datavorous/entrance-exam-dataset huggingface dataset:
Let f be a differentiable function such that x2fx-x=4∫0xt ft dt, f1=23. Then 18 f3 is equal to

Options:
A 210
B 160
C 150
D 180

The correct answer is:
160

Solution:
Given,

x2fx-x=4∫0xt ft dt

Now Differentiating both side w.r.t. x we get,

x2f'x+2xfx-1=4x fx

⇒x2f'x-1=2x fx

⇒dydx-2xy=1x2 Let y=fxdydx=f'x

Which is a linear differential equation,

So, I.F =e∫-2xdx=e-2lnx=1x2

Now solution of the differential equation is given by,

yx2=∫1x4dx+C

⇒yx2=-13x3+C

Now given, f1=23

So, 23=-13+C

⇒C=1

Hence, the function will be,

fx=-13x+x2

So, the required value is given by,

18 f3=18-19+9=160

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let f:ℝ→ℝ be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)-1,∀ x, y∈ℝ. If f'(0)=2, then |f(-2)| is equal to

Options:

The correct answer is:
3

Solution:
Given,

Functional equation fx+y=fx+fy-1

Now taking x=0 & y=0 in above equation we get,

f0+0=f0+f0-1⇒f0=1

Now we know that,

f'x=limh→0fx+h-fxh

⇒f'x=limh→0fx+fh-1-fxh

⇒f'x=limh→0fh-1h

Now let limh→0fh-1h=k

So, the equation becomes f'x=k

Now putting x=0 in above equation we get,

f'0=k⇒k=2 as given f'0=2

So, f'x=2

Now integrate both side we get,

fx=2x+c

Now again taking x=0 we get,

f0=2×0+c⇒c=1 as f0=1

So, fx=2x+1

And f-2=2×-2+1⇒f-2=-3

Hence, f-2=3

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let $f$ be a differentiable function with $\lim _{x \rightarrow \infty} f(x)=0 .$ If $y^{\prime}+y f^{\prime}(x)-f(x) f^{\prime}(x)=0, \lim _{x \rightarrow \infty} y(x)=0$, then $\left(\right.$ where $\left.y^{\prime}=\frac{d y}{d x}\right)$

Options:
A $y+1=e^{f(x)}+f(x)$
B $y-1=e^{f(x)}+f(x)$
C $y+1=e^{-f(x)}+f(x)$
D $y-1=e^{-f f] x}+f(x)$

The correct answer is:
$y+1=e^{-f(x)}+f(x)$

Solution:
Hint:
$\frac{d y}{d x}+f^{\prime}(x) y=f^{\prime}(x) f(x)$
$\Rightarrow y \times e^{f(x)}=\int f^{\prime}(x) f(x) e^{f(x)} d x$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+c \quad[$ Putting $f(x)=0 ; y=0, c=1]$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+1$
$\Rightarrow y=f(x)-1+e^{-f(x)}$
$\Rightarrow y+1=e^{-f(x)}+f(x)$

Topic: Mathematics, Differential Equations, JEE Main
Let f:ℝ→ℝ be a function defined as fx=asinπx2+2-x,a∈ℝ, where t is the greatest integer less than or equal to t. If limx→-1fx exists, then the value of ∫04fxdx is equal to

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

The correct answer is:
-2

Solution:
Given,

fx=asinπx2+2-x,a∈ℝ

Now given limx→-1fx exists,

So limx→-1+asinπx2+2-x=-a+2

And limx→-1-asinπx2+2-x=0+3=3

So, limx→-1fx exist when -a+2=3⇒a=-1

Now,

∫04fxdx=∫01fxdx+∫12fxdx+∫23fxdx+∫34fxdx

⇒∫04fxdx=∫01-sinπx2+2-xdx+∫12-sinπx2+2-xdx+∫23-sinπx2+2-xdx+∫34-sinπx2+2-xdx

=∫010+1dx+∫12-1+0dx+∫230-1dx+∫341-2dx

=1-1-1-1=-2

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let f:ℝ→ℝ be a function defined by fx=4x4x+2 and M=∫faf1−axsin4x1−xdx, N=∫faf1−asin4x1−xdx; a≠12. If αM=βN,α, β∈ℕ, then the least value of α2+β2 is equal to ______

