datavorous/entrance-exam-dataset huggingface dataset:
If $f(x)=\log _{x^{2}}(\log x)$, then $f^{\prime}(x)$ at $x=e$ is

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

The correct answer is:
$\frac{1}{2 e}$

Solution:
We have, $f(x)=\log _{x^{2}}(\log x)$
$\Rightarrow \quad f(x)=\frac{1}{2}\left[\log _{x}(\log x)\right] \quad\left[\because \log _{a^{n}} b=\frac{1}{n} \log _{a} b\right]$
$\Rightarrow \quad f(x)=\frac{1}{2}\left[\frac{\log (\log x)}{\log x}\right] \quad\left(\because \log _{a} b=\frac{\log b}{\log a}\right)$
$\Rightarrow f^{\prime}(x)=\frac{1}{2}\left[\frac{\log (x) \frac{d}{d x}[\log (\log (x))]-\log (\log x) \frac{d}{d x} \log x}{(\log x)^{2}}\right]$
$=\frac{1}{2}\left[\frac{\log (x) \frac{1}{\log x} \times \frac{1}{x}-\left(\log (\log x) \frac{1}{x}\right)}{(\log x)^{2}}\right]$
$=\frac{1}{2}\left[\frac{\frac{1}{x}-\frac{\log (\log (x))}{x}}{(\log x)^{2}}\right]$
$\Rightarrow \quad f^{\prime}(x)=\frac{1}{2}\left[\frac{1-\log \log (x)}{x(\log x)^{2}}\right]$
$\Rightarrow \quad f^{\prime}(e)=\frac{1}{2}\left[\frac{1-\log \log (e)}{e(\log e)^{2}}\right]=\frac{1}{2}\left[\frac{1-0}{e}\right]=\frac{1}{2 e}$

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\log _x(\log x)$, then $f(x)$ at $x=e$ is

Options:
A $e$
B $\frac{1}{e}$
C 1
D None of these

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

Solution:
Step 1: Solve for the first derivative of the given function
The given function is $f\left({ }^x\right)=\log _x(\log x)$.
$\Rightarrow f\left({ }^x\right)=\frac{\log (\log x)}{\log x} \quad\left[\because \log _{\mathrm{a}} b=\frac{\log b}{\log \mathrm{a}}\right]$
Differentiate both sides of the equation with respect to $x$.
$\begin{array}{ll}\Rightarrow \frac{\mathrm{d}}{\mathrm{d} x}[f(x)]=\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{\log (\log x)}{\log x}\right] \\
\Rightarrow f \prime(x)= \frac{\log x \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\log (\log x)]-\log (\log x) \cdot \frac{\mathrm{d}}{\mathrm{d} x}(\log x)}{(\log x)^2} \left[\because \frac{\mathrm{d}}{\mathrm{d x}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)= \frac{\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^2}\right] \\
=\frac{\log x \cdot \frac{1}{\log x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log x)-\log (\log x) \cdot \frac{1}{2}}{(\log x)^2}
\end{array}$
$\begin{aligned} & {\left[\because \frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{g}(\mathrm{x}))]=\mathrm{f}^{\prime}[\mathrm{g}(\mathrm{x})] \cdot \mathrm{g}^{\prime}(\mathrm{x})\right]} \\
& =\frac{\frac{1}{x}-\frac{\log (\log x)}{x}}{(\log x)^2} \\
& {\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\log x)=\frac{1}{\mathrm{x}}\right]} \\
& =\frac{\frac{1-\log (\log x)}{x}}{(\log x)^2} \end{aligned}$
$\begin{aligned}
& =\frac{\frac{1}{x}-\frac{\log (\log x)}{x}}{(\log x)^2} \left[\because \frac{\mathrm{d}}{\mathrm{d x}}(\log \mathrm{x})=\frac{\mathrm{1}}{\mathrm{x}}\right]\end{aligned}$
$\begin{aligned} = & \frac{\frac{1-\log (\log x)}{x}}{(\log x)^2}=\frac{1-\log (\log x)}{x(\log x)^2}\end{aligned}$
Step 2: Solve for the required value
Substitute $x=e$ in the above equation.
$\begin{aligned} \Rightarrow f \prime(e)= & \frac{1-\log (\log e)}{e(\log e)^2} \\
& =\frac{1-\log (1)}{e(1)^2} \\
& =\frac{1-0}{e} =\frac{1}{e}\end{aligned}$
Therefore, the value of $f \prime\left({ }^x\right)$ at $x=e$ is $\frac{1}{e}$.
Hence, option(B) is the correct option.

