datavorous/entrance-exam-dataset huggingface dataset:
If $f(2)=4$ and $f^{\prime}(2)=1$, then
$$
\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}
$$
is equal to

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

The correct answer is:
$2$

Solution:
Given that, $f(2)=4$ and $f^{\prime}(2)=1$
$$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2} \\
= & \lim _{x \rightarrow 2} f(2)-2 \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \\
= & f(2)-2 f^{\prime}(2) \\
= & 4-2(1) \\
= & 2
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Main
If $\overrightarrow{\mathrm{F}}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k}$, then the scalar and vector products of $\vec{F}$ and $\vec{r}$ have the magnitudes respectively as:

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

The correct answer is:
$10, \sqrt{2}$

Solution:
$\begin{aligned}
& \overrightarrow{\mathrm{F}}=2 \hat{i}+\hat{j}-\hat{k} \\
& \vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k} \text { (given) }
\end{aligned}$
The scalar product of $\overrightarrow{\mathrm{F}}$ and $\vec{r}$ is
$\begin{aligned}
\vec{F} . \vec{r} & =(2)(3)+(1)(2)+(-1)(-2) \\
& =6+2+2=10
\end{aligned}$
The vector product of $\vec{F}$ and $\vec{r}$ is
$\begin{aligned}
\overrightarrow{\mathrm{F}} \times \vec{r} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & -1 \\
3 & 2 & -2
\end{array}\right| \\
& =\hat{i}[(1 \times-2)-(2 \times-1)]-\hat{j}[(2 \times-2) \\
& -(-1 \times 3)]+\hat{k}[(2 \times 2)-(3 \times 1)] \\
& =\hat{i}(0)-\hat{j}(-1)+\hat{k}(1)=\hat{j}+\hat{k}
\end{aligned}$
Magnitude $= \sqrt{(1)^2+(1)^2}=\sqrt{2}$

Topic: Physics, Mathematics in Physics, JEE Main
If $f\left(\frac{t+1}{2 t+1}\right)=t+1$, then $\int f(x) d x=$

Options:
A $\frac{x^2}{2}+c$
B $\log (2 x-1)+\frac{1}{2} \log (x+1)+c$
C $\frac{1}{2} \log (2 x-1)+c$
D $\frac{x}{2}+\frac{1}{4} \log (2 x-1)+c$

The correct answer is:
$\frac{x}{2}+\frac{1}{4} \log (2 x-1)+c$

Solution:
We have, $f\left(\frac{t+1}{2 t+1}\right)=t+1$
$$
\begin{aligned}
& \text { Put } \frac{t+1}{2 t+1}=x \Rightarrow t+1=2 t x+x \\
& \Rightarrow \quad 1-x=t(2 x-1) \Rightarrow t=\frac{1-x}{2 x-1} \\
& \therefore \quad f(x)=\frac{1-x}{2 x-1}+1 \Rightarrow f(x)=\frac{1-x+2 x-1}{2 x-1} \\
& \Rightarrow \quad f(x)=\frac{x}{2 x-1} \\
& \therefore \quad \int f(x) d x=\int \frac{x}{2 x-1} d x=\frac{1}{2} \int \frac{2 x}{2 x-1} d x \\
& =\frac{1}{2} \int \frac{2 x-1+1}{2 x-1} d x=\frac{1}{2} \int\left(1+\frac{1}{2 x-1}\right) d x \\
& =\frac{1}{2}\left[x+\frac{1}{2} \log (2 x-1)\right]+c \\
& =\frac{x}{2}+\frac{1}{4} \log (2 x-1)+c \\
&
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
If $f:[-3,4] \rightarrow R, f(x)=2 x$, and $g:[-2,6] \rightarrow R, g(x)=x^2$. Then find function $(f+g)(x)$.

