datavorous/entrance-exam-dataset huggingface dataset:
The value of $\int_a^{a+(\pi / 2)}\left(\sin ^4 x+\cos ^4 x\right) d x$ is

Options:
A Independent of a
B $a\left(\frac{\pi}{2}\right)^2$
C $\frac{3 \pi}{8}$
D $\frac{3 \pi a^2}{8}$

The correct answer is:
$\frac{3 \pi}{8}$

Solution:
Since $\sin ^4 x+\cos ^4 x$ is a periodic function with period $\frac{\pi}{2}$, therefore $\int_a^{a+(\pi / 2)}\left(\sin ^4 x+\cos ^4 x\right) d x$
$=\int_0^{\pi / 2}\left(\sin ^4 x+\cos ^4 x\right) d x$
$=2 \int_0^{\pi / 2} \sin ^4 x d x=\frac{3 \Gamma(5 / 2) \Gamma(1 / 2)}{2 \Gamma\left(\frac{4+0+2}{2}\right)}=\frac{3 \pi}{8}$

Topic: Mathematics, Definite Integration, JEE Main
The value of a+b such that the inequality aâ¤5cosâ¡Î¸+3cosÎ¸+Ï3+3â¤b holds true for all the real values of Î¸ is (equality holds on both sides atleast once for real values of Î¸)

Options:

The correct answer is:
6

Solution:
The given expression is
5cos⁡θ+3cos⁡θcos⁡60o-sin⁡θsin⁡60o+3
=5cos⁡θ+3cos⁡θ12-3sin⁡θ32+3
=cos⁡θ5+32+sin⁡θ-332+3
=132cos⁡θ-332sin⁡θ+3
132cos⁡θ-332sin⁡θ∈-k,k where
k=1322+-3322=1694+274=142=7
⇒132cos⁡θ-332sin⁡θ∈-7,7
⇒132cos⁡θ-332sin⁡θ+3∈-4,10
⇒a=-4 and b=10

⇒a+b=6

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of $\int_{a}^b \frac{x}{|x|} d x, a \lt b \lt 0$ is

Options:
A $-(|a|+|b|)$
B $|b|-|a|$
C $|a|-|b|$
D $|a|+|b|$

The correct answer is:
$|b|-|a|$

Solution:
$\begin{aligned} \int_a^b \frac{x}{|x|} d x=-\int_a^b d x & \\ & (a \lt b \lt 0) \\ & =-(b-a)=|b|-|a|\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
The value of $[\mathbf{a}-\mathbf{b} \mathbf{b}-\mathbf{c} \mathbf{c}-\mathbf{a}]$, where $|\mathbf{a | = 1},| \mathbf{b} \mid=5$ and $|\mathbf{c}|=3$ is

Options:
A 0
B 1
C 2
D 4

The correct answer is:
0

Solution:
Since, \(\vec{a}-\vec{b}+\vec{b}-\vec{c}+\vec{c}-\vec{a}=0\), therefore the vectors \(\vec{a}-\vec{b}, \vec{b}-\vec{c}\) and \(\vec{c}-\vec{a}\) are coplanar. We know that if three vectors are coplanar, then their box product would be equal to zero.
Also try to recollect why this is so.
Hence for our present case, \(\left[\begin{array}{llll}\vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a}\end{array}\right]=0\).

Topic: Mathematics, Vector Algebra, JEE Main
The value of $a \cos \theta+b \sin \theta$ lies between

Options:
A $a + b$ and $a - b$
B $a$ and $b$
C $-\left(a^2+b^2\right)$ and $\left(a^2+b^2\right)$
D $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$

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

Solution:
$\begin{aligned} & \qquad=\sqrt{a^2+b^2} \sin (\theta+\phi) \\ & \text {Since, }-1 \lt \sin (\theta+\phi) \lt 1 \\ & \text {Then }-\sqrt{a^2+b^2} \lt \sin (\theta+\phi) \lt \sqrt{a^2+b^2}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of a for which ax2+sin-1⁡x2-2x+2+cos-1⁡x2-2x+2=0 has a real solution, is

