datavorous/entrance-exam-dataset huggingface dataset:
If $f(x)=3 x-5$, then $f^{-1}(x)$

Options:
A Is given by $\frac{1}{3 x-5}$
B Is given by $\frac{x+5}{3}$
C Does not exist because $f$ is not one-one
D Does not exist because $f$ is not onto

The correct answer is:
Is given by $\frac{x+5}{3}$

Solution:
Let $f(x)=y \Rightarrow x=f^{-1}(y)$.
Hence $f(x)=y=3 x-5 \Rightarrow x=\frac{y+5}{3} \Rightarrow f^{-1}(y)=x=\frac{y+5}{3}$
$\therefore f^{-1}(x)=\frac{x+5}{3}$
Also $f$ is one-one and onto, so $f^{-1}$ exists and is given by $f^{-1}(x)=\frac{x+5}{3}$

Topic: Mathematics, Functions, JEE Main
If $f(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7$, then $\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}$ is

Options:
A
$-\frac{53}{3}$

B
$\frac{53}{3}$

C
$-\frac{55}{3}$

D
$\frac{55}{3}$

The correct answer is:
$\frac{53}{3}$

Solution:
$$
\text { } \begin{aligned}
\text { Let } f^{\prime}(x) & =3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7 \\
f^{\prime}(x) & =30 x^9-56 x^7+30 x^5-63 x^2+6 x \\
f^{\prime}(1) & =30-56+30-63+6 \\
& =66-63-56=-53
\end{aligned}
$$
$$
\begin{aligned}
& \text { Consider } \lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha} \\
& =\lim _{\alpha \rightarrow 0} \frac{f^{\prime}(1-\alpha)(-1)-0}{3 \alpha^2+3} \quad \text { (By using }
\end{aligned}
$$
L'hospital rule)
$$
=\frac{f^{\prime}(1-0)(-1)}{3(0)^2+3}=\frac{-f^{\prime}(1)}{3}=\frac{53}{3}
$$

Topic: Mathematics, Limits, JEE Main, JEE Main 2012 (19 May Online)
If $f(x)=3[x]+\{x+1\}$, where $[x]$ is greatest integer function of $x$ and $\{x\}$ is fractional part function of $x$, then $f(-1.32)=$

Options:
A -4.6
B -2.6
C -7.4
D -3.4

The correct answer is:
-2.6

Solution:
$$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=3[\mathrm{x}]+5\{\mathrm{x}+1\} \\
& \mathrm{x}=-1.32 \Rightarrow[\mathrm{x}]=[-1.32]=-2 \\
& \text { Also } \mathrm{x}+1=-1.32+1=-0.32 \\
& \therefore[\mathrm{x}+1]=[-0.32]=-1 \text { and }\{\mathrm{x}+1\}=0.68 \\
& \therefore \mathrm{f}(\mathrm{x})=3(-2)+5(0.68) \\
& =-6+3.4=-2.6
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Main
If $f(x)=\frac{2 x}{4+3|x|}, x \in R$, then $f^{\prime}(0)=$

Options:
A 0
B $\frac{1}{4}$
C $\frac{1}{2}$
D $\frac{3}{4}$

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

Solution:
We have, $f(x)=\frac{2 x}{4+3|x|}, x \in R$
$$
\begin{aligned}
& = \begin{cases}\frac{2 x}{4+3 x}, & \text { if } x \geq 0 \\
\frac{2 x}{4-3 x}, & \text { if } x < 0\end{cases} \\
\therefore \quad f^{\prime}(x) & = \begin{cases}\frac{(4+3 x) \cdot 2-2 x \cdot 3}{(4+3 x)^2}, & \text { if } x \geq 0 \\
\frac{(4-3 x) \cdot 2-2 x(-3)}{(4-3 x)^2}, & \text { if } x < 0\end{cases}
\end{aligned}
$$
$$
\begin{aligned}
& = \begin{cases}\frac{8+6 x-6 x}{(4+3 x)^2}, & \text { if } x \geq 0 \\
\frac{8-6 x+6 x}{(4-3 x)^2}, & \text { if } x < 0\end{cases} \\
& = \begin{cases}\frac{8}{(4+3 x)^2}, & \text { if } x \geq 0 \\
\frac{8}{(4-3 x)^2}, & \text { if } x < 0\end{cases}
\end{aligned}
$$
Now, $L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{8}{(4-3 x)^2}=\frac{8}{16}=\frac{1}{2}$
$$
\text { and } R f^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \frac{8}{(4+3 x)^2}=\frac{8}{16}=\frac{1}{2}
$$
Since, $\quad L f^{\prime}(0)=R f^{\prime}(0)$
$$
\therefore \quad f^{\prime}(0)=\frac{1}{2}
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f x = 4+ax â 4âax x , â1 â¤x<0 3x+2 xâ8 , 0 â¤x â¤1 is continuous in [ â1,1 ], then the value of a is

