ are radii of given circles and $p_1, p_2$ are the perpendicular distances from centres of circles to common chords.
Hence required length $=2 \sqrt{a^2-\frac{a^4}{a^2+b^2}}=\frac{2 a b}{\sqrt{a^2+b^2}}$.

Topic: Mathematics, Circle, JEE Main
The length of conjugate axis of a hyperbola is greater than the length of transverse axis. Then, the eccentricity e is

Options:
A $=\sqrt{2}$
B $>\sqrt{2}$
C $ < \sqrt{2}$
D $ < \frac{1}{\sqrt{2}}$

The correct answer is:
$>\sqrt{2}$

Solution:
Let the length of conjugate axis $=b$
and the length of transvers axis $=a$ Given that, $b>a$
Now,$b^{2}>a^{2}$
$\therefore$ Eccentricity $(e)=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$e^{2}=\left(\sqrt{1+\frac{b^{2}}{a^{2}}}\right)^{2}>2$
$e>\sqrt{2}$

Topic: Mathematics, Hyperbola, JEE Main
The length of different internodes in a culm of sugarcane is variable because of

Options:
A Intercalary meristem
B Shoot apical meristem
C Position of axillary buds
D Size of leaf lamina at the node below each internode

The correct answer is:
Intercalary meristem

Solution:
Intercalary meristem : They are remains of apical meristem between permanent tissues and internodes; responsible for primary growth of the plant.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2008 (Mains)
The length of elastic string, obeying Hooke's law is $\ell_{1}$ metres when the tension $4 \mathrm{~N}$ and $\ell_{2}$ metres when the tension is $5 \mathrm{~N}$. The length in metres when the tension is $9 \mathrm{~N}$ is $-$

Options:
A $5 \ell_{1}-4 \ell_{2}$
B $5 \ell_{2}-4 \ell_{1}$
C $9 \ell_{1}-8 \ell_{2}$
D $9 \ell_{2}-8 \ell_{1}$

The correct answer is:
$5 \ell_{2}-4 \ell_{1}$

Solution:
Let $\ell_{0}$ be the unstretched length and $\ell_{3}$ be the length under a tension of $9 \mathrm{~N}$. Then
$$
\begin{aligned}
\mathrm{Y} &=\frac{4 \ell_{0}}{\mathrm{~A}\left(\ell_{1}-\ell_{0}\right)}=\frac{5 \ell_{0}}{\mathrm{~A}\left(\ell_{2}-\ell_{0}\right)} \\
&=\frac{9 \ell_{0}}{\mathrm{~A}\left(\ell_{3}-\ell_{0}\right)}
\end{aligned}
$$
These give
$$
\frac{4}{\ell_{1}-\ell_{0}}=\frac{5}{\ell_{2}-\ell_{0}} \Rightarrow \ell_{0}=5 \ell_{1}-4 \ell_{2}
$$
Further, $\frac{4}{\ell_{1}-\ell_{0}}=\frac{9}{\ell_{2}-\ell_{0}}$
Substituting the value of $\ell_{0}$ and solving, we get $\ell_{3}=5 \ell_{2}-4 \ell_{1}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The length of intercept cut by the line 4x+43y-1=0 on the curve y2=4x+34 is equal to

Options:
A 4 units
B 9 units
C 12 units
D 16 units

The correct answer is:
16 units

Solution:
The given line passes through the focus 14,0, so it is a focal chord for the given parabola.

Hence, its slope is tan⁡α=-13
∴ the length of the intercept is the same as the length of the focal chord

Hence, the length of the intercept=4acosec2α=4⋅1⋅22=16 units

Topic: Mathematics, Parabola, JEE Main
The length of latus -rectum of the parabola $x^{2}+2 y=8 x-7$ is

Options:
A 8
B 2
C 6
D 4

The correct answer is:
2

Solution:
We have,
$\begin{array}{l}
x^{2}+2 y=8 x-7 \\
x^{2}-8 x=-2 y-7 \\
x^{2}-8 x+16=-2 y-7+16 \\
(x-4)^{2}=-2 y+9 \\
(x-4)^{2}=-2(y-9 / 2)
\end{array}$
$\therefore$ Length of latus rectum $=2$

