datavorous/entrance-exam-dataset huggingface dataset:
The value of \( \cos 65^{\circ} \cos 55^{\circ} \cos 5^{\circ}= \)

Options:
A \( \frac{\sqrt{3}-1}{4 \sqrt{2}} \)
B \( \frac{\sqrt{3}-1}{8 \sqrt{2}} \)
C \( \frac{\sqrt{3}+1}{4 \sqrt{2}} \)
D \( \frac{\sqrt{3}+1}{8 \sqrt{2}} \)

The correct answer is:
\( \frac{\sqrt{3}+1}{8 \sqrt{2}} \)

Solution:
Given that

cos65°cos55°cos5°

∵ 2cosAcosB=cosA+B+cosA-B

∴ cos60°+cos50°2cos65°

⇒ cos65°4+2cos50°cos65°4

Again using the same formula, we get

⇒ cos65°4+cos115°+cos15°4

Now using cosC+cosD=2cos C+D2cosC-D2, We get

⇒ 2cos90°cos25°+cos15°4

⇒ cos15°4=3+182; cos90°=0

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of $\cos A-\sin A$ when $A=\frac{5 \pi}{4}$, is

Options:
A $\sqrt{2}$
B $\frac{1}{\sqrt{2}}$
C 0
D 1

The correct answer is:
0

Solution:
$\begin{aligned}
& \cos A-\sin A= \cos \frac{5 \pi}{4}-\sin \frac{5 \pi}{4},\left(\text {As } A=\frac{5 \pi}{4}\right) \\
&=-\cos \frac{\pi}{4}+\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=0
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of \( \cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \ldots \cos \frac{32 \pi}{65} \) is

Options:
A \( \frac{1}{32} \)
B \( \frac{1}{64} \)
C \( -\frac{1}{32} \)
D \( -\frac{1}{64} \)

The correct answer is:
\( \frac{1}{64} \)

Solution:
cos⁡π65cos⁡2π65…cos⁡32π65=cos⁡π65.cos⁡2π65…cos⁡25π65
=sin⁡26π6526sin⁡π65=sin⁡65π6564sin⁡π65
=sin⁡π-π6564sin⁡π65=164

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of \( \cos \left(-89^{\circ}\right)+\cos \left(-87^{\circ}\right)+\cos (-85)^{\circ}+\cdots+\cos \left(85^{\circ}\right)+\cos \left(87^{\circ}\right)+\cos \left(89^{\circ}\right)= \)

Options:
A \( \operatorname{cosec} 1^{\circ} \)
B \( \sec 1^{\circ} \)
C \( 2 \sec 1^{\circ} \)
D \( 2 \operatorname{cosec} 1^{\circ} \)

The correct answer is:
\( \operatorname{cosec} 1^{\circ} \)

Solution:
Given: cos-89°+cos-87°+cos-85°+⋯+cos85°+cos87°+cos89°

=2cos1°+cos3°+cos5°+⋯+cos89° as cos(-x)=cosx

Using result of sum of ∑k=0n-1cosa+kd=sin nd 2sind2cos2a+(n − 1)⋅d2

=2sin45°sin1°cos1°+89°2

=2sin45°cos45°sin1°

=sin90°sin1°

=cosec1°

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of $\int \cos \left(\log _e(x)\right) \mathrm{d} x$ is equal to (where $C$ is a constant of integration.)

Options:
A $x[\cos (\log x)-\sin (\log x)]+C$
B $\frac{x}{2}[\sin (\log x)-\cos (\log x)]+C$
C $\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C$
D $x[\cos (\log x)+\sin (\log x)]+C$

The correct answer is:
$\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C$

Solution:
$\begin{aligned} & \int \cos \left(\log _{\mathrm{e}} x\right) \mathrm{d} x \operatorname{let}_{\log _{\mathrm{e}} x=t} \\ & \Rightarrow \mathrm{d} x=e^t \\ & \Rightarrow I=\int \cos t \cdot e^t \mathrm{~d} t=\cos t \cdot e^t+\sin t \cdot e^t-I \quad[\text { Integrating by parts] } \\ & \Rightarrow 2 I=e^t(\cos t+\sin t) \\ & \Rightarrow I=\frac{x}{2}\left\{\cos \left(\log _{\mathrm{e}} x\right)+\sin \left(\log _{\mathrm{e}} x\right)\right\}\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
The value of cos⁡tan-1⁡34+cos⁡tan-1⁡x is equal to