Topic: Mathematics, Differentiation, JEE Main
If fx=loge1-x1+x, x< 1, then f2x1+x2 is equal to

Options:
A fx2
B 2fx2
C –2fx
D 2fx

The correct answer is:
2fx

Solution:
fx=ln1-x1+x

⇒f2x1+x2=ln1-2x1+x21+2x1+x2

=ln1+x2-2x1+x2+2x

=ln1-x1+x2

=2ln1-x1+x

=2ln1-x1+x      ∵x<1

=2fx

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If fx=log⁡sec2⁡xcot2⁡xforx≠0Kforx=0 is continuous at x=0 then K is

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

The correct answer is:
1

Solution:
Since f(x) is continuous at x=0
⇒ f0=limxâ0â¡logâ¡sec2â¡xcotâ¡2x
⇒ k= limxâ0â¡cot2â¡xâlogâ¡(1+tan2â¡x)
=limxâ0â¡logâ¡(1+tan2â¡x)tan2â¡x
⇒ k=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If â«fxlogâ¡ sinâ¡ xdx=logâ¡[logâ¡sinâ¡ x]+c, then fx=

Options:
A cot⁡x
B tanx
C sec⁡x
D cosec⁡x

The correct answer is:
cot⁡x

Solution:
Given â«fxlogâ¡ sinâ¡ xdx=logâ¡ [logâ¡sinâ¡ x]+C
By using anti differentiation method, we will get
ddxlogâ¡ logâ¡sinâ¡ x+c
= 1logâ¡ sinâ¡ xâ1sinâ¡ xâcosâ¡ x
= cotâ¡ xlogâ¡ sinâ¡ x
â´ fx=cotâ¡ x

Topic: Mathematics, Indefinite Integration, JEE Main
If fx=min9-x2,1+x2, âxâ-3,3, then the number of point(s) where fx is non-differentiable is/are

Options:
A 4
B 3
C 2
D 0

The correct answer is:
4

Solution:
Using graph of
y=9-x2⇒x2+y2=9 ,
y=1+x2⇒y2-x2=1
In the figure, we can see the graph of fx
Clearly, fx is non-differentiable at
x=-3, x1, x2, 3 ( 4 points)
because there are sharp points at x1, x2 and at x=3, x=–3 there are vertical tangents.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=maxâ¡xn, sinâ¡Ïx, nâN has maximum points of non-differentiability for xâ(0, 4), then n cannot be

Options:
A 4
B 2
C 5
D 6

The correct answer is:
2

Solution:
fx=maxâ¡xn, sinâ¡Ïx
Thus, for the maximum points of non-differentiability, graphs of y=xn and y=sinâ¡Ïx must intersect at maximum number of points which occurs when n>3.5

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=min⁡x-1, x, x+1, then the value of 6∫-11fxdx is

Options:

The correct answer is:
3

Solution:
∫-11fxdx= shaded area
=212base⋅height
=212⋅1⋅12=12

Topic: Mathematics, Definite Integration, JEE Main
If $f(x)=\sum_{p=1}^7 p^2 \sin ^{-1}\left(\frac{4}{5} \sin (p x)-\frac{3}{5} \cos (p x)\right)$ then the value of $\frac{d f}{d x}$ at $x=1$ is (Given that $\sin ^{-1}(\sin x)=x$ )