Options:
A -2π
B 2π
C -π2
D π2

The correct answer is:
-π2

Solution:
Here, x2-2x+2=x-12+1â¥1
But, -1â¤(x2-2x+2)â¤1
Which is possible only when
x2-2x+2=1
âx=1
Then, a12+sin-1â¡1+cos-1â¡1=0
âa+Ï2+0=0
âa=-Ï2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of a for which both the roots of the equation 1 - a 2 x 2 + 2 a x - 1 = 0 lie between 0 and 1, will always be greater than

Options:

The correct answer is:
2

Solution:
Given equation is 1 - a 2 x 2 + 2 a x - 1 = 0
Its discriminant D=4 and roots are 1 a - 1 ,â¡ 1 a + 1
Given, 0 < 1 a - 1 < 1 ,â¡ 0 < 1 a + 1 < 1
Now, 1 a - 1 > 0 â a > 1 ,â¡
1 a - 1 < 1 â 1 a - 1 - 1 < 0
â 2 - a a - 1 < 0 â a < 1 or â¡ a > 2
â´ a > 2 .... (1)
and 1 a+1 >0âa>â1
and 1 a + 1 < 1 â - a a + 1 < 0
â a < - 1 or â¡ a > 0
â´ a>0 ... (2)
From (1) and (2), a > 2

Topic: Mathematics, Quadratic Equation, JEE Main
The value of \( a \) for which both the roots of the equation \( \left(1-a^{2}\right) x^{2}+2 a x-1=0 \) lie between \( 0 \) and \( 1 \), will always be greater than

Options:

The correct answer is:
2

Solution:
Given equation is 1 - a 2 x 2 + 2 a x - 1 = 0
Its discriminant D=4 and roots are 1 a - 1 ,â¡ 1 a + 1
Given, 0 < 1 a - 1 < 1 ,â¡ 0 < 1 a + 1 < 1
Now, 1 a - 1 > 0 â a > 1 ,â¡
1 a - 1 < 1 â 1 a - 1 - 1 < 0
â 2 - a a - 1 < 0 â a < 1 or â¡ a > 2
â´ a > 2 .... (1)
and 1 a+1 >0âa>â1
and 1 a + 1 < 1 â - a a + 1 < 0
â a < - 1 or â¡ a > 0
â´ a>0 ... (2)
From (1) and (2), a > 2

Topic: Mathematics, Quadratic Equation, JEE Main
The value of ' $a$ ' for which one root of the quadratic equation $\left(a^2-5 a+3\right) x^3+(3 a-1) x+2=0$ is twice as large as the other is

Options:
A
$-\frac{1}{3}$

B
$\frac{2}{3}$

C
$-\frac{2}{3}$

D
$\frac{1}{3}$

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

Solution:
$3 \alpha=\frac{1-3 a}{a^2-5 a+3} \& 2 \alpha^2=\frac{2}{a^2-5 a+3}$
$2\left[\frac{1}{9} \frac{(1-3 a)^2}{\left(a^2-5 a+3\right)^2}\right]=\frac{2}{a^2-5 a+3}$
$\frac{(1-3 a)^2}{\left(a^2-5 a+3\right)}=9$ or $9 a^2-6 a+1$
$=9 a^2-45 a+27$ or $39 a=26$ or $a=\frac{2}{3}$

Topic: Mathematics, Quadratic Equation, JEE Main
The value of ' $a$ ' for which one root of the quadratic equation $\left(a^2-5 a+3\right) x^3+(3 a-1) x+2=0$ is twice as large as the other is

Options:
A
$-\frac{1}{3}$

B
$\frac{2}{3}$

C
$-\frac{2}{3}$

D
$\frac{1}{3}$

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

Solution:
$3 \alpha=\frac{1-3 a}{a^2-5 a+3} \& 2 \alpha^2=\frac{2}{a^2-5 a+3}$
$2\left[\frac{1}{9} \frac{(1-3 a)^2}{\left(a^2-5 a+3\right)^2}\right]=\frac{2}{a^2-5 a+3}$
$\frac{(1-3 a)^2}{\left(a^2-5 a+3\right)}=9$ or $9 a^2-6 a+1$
$=9 a^2-45 a+27$ or $39 a=26$ or $a=\frac{2}{3}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2003
The value of ' \(a\) ' for which the scalar triple product formed by the vectors \(\vec{a}=\hat{i}+a \hat{j}+\hat{k}, \vec{\beta}=\hat{j}+a \hat{k}\) and \(\vec{\gamma}=a \hat{i}+\hat{k}\) is maximum, is