Options:
A $(f+g)(x)=3 x^2+2 \mathrm{x}$
B $(f+g)(x)=2 x+x^2$
C $(f+g)(x)=10 x^2-9$
D none of these

The correct answer is:
$(f+g)(x)=2 x+x^2$

Solution:
$f \longrightarrow:[-3,4] \rightarrow R, f(x)=2 x$, and $g:[-2,6] \rightarrow R, g(x)=x^2$
For $(f+g)(x)$ we consider common domain of $f(x)$ and $g(x)$.
Now $[-3,4] \cap[-2,6] \equiv[-2,4]$
So, $(f+g):[-2,4] \rightarrow R,(f+g)(x)=2 x+x^2$.

Topic: Mathematics, Functions, JEE Main
If f3x-43x+4=x+2, x≠-43, and ∫fxdx=Alog1-x+Bx+C , then the ordered pair A, B is equal to

Options:
A -83,-23
B -83,23
C 83,23
D 83,-23

The correct answer is:
-83,23

Solution:
Given f3x-43x+4=x+2, x≠-43
Let, 3x-43x+4=t
3x-4=3tx+4t
x=4t+43-3t+2
∴ft=10-2t3-3t
⇒fx=2x-103x-3
∫fxdx=∫2x-103x-3dx
=∫2x3x-3dx-10∫dx3x-3
=2x3-83lnx-1+C, where C is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2017 (09 Apr Online)
If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then
$$
(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)
$$
is equal to

Options:
A $f(4 \sqrt{2})$
B $f(3 \sqrt{2})$
C $f(2 \sqrt{2})$
D $f(\sqrt{2})$

The correct answer is:
$f(4 \sqrt{2})$

Solution:
Given, $\quad f(x)=x^2-3$
Now, $\quad f(-1)=(-1)^2-3=-2$
$$
\begin{aligned}
& \Rightarrow \quad f \circ f(-1)=f(-2)=(-2)^2-3=1 \\
& \Rightarrow \quad f \circ f \circ f(-1)=f(1)=1^2-3=-2
\end{aligned}
$$
Now,
$$
f(0)=0^2-3=-3
$$
$$
\begin{aligned}
& \Rightarrow \quad f \circ f(0)=f(-3)=(-3)^2-3=6 \\
& \Rightarrow \quad f \circ f \circ f(0)=f(6)=6^2-3=33
\end{aligned}
$$
Again,
$$
\begin{aligned}
& f(1)=1^2-3=-2 \\
& \Rightarrow \quad f \circ f(1)=f(-2)=(-2)^2-3=1 \\
& \Rightarrow \quad f \circ f \circ f(1)=(1)^2-3=-2 \\
& \therefore f \circ f \circ f(-1)+f \circ f \circ f(0)+f \circ f \circ f(1) \\
& =-2+33-2=29 \\
&
\end{aligned}
$$
Now,
$$
\begin{aligned}
f(4 \sqrt{2}) & =(4 \sqrt{2})^2-3=32-3 \\
& =29
\end{aligned}
$$
Hence, option (a) is correct.

Topic: Mathematics, Functions, JEE Main
If $f(9)=9, f^{\prime}(9)=4$, then $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=$

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

The correct answer is:
$4$

Solution:
Applying L-Hospital's rule,
$\lim _{x \rightarrow 9} \frac{\frac{1}{2 \sqrt{f(x)}} \cdot f(x)}{\frac{1}{2 \sqrt{x}}}=\frac{\frac{f^{\prime}(9)}{\sqrt{f(9)}}}{\frac{1}{\sqrt{9}}}=\frac{\frac{4}{3}}{\frac{1}{3}}=4$

Topic: Mathematics, Limits, JEE Main
If fa+b+1-x=fx, for all x, where a and b are fixed positive real numbers, then 1a+b∫abxfx+fx+1dx is equal to