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

The correct answer is:
-12

Solution:
Given that fx is continuous in the interval [-1, 1], so, it will be continuous at x=0.
Now f0+= -14
f0-= limx â0-â¡2axx (4+ax + 4-ax ) (rationalizing)
= limxâ0-â¡2a4=a2
Now f0+=f(0-) (âµ f(x) is continuous at point 0)
â a2=-14 â a=-12

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx-4x+2=2x+1, x∈R-1,-2, then∫fxdx is equal to

Options:
A 12ln1-x-3x+C
B -12ln1-x-3x+C
C 12ln1-x+3x+C
D -12ln1-x+3x+C

The correct answer is:
-12ln1-x-3x+C

Solution:
We have, fx-4x+2=2x+1

Let x-4x+2=X⇒xX+2X=x-4⇒xX-1=-4-2X⇒x=22+X1-X

Putting x=22+X1-X in the given functional equation, we get

⇒fX=2×22+X1-X+1=8+4X+1-X1-X=3X+91-X=3X+31-X

Replacing X→x, we get

fx=3x+31-x

⇒∫fxdx=3∫x+31-xdx=3∫4-1-x1-xdx=3∫41-xdx-∫dx

=-12ln1-x-3x+C, where C is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2018 (15 Apr)
If $f(x)=4 x^3+3 x^2+3 x+4$, then $x^3 f\left(\frac{1}{x}\right)$ is

Options:
A $f(-x)$
B $\frac{1}{f(x)}$
C $\left(f\left(\frac{1}{x}\right)\right)^2$
D $f(x)$

The correct answer is:
$f(x)$

Solution:
$\begin{aligned} x^3 f\left(\frac{1}{x}\right)= & x^3\left[\frac{4}{x^3}+\frac{3}{x^2}+\frac{3}{x}+4\right] \\ & =4+3 x+3 x^2+4 x^3=f(x)\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If $f(x)=\left\{\begin{array}{lll}4 x+3, & \text { if } & 1 \leq x \leq 2 \\ 3 x+5, & \text { if } & 2 \lt x \leq 4\end{array}\right.$ then $\int_1^4 f(x) d x=$

Options:
A 80
B 20
C -20
D 37

The correct answer is:
37

Solution:
$\begin{aligned} & \int_1^4 f(x) d x=\int_1^2(4 x+3) d x+\int_2^4(3 x+5) d x \\ & =\left|2 x^2+3 x\right|_1^2+\left|\frac{3 x^2}{2}+5 x\right|_2^4=37\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
If $f\left(x^{5}\right)=5 x^{3}$, then $f^{\prime}(x)$ is equal to

Options:
A $\frac{3}{\sqrt[5]{\mathrm{x}^{2}}}$
B $\frac{3}{\sqrt[5]{x}}$
C $\frac{3}{x}$
D $\sqrt[5]{x}$

The correct answer is:
$\frac{3}{\sqrt[5]{\mathrm{x}^{2}}}$

Solution:
Given, $\mathrm{f}\left(\mathrm{x}^{5}\right)=5 \mathrm{x}^{3}$
Let $\quad \mathrm{x}^{5}=\mathrm{y} \Rightarrow \mathrm{x}^{3}=\mathrm{y}^{3 / 5}$
$\therefore \quad \mathrm{f}(\mathrm{y})=5 \mathrm{y}^{3 / 5}$
or $\quad \mathrm{f}(\mathrm{x})=5 \mathrm{x}^{3 / 5}$
On differentiating w.r.t. x, we get
$$
\begin{aligned}
f^{\prime}(x) &=5 \cdot \frac{3}{5} x^{-2} \\
&=\frac{3}{\sqrt[5]{x^{2}}}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If fx=5x+4tanâ¡xx xâ 0k x=0, then fx is continuous at x=0 for