Topic: Mathematics, Parabola, JEE Main
The length of latusrectum of parabola $y^2+8 x-2 y+17=0$ is

Options:
A $2$
B $4$
C $8$
D $16$

The correct answer is:
$8$

Solution:
The given equation of the parabola is
$\Rightarrow \quad \begin{array}{ll} & y^3+8 x-2 y+17=0 \\ \Rightarrow \quad & (y-1)^2+8 x+16=0 \\ & (y-1)^2=-8(x+2)\end{array}$
Length of the latusrectum $=4 a=8$

Topic: Mathematics, Parabola, JEE Main
The length of metallic wire is $\ell_1$ when tension in it is $T_1$. It is $\ell_2$ when the tension is $T_2$. The original length of the wire will be -

Options:
A $\frac{\ell_1+\ell_2}{2}$
B $\frac{\mathrm{T}_2 \ell_1+\mathrm{T}_1 \ell_2}{\mathrm{~T}_1+\mathrm{T}_2}$
C $\frac{\mathrm{T}_2 \ell_1-\mathrm{T}_1 \ell_2}{\mathrm{~T}_2-\mathrm{T}_1}$
D $\frac{\mathrm{T}_1 \ell_1-\mathrm{T}_2 \ell_2}{\mathrm{~T}_2-\mathrm{T}_1}$

The correct answer is:
$\frac{\mathrm{T}_2 \ell_1-\mathrm{T}_1 \ell_2}{\mathrm{~T}_2-\mathrm{T}_1}$

Solution:
Assuming Hooke's law to be valid.
$\begin{aligned} & \mathrm{T} \propto(\Delta \ell) \\ & \mathrm{T}=\mathrm{k}(\Delta \ell)\end{aligned}$
Let, $\ell_0=$ natural length $($ original length)
$\begin{aligned} & \Rightarrow \mathrm{T}=\mathrm{k}\left(\ell-\ell_0\right) \\ & \text { so, } \mathrm{T}_1=\mathrm{k}\left(\ell_1-\ell_0\right) \& \mathrm{~T}_2=\mathrm{k}\left(\ell_2-\ell_0\right) \\ & \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\ell_1-\ell_0}{\ell_2-\ell_0} \Rightarrow \ell_0=\frac{\mathrm{T}_2 \ell_1-\mathrm{T}_1 \ell_2}{\mathrm{~T}_2-\mathrm{T}_1}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The length of metallic wire is l1 when tension in it is T1. It is l2 when the tension is T2. The original length of the wire will be :

Options:
A T2l1+T1l2 T1+T2
B l1+l22
C T2l1-T1l2 T2-T1
D T1l1-T2l2 T2-T1

The correct answer is:
T2l1-T1l2 T2-T1

Solution:
Assuming Hooke's law to be valid

T∝Δℓ

T=kΔℓ

Let,ℓ0= natural length (original length) ⇒T=kℓ-ℓ0

so, T1=kℓ1-ℓ0 & T2=kℓ2-ℓ0

⇒T1T2=ℓ1-ℓ0ℓ2-ℓ0

⇒ℓ0=T2ℓ1-T1ℓ2 T2-T1

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The length of perpendicular drawn from the point (2,3,4) to line 4-x2=y6=1-z3, is

Options:
A 37101
B 27101
C 27103
D 37103

The correct answer is:
37101

Solution:
Let P be the foot of the perpendicular drawn from A(2,3,4) to the given line l.

Then, x-4-2=y6=z-1-3

Now, any point on the line l is given by

x=4-2λ,y=6λ,z=1-3λ

The coordinates of P are (4-2λ,6λ,1-3λ)

The direction ratios of AP are

(4-2λ-2,6λ-3,1-3λ-4)

i.e. (2-2λ,6λ-3,-3-3λ)

And the direction ratios of l are-2,6 and -3.

Given, AP⊥l

∴ -2(2-2λ)+6(6λ-3)-3(-3-3λ)=0

⇒ λ=1349

∴ AP2=(4-2λ-2)2+(6λ-3)2+(1-3λ-4)2

=22-26λ+49λ2

Put λ=1349, we get

AP2=90949⇒AP=37101

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The length of tangent from the point $(5,1)$ to the circle $x^2+y^2+6 x-4 y-3=0$, is

Options:
A $81$
B $29$
C $7$
D $21$

The correct answer is:
$7$

Solution:
Length of tangent is given by $L_T=\sqrt{S_1}=\sqrt{49}=7$.