Options:

The correct answer is:
5

Solution:
Given,

fx=4x4x+2

⇒f1-x=41-x41-x+2

⇒f1-x=44+2·4x=22+4x

⇒fx+f1−x=1

Now, solving M=∫faf1−ax·sin4x1−xdx

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

⇒M=∫faf1−afa+f1-a-x·sin4fa+f1-a-x1−fa+f1-a-xdx

⇒M=∫faf1−a1-x·sin41-xxdx

⇒M=∫faf1−asin41-xxdx-∫faf1−ax·sin41-xxdx

⇒M=N−M

⇒2M=N

So, on comparing we get, α=2, β=1

Hence, α2+β2=22+12=5

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let f:ℝ→ℝ be a function defined by fx=logm 2sinx-cosx+m-2, for some m, such that the range of f is 0,2. Then the value of m is _____ .

Options:
A 5
B 3
C 2
D 4

The correct answer is:
5

Solution:
Given,

f:ℝ→ℝ be a function defined by fx=logm 2sinx-cosx+m-2, for some m,

Also given the range of f is 0,2,

Now we know that,

-2≤sinx-cosx≤2

⇒-2≤2sinx-cosx≤2

(Assuming 2sinx-cosx=k)

⇒-2≤k≤2 …1

Now taking function fx=logmk+m-2

Given, 0≤fx≤2

⇒0≤logmk+m-2≤2

⇒1≤k+m-2≤m

⇒-m+3≤k≤2 …2

Now from equations 1 & 2, we get

-m+3=-2

⇒m=5

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let $f$ be a function from the set of natural numbers to the set of even natural numbers given by $f(x)=2 x$. Then $f$ is

Options:
A one to one but not onto
B onto but not one-one
C both one-one and onto
D neither one-one nor onto

The correct answer is:
both one-one and onto

Solution:
$1 ..... 2$
$2 ..... 4$
$3 ..... 6$
$4 ..... 8$
Domain (x) $\quad$ Co-domain $(2 \mathrm{x})$ For each element in the domain there is only one element in the therefore $\mathrm{f}(\mathrm{x})$ is one-one. No image in the co-domain which has no pre-image in the domain, therefore $\mathrm{f}(\mathrm{x})$ is onto Hence $\mathrm{f}(\mathrm{x})$ both is one-one and onto.

Topic: Mathematics, Functions, NDA, NDA 2013 (Phase 2)
Let $f$ be a non-negative function defined on the interval $[0,1]$. If $\int_0^x \sqrt{1-\left\{f^{\prime}(t)\right\}^2} d t=\int_0^x f(t) d t, \quad 0 \leq x \leq 1$
$$
\text { and } f(0)=0 \text {, then }
$$

Options:
A
$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$
B
$f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$

C
$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$
D
$f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$

The correct answer is:
$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$

Solution:
Given $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t$,
$$
0 \leq x \leq 1
$$
Applying Leibnitz theorem, we get
$$
\begin{array}{rlrl}
& & \sqrt{1-\left(f^{\prime}(x)\right)^2} & =f(x) \\
\Rightarrow & & 1-\left(f^{\prime}(x)\right)^2 & =f^2(x) \\
\Rightarrow & & \left(f^{\prime}(x)\right)^2 & =1-f^2(x) \\
\Rightarrow & & f^{\prime}(x) & =\pm \sqrt{1-f^2(x)} \\
\Rightarrow & & \quad \frac{d y}{d x} & =\pm \sqrt{1-y^2} \\
& \text { where } y=f(x) \Rightarrow & \frac{d y}{\sqrt{1-y^2}}=\pm d x
\end{array}
$$
On integrating both sides, we get
$$
\begin{aligned}
& \sin ^{-1}(y)=\pm x+C \\
& \because \quad f(0)=0 \Rightarrow C=0 \Rightarrow y=\pm \sin x \\
& y=\sin x=f(x) \text { given } f(x) \geq 0 \text { for } \\
& x \in[0,1]
\end{aligned}
$$
It is known that $\sin x < x, \forall x \in R^{+}$
$\therefore \quad \sin \left(\frac{1}{2}\right) < \frac{1}{2} \Rightarrow f\left(\frac{1}{2}\right) < \frac{1}{2}$
and $\sin \left(\frac{1}{3}\right) < \frac{1}{3} \Rightarrow f\left(\frac{1}{3}\right) < \frac{1}{3}$