Options:
A k=9
B k=-1
C no values of k
D k=2

The correct answer is:
no values of k

Solution:
limxâ0-â¡fx=limxâ0-â¡5x+4tanâ¡xx
=limxâ0-â¡5-x+4tanâ¡xx
=limxâ0-â¡-5xx+limxâ0-â¡4tanâ¡xx
=-5+4=-1
and limxâ0+â¡fx=limxâ0+â¡5x+4tanâ¡xx
=limxâ0+â¡5xx+4tanâ¡xx
=5+4=9
limxâ0-â¡fxâ limxâ0+â¡fx
â´fx is not continuous at x=0, for any value of k

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=∫5x8+7x6x2+1+2x72dx, x≥0, and f0=0, then the value of f(1) is

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

The correct answer is:
14

Solution:
We have, fx=∫5x8+7x6x2+1+2x72 dx
=∫5x8+7x6x141x5+1x7+22dx
=∫5x6+7x81x5+1x7+22dx
Put, 1x5+1x7+2=t
⇒-5x6-7x8dx=dt
∴fx=-∫dtt2=1t+c, where c is the constant of integration.
=x7x2+1+2x7+c
Now, f0=0⇒ c=0
∴ f1=112+2+2.17=14

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If $f(x)=\left\{\begin{array}{r}
\frac{\sin 2 x}{5 x}, \text { when } x \neq 0 \\
k, \text { when } x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ will be

Options:
A $1$
B $\frac{2}{5}$
C $-\frac{2}{5}$
D None of these

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

Solution:
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin 2 x}{2 x .5}=\frac{2}{5}=k$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=a+bcosx+csinxx2;x>09;x≤0 is continuous at x=0, then the value of a+b5 is ______. If your answer is K, mark 10K as answer after rounding off to nearest integer.

Options:

The correct answer is:
72

Solution:
For fx to be continuous at x=0
limxâ0fx=f0
limxâ0a+bcos x+csin xx2=9
limxâ0a+b1-x22!+x24!-....+cx-x33!+x55!-....x2=9
limxâ0a+b+cx+-b2x2+-c6x3.....x2=9
for the above limit to satisfy
a+b=0,
c=0 and
-b2=9âb=-18, c=0, a=18

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=a+bcosx+csinxx2;x>09;x≤0 is continuous at x=0, then the value of a+b5 is

Options:

The correct answer is:
7.20

Solution:
For fx to be continuous at x=0
limxâ0fx=f0
limxâ0a+bcos x+csin xx2=9
limxâ0a+b1-x22!+x24!-....+cx-x33!+x55!-....x2=9
limxâ0a+b+cx+-b2x2+-c6x3.....x2=9
for the above limit to satisfy
a+b=0,
c=0 and
-b2=9âb=-18, c=0, a=18

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=a+cos-1x+b:xâ¥1-x:x<1 is differentiable at x=1, then the value of b-a is equal to

Options:
A 0
B 1
C -1
D π2

The correct answer is:
π2

Solution:
Since, f(x) is differentiable, it must be continuous at x=1
∴f1-=f1+=f1
-1=a+cos-11+b ... 1
f'x=−11−x+b2:x≥1−1:x<1
Now, for fx to be differentiable at x=1
f ′ 1 − = f ′ 1 +
-11-1+b2=-1⇒1-1+b2=1⇒b=-1 ... 2
From equation 1 & 2
-1=a+π2⇒a=-π2-1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=a \cos (x+c)+d$, then range of $f(x)$ is

Options:
A $[d+a, d+2 a]$
B $[a-d, a+d]$
C $[d+a, a-d]$
D $[d-a, d+a]$

The correct answer is:
$[d-a, d+a]$

Solution:
$f(x)=a \cos (x+c)+d$ ....(i)
For minimum $\cos (x+c)=-1$
from (i), $f(x)=-a+d=(d-a)$
For maximum $\cos (x+c)=1$
from (i), $f(x)=a+d=(d+a)$

Topic: Mathematics, Functions, JEE Main
If $f(x)=a|\sin x|+b e^{|x|}+c|x|^3$, where $a, b, c \in R$, is differentiable at $x=0$, then