Options:
A ∫a-1b-1fx+1dx
B ∫a-1b-1fxdx
C ∫a+1b+1fxdx
D ∫a+1b+1fx+1dx

The correct answer is:
∫a-1b-1fx+1dx

Solution:
I=1a+b∫abxfx+fx+1dx …1
x→a+b-x
I=1a+b∫aba+b-xfa+b-x+a+b+1-xdx
I=1a+b∫abaa+b-xfx+1+fxdx …2
∵ Keeping x→x+1 in given relation
1+2
2I=∫abfx+1+fxdx
2I=∫abfx+1dx+∫abfxdx
∫abfa+b+1-xdx+∫abfxdx
2I=2∫abfxdx
I=∫abfxdx
Put x=t+1⇒I=∫a-1b-1ft+1dt=∫a-1b-1fx+1dx

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If $f: A \rightarrow B$ and $g: B \rightarrow C$ are two functions such that $g \circ f: A \rightarrow C$ is a bijection, then which one of the following is always true?

Options:
A f and g are bijections
B f is an injection and g is a surjection
C f is a surjection and g is an injection
D f is a bijection but g is not a bijection

The correct answer is:
f is an injection and g is a surjection

Solution:
If $f: A \rightarrow B$ and $g: B \rightarrow C$ are two function such that $g \circ f: A \rightarrow C$ is bijection.
As, $g \circ f$ is one-one for $x, y$
Let
$f(x)=f(y)$
$g \circ(f(x))=g \circ(f(y))$
$$
\Rightarrow \quad g f(x)=g f(y) \Rightarrow x=y
$$
$\therefore f$ is one one.
Now, Let $Z \in C$
As gof is onto for every element $Z \in C$ there exist $y \in A$ that is in domain of gof.
$$
\begin{gathered}
g \circ f(y)=Z \\
g(f(y))=Z
\end{gathered}
$$
$f$ lies in range of $B$ as $f: A \rightarrow B$.
For $\quad a, y \in A, f(y) \in B$, let $f(y)=x$.
For every element $Z \in C$, we have at $X \in B$.
Such that
$$
g(x)=Z \text {. }
$$
$\therefore g$ is onto
Hence, $f$ is an injection and $g$ is a surjection.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 1)
If f:AâB defined as fx=2sinâ¡x-2cosâ¡x+32 is an invertible function, then the correct statement can be

Options:
A A=π4,5π4,B=22,55
B A=-π4,5π4,B=2,52
C A=-π4,3π4,B=2,42
D A=-π4,3π4,B=2,52

The correct answer is:
A=-π4,3π4,B=2,52

Solution:
fx=2212sin⁡x-12cos⁡x+32=22sin⁡x-π4+32
Clearly, for options A and B fx is not injective and for option C fx is not surjective

fx is bijective only for option D

Topic: Mathematics, Functions, JEE Main
If f:AâB defined by fx=sinx-cosx+32 is an invertible function, then the correct statement can be

Options:
A A=π4,5π4,B=32,42
B A=-π4,5π4,B=22,42
C A=-π4,3π4,B=2,42
D A=-π4,3π4,B=22,42

The correct answer is:
A=-π4,3π4,B=22,42

Solution:
fx=212sin⁡x-12cos⁡x+32=2sinx-π4+32
Clearly, for options A and B fx is not injective
and for the option C fx is not surjective
Hence, fx is bijective only for option D

Topic: Mathematics, Functions, JEE Main
If $\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x})$ then $\int_a^b \mathrm{xf}(\mathrm{x}) \mathrm{dx}$ is equal to

Options:
A
$\frac{a+b}{2} \int_a^b f(a+b-x) d x$

B
$\frac{a+b}{2} \int_a^b f(b-x) d x$

C
$\frac{a+b}{2} \int_a^b f(x) d x$

D
$\frac{b-a}{2} \int_a^b f(x) d x$

The correct answer is:
$\frac{a+b}{2} \int_a^b f(a+b-x) d x$

Solution:
$I=\int_a^b x f(x) d x=\int_a^b(a+b-x) f(a+b-x) d x$
$=(a+b) \int_a^b f(a+b-x) d x-\int_a^b x f(a+b-x) d x$
$=(a+b) \int_a^b f(a+b-x) d x-\int_a^b x f(x) d x$
$2 I=(a+b) \int_a^b f(x) d x$
$I=\frac{(a+b)}{2} \int_a^b f(x) d x ; I=\frac{(a+b)}{2} \int_a^b f(a+b-x) d x$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2003
If f:ℝ→ℝ be a continuous function satisfying ∫0π2fsin2xsin x dx+α∫0π4fcos2xcos x dx=0, then the value of α is