Topic: Mathematics, Circle, JEE Main
The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to:

Options:
A $\frac{\sqrt{1691}}{5}$
B $\frac{\sqrt{2009}}{5}$
C $\frac{\sqrt{1741}}{5}$
D $\frac{\sqrt{1541}}{5}$

The correct answer is:
$\frac{\sqrt{1691}}{5}$

Solution:
Correct option is (1) $\frac{\sqrt{1691}}{5}$
Equation of chord with given middle point.
$\mathrm{T}=\mathrm{S}_1$
$\frac{\mathrm{x}}{25}+\frac{\mathrm{y}}{40}=\frac{1}{25}+\frac{1}{100}$
$\frac{8 x+5 y}{200}=\frac{8+2}{200}$
$\mathrm{y}=\frac{10-8 \mathrm{x}}{5}$
$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1 \text { (put in original equation) }$
$\frac{16 \mathrm{x}^2+100+64 \mathrm{x}^2-160 \mathrm{x}}{400}=1$
$4 \mathrm{x}^2-8 \mathrm{x}-15=0$
$\mathbf{x}=\frac{8 \pm \sqrt{304}}{8}$
$\mathrm{x}_1=\frac{8+\sqrt{304}}{8} ; \mathrm{x}_2=\frac{8-\sqrt{304}}{8}$
$\text {similarly, } \mathbf{y}=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$
$\mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5}$
$\text {Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}$
$=\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}$
$=\frac{\sqrt{1691}}{5}$

Topic: Mathematics, Ellipse, JEE Main
The length of the chord of the ellipse x225+y216=1, whose mid point is 1,25, is equal to:

Options:
A 16915
B 20095
C 17415
D 15415

The correct answer is:
16915

Solution:
Equation of chord with given middle point is given by,

T=S1

⇒x25+y40=125+1100

⇒8x+5y200=8+2200

⇒y=10-8x5 ..i

⇒x225+(10-8x)2400=1

⇒16x2+100+64x2-160x400=1

⇒4x2-8x-15=0

⇒x=8±3048

⇒x1=8+3048; x2=8-3048

Similarly, y=10-8±3045=2±3045

⇒y1=2-3045;y2=2+3045

Distance =x1-x22+y1-y22

=4×30464+4×30425

=16915

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The length of the chord of the parabola x2=4y having equation x-2y+42=0 is

Options:
A 63 units
B 82 units
C 211 units
D 32 units

The correct answer is:
63 units

Solution:
Solving the equations of the given parabola and the line, we get,
2y-422=4y⇒y2-10y+16=0

⇒y1-y2=y1+y22-4y1y2=100-64=6
⇒ Length of the chord =AB=x1-x22+y1-y22=2y1-y22+y1-y22=3y1-y2
=63 units

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The length of the chord of the parabola x2=4y having equation x-2y+42=0 is

Options:
A 63 units
B 82 units
C 211 units
D 32 units

The correct answer is:
63 units

Solution:
Solving the equations of the given parabola and the line, we get,
2y-422=4y⇒y2-10y+16=0⇒y1-y2=y1+y22-4y1y2=100-64=6
⇒ Length of the chord =AB=x1-x22+y1-y22=2y1-y22+y1-y22=3y1-y2
=63

Topic: Mathematics, Parabola, JEE Main
The length of the chord of the parabola $y^{2}=4 a x(a>0)$ which passes through the vertex and makes an acute angle $\alpha$ with the axis of the parabola is

Options:
A $\pm 4 a \cot \alpha \operatorname{cosec} \alpha$
B $4 a \cot \alpha \operatorname{cosec} \alpha$
C $-4 a \cot \alpha \operatorname{cosec} \alpha$
D $4 a \operatorname{cosec}^{2} \alpha$