Options:
A
$a=0, b$ and $c$ are any real numbers

B
$c=0, a=0, b$ is any real number

C
$b=0, c=0, a$ is any real number

D
$a=0, b=0, c$ is any real number

The correct answer is:
$a=0, b=0, c$ is any real number

Solution:
$|\sin x|$ and $e^{|x|}$ are not differentiable at $x=0$ and $|x|^3$ is differentiable at $x=0$. $\therefore$ for $f(x)$ to be differentiable at $x=0$, we must have $a=0, b=0$ and $c$ is any real number.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2012 (26 May Online)
If fx=a+tan-1⁡x-b;x≥1x2;x<1 is differentiable at x=1, then 4a-b can be

Options:
A 0
B 1
C -1
D π

The correct answer is:
π

Solution:
Also fx must be continuous at x=1
∴f1-=f1+=f1
12=a+tan-1⁡1-b…i
f ′ x = 1 1+ x−b 2 ; x≥1 1 2 ; x<1
Now for fx to be differentiable at x=1
f ′ 1 − = f ′ 1 +
11+1-b2=12⇒1+1-b2=2

⇒b=2 or 0…ii
From equation i and ii, we get
12=a±π4⇒a=12-π4 or 12+π4
a,b≡12-π4,0 or 12+π4,2
Hence, 4a-b=2-π,π

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)= \begin{cases}a x^2+b x+1 & \text { if }|2 x-3| \geq 2 \\ 3 x+2 & \text {; if } \frac{1}{2}

MathematicsContinuity and DifferentiabilityMHT CETMHT CET 2022 (06 Aug Shift 2)

Options:

A $\frac{23}{5}$
B $\frac{1}{5}$
C $\frac{13}{5}$
D $\frac{31}{5}$

Solution:

1892 Upvotes

Verified Answer

The correct answer is:
$\frac{23}{5}$

$\begin{aligned} & f(x)=\left\{\begin{array}{ccc}a x^2+b x+1 & ; & |2 x-3| \geq 2 \\ 3 x+2 & ; & \frac{1}{2}for continuity at $x=\frac{1}{2}$
for continuity at $x=\frac{5}{2}$
from (1) and (2)
$a=-\frac{4}{5}$ and $b=\frac{27}{5}$
$a+b=\frac{-4+27}{5}=\frac{23}{5}$

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Options:
A $\frac{23}{5}$
B $\frac{1}{5}$
C $\frac{13}{5}$
D $\frac{31}{5}$

The correct answer is:
$\frac{23}{5}$

Solution:
$\begin{aligned} & f(x)=\left\{\begin{array}{ccc}a x^2+b x+1 & ; & |2 x-3| \geq 2 \\ 3 x+2 & ; & \frac{1}{2}for continuity at $x=\frac{1}{2}$
for continuity at $x=\frac{5}{2}$
from (1) and (2)
$a=-\frac{4}{5}$ and $b=\frac{27}{5}$
$a+b=\frac{-4+27}{5}=\frac{23}{5}$

Topic: Mathematics, Continuity and Differentiability, MHT CET, MHT CET 2022 (06 Aug Shift 2)
If fx=ax2+bx+c,f-1>12,f1<-1 and f-3<-12, then

Options:
A a=0
B a<0
C a>0
D Sign of a can not be determined

The correct answer is:
a<0

Solution:
Clearly, parabola is opening downwards âa<0.

Topic: Mathematics, Quadratic Equation, JEE Main
If f(x)=a−(x−3)8/9, then the maximum value of f( x ) is

Options:
A 3
B a−3
C a
D None of these

The correct answer is:
a

Solution:
∵ fx=a-x-38/9
∴ f 'x=0-89x-3-1/9
At x=3, f 'x is not defined.
hence sign scheme of f 'x is

∴fx is maximum at x=3
Hence maximum value of fx is equal to "a"

Topic: Mathematics, Application of Derivatives, JEE Main
If fx=ax+b, if x≤1ax2+c, if 1<x≤2dx2+1x, if x≥2 is differentiable on R, then ad-bc=

Options:
A 0
B 1
C -1
D 2

The correct answer is:
-1

Solution:
Given: fx=ax+b, if x≤1ax2+c, if 1<x≤2dx2+1x, if x≥2 is differentiable on R