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

The correct answer is:
-2

Solution:
Given,

∫0π2fsin 2xsin x dx+α∫0π4fcos 2xcos x dx=0

Now, let I=∫0π2fsin 2x.sin x dx

Now using property ∫abfxdx=∫acfxdx+∫cbfxdx we get,
⇒I=∫0π4fsin 2x sin x dx+∫π4π2fsin 2x.sin x dx

Now using property ∫abfxdx=∫abfa+b-xdx we get,

⇒I=∫0π4fcos 2x sin π4-xdx+∫0π4fsin 2π4+x sin π4+xdx

⇒I=∫0π4fcos 2x 12cos x-12 sin xdx+∫0π4fcos 2x 12cos x+12 sin xdx
⇒I=∫0π4fcos 2x2 cos xdx

So, the putting the value of I in ∫0π2fsin 2xsin x dx+α∫0π4fcos 2xcos x dx=0 we get, α=-2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If fθ=cos2⁡θcos⁡θsin⁡θ-sin⁡θcos⁡θsin⁡θsin2⁡θcos⁡θsin⁡θ-cos⁡θ0 then, fπ6+fπ3+fπ2+f2π3+f5π6+fπ+.…….+f53π6 is equal to

Options:

The correct answer is:
53

Solution:
fθ=cos2⁡θcos2⁡θ-cos⁡θsin⁡θ-cos⁡θsin⁡θ-sin⁡θ-cos2⁡θsin⁡θ-sin3⁡θ
=cos4⁡θ+sin4⁡θ+2cos2⁡θsin2⁡θ=cos2⁡θ+sin2⁡θ2=1
fπ6=f2π6=f3π6=.…….=f53π6=1

Topic: Mathematics, Determinants, JEE Main
If f denotes the ratio of the number of nuclei decayed Nd to the number of nuclei at t=0, N0 then for a collection of radioactive nuclei, the rate of change of f with respect to time is given as:

[λ is the radioactive decay constant]

Options:
A -λ1-e-λt
B λ1-e-λt
C λe-λt
D -λe-λt

The correct answer is:
λe-λt

Solution:
N=N0e-λt
Nd=N0-N
Nd=N01-e-λ
NdN0=f=1-e-λt
dfdt=λe-λt

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (27 Jul Shift 1)
If $F_e$ and $F_e$ are the focal length of the objective and eye-piece respectively of a telescope, then its magnifying power will be

Options:
A $F_o+F_e$
B $F_o \times F_e$
C $F_o / F_e$
D $\frac{1}{2}\left(F_o+F_e\right)$

The correct answer is:
$F_o / F_e$

Solution:
Correct option is $3$
$m=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}$

Topic: Physics, Ray Optics, JEE Main
If f& g are differentiable functions in [0, 1] satisfying f0=2=g1, g0=0 & f1=6, then for some c∈]0, 1[

Options:
A f'c=g'c
B f'c=2g'c
C 2f'c=g'c
D 2f'c=3g'c

The correct answer is:
f'c=2g'c

Solution:
Let hx=fx-2gx clearly h0=h1
Also hx is continuous and differentiable.
Hence hx satisfy all conditions of Rolle's theorem in 0, 1. According to Rolle's theorem there exist atleast one c∈0, 1 such that h'x=0⇒f'x-2g'x=0
⇒f'c=2g'c.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (06 Apr)
If f is a function defined as f(x)=x2âx+5,âf:(12,ââ)â(194,ââ), and g(x) is its inverse function, then g'7 is equal to