Options:
A 0
B 628
C 1140
D 784

The correct answer is:
784

Solution:
Given
$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \sin ^{-1}\left(\frac{4}{5} \sin (\mathrm{px})-\frac{3}{5} \cos (\mathrm{px})\right) \\
& \Rightarrow \sin ^{-1}(\sin (\mathrm{px}) \cos \alpha-\cos (\mathrm{px}) \sin \alpha \\
& \left\{\begin{array}{l}
\text { Let } \sin (\alpha)=\frac{3}{5} \\
\cos (\alpha)=\frac{4}{5}
\end{array}\right\} \\
& \Rightarrow \sin ^{-1}(\sin (\mathrm{px}-\alpha)) \\
& =(\mathrm{px}-\alpha) \\
& \text { now } \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2(\mathrm{px}-\alpha)
\end{aligned}$
$\begin{aligned}
& =x \sum_{\mathrm{p}=1}^7 \mathrm{p}^3-\alpha \sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \\
& \mathrm{f}(\mathrm{x})=\mathrm{x}\left[\frac{7(7+1)}{2}\right]^2-\alpha\left(\frac{7(7+1)(14+1)}{6}\right) \\
& \mathrm{f}^{\prime}(\mathrm{x})=(28)^2=784
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $f(x)=p x^3+q x^2+r x+t$ attains local minimum and local maximum values at $x=-2$ and $x=2$ respectively and $p$ is a root of $9 x^2-1=0$, then $\mathrm{p}+\mathrm{q}+\mathrm{r}=$

Options:
A $\frac{4}{3}$
B 4
C $\frac{11}{3}$
D $\frac{13}{3}$

The correct answer is:
$\frac{11}{3}$

Solution:
$\because f^{\prime}(x)=3 p x^2+2 q x+r$ ...(i)
Since, $f(x)$ attain local minimum and maximum at $x=-2$ and $x=2$ respectively.
$\therefore f^{\prime}(x)=k(x+2)(x-2)=k\left(x^2-4\right)$ ...(ii)
From (i) and (ii)
$\begin{aligned} & k\left(x^2-4\right)=3 p x^2+2 q x+r \\ & \Rightarrow \quad k x^2-4 k=3 p x^2+2 q x+r\end{aligned}$
Comparing both sides, we get
$k=3 p, q=0$ and $r=-4 k$
Now, $f^{\prime \prime}(x)=6 p x+2 q=6 p x \quad\{\because q=0\}$
$\Rightarrow f^{\prime \prime}(-2)=-12 p>0 \quad\{\because x=-2$ is local minimum $\}$
$\Rightarrow p < 0$
$\because p$ is a root of $9 x^2-1=0 \Rightarrow x= \pm \frac{1}{3}$
$\therefore p < 0 \Rightarrow p=-\frac{1}{3}$
So, $k=3 \times\left(-\frac{1}{3}\right)=-1$
$\Rightarrow r=-4 k=-4(-1)=4$
$\therefore p+q+r=-\frac{1}{3}+0+4=\frac{11}{3}$.

Topic: Mathematics, Application of Derivatives, JEE Main
If fx=px+qx≤2x2-5x+62<x<3ax2+bx+1x≥3 is differentiable everywhere, then p+q+1a+1b is equal to

Options:
A 7110
B 5110
C 335
D 315

The correct answer is:
7110

Solution:
∵ Continuous at x=2

⇒p2+q=22-52+6⇒q=-2p…i
Continuous at x=3

⇒a29+b3+1=32-53+6⇒9a2+3b+1=0…ii
f'x=px≤22x−52<x<32a2x+bx≥3
∵ Differentiable at x=2

⇒p=22-5⇒p=-1…..iii
∵ Differentiable at x=3

⇒2a23+b=23-5⇒6a2+b=1……iv
∴p=-1,q=4,a=23,b=-53 {from equations i,ii,iii and iv}
p+q+1a+1b=1+4+32+35=7110

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=px+q:x≤2x2-5x+6:2<x<3ax2+bx+1:x≥3 is differentiable everywhere,

then p+q+1a+1b is equal to

Options:

The correct answer is:
5.85

Solution:
Continuity at x=2⇒p2+q=22-52+6⇒q=-2p
Continuity at x=3⇒a9+b3+1=0
Differentiable at x=2⇒p=22-5⇒p=-1
Differentiable at x=3 ⇒2a3+b=23-5⇒6a+b=1
∴p=-1, q=2, a=49, b=-53
p+q+1a+1b=1+2+94+35