Options:
A 45+x1+x2
B 35+11+x2
C 45+11+x2
D 35+x1x2

The correct answer is:
45+11+x2

Solution:
cos⁡tan-1⁡34+cos⁡tan-1⁡x
=cos⁡cos-1⁡45+cos⁡cos-1⁡11+x2
=45+11+x2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of ∫cos⁡xsin2⁡xsin⁡x+cos⁡xdx is equal to

Options:
A log⁡1+tan⁡xtan⁡x-cot⁡x+C
B log⁡tan⁡x1+tan⁡x+C
C log⁡tan⁡x1+tan⁡x-tan⁡x+C
D log⁡tan⁡x1+tan⁡x+cot⁡x+C

The correct answer is:
log⁡1+tan⁡xtan⁡x-cot⁡x+C

Solution:
Let, I= ∫cos⁡xsin2⁡xsin⁡x+cos⁡xdx
On dividing numerator and denominator by cos3⁡x, and substituting tanx=t, sec2xdx=dt, we get,
I= ∫dtt21+t
= ∫-1t+1t2+11+tdt (using partial fraction)
⇒I=− ∫ 1 t dt + ∫ t −2 dt+ ∫ 1 1+t dt
=-log⁡t-1t+log⁡1+t+C
= -1tan⁡x+log⁡1+tan⁡xtan⁡x+C
=-cot⁡x+log⁡1+tan⁡xtan⁡x+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of cot⁡70o+4cos⁡70o is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
cot⁡70o+4cos⁡70o
=cos⁡70°+4cos⁡70osin⁡70osin⁡70o=sin⁡20°+2×2sin⁡20°cos⁡20°cos⁡20°
=sin⁡20+2sin⁡40ocos⁡20o
=(sin⁡20+sin⁡40o)+sin⁡40ocos⁡20o
=2×12cos⁡10o+cos⁡50ocos⁡20o
=cos⁡10o+cos⁡50ocos⁡20o=2cos⁡30o×cos⁡20ocos⁡20o
=2×32=3

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of \( \cot 70^{\circ}+4 \cos 70^{\circ} \) is

Options:
A \( \frac{1}{\sqrt{3}} \)
B \( \sqrt{3} \)
C \( 2 \sqrt{3} \)
D \( \frac{1}{2} \)

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

Solution:
cot70°+4cos70°

=cos70°+4sin70°·cos70°sin70°

Using 2sinAcosA=sin2A, we get

=cos70°+2sin140°sin70°
=cos90°-70°+2sin180°-40°sin70°

Using, cos90°-θ=sinθ, sin180°-θ=sinθ, we get

=sin20°+2sin40°sin70°

=sin20°+sin40°+sin40°sin70°

Now, using sinC+sinD=2sinC+D2cosC-D2, we get

=2sin30°cos10°+sin40°sin70°

=2×12×cos90°-10°+sin40°sin70°

Again, using cos90°-θ=sinθ, we get

=sin80°+sin40°sin70°

Again, using sinC+sinD=2sinC+D2cosC-D2, we get

=2sin60°cos20°sin70°

=2×32×cos90-70°sin70°

=3·sin70°sin70°=3.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)$ is

Options:
A
$\frac{6}{17}$

B
$\frac{3}{17}$

C
$\frac{4}{17}$

D
$\frac{5}{17}$

The correct answer is:
$\frac{6}{17}$

Solution:
$$
\begin{aligned}
& \text { Let } E=\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right) \\
& \Rightarrow E=\cot \left(\tan ^{-1} \frac{17}{6}\right)=\frac{6}{17} .
\end{aligned}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2008
The value of cot⁡∑n=119cot-1⁡1+∑p=1n2p is:

Options:
A 2119
B 1921
C 2322
D 2223

The correct answer is:
2119

Solution:
We have, cot⁡∑n=119cot-1⁡1+∑p=1n2p
=cot∑n=119tan-111+nn+1
=cot∑n=119tan-1n+1-n1+nn+1
=cot⁡∑n=119(tan-1n+1-tan-1n)
=cot⁡tan-120-tan-11
=1tan(tan-120-tan-11)
=120-11+(20).1=2119

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The value of current i1 flowing from A to C in the circuit diagram is :

Options:
A 5 A
B 2.25 A
C 0.75 A
D 1 A

The correct answer is:
1 A

Solution:
The Diagram shown in the question represents the balanced wheatstone bridge. It satisfy the condition of balanced wheat stone bridge.
Multiplication of dimensionally opposite resistance is same.
2×2=2×2⇒4=4
When the circuit is balanced wheat stone bridge then current through the branch BD is zero, so it can be removed.

Now, after removing BD wire voltage across AC is 8 V.

So, current will be,

i2=84+4=1 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The value of current in the 6 Ω resistance is:

Options:
A 4 A
B 8 A
C 10 A
D 6 A

The correct answer is:
10 A

Solution:
Applying KCL at point P,

V-06+V-905+V-14020=0
⇒10 V+12 V-1080+3 V-420=0
⇒V=60
∴ Current in 6 Ω=V-06=10 A
Hence, option 3.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The value of ∫ dx x( x n +1 ) is equal to

Options:
A 1 n log e ( x n x n +1 )+c
B 1 n log e ( x n +1 x n )+c
C log e ( x n x n +1 )+c
D None of these

The correct answer is:
1 n log e ( x n x n +1 )+c

Solution:
I= ∫ dx ( 1+ 1 x n )× x n+1
put 1+ 1 x n =t
− n x n+1 .dx=dt
I= ∫ − dt n t =− 1 n ln| t |+c
I=− 1 n ln( 1+ 1 x n )+c
I= 1 n ln( x n x n +1 )+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of det $A$, where $A=\left(\begin{array}{ccc}1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1\end{array}\right)$, lies

Options:
A in the closed interval [1,2]
B in the closed interval [0,1]
C in the open interval (0,1)
D in the open interval (1, 2)

The correct answer is:
in the closed interval [1,2]

Solution:
We have, $|A|=\left|\begin{array}{ccc}1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1\end{array}\right|$
$=1[1-(-\cos \theta)(\cos \theta)]$
$-\cos \theta[-\cos \theta+\cos \theta]+0\left(\cos ^{2} \theta+1\right)$
$=1+\cos ^{2} \theta$
Now, we know that $-1 \leq \cos \theta \leq 1$
$\Rightarrow \quad 0 \leq \cos ^{2} \theta \leq 1$
$\Rightarrow \quad 1 \leq 1+\cos ^{2} \theta \leq 2$
$\Rightarrow \quad 1 \leq|A| \leq 2$
$\therefore \quad|A| \in[1,2]$

Topic: Mathematics, Determinants, JEE Main
The value of $\lambda$ does the line $\mathrm{y}=\mathrm{x}+\lambda$ touches the ellipse $9 x^{2}+16 y^{2}=144$ is/are

Options:
A $\pm 2 \sqrt{2}$
B $2 \pm \sqrt{3}$
C $\pm 5$
D $5 \pm \sqrt{2}$

The correct answer is:
$\pm 5$

Solution:
$\because$ Equation of ellipse is $9 x^{2}+16 y^{2}=144$ or $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ then we get $a^{2}=16$ and $b^{2}=9$ and comparing the line $\mathrm{y}=\mathrm{x}+\lambda$ with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ $\therefore \mathrm{m}=1$ and $\mathrm{c}=\lambda$
If the line $y=x+\lambda$ touches the ellipse $9 x^{2}$ $+16 y^{2}=144$, then $c^{2}=a^{2} m^{2}+b^{2}$
$$
\begin{array}{l}
\Rightarrow \lambda^{2}=16 \times 1^{2}+9 \Rightarrow \lambda^{2}=25 \\
\therefore \lambda=\pm 5
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
The value of $\int \frac{e^x}{e^x+1} d x$ is