So fx should be continuous on R

Now continuity at x=1

f1-=f1+=f1

⇒a+b=a+c⇒b=c ...i

Now continuity at x=2

f2-=f2+=f2

⇒4a+c=4d+12=8a+2c=4d+1 ...ii

Now

f'x=a, if x≤12ax, if 1<x≤2dx2-1x2, if x≥2

It is differentiable on R

So f'1-=f'1+⇒a=2a⇒a=0 ...iii

f'2-=f'2+⇒4a=4d-14⇒d=14 ...iv

Using Equations ii, iii & iv

c=b=1

Now ad-bc=0-1=-1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)$ and $g(x)$ are twice differentiable. functions on (0,3)
satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x), f^{\prime}(1)=4, g^{\prime}(1)=6, \quad f(2)=3$
$g(2)=9,$ then $f(1)-g(1)$ is

Options:
A 4
B -4
C 0
D -2

The correct answer is:
-4

Solution:
According to question, $f^{\prime \prime}(x)=g^{\prime \prime}(x)$
Integrating w.r.t. $x$, we get $\begin{array}{ll} & f^{\prime}(x)=g^{\prime}(x)+c_{1} \\ \text { Put } & x=1 \Rightarrow f^{\prime}(1)=g^{\prime}(1)+C_{1} \\ \Rightarrow & 4=6+C_{1} \\ \therefore & C_{1}=-2 \\ \therefore \quad & f^{\prime}(x)=g^{\prime}(x)-2\end{array}$
Again, integrating w.r.t. $x$, we get $f(x)=g(x)-2 x+C_{2}$
$\Rightarrow \quad 3=9-4+C_{2} \Rightarrow C_{2}=-2$
$\therefore \quad f(x)=g(x)-2 x-2$
Put $x=1,$ we get $f(1)-g(1)=-2(1)-2=-4$

Topic: Mathematics, Indefinite Integration, JEE Main
If $f(x)=b e^{a x}+a e^{b x}$, then $f^{\prime \prime}(0)$ is equal to

Options:
A 0
B $2 \mathrm{ab}$
C $a b(a+b)$
D ab

The correct answer is:
$a b(a+b)$

Solution:
Given,
$$
f(x)=b e^{a x}+a e^{b x}
$$
On differentiating w.r.t. $x$, we get
$$
f^{\prime}(x)=a b e^{a x}+a b e^{b x}
$$
Again, differentiating, we get
$$
$$
\begin{aligned}
\mathrm{f}^{\prime \prime}(\mathrm{x}) &=\mathrm{a}^{2} b \mathrm{e}^{\mathrm{ax}}+\mathrm{ab}^{2} \mathrm{e}^{\mathrm{bx}} \\
\Rightarrow \quad \mathrm{f}^{\prime \prime}(0) &=\mathrm{a}^{2} \mathrm{~b}+\mathrm{ab}^{2} \\
&=\mathrm{ab}(\mathrm{a}+\mathrm{b})
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If f(x)=cos-1sinx4cos2x-1, then 1πf'π3⋅fπ10 is

Options:

The correct answer is:
0.60

Solution:
fx=cos-1sin⁡x41-sin2x-1
fx=cos-1sin⁡x3-4sin2x=cos-13sin⁡x-4sin3x
fx=cos-1sin⁡3x=π2-sin-1sin⁡3x
In the neighbourhood of x=π3⇒fx=π2-π-3x
fx=3x-π2
∴f'π3=3
In the neighbourhood of x=π10⇒fx=π2-3x
fπ10=π2-3π10=π5

Topic: Mathematics, Differentiation, JEE Main
If fx=cos-1x32-1-x-x2+x3, â0â¤xâ¤1, then the minimum value of fx is

Options:

The correct answer is:
0

Solution:
fx=cos-1xx-1-x1-x2
=cos-1xx-1-x21-x2
=cos-1x+cos-1x

At x=1,

minimum value of f(x)=0

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $f(x)=\cos ^2 x+\sec ^2 x$, then

Options:
A $f(x) \lt 1$
B $f(x)=1$
C $1 \lt f(x) \lt 2$
D $f(x) \geq 2$

The correct answer is:
$f(x) \geq 2$

Solution:
Since $\left(x-\frac{1}{x}\right)^2 \geq 0, \forall x \in R$,
we have $x^2+\frac{1}{x^2} \geq 2$ and
Hence, $f(x)=\cos ^2 x+\frac{1}{\cos ^2 x} \geq 2$.