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

The correct answer is:
\(-\frac{1}{\sqrt{3}}\)

Solution:
Hint : \(\Delta=1+\mathrm{a}^3-\mathrm{a}\) For \(\max ; \frac{\mathrm{d} \Delta}{\mathrm{da}}=0\)
\(\frac{\mathrm{d} \Delta}{\mathrm{da}}=3 \mathrm{a}^2-1 \quad \mathrm{a}= \pm \frac{1}{\sqrt{3}}\)
\(\frac{\mathrm{d}^2 \Delta}{\mathrm{da}^2}=6 \mathrm{a} < 0\) for \(\mathrm{a}=-\frac{1}{\sqrt{3}}\)

Topic: Mathematics, Vector Algebra, JEE Main
The value of a for which the sum of cubes of the roots of the equation, x2-ax+2a-3=0 ∀a∈12,4 attains its minimum value is

Options:
A greater than 4
B less than 2
C greater than 32
D less than 12

The correct answer is:
greater than 32

Solution:
Let a,β be the roots,
Then, α3+β3=α+β3-3αβα+β
=a3-32a-3a
=a3-6a2+9a
=E(say)
∴dEda=3a2-12a+9
=3a-1a-3
Thus, E attains its minimum value at ‘a’=3

Topic: Mathematics, Quadratic Equation, JEE Main
The value of $a$ for which the system of equations
$a^3 x+(a+1)^3 y+(a+2)^3 z=0$,
$\begin{array}{r}a x+(a+1) y+(a+2) z=0 \\ x+y+z=0\end{array}$ has a non zero solution is

Options:
A -1
B 0
C 1
D None of these

The correct answer is:
-1

Solution:
The system will have a non-zero solution, if
$\begin{gathered}\Delta=\left|\begin{array}{ccc}a^3 & (a+1)^3 & (a+2)^3 \\ a & a+1 & a+2 \\ 1 & 1 & 1\end{array}\right|=0 \\ \Rightarrow\left|\begin{array}{ccc}a^3 & 3 a^2+3 a+1 & 3(a+1)^2+3(a+1)+1 \\ a^2 & 1 & 1 \\ 1 & 0 & 0\end{array}\right|=0\end{gathered}$
$\Rightarrow 3 a^2+3 a+1-\left\{3(a+1)^2+3(a+1)+1\right\}$ $\begin{aligned} & c_2 \rightarrow c_2-c_1 \\ & \text { by } c_3 \rightarrow c_3-c_2\end{aligned}$
$\Rightarrow-6(a+1)=0 \Rightarrow a=-1$ (expanding along ${ }^{R_3}$)

Topic: Mathematics, Determinants, JEE Main
The value of ${ }^{\prime} a$ for which the system of equations
$$
\begin{array}{r}
a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0 \\
a x+(a+1) y+(a+2) z=0 \\
x+y+z=0
\end{array}
$$
has a non-zero solution is

Options:
A 1
B 0
C $-1$
D None of these

The correct answer is:
$-1$

Solution:
For non-zero solution,
$$
\left|\begin{array}{ccc}
a^{3} & (a+1)^{3} & (a+2)^{3} \\
a & (a+1) & (a+2) \\
1 & 1 & 1
\end{array}\right|=0
$$
$$
\left.\begin{array}{l}
\Rightarrow \quad-\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & (a+1) & (a+2) \\
a^{3} & (a+1)^{3} & (a+2)^{3}
\end{array}\right|=0 \\
\Rightarrow & -(a-a-1)(a+1-a-2)(a+2-a) \\
& \quad \quad \times(a+a+1+a+2)=0 \\
\Rightarrow & -2(3 a+3)=0 \\
\Rightarrow & \quad a=-1 \\
& {\left[\begin{array}{ccc}
1 & 1 & 1 \\
x & y & z \\
x^{3} & y^{3} & z^{3}
\end{array} \mid=\right.} \\
(x-y)(y-z)(z-x)(x+y+z)
\end{array}\right]
$$

Topic: Mathematics, Determinants, JEE Main
The value of a for which twice the sum of the cubes of the roots of the equation a=x2-3x-2 attains its minimum value is (where, a∈0,π)