Options:
A $e^x+c$
B $\left(e^x+1\right)+c$
C $\log \left(e^x+1\right)+c$
D None of these

The correct answer is:
$\log \left(e^x+1\right)+c$

Solution:
\(\begin{aligned}
& \Rightarrow \int \frac{e^x}{e^x+1} \cdot d x \\
& \text {Let, } e^x+1=t \\
& \text {So, } d x \cdot e^x=d t \\
& \Rightarrow \int \frac{d t}{t} \Rightarrow \log (t)+C \\
& \Rightarrow \log \left(e^x+1\right)+c
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, JEE Main
The value of ∫exx1+e2xdx is equal to (where, C is the constant of integration)

Options:
A tan-1⁡2ex+C
B lnâ¡1+ex1-ex+C
C 2tan-1â¡ex+C
D tan-1â¡xex+C

The correct answer is:
2tan-1⁡ex+C

Solution:
Let, I=â«exx1+e2xdx
Substituting ex=t
ex2xdx=dt
I=2â«dt1+t2
=2tan-1â¡t+C
I=2tan-1â¡ex+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of equilibrium constant of the reaction
$$
\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{I}_2 \text { is } 8.0
$$
The equilibrium constant of the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ will be

Options:
A $\frac{1}{8}$
B $\frac{1}{16}$
C $\frac{1}{64}$
D 16

The correct answer is:
$\frac{1}{64}$

Solution:
$$
\mathrm{HI} \rightleftharpoons \frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{I}_2, \mathrm{~K}_1=8.0
$$
or $2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2, \mathrm{~K}_2=64$
$$
\text { or } \mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}, \mathrm{K}_3=\frac{1}{64}
$$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2008 (Mains)
The value of f0 so that the function f x = 1 - cos 1 - cos x x 4 is continuous everywhere is k, then value of 10âk is

Options:

The correct answer is:
1.25

Solution:
If fx is continuous at x=0
fx= lim x â 0 f x
= lim x â 0 1 - cos 1 - cos x x 4 Ã 1 + cos 1 - cos x 1 + cos 1 - cos x
= lim x â 0 sin 2 1 - cos x x 4 Â· 1 + cos 1 - cos x Â· 1 - cos x 2 1 - cos x 2
= lim x â 0 sin 1 - cos x 1 - cos x 2 Ã lim x â 0 1 - cos x x 2 2 Ã lim x â 0 1 1 + cos 1 - cos x
= 1 2 Ã 1 4 Ã 1 2 = 1 8
â´âk= 1 8 â10âk= 5 4 =1.25

Topic: Mathematics, Continuity and Differentiability, JEE Main
The value of f(0), so that the function fx=2x-sin-1x2x+tan-1x is continuous at each point in its domain, is

Options:
A 13
B -13
C 23
D -23

The correct answer is:
13

Solution:
Since, x,sin-1x,tan-1x are continuous functions, so the
function f is clearly continuous at each point of its domain except possibly at x=0.

So, for f to be continuous at x=0

f0=limx→0fx=limx→02-sin-1xx2+tan-1xx=13

∵limx→0sin-1xx=1,limx→0tan-1xx=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
The value of f0, such that fx=1x21-cossinx can be made continuous at x=0, is

Options:
A 12
B 2
C 14
D 4

The correct answer is:
12

Solution:
limxâ0â¡1-cossinxx2=limxâ0â¡2sin2â¡sinx2x2 limxâ02.sinsinx2sinx22.sinx2x2=12â¡
Thus, for continuity at x=0,

f(0)=12

Topic: Mathematics, Continuity and Differentiability, JEE Main
The value of f0 such that the function fx=1+2x3-1+x4x is continuous at x=0, is

Options:
A 112
B 512
C 0
D 912

The correct answer is:
512

Solution:
For fx to be continuous at x=0, we must have
f0=limx→01+2x3-1+x4x
=limx→01+23x+13-232!4x2+.....x-1+14x+14-342!x2+.....x
=limx→0x512+-49+332x+terms containing x2 and higher powersx
=512