The correct answer is:
$4 a \cot \alpha \operatorname{cosec} \alpha$

Solution:
Hint:
Equation of OP:-
$y=x \tan \alpha$
Solving with $y^{2}=4 a x$, we get :
$x^{2} \tan ^{2} \alpha=4 a x \quad \Rightarrow x=4 a \cot ^{2} \alpha$
Substituting, $y=4 a \cot \alpha$
$\therefore \mathrm{P} \equiv\left(4 \mathrm{a} \cot ^{2} \alpha, 4 \mathrm{a} \cot \alpha\right)$
So, $\overline{\mathrm{OP}}=\sqrt{16 \mathrm{a}^{2} \cot ^{4} \alpha+16 \mathrm{a}^{2} \cot ^{2} \alpha}=4 \mathrm{a} \cot \alpha \operatorname{cosec} \alpha\left(\right.$ as $0^{\circ} < \alpha < 90^{\circ}$, so $\left.\cot \alpha>0, \operatorname{cosec} \alpha>0\right)$

Topic: Mathematics, Parabola, JEE Main
The length of the chord of the parabola $y^2=4 a x$ which passes through the vertex and makes an angle $\theta$ with the axis of the parabola, is

Options:
A $4 a \cos \theta \operatorname{cosec}^2 \theta$
B $4 a \cos ^2 \theta \operatorname{cose} \theta$
C $a \cos \theta \operatorname{cosec}^2 \theta$
D $\operatorname{acos}^2 \theta \operatorname{cose} \theta$

The correct answer is:
$4 a \cos \theta \operatorname{cosec}^2 \theta$

Solution:
$y=x \tan \theta$ will be equation of chord. The points of intersection of chord and parabola are
$(0,0),\left(\frac{4 a}{\tan ^2 \theta}, \frac{4 a}{\tan \theta}\right)$
$\begin{aligned}& \text {Hence length of chord }=4 a \sqrt{\left(\frac{1}{\tan ^2 \theta}\right)^2+\frac{1}{\tan ^2 \theta}} \\& =\frac{4 a}{\tan \theta} \sqrt{\frac{1+\tan ^2 \theta}{\tan ^2 \theta}}=4 a \operatorname{cosec}^2 \theta \cos \theta\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The length of the chord y=3x-23 intercepted by the parabola y2=4x-1 is equal to

Options:
A 43 units
B 83 units
C 163 units
D 43 units

The correct answer is:
163 units

Solution:
Focus of y2=4x-1 is 2,0 which satisfies the equation y=3x-23.
Hence, line y=3x-23 is a focal chord.
Now, the length of the focal chord equals to 4a cosec2θ where a=1 and tan⁡θ=3 or θ=60o
⇒ length of chord =4×cosec260o
=4×43=163 units

Topic: Mathematics, Parabola, JEE Main
The length of the common chord of the circles of radii 15 and 20, whose centres are 25 unit of distance apart, is

Options:
A 12
B 16
C 24
D 25

The correct answer is:
24

Solution:
Given, $r_1=15$ unit
$r_2=20 \text { unit }$
$C_1 C_2=25$ unit
Let $\angle A C_2 D=\theta$
Then, in right angled triangle $A D C_2$,
$A D=r_2 \sin \theta$
$\Rightarrow \quad \frac{A D}{r_2}=\sin \theta$ ...(i)
Now, in right angled triangle $A D C_1$
$A D=r_1 \sin (90-\theta)$
$\Rightarrow \quad \frac{A D}{r_1}=\cos \theta$ ...(ii)
On squaring and adding Eqs. (i) and (ii), we get
$A D^2\left[\frac{1}{r_1^2}+\frac{1}{r_2^2}\right]=1$
$\Rightarrow \quad A D^2\left(\frac{r_1^2+r_2^2}{r_1^2 r_2^2}\right)=1$
$\Rightarrow \quad A D^2=\frac{r_1^2 r_2^2}{r_1^2+r_2^2}$
$\Rightarrow \quad A D^2=\frac{225 \times 400}{225+400}$
$\Rightarrow \quad A D=\frac{15 \times 20}{25}=12$
Thus, length of common chord $=2 A D$
$=2 \times 12$
$=24$ unit

Topic: Mathematics, Circle, JEE Main
The length of the diameter of the circle which touches the $x$-axis at the point $(1,0)$ and passes through the point $(2,3)$ is