Options:
A greater than 4
B less than 2
C greater than 74
D less than 1

The correct answer is:
greater than 74

Solution:
The given equation is x2-ax+2a-3=0

Let α,β be the roots. Then,

2α3+β3=2α+β3-3αβα+β=2a3-32a-3a
=2a3-6a2+9a
=2E (say)
∴dEda=3a2-12a+9
=3a-1a-3
Thus, E attains its minimum value at a=3

Topic: Mathematics, Application of Derivatives, JEE Main
The value of a such that the area bounded by the curve y=x2+2ax+3a2, the coordinate axes and the line x=1, attains its least value is equal to

Options:
A 16
B -16
C 13
D -13

The correct answer is:
-16

Solution:
y=x2+2ax+3a2 is positive for all x∈R
Area =∫01x2+2ax+3a2dx
=13+2a12+3a2
=13+a+3a2
Area =9a2+3a+13
Which is minimum at a=-16

Topic: Mathematics, Area Under Curves, JEE Main
The value of a, such that the volume of the parallelopiped formed by the vectors i^+j^+k^, j^+ak^ and ai^+k^ becomes minimum, is

Options:
A 3
B -13
C 12
D -2

The correct answer is:
12

Solution:
Volume of parallelopiped =aâ bâ câ

Here, the parallelopiped is formed by the vectors i^+j^+k^, j^+ak^ and ai^+k^
So, the volume, V =11101aa01=1+a2-a

=34+a-122 cubic units

Hence, the volume of the parallelopiped is minimum when a=12

Topic: Mathematics, Vector Algebra, JEE Main
The value of acceleration due to gravity at a height of \( 10 \mathrm{~km} \) from the surface of earth is \( x \). At what depth inside the earth is the value of the acceleration due to gravity has the same value \( x ? \)

Options:
A \( 10 \mathrm{~km} \)
B \( 15 \mathrm{~km} \)
C \( 5 \mathrm{~km} \)
D \( 20 \mathrm{~km} \)

The correct answer is:
\( 20 \mathrm{~km} \)

Solution:
gh=g1-2hR

gd=g1-dR⇒gh=gd

g1-2hR=g1-dR⇒d=2h=2×10=20 km

Topic: Physics, Gravitation, JEE Main
The value of acceleration due to gravity at Earth's surface is 9.8 m s-2 . The altitude above its surface at which the acceleration due to gravity decreases to 4.9 m s-2, is close to: (Radius of earth =6.4×106 m )

Options:
A 1.6×106 m
B 2.6×106 m
C 6.4×106 m
D 9.0×106 m

The correct answer is:
2.6×106 m

Solution:
Acceleration due to gravity g=GMR2 .......(1)
⇒g2=GM(R+h)2 ............(2)
From (1) and (2),
GM2R2=GM(R+h)2
R+h=2R
h=0.414×6.4×106
=2.6×106 m

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The value of b>3 for which 12∫3b1x2-1x2-4dx=loge4940, is equal to _____.

Options:

The correct answer is:
6

Solution:
Given 12×∫3b1dxx2-4x2-1=ln4940 & 4 b>3

⇒ By Using partial fraction

1x2-4x2-1=131x2-4-1x2-1

⇒123∫3b1x2-4-1x2-1=ln4940

⇒ 414logx-2x+2-12logx-1x+13b=ln4940

⇒ logb-2b+12b+2b-12×54=log4940

⇒ b-2b+12b+2b-12×54=4940

⇒b-2b+12b+2b-12=4950 ...i

Comparing by hit and trial and taking b+12=49⇒b+1=7⇒b=6

Now putting b=6 in equation (i) to corsscheck,

we get 4×728×52=4950 which is true

So b=6 is correct option,

Hence b=6

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The value of $b$ and $c$ for which the identity $f(x+1)-f(x)=8 x+3$ is satisfied, where $f(x)=b x^2+c x+d$, are