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=sin-12×3x1+9x, then f'-12 is equal to

Options:
A 3 loge3
B -3 loge3
C -3 loge3
D 3 loge3

The correct answer is:
3 loge3

Solution:
Given fx=sin-12×3x1+9x
Let 3x=tant⇒t=tan-13x
So, f(x)=sin-12tan⁡t1+tan2t
We know that, sin(2t)=2tan(t)1+tan2(t)
⇒f(x)=sin-1sin2t
∴f(x)=2t=2tan-13x
⇒dfxdx=21+(3x)2 ×3x.loge3
At x=12, dfdx=21+3122×312.loge3
=12×3×loge3
=3×loge3

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\frac{\cos x}{\sin ^2 x+\cos ^4 x}$, for $x \in R$, then $f(2002)$ is equal to

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

The correct answer is:
$1$

Solution:
We have,
$$
f(x)=\frac{\cos ^2 x+\sin ^4 x}{\sin ^2 x+\cos ^4 x}
$$
$\begin{aligned} & =\frac{1-\sin ^2 x+\sin ^4 x}{1-\cos ^2 x+\cos ^4 x} \\ & =\frac{1-\sin ^2 x\left(1-\sin ^2 x\right)}{1-\cos ^2 x\left(1-\cos ^2 x\right)} \\ & =\frac{1-\sin ^2 x \cos ^2 x}{1-\cos ^2 x \sin ^2 x}=1 \\ \therefore \quad f(2002) & =1\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If fx=sin2xcx+x4x+a2-a; xâ 0, a<0, câ 0b; x=0, bâ 0

and fx is continuous at x=0, then the value of bc is equal to

Options:

The correct answer is:
2

Solution:
∵fx is continuous at x=0
∴limx→0fx=f0
limx→0sin⁡2xcx+x4x+a2-ax+a2+ax+a2+a=b
limx→02c+xx+a2+a4x+a2-a2=b⇒2c+0=b (as a<0)
⇒bc=2

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=sina+2x+sinxx;x<0b;x=0x+3x21/3-x1/3x1/3;x>0 is continuous at x=0 , then a+2b is equal to:

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

The correct answer is:
0

Solution:
f0-=limx→0-sinaxcos2xx+sin2xcosaxx+sinxx

=limx→0-acos2x+2cosax+1

=a+3

Also, f0=b
f0+=limx→0x13x131+3x13-1x , this is 00form

Apply LH Rule

⇒f0+=1

∴ a=-2

b=1

∴a+2b=0

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If fx=sin(p+1)x+sinxx,x<0q,x=0x+x2-xx3/2,x>0 is continuous at x=0 , then the ordered pair (p, q) is equal to:

Options:
A -32,-12
B -12,32
C 52,12
D -32,12

The correct answer is:
-32,12

Solution:
∵ fx=sin⁡p+1x +sinxxqx2+x-xx3/2 :x<0:x=0:x>0

∵f is continuous at x=0

L.H.L=

limh→0 sinp+1-h+sin(-h)-h=limh→0sinp+1h+sinhh

=limh→0sinp+1hhp+1p+1+limh→0sinhh=p+1+1

=p+2

R.H.L= limh→0 h2+h- hh32× h2+h+hh2+h+h=limh→0h2+h-hh32 h12h+1+1

limh→011+h+1=12

⇒L.H.L.=R.H.L.=f0⇒p+2=12=q
=p=-32 & q=12

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If fx=sinx+cosx and gx=x2-1, then gfx is invertible in the domain

Options:
A 0,π2
B -π4,π4
C -π2,π2
D 0,π

The correct answer is:
-π4,π4

Solution:
Given, fx=sinx+cosx
gx=x2-1
gfx=fx2-1=sinx+cosx2-1=sin2x
But sin2x is invertible in a case when -Ï2â¤2xâ¤Ï2
â -Ï4â¤xâ¤Ï4

Topic: Mathematics, Functions, JEE Main
If fx satisfies fx+f3-x=3 âxâR, then the value of integral I=â«3494ffxdx is equal to

Options:
A 3
B 6
C 94
D 92

The correct answer is:
94

Solution:
Applying a+b-x property and adding, we get,
2I=â«3494ffx+ff3-xdx
=â«3494ffx+f3-fxdx
=â«3494fÎ»+f3-Î»dx
=â«34943dx
â2I=394-34
âI=94

Topic: Mathematics, Definite Integration, JEE Main
If $f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right)$, then $f^{\prime}\left(-\frac{1}{2}\right)$ equals.