Options:
A - 1 1 3
B 1 1 3
C 1 3
D - 1 3

The correct answer is:
1 3

Solution:
fâ¡x=x2-x+5

f'(x)=2x-1
â for y = 7 , x > 1 2
â´ 7 = x 2 - x + 5
â x 2 - x - 2 = 0
â x - 2 x + 1 = 0
â x = - 1, 2
â x=2 (as x=- 1 is rejected)
If fx & g x are inverse of each other, gfx=x
âg'fx .f'x=1
substituting x=2 in this equation,
we get, g'(f(2)).f'(2) = 1
âg'(f(2))=1f'(2)
g'(7)=12(2)-1
=13 (at x=2)

Topic: Mathematics, Functions, JEE Main
If \( f \) is a function such that \( 2 f(x)+f(2-x)=x^{2} \) then find value of \( f(4) \).

Options:

The correct answer is:
9.33

Solution:
Given, 2fx+f2-x=x2 ...i

Substitute x by 2-x in equation i we get,

2f2-x+fx=2-x2 ...ii

Now, i×2-ii we get,

3fx=2x2-2-x2

⇒fx=132x2-2-x2
Then, f4=13242-2-42

=1332-4=283=9.33

Topic: Mathematics, Functions, JEE Main
If $f$ is a real- valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and $f(0)=0$, then $f(1)$ equal

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

The correct answer is:
$0$

Solution:
$\begin{aligned} & \lim _{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right| \leq \lim _{x \rightarrow y}|x-y| \text { or }\left|f^{\prime}(x)\right| \leq 0 \\ & \quad \Rightarrow f^{\prime}(x)=0 \Rightarrow f(x) \text { is constant, As } f(0)=0 \\ & \quad \therefore f(1)=0 .\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and $f(0)=0$, then $f(1)$ equals

Options:
A
-1

B
0

C
2

D
1

The correct answer is:
0

Solution:
$$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& \left|f^{\prime}(x)\right|=\lim _{h \rightarrow 0}\left|\frac{f(x+h)-f(x)}{h}\right| \leq \lim _{h \rightarrow 0}\left|\frac{(h)^2}{h}\right| \\
& \Rightarrow\left|f^{\prime}(x)\right| \leq 0 \Rightarrow f^{\prime}(x)=0 \quad \Rightarrow f(x)=\text { constant } \\
& \text { As } f(0)=0 \Rightarrow f(1)=0 .
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
If $f$ is a relation from set of positive real number to the set of positive real numbers defined by $f(x)=3 x^2-2$, then $f$ is

Options:
A one-one but not onto
B onto but not one-one
C a bijection
D not a function

The correct answer is:
not a function

Solution:
(w) $f: R_{+} \rightarrow R_{+}$
$$
f(x)=3 x^2-2
$$
As
$$
\begin{aligned}
& x \in(0, \infty) \\
& 3 x^2-2 \in(-2, \infty)
\end{aligned}
$$
But co-domain of $f$ is $R_{+}$.
Hence, $f$ is not a function.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
If $f$ is an exponential function and $g$ is a logarithmic function, then $f \circ g(1)$ will be

Options:
A $e$
B $\log _e e$
C $0$
D $2 e$

The correct answer is:
$\log _e e$

Solution:
$(f \circ g)(x)=f\{g(x)\}=f(\log 1)=e^0=1=\log _e e$

Topic: Mathematics, Functions, JEE Main
If $f$ is continuous function, then

Options:
A $\int_{-2}^2 f(x) d x=\int_0^2[f(x)-f(-x)] d x$
B $\int_{-3}^5 2 f(x) d x=\int_{-6}^{10} f(x-1) d x$
C $\int_{-3}^5 f(x) d x=\int_{-4}^4 f(x-1) d x$
D $\int_{-3}^5 f(x) d x=\int_{-2}^6 f(x-1) d x$

The correct answer is:
$\int_{-3}^5 f(x) d x=\int_{-2}^6 f(x-1) d x$

Solution:
Since, $f$ is continuous function. Let $x=t-1$
$\therefore d x=d t$. When $x=-3 \rightarrow 5$, then $t=-2 \rightarrow 6$

Topic: Mathematics, Definite Integration, JEE Main
If $f$ is continuous on its domain $\mathrm{D}$, then $|\mathrm{f}|$ is also continuous on $\mathrm{D}$.