Options:
A $b=2, c=1$
B $b=4, c=-1$
C $b=-1, c=4$
D $b=-1, c=1$

The correct answer is:
$b=4, c=-1$

Solution:
$\begin{aligned}
& \text { Given: } f(x+1)-f(x)=8 x+3, f(x)=b x^2+c x+d \\
& \Rightarrow \quad f(x+1)-f(x)=8 x+3 \\
& \Rightarrow b(x+1)^2+c(x+1)+d-b x^2-c x-d=8 x+3 \\
& \Rightarrow \quad b\left[(x+1)^2-x^2\right]+c=8 x+3 \\
& \Rightarrow \quad b\left[x^2+1+2 x-x^2\right]+c=8 x+3 \\
& \Rightarrow \quad b[1+2 x]+c=8 x+3 \\
& \Rightarrow \quad b+c+2 b x=8 x+3 \\
&
\end{aligned}$
Now compare like terms
$\begin{aligned}
& \Rightarrow \quad 2 \mathrm{~b}=8 \\
& \therefore \quad \mathrm{b}=4 \\
& \Rightarrow \quad \mathrm{b}+\mathrm{c}=3 \\
& \Rightarrow \quad 4+\mathrm{c}=3 \\
& \Rightarrow \quad \mathrm{c}=3-4 \\
& \therefore \quad \mathrm{c}=-1
\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The value of $b$ for which the equations $x^2+b x-1=0, x^2+x+b=0$ have one root in common is

Options:
A
$-\sqrt{2}$

B
$-i \sqrt{3}$

C
$i \sqrt{5}$

D
$\sqrt{2}$

The correct answer is:
$-i \sqrt{3}$

Solution:
If $\quad a_1 x^2+b_1 x+c_1=0$
and $a_2 x^2+b_2 x+c_2=0$
have a common real root, then
$$
\begin{aligned}
\Rightarrow \quad\left(a_1 c_2-a_2 c_1\right)^2=\left(b_1 c_2-\right. & \left.b_2 c_1\right) \\
& \left(a_1 b_2-a_2 b_1\right)
\end{aligned}
$$
$\left.\therefore \begin{array}{l}x^2+b x-1=0 \\ x^2+x+b=0\end{array}\right\}$ have a common root.
$\Rightarrow \quad(1+b)^2=\left(b^2+1\right)(1-b)$
$\Rightarrow b^2+2 b+1=b^2-b^3+1-b$
$\Rightarrow \quad b^3+3 b=0 \Rightarrow b\left(b^2+3\right)=0$
$\Rightarrow \quad b=0, \pm \sqrt{3} i$

Topic: Mathematics, Quadratic Equation, JEE Main
The value of $b$ such that scalar product of the vectors $(\mathbf{i}+\mathbf{j}+\mathbf{k})$ with the unit vector parallel to the sum of the vectors $(2 \mathbf{i}+4 \mathbf{j}-\mathbf{5})$ and $(b \mathbf{i}+\mathbf{j}+3 \mathbf{k})$ is 1 , is

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

The correct answer is:
1

Solution:
$\begin{aligned}
& \text { Parallel vector }=(2+b) \mathbf{i}+6 \mathbf{j}-2 \mathbf{k} \\
& \text { Unit vector }=\frac{(2+b) \mathbf{i}+6 \mathbf{j}-2 \mathbf{k}}{\sqrt{b^2+4 b+44}}
\end{aligned}$
According to the condition,
$\begin{aligned}
& \qquad 1=\frac{(2+b)+6-2}{\sqrt{b^2+4 b+44}} \\
& \text { According to the condition, } \\
& \Rightarrow b^2+4 b+44=b^2+12 b+36 \Rightarrow 8 b=8 \Rightarrow b=1 .
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
The value of binding energy per nucleon of 20 40 Ca nucleus is

Given:

Mass of 20 40 Ca nucleus =39.962589âu

Mass of proton =1.007825âu

Mass of neutron =1.008665âu

and 1 u=931 MeV C-2

Options:
A 18.32 MeV
B 8.55 MeV
C 9.94 MeV
D 14.72 MeV

The correct answer is:
8.55 MeV

Solution:
The nucleus 20 40 Ca contains 20 protons and 20 neutrons.

Mass of 20 protons =20Ã1.007825

=20.1565âu

Mass of 20 neutrons =20Ã1.008665

=20.1733âu

Total mass =40.3298âu

___________

Mass 20 40 Ca nucleus =39.962589âu

Mass defect, Îm =0.367211âu

Binding energy=0.367211Ã931=341.87 MeV

Binding energy per nucleon=341.8740=8.547âMeV

Topic: Physics, Nuclear Physics, JEE Main
The value of -C115+2·C215-3·15C3+.....-15·C1515+C114+C314+C514+....+C1114 is equal to

Options:
A 214
B 213-13
C 216-1
D 213-14

The correct answer is:
213-14

Solution:
-C115+2·C215-3·C315+…-15·C1515+C114+C314+…+C1114

=∑r=115-1r·r·15Cr+C114+C314+…+C1114+C1314-C1314

=∑r=115-1r·15·Cr-114+213-14

=15-C014+C114……..-C1414+213-14

=213-14

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The value of c, in the Lagrange’s mean value theorem for the function fx=x3-4x2+8x+11, when x∈0,1, is

Options:
A 4-53
B 4-73
C 23
D 7-23

The correct answer is:
4-73

Solution:
As fx, is polynomial function, so it is continuous and differentiable in 0,1.