Topic: Mathematics, Trigonometric Equations, JEE Main
If f( x )=cos( lnx), then f(x)⋅f( y )− 1 2 [ f( x y )+f( xy ) ] is equal to

Options:
A −1
B 1 2
C −2
D 0

The correct answer is:
0

Solution:
Given, f( x )=cos( lnx )
Hence, f(x)âf( y )â 1 2 [ f( x y )+f( xy ) ]
=cos(lnx)âcos(lny)â 1 2 [cos(lnxâlny)+cos(lnx+lny)]
=cos(lnx)âcos(lny)â 1 2 [(2cos(lnx)âcos(lny)]
= cos( lnx )âcos( lny )âcos( lnx )âcos( lny )=0

Topic: Mathematics, Functions, JEE Main
If $f(x)=\cos (\log x)$, then the value of $f(x) \cdot f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]$

Options:
A $1$
B $-1$
C $0$
D $\pm 1$

The correct answer is:
$0$

Solution:
$f(x)=\cos (\log x)$
Now let
$y=f(x) . f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]$
$\begin{aligned}
\Rightarrow \quad y= & \cos (\log x) \cdot \cos (\log 4)-\frac{1}{2}\left[\cos \log \left(\frac{x}{4}\right)+\cos (\log 4 x)\right] \\
\Rightarrow y= & \cos (\log x) \cos (\log 4) -\frac{1}{2}[\cos (\log x-\log 4)+\cos (\log x+\log 4)] \\
\Rightarrow y= & \cos (\log x) \cos (\log 4)-\frac{1}{2}[2 \cos (\log x) \cos (\log 4)] \\
\Rightarrow y= & 0 .
\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If $f(x)=\cos (\log x)$, then the value of $f(x) \cdot f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]$

Options:
A $1$
B $-1$
C $0$
D $\pm 1$

The correct answer is:
$0$

Solution:
$f(x)=\cos (\log x)$
Now let
$y=f(x) . f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]$
$\Rightarrow y=\cos (\log x) \cdot \cos (\log 4)-\frac{1}{2}\left[\cos \log \left(\frac{x}{4}\right)+\cos (\log 4 x)\right]$
$\Rightarrow y=\cos (\log x) \cos (\log 4)$
$-\frac{1}{2}[\cos (\log x-\log 4)+\cos (\log x+\log 4)]$
$\Rightarrow y=\cos (\log x) \cos (\log 4)-\frac{1}{2}[2 \cos (\log x) \cos (\log 4)]$
$\Rightarrow y=0$.

Topic: Mathematics, Functions, JEE Main
If $f(x)=\cos x, 0 \leq x \leq \frac{\pi}{2}$, then the real number ' $c$ ' of the mean value theorem is

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\sin ^{-1}\left(\frac{2}{\pi}\right)$
D $\cos ^{-1}\left(\frac{2}{\pi}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{2}{\pi}\right)$

Solution:
We know that
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
$\Rightarrow f^{\prime}(c)=\frac{0-1}{\pi / 2}=-\frac{2}{\pi}$
But $f^{\prime}(x)=-\sin x \Rightarrow f^{\prime}(c)=-\sin c$
From (i) and (ii), we get
$-\sin c=-\frac{2}{\pi} \Rightarrow c=\sin ^{-1}\left(\frac{2}{\pi}\right)$

Topic: Mathematics, Application of Derivatives, JEE Main
If fx=cosâ¡x-2ax+b is a function which increases for all x, then the maximum value of 2a+1 is

Options:

The correct answer is:
0

Solution:
f x =cos x â2ax+b,ââf is differentiable everywhere.
f'x=âsinxâ2a,

f'xâ¥0 for all x

â-sinx-2aâ¥0 âsinxâ¤-2a

f increases for all x

â-2aâ¥1â2a+1â¤0

Hence, the maximum value of 2a+1=0

Topic: Mathematics, Application of Derivatives, JEE Main
If $f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}\right|$, then $\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}$

Options:
A
Exists and is equal to $-2$

B
Does not exist

C
Exist and is equal to 0

D
Exists and is equal to 2

The correct answer is:
Exists and is equal to $-2$

Solution:
$\begin{aligned} & f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}\right| \\=& \cos x\left(x^2-2 x^2\right)-x(2 \sin x-2 x \tan x) \\=&-x^2 \cos x-2 x \sin x+2 x^2 \tan x \\\left.+2 x \sin x-x^2 \tan x\right) \\=& x^2 \tan x-x^2 \tan x \\ \Rightarrow & f^{\prime}(x)=2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right) \\ \therefore & \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x} \\=& \lim _{x \rightarrow 0} \frac{2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right)}{x} \\=& \lim _{x \rightarrow 0}(\tan x-\cos x)+x\left(\sec ^2 x+\sin x\right) \\=& 2(0-1)+0=-2 \\ \text { So, } \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}=-2 \end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If fx=cos⁡x.cos⁡2x.cos⁡4x .cos⁡8x .cos⁡16x, then f'π4 is (where  f'(x)=ddxf(x))