Topic: Mathematics, Definite Integration, JEE Main
Let f be a non-negative function in [0,1] and twice differentiable in (0,1). If ∫0x1-f'(t)2 dt=∫0xf(t)dt,0≤x≤1 and f(0)=0, then limx→01x2∫0xf(t)dt:

Options:
A does not exist
B equals 0
C equals 1
D equals 12

The correct answer is:
equals 12

Solution:
Newton - Leibnitz rule

ddx∫fxgxhxdx = g'xhg(x) - f'xhf(x)

Given that

∫0x1-f'(t)2dt=∫0xf(t)dt

Differentiate w.r.t. x

1-f'(x)2=f(x)

Squaring on both sides
⇒1-f'(x)2=(f(x))2

⇒f'(x)2=1-(f(x))2

fx = y ⇒f'x = dydx
⇒dydx2=1-y2 ⇒dydx=±1-y2

⇒dy1-y2=±dx

By using Variable Separable form

sin-1y=±x+c

When x = 0 , y = 0 ⇒ c = 0.

∴sin-1y=x⇒y=sinx

Now limx→01x2∫0xsintdt

It is in the form of 00 so apply L hospital's rule

=limx→0ddx∫0xsintdtdx2dx

= 12limx→0sinxx

=12

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Let f be a polynomial function such that f3x=f′x. f′′x, for all x∈R. Then :

Options:
A f2+fâ²2=28
B fâ²â²2-fâ²2=0
C f2-f′2+f′′2=Â 10
D f′′2-f2=4

The correct answer is:
f′′2-f′2=0

Solution:
Degree of fx will be 3
f(x)=ax3+bx2+cx+d
f3x=27ax3+9bx2+3cx+d
f'(x)=3ax2+2bx+c
f′′x=6ax+2b
f3x=f′x f′′x
Comparing the coefficient, we get
27a=18a2 ⇒a= 3 2
Also b=0, c=0, d=0
fx=32x3, f( 2 )=12
⇒ f′x=92x2, f′′x=9x
Hence, f'(2)=18, f''(2)=18
Hence, f''2−f'(2)=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2017 (09 Apr Online)
Let f be a real valued function, defined onR--1,1 and given by fx=3logex-1x+1-2x-1. Then in which of the following intervals, function fx is increasing?

Options:
A

-∞,-1∪12,∞-1

B -∞,∞--1,1
C -1,12
D -∞,12--1

The correct answer is:

-∞,-1∪12,∞-1

Solution:
fx=3logex-1-3logex+1-2x-1

f'x=3x-1-3x+1+2x-12

f'x=42x-1x-12x+1

f'x≥0

⇒ x∈-∞,-1∪12,1∪1,∞ or -∞,-1∪12,∞-1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let $f$ be a real-valued differentiable function on $R$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$, then the value of $f(-3)$ is equal to

Options:

The correct answer is:
9

Solution:
The equation of the tangent at $(x, y)$ to the given curve $y=f(x)$ is
$$
\begin{gathered}
Y-y=\frac{d y}{d x}(X-x) \\
Y \text {-intercept }=y-x \frac{d y}{d x}
\end{gathered}
$$
According to the question
$$
x^3=y-x \frac{d y}{d x}
$$
$$
\Rightarrow \quad \frac{d y}{d x}-\frac{y}{x}=-x^2
$$
which is linear in $x$.
$$
\text { IF }=e^{\int \frac{-1}{x} d x}=\frac{1}{x}
$$
$\therefore$ Required solution is
$$
\begin{aligned}
& y \cdot \frac{1}{x}=\int-x^2 \cdot \frac{1}{x} d x \\
& \Rightarrow \quad \frac{y}{x}=\frac{-x^2}{2}+c \\
& \Rightarrow \quad y=\frac{-x^3}{2}+c x \\
& \text { At } x=1, y=1 \text {, } \\
& 1=\frac{-1}{2}+c \\
& \Rightarrow \quad c=\frac{3}{2} \\
&
\end{aligned}
$$
Now, $\begin{aligned} f(-3) & =\frac{27}{2}+\frac{3}{2}(-3) \\ & =\frac{27-9}{2}=9\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
Let f be a twice differentiable function defined on R such that f0=1, f'0=2 and f'x≠0 for all x∈R. If fxf'xf'xf''x=0, for all x∈R, then the value of f1 lies in the interval

Options:
A 9,12
B 3,6
C 0,3
D 6,9

The correct answer is:
6,9

Solution:
We have, fxf'xf'xf''x=0

⇒fxf''x-f'x2=0

⇒f''xf'x=f'xfx

On integrating both side, we get

lnf'x=lnfx+lnc

⇒f'x=cfx

⇒f'xfx=c

Again integrating, we get

ln fx=cx+k1

⇒fx=kecx

Since, f0=1=k

Therefore, f'0=c=2

Now, fx=e2x

Hence, f1=e2∈6,9

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let f be a twice differentiable function on R. If f'0=4 and fx+∫0xx-tf'tdt =e2x+e-2xcos2x+2ax, then 2a+15a2 is equal to _______.

Options:

The correct answer is:
8

Solution:
Given,

fx+∫0xx-tf'tdt=e2x+e-2xcos2x+2xa ⋯i

Putting x=0 both side we get, f0=2 ⋯ii

On differentiating equation i w.r.t. x we get:

f'x+∫0xf'tdt+xf'x-xf'x=2e2x-e-2xcos2x-2e2x+e-2xsin2x+2a

⇒f'x+fx-f0=2e2x-e-2xcos2x-2e2x+e-2xsin2x+2a

Replace x by 0 we get:

⇒4=2a ⇒a=12

Now putting the value of a in2a+15·a2

We get, 25·122=23=8

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Let $\mathrm{f}$ be an odd function defined on the set of real numbers such that for $\mathrm{x} \geq 0$, $f(x)=3 \sin x+4 \cos x$.
Then $\mathrm{f}(\mathrm{x})$ at $\mathrm{x}=-\frac{11 \pi}{6}$ is equal to:

Options:
A
$\frac{3}{2}+2 \sqrt{3}$

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

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

D
$-\frac{3}{2}-2 \sqrt{3}$

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

Solution:
Given $\mathrm{f}$ be an odd function $f(x)=3 \sin x+4 \cos x$
Now,
$$
\begin{array}{r}
f\left(\frac{-11 \pi}{6}\right)=3 \sin \left(\frac{-11 \pi}{6}\right)+4 \cos \left(\frac{-11 \pi}{6}\right) \\
f\left(\frac{-11 \pi}{6}\right)=3 \sin \left(-2 \pi+\frac{\pi}{6}\right)+4 \cos \left(-2 \pi+\frac{\pi}{6}\right) \\
f\left(\frac{-11 \pi}{6}\right)=3 \sin \left\{-\left(2 \pi-\frac{\pi}{6}\right)\right\}+4 \cos \left\{-\left(2 \pi-\frac{\pi}{6}\right)\right\} \\
\left\{\text { For odd functions } \begin{array}{l}
\sin (-\theta)=-\sin \theta \\
\text { and } \cos (-\theta)=-\cos \theta
\end{array}\right\}
\end{array}
$$
$$
\begin{aligned}
\therefore f\left(\frac{-11 \pi}{6}\right)=-3 \sin \left(2 \pi-\frac{\pi}{6}\right)-4 \cos \left(2 \pi-\frac{\pi}{6}\right) \\
\Rightarrow f\left(\frac{-11 \pi}{6}\right) &=+3 \sin \left(\frac{\pi}{6}\right)-4 \cos \frac{\pi}{6} \\
\Rightarrow f\left(\frac{-11 \pi}{6}\right) &=3 \times \frac{1}{2}-4 \times \frac{\sqrt{3}}{2} \\
\text { or } f\left(\frac{-11 \pi}{6}\right) &=\frac{3}{2}-2 \sqrt{3}
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Main, JEE Main 2014 (11 Apr Online)
Let f be any continuous function on 0, 2 and twice differentiable on 0, 2. If f0=0, f1=1 and f2=2, then :