Here f0=11, f1=1-4+8+11=16

f'x=3x2-8x+8

∴f.c=f1-f01-0=16-111=3c2-8c+8

3c2-8c+3=0

c=8±276=4±73

∴c=4-73∈0,1.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The value of $\frac{\sin 85^{\circ}-\sin 15^{\circ}}{\cos 65^{\circ}}=$

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

The correct answer is:
1

Solution:
$\frac{\sin 85^{\circ}-\sin 35^{\circ}}{\cos 65^{\circ}}$
$\begin{aligned}
&=\frac{2 \cos \left(\frac{85^{\circ}+35^{\circ}}{2}\right) \sin \left(\frac{85^{\circ}-35^{\circ}}{2}\right)}{\cos 65^{\circ}} \\
&=\frac{2 \cos 60^{\circ} \sin 25^{\circ}}{\cos 65^{\circ}} \\
&=\frac{2 \times \frac{1}{2} \times \sin 25^{\circ}}{\cos \left(90^{\circ}-25^{\circ}\right)} \\
&=\frac{\sin 25^{\circ}}{\sin 25^{\circ}}=1
\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of $c$ of Lagrange's mean value theorem for $\mathrm{f}(x)=\sqrt{25-x^2}$ on $[1,5]$ is

Options:
A $\sqrt {15}$
B $5$
C $\sqrt{10}$
D $1$

The correct answer is:
$\sqrt {15}$

Solution:
$\begin{aligned}
& \mathrm{f}(x)=\sqrt{25-x^2} \\
& \therefore \quad \mathrm{f}^{\prime}(x=\mathrm{c})=\frac{-2 \mathrm{c}}{2 \sqrt{25-\mathrm{c}^2}} \\
&=\frac{-\mathrm{c}}{\sqrt{25-\mathrm{c}^2}}
\end{aligned}$
Applying Lagrange's mean value theorem, we get
$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(1)-\mathrm{f}(5)}{1-5} \\
& \therefore \quad \frac{\mathrm{c}}{\sqrt{25-\mathrm{c}^2}}=\frac{\sqrt{25-1}-\sqrt{25-5^2}}{1-5} \\
& \therefore \quad \frac{-\mathrm{c}}{\sqrt{25-\mathrm{c}^2}}=\frac{-\sqrt{24}}{4}
\end{aligned}$
$\begin{array}{ll}
\therefore & 4 \mathrm{c}=\sqrt{24} \cdot \sqrt{25-\mathrm{c}^2} \\
\therefore & 16 \mathrm{c}^2=24\left(25-\mathrm{c}^2\right) \\
\therefore & \mathrm{c}^2=15 \\
\therefore & \mathrm{c}= \pm \sqrt{15}
\end{array}$
Since $c=-\sqrt{15}$ does not lie in $[1,5]$ $c=\sqrt{15}$

Topic: Mathematics, Application of Derivatives, JEE Main
The value of $C_V$ for one mole of neon gas is

Options:
A $\frac{1}{2} R$
B $\frac{3}{2} R$
C $\frac{5}{2} R$
D $\frac{7}{2} R$

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

Solution:
This is the value for monoatomic gases.
Neon gas is monoatomic and for monoatomic gases $C_v=\frac{3}{2} R$

Topic: Physics, Kinetic Theory of Gases, JEE Main
$\text { The value of } \cos ^{-1}\left(\cos \frac{5 \pi}{3}\right)+\sin ^{-1}\left(\cos \frac{5 \pi}{3}\right)$is

Options:
A $\frac{\pi}{2}$
B $\frac{5 \pi}{3}$
C $\frac{10 \pi}{3}$
D $0$

The correct answer is:
$\frac{\pi}{2}$

Solution:
$\cos ^{-1}\left[\cos \frac{5 \pi}{3}\right]+\sin ^{-1}\left[\frac{\cos 5 \pi}{3}\right]=\frac{\pi}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $\cos ^{-1}\left(\cos \left(\frac{7 \pi}{6}\right)\right)$ is