Options:
A 2
B 12
C 1
D None of these

The correct answer is:
2

Solution:
fx=2sin⁡x.cos⁡x.cos⁡2x.cos⁡4x.cos⁡8x.cos⁡16x2sin⁡x
=sin⁡32x25sin⁡x
∴f'x=132.32cos⁡32x.sin⁡x-cos⁡x.sin⁡32xsin2⁡x
∴f'π4=32.12-12.032.122=2

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=(\cos x)(\cos 2 x) \ldots(\cos n x)$ then $f^{\prime}(x)+\sum_{r=1}^n(r \tan r x) f(x)$ is equal to

Options:
A $f(x)$
B 0
C $-f(x)$
D $2 f(x)$

The correct answer is:
0

Solution:
$f(x)=(\cos x)(\cos 2 x) \ldots(\cos n x)$
$f^{\prime}(x)=-\sin x \cdot \cos 2 x \ldots \cos n x$ $+\cos x \frac{d}{d x}\{\cos 2 x \cdot \cos 3 x \ldots \cos n x\}$
$f^{\prime}(x)=-\sin x \cdot \cos 2 x \ldots \cos n x$ $+\cos x[-2 \sin 2 x \cdot \cos 3 x \ldots \cos n x$ $\left.+\cos 2 x \frac{d}{d x} \cos 3 x \cdot \cos 4 x \ldots \cos n x\right]$
$f^{\prime}(x) \Rightarrow-(\sin x \cdot \cos 2 x \ldots \cos n x)$ $-(2 \cos x \cdot \sin 2 x \cdot \cos 3 x \ldots \cos n x)$ $+\cos x \cdot \cos 2 x \frac{d}{d x}(\cos 3 x \cdot \cos 4 x \cdot \cos n x)$
$f^{\prime}(x) \Rightarrow-(\sin x \cdot \cos 2 x \ldots \cos n x)$ $-(2 \cos x \cdot \sin 2 x \ldots \cos n x)$ $-(3 \cos x \cdot \cos 2 x \cdot \sin 3 x \ldots \cos n x)$ $-(n \cos x \cdot \cos 2 x \ldots \sin n x)$
So,
$\Rightarrow f^{\prime}(x)+\sum_{r=1}^n(r \tan r x) f(x)$
$=f^{\prime}(x)+\{\tan x+2 \tan 2 x+3 \tan 3 x$ $+\ldots+n \tan n x\} f(x)$
$=f^{\prime}(x)+f(x) \tan x+2 f(x) \tan 2 x$ $+\ldots+n f(x) \tan n x$
$=f^{\prime}(x)+[(\sin x \cdot \cos 2 x \ldots \cos n x)$ $+(2 \cos x \cdot \sin 2 x \ldots \cos n x)$ $+\ldots+(n \cos x \cdot \cos 2 x \ldots \sin n x)]$
$=f^{\prime}(x)-f^{\prime}(x)$
$\Rightarrow 0$
Hence, $f^{\prime}(x)+\sum_{r=1}^n(r \tan r x) f(x)=0$

Topic: Mathematics, Differentiation, JEE Main
If fx=cos⁡x+sin⁡x and gx=x2-1, then gfx is injective in the interval

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

The correct answer is:
–π4,π4

Solution:
gfx=sin⁡x+cos⁡x2-1=sin⁡2x
Now, sin⁡2x is injective if 2x∈nπ-π2,nπ+π2
⇒x∈nπ2-π4,nπ2+π4 ∀n∈Z

Topic: Mathematics, Functions, JEE Main
If fx=cosxx12sinxx22xtanxx1, then limx→0⁡f′xx

Options:
A does not exist
B exists and is equal to -2
C exists and is equal to 0
D exists and is equal to 2