∴ f0=512

Topic: Mathematics, Continuity and Differentiability, JEE Main
The value of α for which 4α∫-12e-αxdx=5 , is

Options:
A loge2
B loge32
C loge2
D loge43

The correct answer is:
loge2

Solution:
4α∫-10eαxdx+∫02e-αxdx=5

⇒4αeαxα-10+e-αx-α​02=5

⇒4α1-e-αα-e-2α-1α=5

⇒42-e-α-e-2α=5

Put e-α=t

⇒4t2+4t-3=0⇒2t+32t-1=0

⇒e-α=12⇒α=ln2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The value of $\lambda$ for which the curve $(7 x+5)^{2}+(7 y+3)^{2}=\lambda^{2}(4 x+3 y-24)^{2}$ representa a parabola is

Options:
A $\pm \frac{6}{5}$
B $\pm \frac{7}{5}$
C $\pm \frac{1}{5}$
D $\pm \frac{2}{5}$

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

Solution:
Given curve is
$$
\begin{array}{l}
(7 x+5)^{2}+(7 y+3)^{2}=\lambda^{2}(4 x+3 y-24)^{2} \\
\Rightarrow 49 x^{2}+25+70 x+49 y^{2}+9+42 y \\
=\lambda^{2}\left(16 x^{2}+9 y^{2}+576+24 x y-144 y-192 x\right) \\
\Rightarrow \left(49-16 \lambda^{2}\right) x^{2}+\left(49-9 \lambda^{2}\right) y^{2}+\left(70+192 \lambda^{2}\right) x \\
+\left(42+144 \lambda^{2}\right) y-24 \lambda^{2} x y+\left(25-576 \lambda^{2}\right)=0
\end{array}
$$
On comparing with $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0,$ we get
$a=49-16 \lambda^{2}, b=\left(49-9 \lambda^{2}\right), h=-12 \lambda^{2}$
Condition for parabola is ab $=h^{2}$.
$$
\begin{array}{l}
\therefore \quad\left(49-16 \lambda^{2}\right)\left(49-9 \lambda^{2}\right)=\left(-12 \lambda^{2}\right)^{2} \\
\Rightarrow \quad 2401-441 \lambda^{2}-784 \lambda^{2}+144 \lambda^{4}=144 \lambda^{4} \\
\Rightarrow \quad 2401-1225 \lambda^{2}=0 \\
\Rightarrow \quad \lambda^{2}=\frac{2401}{1225}
\end{array}
$$
$\Rightarrow \quad \lambda^{2}=\frac{(49)^{2}}{(35)^{2}}$
$\Rightarrow \quad \lambda=\pm \frac{49}{35}$
$\Rightarrow \quad \lambda=\pm \frac{7}{5}$

Topic: Mathematics, Parabola, JEE Main
The value of $\lambda$, for which the line $2 x-\frac{8}{3} \lambda y=-3$ is a normal to the conic $x^2+\frac{y^2}{4}=1$ is

Options:
A $\frac{\sqrt{3}}{2}$
B $\frac{1}{2}$
C $-\frac{\sqrt{3}}{2}$
D $\frac{3}{8}$

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

Solution:
Equation of normal to ellipse $x^2+\frac{y^2}{4}=1$ is $\frac{x}{\cos \theta}-\frac{2 y}{\sin \theta}=1-4$ (i)
Comparing with $2 x-\frac{8}{3} \lambda y=-3$
$\cos \theta=\frac{1}{2}, \lambda=\frac{3}{4 \sin \theta}=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Ellipse, JEE Main
The value of $\lambda$ for which the lines $3 x+4 y=5, \quad 5 x+4 y=4$ and $\lambda x+4 y=6$ meet at a point is