Options:
A
$\sqrt{3} \log _e \sqrt{3}$

B
$-\sqrt{3} \log _e \sqrt{3}$

C
$-\sqrt{3} \log _e 3$

D
$\sqrt{3} \log _e 3$

The correct answer is:
$\sqrt{3} \log _e \sqrt{3}$

Solution:
Since $f(x)=\sin \left(\frac{2 \times 3^x}{1+9^x}\right)$
Suppose $3^x=\tan t$
$$
\begin{aligned}
&\Rightarrow f(x)=\sin ^{-1}\left(\frac{2 \tan t}{1+\tan ^2 t}\right) \\
&=\sin ^{-1}(\sin 2 t)=2 t=2 \tan ^{-1}(3 x) \\
&\text { So, } f^{\prime}(x)=\frac{2}{1+\left(3^x\right)^2} \times 3^x \cdot \log _e 3 \\
&\therefore \quad f^{\prime}\left(-\frac{1}{2}\right)=\frac{2}{1+\left(3^{-\frac{1}{2}}\right)^2} \times 3^{-\frac{1}{2}} \cdot \log _e 3 \\
&=\frac{1}{2} \times \sqrt{3} \times \log _e 3=\sqrt{3} \times \log _e \sqrt{3}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, then $f(x)$ is
differentiable on

Options:
A $[-1,1]$
B $R-\{-1,1\}$
C $R-(-1,1)$
D None of these

The correct answer is:
$R-\{-1,1\}$

Solution:
Given, $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
On differentiating w.r.t. $x$, we get $\begin{aligned} f^{\prime}(x) &=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)} \\ &=\frac{1+x^{2}}{\sqrt{\left(1-x^{2}\right)^{2}}} \times \frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}} \end{aligned}$
$=\frac{2}{1+x^{2}} \times \frac{1-x^{2}}{\left|1-x^{2}\right|}$
$=\left\{\begin{array}{l}\frac{2}{1+x^{2}}, \text { if }|x| < 1 \\ -\frac{2}{1+x^{2}}, \text { if }|x|>1\end{array}\right.$
$\therefore f^{\prime}(x)$ does not exist for $|x|=1, i e, x=\pm 1$
Hence, $f(x)$ is differentiable on $R-\{-1,1\}$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\sin ^2 x$ and the composite function $g\{f(x)\}=\sin x \mid$, then the function $g(x)$ is equal to

Options:
A $\sqrt{x-1}$
B $\sqrt{x}$
C $\sqrt{x+1}$
D $-\sqrt{x}$

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

Solution:
$\begin{aligned} & (g \circ f)(x)=|\sin x| \text { and } f(x)=\sin ^2 x \\ & \Rightarrow g\left(\sin ^2 x\right)=|\sin x| ; \therefore g(x)=\sqrt{x}\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If fx=sincos-11-22x1+22x and its first derivative with respect to x is -baloge2 when x=1, where a and b are integers, then the minimum value of a2-b2 is _______.

Options:

The correct answer is:
481

Solution:
Given fx=sincos-11-22x1+22x at x=1;22x=2x2

For sincos-11-2x21+2x2

Let tan-12x=θ;θ∈-π2,π2

∴sincos-1cos2θ=sin2θ

If x>1⇒π2>θ>π4∴ π>2θ>π2

=2sinθcosθ=2tanθ1+tan2θ

Hence, fx=2·2x1+22x

∴f'x=1+22x2.2xln2-22x·2·ln2·2·2x1+22x

∴f'1=20ln2-32ln225=-1225ln2

So, a=25, b=12.