Options:
A
$\frac{10}{3}$

B
$\frac{3}{5}$

C
$\frac{6}{5}$

D
$\frac{5}{3}$

The correct answer is:
$\frac{10}{3}$

Solution:
Let $(h, k)$ be centre.
$\begin{array}{ll}
(h-1)^2+(k-0)^2=k^2 & \Rightarrow h=1 \\
(h-2)^2+(k-3)^2=k^2 & \Rightarrow k=\frac{5}{3}
\end{array}$
$\therefore$ diameter is $2 \mathrm{k}=\frac{10}{3}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2012 (Offline)
The length of the focal chord of the parabola y2=4x at a distance of 0.4 units from the origin is

Options:
A 22 units
B 23 units
C 24 units
D 25 units

The correct answer is:
25 units

Solution:
Let, PQ is a focal chord and S is the focus
Now, OS=1
⇒ In ΔOSR (R is the foot of the perpendicular from O on PQ)
⇒cosecθ=10.4=2.5
Length of PQ is equal to the length of latus rectum×cosec2θ
⇒PQ=42.52=4×6.25
=25

Topic: Mathematics, Parabola, JEE Main
The length of the intercept cut by the line 4x+43y-1=0 between the curve y2=x is equal to

Options:
A 4
B 9
C 12
D 16

The correct answer is:
4

Solution:
Given line is a focal chord for the parabola and its slope is tanâ¡Î±=-13
â´ Length of the focal chord =4acosec2Î±=4â14â22=4

Topic: Mathematics, Parabola, JEE Main
The length of the latus rectum of a parabola whose focal chord $P S Q$ is such that $P S=3$ and $Q S=2$ is

Options:
A $\frac{24}{5}$
B $\frac{12}{5}$
C $\frac{6}{5}$
D $\frac{12}{10}$

The correct answer is:
$\frac{24}{5}$

Solution:
We know that the length of the semi latus rectum is the harmonic mean of focal radii, so length of the latus rectum
$$
=2\left(\frac{2(P S)(Q S)}{P S+Q S}\right)=2 \times \frac{2 \times 3 \times 2}{3+2}=\frac{24}{5}
$$

Topic: Mathematics, Parabola, JEE Main
The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S(>R) respectively from the origin, is :

Options:
A 2( S-R)
B 2(S+R)
C 4(S-R)
D 4(S+R)

The correct answer is:
4(S-R)

Solution:
A parabola, whose vertex and focus are on the positive x- axis at a distance R and S(>R) respectively from the origin.

y2=4a(x-R)
y2=4(S-R)(x-R)
Length of Latus rectum=4(S-R)

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The length of the latus rectum of the curve represented by x=3cosâ¡t+sinâ¡t and y=4cosâ¡t-sinâ¡t, is

Options:
A 92 units
B 92 units
C 3223 units
D 92 units

The correct answer is:
92 units

Solution:
x3=cosâ¡t+sinâ¡t and y4=cosâ¡t-sinâ¡t
x32+y42=2
x2322+y422=1
Hence, the length of the latus rectum =232242=92

Topic: Mathematics, Ellipse, JEE Main
The length of the latus rectum of the ellipse $5 x^2+9 y^2=45$ is

Options:
A $\sqrt{5} / 4$
B $\sqrt{5} / 2$
C $5 / 3$
D $10 / 3$

The correct answer is:
$10 / 3$

Solution:
Here the ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$.
Latus rectum $=\frac{2 b^2}{a}=\frac{2 \cdot 5}{3}=\frac{10}{3}$.

Topic: Mathematics, Ellipse, JEE Main
The length of the longest interval, in which the function $3 \sin x-4 \sin ^3 x$ is increasing, is

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

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

Solution:
$3 \sin x-4 \sin ^3 x=\sin 3 x$
It is increasing, when $-\pi / 2 \leq 3 x \leq \pi / 2$ i.e., $-\pi / 6 \leq x \leq \pi / 6$
The length of interval $=\left|\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)\right|=\frac{\pi}{3}$

Topic: Mathematics, Application of Derivatives, JEE Main
The length of the longest interval in which the function fx=x3-3a2x+4 is decreasing is ∀a>0

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

The correct answer is:
2a

Solution:
f ′ x =3 x 2 −3 a 2
=3x-ax+a
∴fx decreases in x∈-a,a
i.e. length =2a

Topic: Mathematics, Application of Derivatives, JEE Main
The length of the longest interval in which the function y=sin⁡2x-2sin⁡x increases for x∈0,π is