Options:
A 2
B 1
C 4
D 3

The correct answer is:
1

Solution:
Given lines are $3 x+4 y=5, \quad 5 x+4 y=4, \quad \lambda x+4 y=6$. These lines meet at a point if the point of intersection of first two lines lies on the third line.
From $3 x+4 y=5$ and $5 x+4 y=4$
We get $x=\frac{-1}{2}, y=\frac{13}{8}$
This lies on $\lambda x+4 y=6$, if
$\begin{aligned}
& \text { This lies on } \lambda x+4 y=6, \text { if } \lambda\left(-\frac{1}{2}\right)+4\left(\frac{13}{8}\right)=6 \\
& \Rightarrow \lambda=1 .
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
The value of λ for which the set x,y:x2+y2-6x+4y≤12∩x,y:4x+3y≤λ contains only one point is

Options:
A 31
B –31
C 19
D –19

The correct answer is:
–19

Solution:
Given line L:4x+3y=λ
Circle S:x2+y2-6x+4y=12

Centre 3,–2, Radius =5
Now, the line will be tangent to the circle
∴ p=r
43+3-2-λ5=5⇒λ=31,–19
So, for only one point of intersection inequality, 4x+3y≤λ must not satisfy the centre of the circle
∴λ=-19

Topic: Mathematics, Circle, JEE Main
The value of $\lambda$ for which the straight line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ may lie on the plane $x-2 y=0,$ is

Options:
A 2
B 0
C $-\frac{1}{2}$
D there is no such $\lambda$

The correct answer is:
there is no such $\lambda$

Solution:
If a line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ lies
on a plane $a_{2} x+b_{2} y+c_{2} z=d,$ then
(i) $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
(ii) $a_{2} x_{1}+b_{2} y_{1}+c_{2} z_{1}=d$
According to the question. $3(1)+(2+\lambda)(-2)+(-1)(0)=0$
$\Rightarrow \quad 3-4-2 \lambda=0$
$\Rightarrow \quad \lambda=-1 / 2$
Also, $\quad \lambda-2=0$
$\Rightarrow \quad \lambda=2$
So. there is no such value of $\lambda$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The value of $\alpha$ for which the sum of the squares of the roots of the equation $x^2-(a-2) x-a-1=0$ assume the least value is

Options:
A
1

B
0

C
3

D
2

The correct answer is:
1

Solution:
$$
\begin{aligned}
& x^2-(a-2) x-a-1=0 \\
& \Rightarrow \alpha+\beta=a-2 \\
& \alpha \beta=-(a+1) \\
& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\
& \quad=a^2-2 a+6=(a-1)^2+5 \\
& \Rightarrow a=1 .
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2005
The value of θ for which the system of equations

sin⁡3θx-2y+3z=0, (cos⁡2θ x+8y-7z=0 and 2x+14y-11z=0 has a non-trivial solution, is (here, n∈Z)

Options:
A nπ
B nπ + -1n π/3
C nπ + -1n π/8
D None of these

The correct answer is:
nπ

Solution:
The system of equations has a non-trivial solution if and only if sinâ¡3Î¸-23cosâ¡2Î¸8-7214-11 = 0
Applying R2â R2 + 4R1, R3â R3 + 7R1, we get,
sinâ¡3Î¸-23cosâ¡2Î¸+4sinâ¡3Î¸052+7sinâ¡3Î¸010 = 0
Expanding along C2, we get,
2cosâ¡2Î¸ + 4sinâ¡3Î¸- 2 + 7sinâ¡3Î¸=0
â 2- 2 cosâ¡2Î¸- sinâ¡3Î¸=0
â4sin2â¡Î¸- 3sinâ¡Î¸- 4sin3â¡Î¸=0
âsinâ¡Î¸ 4sin2â¡Î¸ + 4sinâ¡Î¸- 3=0
âsinâ¡Î¸ 2sinâ¡Î¸- 1 2sinâ¡Î¸ + 3=0
âsinâ¡Î¸=0 or sinâ¡Î¸=1/2 .
[ sinâ¡Î¸=-3/2 is not possible]
â´ For, Î¸=nÏ the system of equations has a non-trivial solution.