Options:
A f"x>0 for all x∈0, 2
B f'x=0 for some x∈0, 2
C f"x=0 for some x∈0, 2
D f"x=0 for all x∈0, 2

The correct answer is:
f"x=0 for some x∈0, 2

Solution:
Given f0=0, f1=1 and f2=2

Let, hx=fx-x

Clearly hx, will be continuous and twice differentiable on 0, 2.

Now, h0=h1=h2=0

By Rolle's mean value theorem in 0, 1, we get

h'c1=0

⇒f'c1-1=0

⇒f'c1=1, where c1∈0, 1

Also, in the interval 1, 2

h'c2=0

⇒f׀c2-1=0

⇒f'c2=1, where c2∈1, 2

Now, use Rolle's theorem on c1, c2 for h'x,

We have, h''(c)=0

⇒f''(c)=0, where c∈c1, c2

Hence, f''(c)=0 for some x∈0, 2.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let f be any function continuous on a,b and twice differentiable on a,b . If all x∈a,b,f'x>0 and f''x<0 , then for any c∈a,b,fc-fafb-fc

Options:
A b+ab-a
B 1
C b-cc-a
D c-ab-c

The correct answer is:
c-ab-c

Solution:
Let’s use LMVT for x∈a,c

fc-f(a)c-a=f'α,α∈a,c

Also use LMVT for x∈c,b

fb-f(c)b-c=f'β,β∈c,b

∵f''x<0⇒f'x is decreasing

f'α>f'β

fc-f(a)c-a>fb-f(c)b-c

fc-f(a)fb-f(c)>c-ab-c ( ∵f(x) is increasing)

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Let f be any function defined on R and let it satisfy the condition: fx-fy≤x-y2,∀x,y∈R. If f0=1, then :

Options:
A fx=0,∀x∈R
B fx can take any value in R
C fx<0,∀x∈R
D fx>0,∀x∈R

The correct answer is:
fx>0,∀x∈R

Solution:
fx-fyx-y≤x-y

Let x-y=h

⇒x=y+h

limx→0fy+h-fyh≤0

⇒f'y≤0⇒f'y=0

⇒fy=k (constant)

And f0=1 given

So, fy=1⇒fx=1

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Let f:ℝ→ℝ be defined as fx=ex,x<0aex+x-1,0≤x<1b+sinπx,1≤x<2e-x-c,x≥2
where a,b,c∈ℝ and t denotes greatest integer less than or equal to t. Then, which of the following statements is true?

Options:
A There exists a,b,c∈ℝ such that f is continuous of ℝ.
B If f is discontinuous at exactly one point, then a+b+c=1
C If f is discontinuous at exactly one point, then a+b+c≠1.
D f is discontinuous at atleast two points, for any values of a,b and c.

The correct answer is:
If f is discontinuous at exactly one point, then a+b+c≠1.