Options:
A $\frac{5 \pi}{6}$
B $\frac{\pi}{3}$
C $\frac{7 \pi}{6}$
D $\frac{\pi}{6}$

The correct answer is:
$\frac{5 \pi}{6}$

Solution:
Let $\alpha$ and $\beta$ are roots of quadratic equation Given, $\frac{\alpha+\beta}{2}=34$ $\alpha+\beta=68$
and $\sqrt{\alpha \beta}=16$ $\alpha \beta=256$
$\therefore$ Required quadratic equation is, $x^{2}-(\alpha+\beta) x+\alpha \beta=0$ $x^{2}-68 x+256=0$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of cos⁡π22⋅cos⁡π23⋅…⋅cos⁡π210⋅sin⁡π210 is:

Options:
A 11024
B 1512
C 12
D 1256

The correct answer is:
1512

Solution:
Given cosπ22·cosπ23·...·cosπ210·sinπ210
=sin29.π21029.sinπ210.sinπ210

Using cosA.cos2A.cos22A.cos23A.cos24A.....cos2n-1A=sin2nA2nsinA

=129

=1512

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of cos⁡π22⋅cos⁡π23⋅…⋅cos⁡π210⋅sin⁡π210 is:

Options:
A 11024
B 1512
C 12
D 1256

The correct answer is:
1512

Solution:
Given cosπ22·cosπ23·...·cosπ210·sinπ210
=sin29.π21029.sinπ210.sinπ210

Using cosA.cos2A.cos22A.cos23A.cos24A.....cos2n-1A=sin2nA2nsinA

=129

=1512

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The value of cos2π7+cos4π7+cos6π7 is equal to

Options:
A -1
B -12
C -13
D -14

The correct answer is:
-12

Solution:
Given, cos2π7+cos4π7+cos6π7

=12sinπ72cos2π7sinπ7+2cos4π7sinπ7+2cos6π7sinπ7

=12sinπ7sin3π7-sinπ7+sin5π7-sin3π7+sin7π7-sin5π7

Since, 2cosAsinB=sinA+B-sinA-B

=12sinπ7-sinπ7+sinπ

=-12, ∵sinπ=0

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of cos2π7+cos4π7+cos6π7 is equal to

Options:
A -1
B -12
C -13
D -14

The correct answer is:
-12

Solution:
Given, cos2π7+cos4π7+cos6π7

=12sinπ72cos2π7sinπ7+2cos4π7sinπ7+2cos6π7sinπ7

=12sinπ7sin3π7-sinπ7+sin5π7-sin3π7+sin7π7-sin5π7

Since, 2cosAsinB=sinA+B-sinA-B

=12sinπ7-sinπ7+sinπ

=-12, ∵sinπ=0

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The value of cos210°–cos10° cos50°+cos250° is

Options:
A 34
B 34+cos⁡20°
C 32
D 32(1+cos⁡20°)

The correct answer is:
34

Solution:
cos210°+ cos250°-cos10°cos50°

Using, cos2x=2cos2x-1, ⇒cos2x=1+cos2x2, we get

=121+cos20°+1+cos100°-2cos50°cos10°

=122+cos100°+cos20°-cos60°-cos40°

=122+cos100°+cos20°-12-cos40°

Using, cosC+cosD=2cosC+D2cosC-D2, we get

=1232+2cos⁡60°cos⁡40°-cos⁡40°

=1232+2×12×cos40°-cos40°

=1232+cos40°-cos40°

=12×32=34.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The value of cos⁡2π63cos⁡3π70cos⁡4π77cos⁡π72cos⁡π40cos⁡3π881cos⁡π90cos⁡2π99 is equal to