Then, a2-b2=252-122

=625-144

=481

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If $f^{\prime}(x)=\sin (\log x)$ and $y=f\left(\frac{2 x+3}{3-2 x}\right)$, then $\frac{d y}{d x}$ equals

Options:
A
$\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$

B
$\frac{12}{\left(3-2 x^2\right)}$

C
$\frac{12}{\left(3-2 x^2\right)} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$

D
$\frac{12}{\left(3-2 x^2\right.} \cos \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$

The correct answer is:
$\frac{12}{\left(3-2 x^2\right)} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$

Solution:
Let $f^{\prime}(x)=\sin [\log x]$ and $y=f\left(\frac{2 x+3}{3-2 x}\right)$
Now, $\frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$
$$
\begin{aligned}
& =\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \frac{[(6-4 x-)-4 x-6]}{\left(3-2 x^2\right)} \\
& =\frac{12}{\left(3-2 x^2\right.} \cdot \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2012 (12 May Online)
If $f(x)=\sin (\sin x)$ and $f^{\prime \prime}(x)+\tan x f^{\prime}(x)+g(x)$ $=0$, then $g(x)$ is :

Options:
A
$\cos ^2 x \cos (\sin x)$

B
$\sin ^2 x \cos (\cos x)$

C
$\sin ^2 x \sin (\cos x)$

D
$\cos ^2 x \sin (\sin x)$

The correct answer is:
$\cos ^2 x \sin (\sin x)$

Solution:
$$
\begin{aligned}
& \text { } f(x)=\sin (\sin x) \\
& \Rightarrow f^{\prime}(x)=\cos (\sin x) \cdot \cos x \\
& \Rightarrow f^{\prime \prime}(x)=-\sin (\sin x) \cdot \cos ^2 x+\cos (\sin x) \text {. } \\
& (-\sin x) \\
& =-\cos ^2 x \cdot \sin (\sin x)-\sin x \cdot \cos (\sin x) \\
& \text { Now } f^{\prime \prime}(x)+\tan x . f^{\prime}(x)+g(x)=0 \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sin x)+\sin x \cdot \cos (\sin x) \\
& -\tan x \cdot \cos x \cdot \cos (\sin x) \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sin x) \text {. } \\
&
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2013 (23 Apr Online)
If fx=sin⁡x, gx=cos⁡x and hx=coscos⁡x, then the integral Ι=∫fgx.fx.hxdx simplifies to -λsin2⁡cos⁡x+C (where, C is the constant of integration). The value of λ is equal to

Options:

The correct answer is:
0.50

Solution:
Given integral, Î=â«sincosâ¡x.sinâ¡x.coscosâ¡xdx
Let, sincosâ¡x=tâ-coscosâ¡x.sinâ¡xdx=dt
â´Î=â«-tdt=-t22+C
=C-sin2cosâ¡x2

Topic: Mathematics, Indefinite Integration, JEE Main
If fx=tan-12x1+22x+1, then ∑r=09fr is

Options:
A tan-1⁡1024
B tan-1⁡10231024
C tan-1⁡10231025
D None of these

The correct answer is:
tan-1⁡10231025

Solution:
Let, y=tan-12x+1-2x1+2x⋅2x+1=tan-1⁡2x+1-tan-1⁡2x
∑r=09fr=tan-1⁡210-tan-1⁡1=tan-1⁡10231025

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If fx=tan-1lne2/x3lne2x3+tan-1lne2x3lne/x6 ∀x≥e, then the incorrect statement is

Options:
A fx is a constant function
B fx≥0
C fx is an even function
D fx≥π

The correct answer is:
fx is an even function

Solution:
We have,
fx=tan-12-3ln⁡x2+3ln⁡x+tan-12+3ln⁡x1-6ln⁡x
=tan-12-tan-13ln⁡x+tan-12+tan-13ln⁡x+π
Using, tan-1A-tan-1B=tan-1A-B1+AB and

tan-1A+tan-1B=tan-1A+B1-AB+π ∵AB>1, we get,
fx=tan-12+tan-12+π= constant

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If fx=tan-1x2+4x+sin-1x2+4x+1, then

Options:
A domain of fx contains 3 integers only
B range of fx has two elements only
C fx is a constant function ∀x∈R
D fx contains only two elements in its domain