Options:
A
True

B
False

The correct answer is:
True

Solution:
No Solution Available

Topic: Mathematics, Continuity and Differentiability
If $f$ is linear function and $f(2)=4, f(-1)$ $=3$, then find $f(x)$.

Options:
A $f(x)=3 x+72$
B $f(x)=\frac{x+10}{3}$
C $f(x)=10-9 x$
D none of these

The correct answer is:
$f(x)=\frac{x+10}{3}$

Solution:
Let linear function is $f(x)=\mathrm{ax}+\mathrm{b}$
Given $f(2)=4 \Rightarrow 2 a+b=4$
Also $f(-1)=3 \Rightarrow-a+b=3$
Solving (1) and (2) we get $a=\frac{1}{3}$ and $b=\frac{10}{3}$
Hence, $f(x)=\frac{x+10}{3}$

Topic: Mathematics, Functions, JEE Main
If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\vec{r}$ and $\vec{\tau}$ be the torque of this force about the origin, then

Options:
A $\overrightarrow{\mathrm{r}} \cdot \vec{\tau} \neq 0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$
B $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}>0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau} < 0$
C $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$
D $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau} \neq 0$

The correct answer is:
$\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$

Solution:
Key Idea Torque is an axial yector ie, its direction is always perpendicular to the plane containing vectors $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{F}}$.
$$
\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}
$$
Torque is perpendicular to both $\vec{r}$ and $\vec{F}$
$$
\therefore \quad \begin{aligned}
\vec{\tau} \cdot \overrightarrow{\mathrm{r}} & =0 \\
\overrightarrow{\mathrm{F}} \cdot \vec{\tau} & =0
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\overrightarrow{\mathrm{r}}$ and $\vec{\tau}$ be the torque of this force about the origin, then :

Options:
A $\vec{r} \cdot \vec{\tau}=0$ and $\vec{F} \cdot \vec{\tau} \neq 0$
B $\overrightarrow{\mathrm{r}} \cdot \vec{\tau} \neq 0$ and $\vec{F} \cdot \vec{\tau}=0$
C $\vec{r} \cdot \vec{\tau}>0$ and $\vec{F} \cdot \vec{\tau} < 0$
D $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\vec{F} \cdot \vec{\tau}=0$

The correct answer is:
$\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\vec{F} \cdot \vec{\tau}=0$

Solution:
$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$
$\vec{\tau}$ is perpendicular to $\overrightarrow{\mathrm{r}}$ and $\vec{F}$.

Topic: Physics, Rotational Motion, JEE Main
If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\overrightarrow{\mathrm{r}}$ and $\vec{\tau}$ be the torque of this force about the origin, then :

Options:
A $\vec{r} \cdot \vec{\tau}=0$ and $\vec{F} \cdot \vec{\tau} \neq 0$
B $\overrightarrow{\mathrm{r}} \cdot \vec{\tau} \neq 0$ and $\vec{F} \cdot \vec{\tau}=0$
C $\vec{r} \cdot \vec{\tau}>0$ and $\vec{F} \cdot \vec{\tau} < 0$
D $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\vec{F} \cdot \vec{\tau}=0$

The correct answer is:
$\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\vec{F} \cdot \vec{\tau}=0$

Solution:
$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$
$\vec{\tau}$ is perpendicular to $\overrightarrow{\mathrm{r}}$ and $\vec{F}$.