Solution:
Given fx=ex,x<0aex+x-1,0≤x<1b+sinπx,1≤x<2e-x-c,x≥2

i.e. fx=0,x<0aex-1,0≤x<1b,x=1b-1,1<x<2-c,x≥2

Here, we know fx is discontinuous at x=1 as b≠b-1

For fx to be continuous at x=0; a=1

For fx to be continuous at x=2; b+c=1

Hence, a+b+c=2

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Let $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ be defined by $f(x)=x^3+1$
Statement 1: The function fhas a local extremum at $x=0$
Statement 2: The function $f$ is continuous and differentiable on $(-\infty, \infty)$ and $f^{\prime}(0)=0$

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1.

D
Statement 1 is false, Statement 2 is true.

The correct answer is:
Statement 1 is false, Statement 2 is true.

Solution:
Let $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ be defined by $f(x)=x^3+1$.
Clearly, $f(x)$ is symmetric along $y=1$ and it has neither maxima nor minima.
$\therefore$ Statement - 1 is false.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (26 May Online)
Let $f$ be differentiable for all $x$. If $f(1)=-2$ and $f^{\prime}(x) \geq 2$ for $x \in[1,6]$, then

Options:
A $f(6) \lt 5$
B $f(6)=5$
C $f(6) \geq 8$
D $f(6) \lt 8$

The correct answer is:
$f(6) \geq 8$

Solution:
As $f(1)=-2$ and $f^{\prime}(x) \geq 2 \forall x \in[1,6]$
Applying lagrange's mean value theorem,
$\begin{aligned} & \frac{f(6)-f(1)}{5}=f^{\prime}(c) \geq 2 \\ & \Rightarrow f(6) \geq 10+f(1) \Rightarrow f(6) \geq 10-2 \Rightarrow f(6) \geq 8\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f$ be differentiable for all $x$. If $f(1)=-2$ and $f^{\prime}(x) \geq 2$ for $x \in[1,6]$, then

Options:
A
$f(6) \geq 8$

B
$f(6) < 8$
C
$f(6) < 5$
D
$f(6)=5$

The correct answer is:
$f(6) \geq 8$

Solution:
As $f(1)=-2 \quad \& f^{\prime}(x) \geq 2 \quad \forall x \in[1,6]$
Applying Lagrange's mean value theorem
$$
\begin{aligned}
& \frac{f(6)-f(1)}{5}=f^{\prime}(c) \geq 2 \\
& \Rightarrow f(6) \geq 10+f(1) \\
& \Rightarrow f(6) \geq 10-2 \\
& \Rightarrow f(6) \geq 8 .
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
Let $\vec{F}$ be the force acting on a particle having position vector $\vec{r}$ and $\vec{T}$ be the torque of this force about the origin. Then

Options:
A
$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{T}}=0$ and $\overrightarrow{\mathrm{F}} . \overrightarrow{\mathrm{T}} \neq 0$

B
$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{T}} \neq 0$ and $\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{T}}=0$

C
$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{T}} \neq 0$ and $\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{T}} \neq 0$

D
$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{T}}=0$ and $\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{T}}=0$

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

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2003
Let $\mathrm{f}$ be the function defined by $f(x)=\left\{\begin{array}{ll}\frac{x^{2}-1}{x^{2}-2|x-1|-1}, & x \neq 1 \\ 1 / 2, & x=1\end{array}\right.$

Options:
A The function is continuous for all values of x
B The function is continuous only for $x>1$
C The function is continuous at $\mathrm{x}=1$
D The function is not continuous at $x=1$

The correct answer is:
The function is not continuous at $x=1$

Solution:
For $x<1, f(x)=\frac{x^{2}-1}{x^{2}+2 x-3}=\frac{x+1}{x+3}$

$\therefore \lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=\frac{1}{2}$

For $x>1, f(x)=\frac{x^{2}-1}{x^{2}-2 x+1}=\frac{x+1}{x-1}$

$\therefore \lim _{x \rightarrow 1^{+}} f(x)=\infty$

$\therefore$ The function is not continuous at $\mathrm{x}=1$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f,âg and h are differentiable functions. If f 0 =1;âg 0 =2;âh 0 =3 and the derivatives of their pair wise products at x=0 are fg'0=6;âgh'0=4 and hf'0=5, then the value of fgh'0 is