The correct answer is:
fx contains only two elements in its domain

Solution:
∵ x2+4x≥0 for the first term
and 0≤x2+4x+1≤1 for the second term
∴ by both the results, there is only one possibility
x2+4x=0
⇒x=0,-4
∴ fx=tan-1⁡0+sin-1⁡1=π2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $f(x)=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$, then $f^{\prime}\left(\frac{\pi}{3}\right)=$

Options:
A $\frac{1}{2(1+\cos x)}$
B $\frac{1}{2}$
C $\frac{1}{4}$
D None of these

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

Solution:
$\begin{aligned} & f(x)=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\tan ^{-1}\left[\tan \frac{x}{2}\right]=\frac{x}{2} \\ & \Rightarrow f(x)=\frac{1}{2} . \text { Hence } f^{\prime}\left(\frac{\pi}{3}\right)=\frac{1}{2} .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If f'x=tan-1⁡sec⁡x+tan⁡x,-π2<x<π2 and f0=0 , then f1 is equal to:

Options:
A π+14
B 14
C π-14
D π+24

The correct answer is:
π+14

Solution:
f'x=tan-1⁡sec⁡x+tan⁡x=tan-1⁡1+sin⁡xcos⁡x=tan-1⁡1-cos⁡π2+xsin⁡π2+x

⇒f'(x)=tan-1⁡2sin2⁡π4+x22sin⁡π4+x2cos⁡π4+x2

⇒f'xdx=π4+x2dx

⇒fx=π4x+x24+c

∵f0=0⇒c=0

So, f1=π+14

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If fx=tan1°x+loge123x loge1234-tan1°, x>0, then the least value of ffx+ff4x is

Options:
A 0
B 8
C 2
D 4

The correct answer is:
2

Solution:
Given,

fx=tan1°x+loge123x loge1234-tan1°, x>0

Now, let tan1°=a, loge123=b and loge1234=c

So, fx=ax+bcx-a

⇒ffx=afx+bcfx-a

⇒ffx=aax+bcx-a+bcax+bcx-a-a

⇒ffx=a2x+ab+bcx-aacx+bc-acx-a=a2x+bcxbc+a2=x

⇒ffx+ff4x=x+4x

∵ x>0, then least value =x·4x=2

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (10 Apr Shift 1)
If $f(x)=\left\{\begin{array}{l}\tan ^{-1} x \text {, when }|x| \leq 1 \\ \frac{1}{2}(|x|-1) \text { when }|x|>1\end{array}\right.$, then the domain of $\frac{d}{d x} f(x)$ is

Options:
A $\mathrm{R}-\{-1,1\}$
B $\mathrm{R}-(-1,1)$
C $\mathrm{R}-[-1,1]$
D $\mathrm{R}-\{-1\}$

The correct answer is:
$\mathrm{R}-\{-1,1\}$

Solution:
$f(x)= \begin{cases}\frac{1}{2}(-x-) & \text { if } x < -1 \\ \tan ^{-1} x & \text { if }-1 \leq x \leq 1 \\ \frac{1}{2}(x-1) & \text { if } x>1\end{cases}$
Since $\int(-1)=\frac{-\pi}{4} ; \int(1)=\frac{\pi}{4}$
$\lim _{x \rightarrow-1-} \int(x)=0$ and $\lim _{x \rightarrow 1+} \int(x)=0$
So, if is not continuous at $-1,1$, hence not differentiable at $-1,1$ Also
$f^{\prime}(x)= \begin{cases}\frac{-1}{2} & \text { if } x < -1 \\ \frac{1}{1+x^2} & \text { if }-1 < x < 1 \\ \frac{1}{2} & \text { if } x>1\end{cases}$
Thus the domain of $\int(x)$ is $R-\{-1,1\}$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 1)
If fx=tan⁡x2, x∈0,π3, then f ′ π 4 (where, . is the greatest integer function)

Options:
A is equal to 1
B is equal to 0
C does not exist
D None of these

The correct answer is:
does not exist

Solution:
LHD: f'π4=limh→0+fπ4−h−fπ4−h
=limh→0+⁡tan⁡π4-h2⁡-tan⁡π42-h
=limh→0+⁡0-1-h→∞
∴ f'π4 does not exist.