Topic: Physics, Rotational Motion, NEET, NEET 2009 (Mains)
If Fk=1+sinπ2k1+sin⁡k-1π2k1+sin⁡2k+1π2k1+sin⁡3k-1π2k, then the value of F1+F2+F3 is equal to

Options:
A 316
B 14
C 516
D 716

The correct answer is:
716

Solution:
F1=1+sinπ21+sin01+sin3π21+sin⁡π
=2×1×0×1=0
F2=1+sinπ41+sinπ41+sin5π41+sin5π4
=1+1221-122=1-122=14
F3=1+sinπ61+sinπ31+sin7π61+sin4π3
=1+121+321-121-32
=1-141-34=34×14=316
⇒F1+F2+F3=0+14+316=716

Topic: Mathematics, Trigonometric Equations, JEE Main
If Fk=1+sinπ2k1+sin⁡k-1π2k1+sin⁡2k+1π2k1+sin⁡3k-1π2k, then the value of F1+F2+F3 is equal to

Options:
A 316
B 14
C 516
D 716

The correct answer is:
716

Solution:
F1=1+sinπ21+sin01+sin3π21+sin⁡π
=2×1×0×1=0
F2=1+sinπ41+sinπ41+sin5π41+sin5π4
=1+1221-122=1-122=14
F3=1+sinπ61+sinπ31+sin7π61+sin4π3
=1+121+321-121-32
=1-141-34=34×14=316
⇒F1+F2+F3=0+14+316=716

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If fk-x+fx=sinâ¡x, then the value of integral Î=â«0kfxdx is equal to

Options:
A cos⁡k
B 2cos2k2
C sin2k2
D sin⁡k

The correct answer is:
sin2k2

Solution:
Applying property of a+bâx and adding, we get,

2Î=â«0kfx+fk-xdx=â«0ksinâ¡xdx=1-cosâ¡k

Topic: Mathematics, Definite Integration, JEE Main
If fn+1=2fn+12 for n=1,2,3…….. and f1=2, then f10110 is equal to

Options:

The correct answer is:
5.2

Solution:
fn+1=fn+12
⇒fn+1-fn=12
⇒f1,f2,f3…….. are in AP with common difference 12.
f101=f1+100d
=2+100×12=52
⇒f10110=5.2

Topic: Mathematics, Sequences and Series, JEE Main
If $f(n)=\frac{1}{n}[(n+1)(n+2)(n+3) \ldots(2 n)]^{\frac{1}{n}}$. then $\lim _{n \rightarrow \infty} f(n)=$

Options:
A $\frac{4}{e}$
B $\log \left(\frac{4}{e}\right)$
C $\frac{2}{e}$
D $\log \left(\frac{2}{e}\right)$

The correct answer is:
$\frac{4}{e}$

Solution:
$$
\begin{aligned}
& \text { Let } A=\lim _{n \rightarrow \infty} \frac{1}{n}[(n+1)(n+2)(n+3) \ldots(2 n)]^{\frac{1}{n}} \\
& =\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \ldots\left(1+\frac{n}{n}\right)\right]^{\frac{1}{n}} \\
& \therefore \log A=\lim _{n \rightarrow \infty} \sum_{\Gamma}^n \frac{1}{n} \log \left(1+\frac{r}{n}\right) \\
& =\int_0^1 \log (1+x) d x=[x \log (1+x)]_0^2-\int_0^1 \frac{x}{1+x} d x \\
& =\log 2-\int_0^1\left(1-\frac{1}{1+x}\right) d x \\
& =\log 2-[x-\log (1+x)]_0^2 \\
& =\log 2-[1-\log 2] \\
& =\log 2-1+\log 2=\log 4-1 \\
& \therefore \quad \log A=\log 4-1 \\
& \Rightarrow \log \left(\frac{A}{4}\right)=-1 \Rightarrow \frac{A}{4}=e^1 \\
& \Rightarrow \quad A=\frac{4}{e} \Rightarrow \lim _{n \rightarrow \infty} f(n)=\frac{4}{e} \\
&
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
If fn=tantan-111+2+tan-111+6+tan-111+12+...+tan-111+n(n+1)
then f2021=

Options:
A 20202022
B 20222024
C 20212023
D 20192021

The correct answer is:
20212023

Solution:
Given fn=tantan-111+2+tan-111+6+tan-111+12+...+tan-111+n(n+1)

⇒fn=tantan-12-11+2×1+tan-13-21+3×2+tan-14-31+4×3+...+tan-1n+1-n1+n(n+1)

Using tan-1a-tan-1b=tan-1a-b1+ab,

⇒fn=tantan-12-tan-11+tan-13-tan-12+tan-14-tan-13+...+tan-1n+1-tan-1n

⇒fn=tantan-1n+1-tan-11

⇒fn=tantan-1n+1-11+1n+1

⇒fn=tantan-1nn+2=nn+2.