Options:
A π2
B π3
C 2π3
D π6

The correct answer is:
π3

Solution:
dydx=2cosâ¡2x-2cosx
=22cos2x-cosâ¡x-1
=22cosâ¡x+1cosâ¡x-1
â´ y increases in 2Ï3,Ï

Topic: Mathematics, Application of Derivatives, JEE Main
The length of the minor axis of the ellipse 5x-102+5y+152=3x-4y+724 is

Options:
A 103 units
B 103 units
C 203 units
D 53 units

The correct answer is:
53 units

Solution:
x-22+y+32=12ââ3x-4y+752 is an ellipse, whose focus is 2,-3, directix 3x-4y+7=0 and eccentricity is 12
Length of â¥ from focus to directrix is 3Ã2-4Ã-3+75=5
ae-ae=5â2a-a2=5âa=103
So, length of major axis is 203

and length of minor axis, b=53

Topic: Mathematics, Ellipse, JEE Main
The length of the normal at point ' $t$ ' of the curve $x=a(t+\sin t), y=a(1-\cos t)$ is

Options:
A a sint
B $2 a \sin ^3(t / 2) \sec (t/ 2)$
C $2 a \sin (t / 2) \tan (t / 2)$
D $2 a \sin (t / 2)$

The correct answer is:
$2 a \sin (t / 2) \tan (t / 2)$

Solution:
$\begin{aligned} & x=a(t+\sin t), y=a(1-\cos t) \\ & \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a(\sin t)}{a(1+\cos t)}=\tan \frac{t}{2}\end{aligned}$
$\begin{aligned} & \qquad y \sqrt{1+\left(\frac{d y}{d x}\right)^2} \\ & \text { Length of the normal }=\sqrt{\tan ^2(t / 2)}=a(1-\cos t) \sec (t / 2) \\ & =a(1-\cos t) \sqrt{1+\tan ^2}=2 a \sin (t / 2) \tan (t / 2) .\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
The length of the normal chord to the parabola $y^2=4 x$, which subtends right angle at the vertex is

Options:
A $6 \sqrt{3}$
B $3 \sqrt{3}$
C $2$
D $1$

The correct answer is:
$6 \sqrt{3}$

Solution:
Normal at $P\left(t_1^2, 2 t_1\right)$ on the parabola $y^2=4 x \ldots(1)$
Meets it again at the point $Q\left(t_2^2, 2 t_2\right)$,
where $t_2=-t_1-\frac{2}{t_1} \ldots(2)$
If $P Q$ subtends a right angle at the vertex $(0,0)$ then $($ Slope of $O P)($ Slope of $O Q)=-1$
$\Rightarrow \frac{2 t_1}{t_1^2} \cdot \frac{2 t_2}{t_2^2}=-1 \Rightarrow t_2=-\frac{4}{t_1} \ldots(3)$
From (ii) and (iii)
$-t_1-\frac{2}{t_1}=-\frac{4}{t_1} \Rightarrow-t_1=-\frac{2}{t_1}$
$\Rightarrow t_1^2=2 \Rightarrow t_1= \pm \sqrt{2} ; \therefore t_2=\pm 2 \sqrt{2}$
$\therefore P$ and $Q$ are $(2 \pm 2 \sqrt{2})$ and $(8,84 \sqrt{2})$
$\therefore P Q=\sqrt{(8-2)^2+(\pm 4 \sqrt{2} \pm 2 \sqrt{2})^2}=\sqrt{36+72}$
$=\sqrt{108}=6 \sqrt{3 .}$

Topic: Mathematics, Parabola, JEE Main
The length of the normal chord which subtends an angle of 90o at the vertex of the parabola y2=4x is

Options:
A 63 units
B 72 units
C 82 units
D 92 units

The correct answer is:
63 units

Solution:
For the normal chord, t2=-t1-2t1 â¦ 1
Also, the chord subtends an angle of 90o at the vertex
â´t1t2=-4 â¦ 2

From 1, t1t2=-t12-2 or -4=t12-2

â´t12=2
â´t2=-4t1=-42=-22 or t22=8
Now, PQ2=a2t12-t222+4a2t1-t22
=1â2-82+42+222=36+72
PQ2=108 or PQ=63 units