Topic: Mathematics, Limits, JEE Main
If $f(x)=\sqrt{\tan x}$ and $g(x)=\sin x \cdot \cos x$, then $\int \frac{f(x)}{g(x)} d x$ is equal to (where $C$ is a constant of integration)

Options:
A $\sqrt{\tan x}+C$
B $\frac{1}{2} \sqrt{\tan x}+C$
C $\frac{3}{2} \sqrt{\tan x}+C$
D $2 \sqrt{\tan x}+C$

The correct answer is:
$2 \sqrt{\tan x}+C$

Solution:
$f(x)=\sqrt{\tan x}, g(x)=\sin x \cdot \cos x$
Now,
$\begin{aligned} & \int \frac{f(x)}{g(x)} d x=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x=\int \frac{\sqrt{\tan x \cdot \sec ^2 x}}{\tan x} d x=\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x \\ & =2 \sqrt{\tan x}+C\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
If $f(x)=\left\{\begin{array}{r}\frac{1}{x} \sin x^2, x \neq 0 \\ 0, x=0\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 0^+} f(x) \neq 0$
B $\lim _{x \rightarrow 0^{-}} f(x) \neq 0$
C $f(x)$ is continuous at $x=0$
D None of these

The correct answer is:
$f(x)$ is continuous at $x=0$

Solution:
$f(0)=0, \lim _{x \rightarrow 0+} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} x\left[\frac{\sin x^2}{x^2}\right]=0$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{array}{r}x-1, x \lt 0 \\ \frac{1}{4}, x=0 \\ x^2, x\gt0\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 0^{+}} f(x)=1$
B $\lim _{x \rightarrow 0^{-}} f(x)=1$
C $f(x)$ is discontinuous at $x=0$
D None of these

The correct answer is:
$f(x)$ is discontinuous at $x=0$

Solution:
Clearly from curve drawn of the given function $f(x)$ is discontinuous at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=x+1;x>10 ;x=17-3x;x<1 then fâ²0 equals to

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

The correct answer is:
- 3

Solution:
As for xÂ <Â 1,Â fÂ x=7-Â 3xâÂ f′x=-Â 3âÂ f′0=-Â 3

Topic: Mathematics, Differentiation, JEE Main
If fx=x–1x–2x–3x–4x–5, then the value of f ′ 5 is equal to

Options:
A 0
B 120
C 24
D 5

The correct answer is:
24

Solution:
f ′ x = x–2 x–3 x–4 x–5 + x−1 x−3 x−4 x−5
+......+x-1x-2x-3x-4
Putting x=5, we get,
f ′ 5 =0+0+0+0+ 5−1 5−2 5−3 5−4
f ′ 5 =24

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then, for what value of $\alpha$ is $f(f(x))=x$

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

The correct answer is:
$-1$

Solution:
$\begin{aligned}
& f(f(x))=\frac{\alpha f(x)}{f(x)+1}=\frac{\alpha\left(\frac{\alpha x}{x+1}\right)}{\left(\frac{\alpha x}{x+1}+1\right)}=\frac{\alpha^2 \cdot x}{\alpha x+x+1} \\
& \quad x=\frac{\alpha^2 \cdot x}{(\alpha+1) x+1} \text { or } x\left((\alpha+1) x+1-\alpha^2\right)=0
\end{aligned}$
or $(\alpha+1) x^2+\left(1-\alpha^2\right) x=0$. This should hold for all $x$.
$\Rightarrow \alpha+1=0,1-\alpha^2=0, \quad \therefore \alpha=-1 \text {. }$

Topic: Mathematics, Functions, JEE Main
If $f(x)=x^2-1$ and $g(x)=3 x+1$, then $(g \circ f)(x)=$

Options:
A $x^2-1$
B $2 x^2-1$
C $3 x^2-2$
D $2 x^2+2$

The correct answer is:
$3 x^2-2$

Solution:
$(g \circ f)(x)=3\left(x^2-1\right)+1=3 x^2-2$

Topic: Mathematics, Functions, JEE Main