Topic: Mathematics, Determinants, JEE Main
The value of ' $\lambda$ ' for which the two vectors $\vec{a}=5 \hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ are perpendicular to each other is

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

The correct answer is:
3

Solution:
Hints : For $\vec{a} \perp \vec{b}$
$\begin{aligned}
& \vec{a} \cdot \vec{b}=0 \\
& \text { i.e., } 5-2 \lambda+1=0 \\
& \lambda=3
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
The value of $\lambda$ for which the vectors $2 \lambda \mathbf{i}+\mathbf{j}-\mathbf{k}$ and $2 \mathbf{j}+\mathbf{k}$ are perpendicular, is

Options:
A None
B -1
C 1
D Any value

The correct answer is:
None

Solution:
Since $(2 \lambda \mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(2 \mathbf{j}+\mathbf{k})=1$ for every $\lambda$, therefore the vectors will not be perpendicular for any $\lambda$.

Topic: Mathematics, Vector Algebra, JEE Main
The value of $g$ at a place decreases by $2 \%$. The barometric height of mercury

Options:
A Increases by $2 \%$
B Decrease by $2 \%$
C Remains unchanged
D Sometimes increases and sometimes decreases

The correct answer is:
Increases by $2 \%$

Solution:
$h=\frac{P}{\rho g} \therefore h \propto \frac{1}{g}$ ( $P$ and $\rho$ are constant)
If value of $g$ decreased by $2 \%$ then $h$ will increase by $2 \%$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The value of $g$ on the earth's surface is $980 \mathrm{~cm} / \mathrm{sec}^2$. Its value at a height of $64 \mathrm{~km}$ from the earth's surface is

Options:
A $960.40 \mathrm{~cm} / \mathrm{sec}^2$
B $984.90 \mathrm{~cm} / \mathrm{sec}^2$
C $982.45 \mathrm{~cm} / \mathrm{sec}^2$
D $977.55 \mathrm{~cm} / \mathrm{sec}^2$
(Radius of the earth $R=6400$ kilometers)

The correct answer is:
$960.40 \mathrm{~cm} / \mathrm{sec}^2$

Solution:
$\frac{g^{\prime}}{g}=\left(\frac{R}{R+h}\right)^2=\left(\frac{6400}{6400+64}\right)^2 \Rightarrow g^{\prime}=960.40 \mathrm{~cm} / \mathrm{s}^2$

Topic: Physics, Gravitation, JEE Main
The value of gravitational acceleration g at a height h above the earth's surface is g4 then (R = radius of earth)

Options:
A h=R
B h=R2
C h=R3
D h=R4

The correct answer is:
h=R

Solution:
gâ²=gRR+h2
When gâ²=g4 then.
g4=gÃRR+h2â12=RR+h
â´ 2R=R+hâR=h

Topic: Physics, Gravitation, JEE Main
The value of \( \hat{i} \times(\hat{j} \times \hat{k})+\hat{j} \times(\hat{k} \times \hat{i})+\hat{k} \times(\hat{i} \times \hat{j}) \) is

Options:
A \( 0 \)
B \( \hat{\mathrm{i}} \)
C \( \hat{j} \)
D \( \widehat{\mathrm{k}} \)

The correct answer is:
\( 0 \)

Solution:
We have,

i^×j^×k^+j^×k^×i^+k^×i^×j^

=i^·k^j^-i^·j^k^+j^·i^k^-j^·k^i^+k^·j^i^-k^·i^j^

=0+0+0=0.

Topic: Mathematics, Vector Algebra, JEE Main
The value of $I=\int_0^{\pi / 2} \frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2 x}} d x$ is

Options:
A
0

B
1

C
2

D
3

The correct answer is:
2

Solution:
$\int_0^{\frac{\pi}{2}} \frac{(\sin x+\cos x)^2}{\sqrt{(\sin x+\cos x)^2}} d x=\int_0^{\frac{\pi}{2}}(\sin x+\cos x) d x=|-\cos x+\sin x|_0^{\frac{\pi}{2}}=2$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2004
The value of I=â«-11xsinâ¡(Ïx)dx is (where [Â·] denotes the greatest integer function)