Topic: Mathematics, Parabola, JEE Main
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines r→=i^+j^+λi^+2j^-k^ and r→=(i^+j^)+μ(-i^+j^-2k^) is

Options:
A 13
B 3
C 3
D 13

The correct answer is:
3

Solution:
n→=ijk12-1-11-2=-3i^+3j^+3k^
Equation of plane
-3x-1+3y-1+3z=0
⇒x-y-z=0
d(2,1,4)=|2-1-4|12+12+12=3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The length of the perpendicular from 1,0,2 on the line x+13=y-2-2=z+1-1 is

Options:
A 362 units
B 635 units
C 32 units
D 23 units

The correct answer is:
362 units

Solution:
Direction ratios of PR→3λ-2,-2λ+2,-λ-3
Direction Ratios of the line QR→3,-2,-1
Since, they are perpendicular
33λ-2-2-2λ+2-1-λ-3=0
9λ-6+4λ-4+λ+3=0
14λ=7
λ=12
R→12,1,-32
PR=1-122+0-12+2+322
=14+44+494
=544
=362 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The length of the perpendicular from the origin, on normal to the curve, x2+2xy-3y2=0, at the point 2, 2, is.

Options:
A 2
B 42
C 2
D 22

The correct answer is:
22

Solution:
x2+2xy-3y2=0

Differentiate the above curve,

2x+2xy'+y-6yy'=0

So, at 2, 2, y'=1.

So, equation of normal to the curve at 2, 2 is x+y=4.

Thus, perpendicular distance from origin on the normal is

0+0-412+12

=42=22.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The length of the perpendicular from the point 1,-2,5 on the line passing through 1,2,4 and parallel to the line x+y-z=0=x-2y+3z-5 is:

Options:
A 212
B 92
C 732
D 1

The correct answer is:
212

Solution:
Given line is parallel to the line x+y-z=0=x-2y+3z-5 is:

So, direction ratio of the line =i^j^k^11-11-23=i^-4j^-3k^

So, equation of line passing through 1,2,4 is

r→=i^+2j^+4k^+λi^-4j^-3k^

Let A1,2,4,Q1,-2,5 and P be 1+λ,2-4λ,4-3λ

Given PQ→ and line are perpendicular,

So, PQ→·i^-4j^-3k^=0

⇒λi^+4-4λj^-1+3λk^.i^-4j^-3k^=0

⇒λ-16+16λ+3+9λ=0

⇒λ=1326=12

Hence PQ→≡12,2,-52

Now PQ=212

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The length of the perpendicular from the point (2,-1, 4) on the straight line x+3 10=y-2-7=z1 is

Options:
A greater than 3 but less than 4
B greater than 4
C less than 2
D greater than 2 but less than 3

The correct answer is:
greater than 3 but less than 4

Solution:
Let P2,-1,4

Now, any point Q on line x+3 10=y-2-7=z1=λ is

Q=10λ-3, -7λ+2 ,λ

Direction Ratio of PQ

=10λ-3-2, -7λ+2+1, λ-4

=10λ-5, -7λ+3, λ-4

∵PQ is perpendicular to given line

∴1010λ-5-7-7λ+3+ λ-4=0

⇒150 λ=75 ⇒λ=12

∴Q≡2,-32,12

Distance PQ=14+494=52≈3.53 which lies in 3,4.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The length of the perpendicular from the point \( (2,-1,4) \) on the straight line \( \frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1} \) is

Options:
A greater than \( 3 \) but less than \( 4 \)
B greater than \( 4 \)
C less than \( 2 \)
D greater than \( 2 \) but less than \( 3 \)

The correct answer is:
greater than \( 3 \) but less than \( 4 \)

Solution:
Let P2,-1,4

Now, any point Q on line x+3 10=y-2-7=z1=λ is

Q=10λ-3, -7λ+2 ,λ

Direction Ratio of PQ

=10λ-3-2, -7λ+2+1, λ-4

=10λ-5, -7λ+3, λ-4

∵PQ is perpendicular to given line

∴1010λ-5-7-7λ+3+ λ-4=0

⇒150 λ=75 ⇒λ=12

∴Q≡2,-32,12

Distance PQ=14+494=52≈3.53 which lies in 3,4.

Topic: Mathematics, Three Dimensional Geometry, JEE Main