Options:
A 0
B 1
C 2cos⁡π9
D cos⁡π11

The correct answer is:
0

Solution:
The given determinant =cos⁡π7-π9cos⁡π7-π10cos⁡π7-π11cos⁡π8-π9 cos⁡π8-π10 cos⁡π8-π11cos⁡π9-π9cos⁡π9-π10cos⁡π9-π11
=cos⁡π7cos⁡π9+sin⁡π7sin⁡π9 cos⁡π7cos⁡π10+sin⁡π7sin⁡π10 cos⁡π7cos⁡π11+sin⁡π7sin⁡π11cos⁡π8cos⁡π9+sin⁡π8sin⁡π9cos⁡π8cos⁡π10+sin⁡π8sin⁡π10cos⁡π8cos⁡π11+sin⁡π8sin⁡π11cos⁡π9cos⁡π9+sin⁡π9sin⁡π9cos⁡π9cos⁡π10+sin⁡π9sin⁡π10cos⁡π9cos⁡π11+sin⁡π9sin⁡π11
=cos⁡π7sin⁡π70cos⁡π8sin⁡π80cos⁡π9sin⁡π90cos⁡π9sin⁡π90cos⁡π10sin⁡π100cos⁡π11sin⁡π110=0

Topic: Mathematics, Determinants, JEE Main
The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$ is

Options:
A $-\frac{2 \sqrt{6}}{5}$
B $-2 \sqrt{6}$
C $-\frac{\sqrt{6}}{5}$
D None of these

The correct answer is:
$-\frac{2 \sqrt{6}}{5}$

Solution:
$\cos \left[2 \cos ^{-1} x+\sin ^{-1} x\right]$

$=\cos \left[\cos ^{-1} x+\cos ^{-1} x+\sin ^{-1} x\right]$

$=\cos \left[\cos ^{-1} x+\frac{\pi}{2}\right]=-\sin \left[\cos ^{-1} x\right]$

$=-\sin \left[\sin ^{-1} \sqrt{1-x^{2}}\right]=-\sqrt{1-x^{2}}$

$=-\sqrt{1-\left(\frac{1}{5}\right)^{2}}=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of cos3π8.cos3π8+sin3π8.sin3π8 is:

Options:
A 12
B 122
C 12
D 14

The correct answer is:
122

Solution:
⇒cos3π8cosπ2-π8+sin3π8sinπ2-π8

⇒cos3π8sinπ8+sin3π8cosπ8

⇒sinπ8cosπ8sin2π8+cos2π8

⇒12sinπ4

⇒122

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The value of cos⁡π11cos⁡2π11cos⁡3π11……cos⁡11π11 is equal to

Options:
A -132
B 1512
C 11024
D -12048

The correct answer is:
11024

Solution:
Required expression =cos⁡π11cos⁡2π11cos⁡3π11cos⁡4π11cos⁡5π112
=cos⁡π11cos⁡2π11cos⁡8π11cos⁡4π11cos⁡5π112
=sin⁡16π1116sin⁡π11×cos⁡5π112
=2sin⁡5π11cos⁡5π1132sin⁡π112
=sin⁡10π1132×sin⁡π112=1322=11024

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$ is :

Options:
A $\frac{1}{16}$
B $\frac{1}{8}$
C $\frac{1}{12}$
D $\frac{1}{4}$

The correct answer is:
$\frac{1}{16}$

Solution:
$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$
$=\cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ} \cos 168^{\circ}$
$=-\cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}$
$=-\frac{16 \sin 12^{\circ}}{16 \sin 12^{\circ}} \cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}$
$=\frac{-8 \sin 24^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}}$
$=\frac{-4 \sin 48^{\circ} \cos 48^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}}$
$=-\frac{2 \sin 96^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}}$
$=-\frac{\sin 192^{\circ}}{16 \sin 12^{\circ}}=\frac{\sin 12^{\circ}}{16 \sin 12^{\circ}}=\frac{1}{16}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of $\cos 255^{\circ}+\sin 195^{\circ}$ is

Options:
A
$\frac{\sqrt{3}-1}{2 \sqrt{2}}$

B
$\frac{\sqrt{3}-1}{\sqrt{2}}$

C
$-\frac{\sqrt{3}-1}{\sqrt{2}}$

D
$\frac{\sqrt{3}+1}{\sqrt{2}}$

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

Solution:
Consider $\cos 255^{\circ}+\sin 195^{\circ}$
$$
\begin{aligned}
& =\cos \left(270^{\circ}-15^{\circ}\right)+\sin \left(180^{\circ}+15^{\circ}\right) \\
& =-\sin 15^{\circ}-\sin 15^{\circ} \\
& =-2 \sin 15^{\circ}=-2\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)=-\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2012 (26 May Online)