Options:
A π
B 2π
C 0
D -π

The correct answer is:
0

Solution:
I=2 â« 0 1 xsin(Ïx) dx
Now, xsin(Ïx)â 0,1
as xâ 0,1
â´ xsin(Ïx) =0
â´I=2â«010dx=0

Topic: Mathematics, Definite Integration, JEE Main
The value of ∑i=1nCn+1i-Cni is equal to

Options:
A 2n
B 2n+1
C 3⋅2n
D 2n-1

The correct answer is:
2n-1

Solution:
Cn+1i-Cni
=Cni+Cni-1-Cni
=Cni-1
⇒ Sum =C0n+C1n+…+Cn-1n
=2n-Cnn
=2n-1

Topic: Mathematics, Binomial Theorem, JEE Main
The value of $I=\int_{-\pi / 2}^{\pi / 2}|\sin x| d x$ is

Options:
A 0
B 2
C -2
D $-2 < $ I $ < 2$

The correct answer is:
2

Solution:
Hints: $I=2 \int_0^{\frac{\pi}{2}} \sin x d x=2(1)=2$

Topic: Mathematics, Definite Integration, JEE Main
The value of $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1=$

Options:
A $-1$
B $-2$
C $-3$
D $-4$

The correct answer is:
$-2$

Solution:
$\begin{aligned} & \frac{i^{584}\left(i^8+i^6+i^4+i^2+1\right)}{i^{574}\left(i^8+i^6+i^4+i^2+1\right)}-1=\frac{i^{584}}{i^{574}}-1 \\ & \quad=i^{10}-1=-1-1=-2\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The value of indefinite integral: \( \int \sqrt{\frac{\sin x-\sin ^{3} x}{1-\sin ^{3} x}} d x \) equals: (where \( c \) is constant of integration)

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

The correct answer is:
\( \frac{2}{3} \sin ^{-1}\left(\sin ^{\frac{3}{2}} x\right)+c \)

Solution:
Let

I=∫sinx-sin3x1-sin3x dx

⇒I=∫sinx1-sin2x1-sin3x dx

⇒I=∫sinxcosx1-sin3x dx

Put sin32x=t⇒32sinxcosxdx=dt

⇒I=23∫dt1-t2

⇒I=23sin-1t+c

⇒I=23sin-1sin32x+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of \( \int \frac{f(x) \phi^{\prime}(x)+\phi(x) f^{\prime}(x)}{(f(x) \cdot \phi(x)+1) \sqrt{\phi(x) \cdot f(x)-1}} d x \) is (Where \( C \) is the constant of integration)

Options:
A \( \cos ^{-1} \sqrt{f(x)^{2}-\phi(x)^{2}} \)
B \( \tan ^{-1}[f(x) \phi(x)] \)
C \( \sin ^{-1} \sqrt{\frac{f(x)}{\phi(x)}} \)
D None of these

The correct answer is:
None of these

Solution:
Let I=∫f(x)ϕ'(x)+ϕ(x)f'(x)f(x)·ϕ(x)+1 ϕ(x)·f(x)-1dx

Let ϕ(x)·f(x)-1=t2

⇒(ϕ(x)·f'(x)+f(x)ϕ'(x))dx=2tdt

∴ I=∫2tdt(t2+2)t2=∫2dtt2+(2)2

=2·12tan-1t2+C

=2tan-1ϕ(x) f(x)-12+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of \(\int_0^\pi \sin ^{50} x \cos ^{49} x d x\) is

Options:
A 0
B \(\pi / 4\)
C \(\pi / 2\)
D 1

The correct answer is:
0

Solution:
Hints: \(I=\int_0^\pi \sin ^{50} x \cdot \cos ^{49} x d x \int_0^4 f(x)=\int_0^4 f(a-x)\)
\(\begin{aligned}
& I=\int_0^\pi \sin ^{50} x\left(-\cos ^{49}(x)\right)=-\int_0^\pi \sin ^{50} x \cdot \cos ^{49} x \\
& =I=-I \\
& I=0
\end{aligned}\)

Topic: Mathematics, Definite Integration, JEE Main