The correct answer is:
exists and is equal to -2

Solution:
Given fx=cosxx12sinxx22xtanxx1

=cosxx2-2x2-2sinxx-x+tanx2x2-x2

=-x2cosx+tanx.x2

=x2tanx-cosx

⇒ Limx→0f'xx=Limx→02xtanx-cosx+x2sec2x+sinxx

=Limx→02tanx-cosx+xsec2x+sinx

=-2

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (15 Apr)
If $f(x)=\int \operatorname{cosec}^5 x d x$, then $f\left(\frac{\pi}{4}\right)=$

Options:
A $-\frac{1}{4}[3 \sqrt{2}-5 \log (\sqrt{2}+1)]+c$
B $-\frac{1}{8}[5 \sqrt{2}-3 \log (\sqrt{2}+1)]+c$
C $-\frac{1}{8}[7 \sqrt{2}+3 \log (\sqrt{2}+1)]+c$
D $\frac{1}{8}[5 \sqrt{2}+\log (\sqrt{2}+1)]+c$

The correct answer is:
$-\frac{1}{8}[7 \sqrt{2}+3 \log (\sqrt{2}+1)]+c$

Solution:
$\begin{aligned} & f(x)=\int \operatorname{cosec}^5 x d x=\int \operatorname{cosec}^2 x \cdot \operatorname{cosec}^3 x d x \\ & =\operatorname{cosec}^3 x(-\cos x)-\int-\cot x \cdot 3 \operatorname{cosec}^2 x \\ & \quad \quad(-\operatorname{cosec} x \cot x) d x \\ & =-\cot x \operatorname{cosec}^3 x-3 \int \cot ^2 x \operatorname{cosec}^3 x d x \\ & =-\cot x \operatorname{cosec}^3 x-3 \int\left(\operatorname{cosec}^2 x-1\right) \operatorname{cosec}^3 x d x \\ & =-\cot x \operatorname{cosec}^3 x-3 \int \operatorname{cosec}^5 x d x+3 \int \operatorname{cosec}^3 x d x \\ & \quad f(x)=-\cot x \operatorname{cosec}^3 x-3 I+3 I_1 \\ & 4 f(x)=\cot x \operatorname{cosec}^3 x+3 I_1 \\ & \quad I_1=\int \operatorname{cosec}{ }^3 x d x \\ & =\int \operatorname{cosec} x \cdot \operatorname{cosec}^2 x d x\end{aligned}$
$\begin{aligned} & =\operatorname{cosec} x(-\cot x)-\int-\cot x \cdot(-\operatorname{cosec} x \cot x) d x \\ & =-\operatorname{cosec} x \cot x-\int \cot ^2 x \operatorname{cosec} x d x \\ & =-\operatorname{cosec} x \cot x-\int\left(\operatorname{cosec}^2 x-1\right) \operatorname{cosec} x d x \\ & =-\operatorname{cosec} x \cot x-\int \operatorname{cosec}^3 x d x+\int \operatorname{cosec} x d x \\ & 2 I_1=-\operatorname{cosec} x \cot x+\log (\operatorname{cosec} x-\cot x) \\ & f(x)=\frac{1}{4}\left[-\cot x \operatorname{cosec}^3 x+\frac{3}{2}(-\operatorname{cosec} x \cot x]\right. \\ & \quad+\log (\operatorname{cosec} x-\cot x)] \\ & f(\pi / 4)=\frac{1}{4}\left[-2 \sqrt{2}+\frac{3}{2}(-\sqrt{2}+\log (\sqrt{2}-1)]\right. \\ & =\frac{1}{8}[-4 \sqrt{2}-3 \sqrt{2}+3 \log (\sqrt{2}-1)] \\ & =-\frac{1}{8}[7 \sqrt{2}-3 \log (\sqrt{2}-1)] \\ & =-\frac{1}{8}[7 \sqrt{2}+3 \log (\sqrt{2}+1)]\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
If ∫fxdx=3fx2+c (where c is the constant of integration) and f1=16, then f6π is equal to

Options:
A π2
B π
C π3
D π6

The correct answer is:
π

Solution:
Differentiating the given integral with respect to âxâ, we get,
f x =6f x â f â² x or 1 6 = f â² x
On integrating, we get,
â« 1 6 dx=â« f â² x dx
âx6=fx+c
As f1=16âc=0
âfx=x6
âf6Ï=Ï

Topic: Mathematics, Indefinite Integration, JEE Main