⇒f2021=20212023.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $f_n(x), g_n(x), h_n(x), n=1,2,3$ are polynomials in $x$ such that $f_n(a)=g_n(a)=h_n(a), n=1,2,3$ and $F(x)=\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| . \quad$ Then $F^{\prime}(a)$ is equal to

Options:
A 0
B $f_1(a) g_2(a) h_3(a)$
C 1
D None of these

The correct answer is:
0

Solution:
We have $F(x)=\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right|$
$\begin{aligned} \therefore F^{\prime}(x)=\left|\begin{array}{lll}f'_1(x) & f'_2(x) & f'_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| & +\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1^{\prime}(x) & g'_2(x) & g_3^{\prime}(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| +\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1^{\prime}(x) & h_2^{\prime}(x) & h_3^{\prime}(x)\end{array}\right|\end{aligned}$
$\Rightarrow F(a)=0\left(\right.$ since $\left.f_n(a)=g_n(a)=h_n(a), \quad n=1,2,3\right)$
Therefore two rows in each determinant become identical on putting $x=a$.

Topic: Mathematics, Differentiation, JEE Main
If f:N→Z defined as fn=n-12:if n is odd-n2:if n is even and g:N→N defined as gn=n--1n, then fog is (where, N is the set of natural numbers and Z is the set of integers)

Options:
A one-one and onto
B one-one and into
C many-one and onto
D many-one and into

The correct answer is:
one-one and onto

Solution:
fog: N→Z
fg1=f2=-1, fg3=f4=-2
fg2=f1=0, fg4=f3=1
Clearly, fgn=-1 n=1-2 n=3-3 n=5: :: : and fgn=0 n=21n=42n=6::
∴ fog is one-one & onto

Topic: Mathematics, Functions, JEE Main
If f: Râ0,Ï2, fx=sin-140x2+x+Î» is a surjective function, then the value of Î» is equal to

Options:

The correct answer is:
40.25

Solution:
For fx to be surjective,
0<40x2+x+λ≤1⇒∞>x2+x+λ40≥1
Now, x2+x+λ≥40
x2+x+λ=x2+x+14+λ-14
=x+122+λ-14
Now, for x2+x+λ to be minimum 40
λ-14=40⇒λ=40.25

Topic: Mathematics, Functions, JEE Main
If f:RâÏ3,Ï defined by fx=cos-1Î»-x2x2+2 is a surjective function, then the value of Î» is equal to

Options:
A 0
B 3
C 2
D 1

The correct answer is:
1

Solution:
Range of fx∈π3,π
π3≤cos-1λ-x2x2+2<π
12≥λ-x2x2+2>-1⇒12≥λ+2-x2+2x2+2>-1
12≥λ+2x2+2-1  > -1⇒32≥λ+2x2+2>0 … 1
Now, 2≤x2+2<∞
12≥1x2+2>0 … 2
On comparing 1 and 2, we get,
λ+2=3⇒λ=1

Topic: Mathematics, Functions, JEE Main
If f:R→π3,π defined by fx=cos-1λ-x2x2+3 is a surjective function, then λ is equal to

Options:

The correct answer is:
1.5

Solution:
Range of f(x)∈π3,π
π3≤cos-1λ-x2x2+3<π
⇒12≥λ-x2x2+3>-1⇒12≥λ+3-x2+3x2+3>-1
12≥λ+3x2+3-1>-1⇒32≥λ+3x2+3>0      …….i
Now, 3≤x2+3<∞
⇒13≥1x2+3>0…….ii
On comparing i and ii, we get,
λ+3=92⇒λ=32

Topic: Mathematics, Functions, JEE Main