Options:

The correct answer is:
16

Solution:
Let, y=fgh
dydx=f'gh+fg'h+fgh'
=122f'gh+2fg'h+2fgh'
=12hf'g+g'f+gf'h+fh'+fg'h+gh'
=12h.fg'+g.fh'+f.gh'
fgh'0=12h0fg'0+g0fh'0+f0gh'0
= 1 2 3 × 6 + 2 × 5 + 1 × 4
= 1 2 1 8 + 1 0 + 4 = 3 2 2 = 1 6

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f,g and h be the real valued functions defined on ℝ as fx=xx,x≠01,x=0, gx=sinx+1x+1,x≠-11,x=-1and hx=2x-fx, where x is the greatest integer ≤x. Then the value of limx→1ghx-1 is

Options:
A 1
B sin1
C -1
D 0

The correct answer is:
1

Solution:
Given,

Function fx=xx,x≠01,x=0, gx=sinx+1x+1,x≠-11,x=-1 and hx=2x-fx

To find limx→1ghx-1

Taking L.H.L, we take x=1-h

limh→0-g(h(1-h-1))

⇒limh→0-g2-h--h-h

⇒limh→0-g(2(-1)+1)

⇒limh→0-g-1=1

Now taking R.H.L, we take x=1+h

limh→0+g(h(1+h-1))

=limh→0+g2h-f(h)

=limh→0+g(0-1)

⇒limh→0+1=1

Topic: Mathematics, Limits, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let $f, g: \mathbf{R} \rightarrow \mathbf{R}$ be defined as :
$f(x)=|x-1| \text { and } g(x)= \begin{cases}\mathrm{e}^x, \text { MARA } & x \geq 0 \\ x+1, & x \leq 0\end{cases}$
Then the function $f(g(x))$ is

Options:
A neither one-one nor onto.
B one-one but not onto.
C onto but not one-one.
D both one-one and onto.

The correct answer is:
neither one-one nor onto.

Solution:
$\begin{aligned} & f(g(x))=|g(x)-1| \\ & f o g\left[\begin{array}{cc}\left|e^x-1\right| & x \geq 0 \\ |x+1-1| & x \leq 0\end{array}\right. \\ & \text { fog }\left[\begin{array}{cc}e^x-1 & x \geq 0 \\ -x & x \leq 0\end{array}\right.\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let $\mathrm{f}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be two functions defined by $f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0, & , x=0\end{cases}$ and $g(x)=x f(x)$
Statement I: $f$ is a continuous function at $\mathrm{x}=0$.
Statement II: $g$ is a differentiable function at $x=0$.

Options:
A
Both statement I and II are false.

B
Both statement I and II are true.

C
Statement $\mathrm{I}$ is true, statement $\mathrm{II}$ is false.

D
Statement $\mathrm{I}$ is false, statement $\mathrm{II}$ is true.

The correct answer is:
Both statement I and II are true.

Solution:
$$
\begin{aligned}
&f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right), & x \neq 0 \\
0, & x=0\end{cases} \\
&\text { and } g(x)=x f(x) \\
&\text { For } f(x) \\
&\mathrm{LHL}=\lim _{h \rightarrow 0^{-}}\left\{-h \sin \left(-\frac{1}{h}\right)\right\}
\end{aligned}
$$
$=0 \times$ a finite quantity between $-1$ and 1
$$
=0
$$
$$
\mathrm{RHL}=\lim _{h \rightarrow 0^{+}} h \sin \frac{1}{h}=0
$$
Also, $f(0)=0$
Thus $\mathrm{LHL}=\mathrm{RHL}=f(0)$
$\therefore f(x)$ is continuous at $x=0$
$$
g(x)= \begin{cases}x^2 \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}
$$
For $g(x)$
$$
\mathrm{LHL}=\lim _{h \rightarrow 0^{-}}\left\{-h^2 \sin \left(\frac{1}{h}\right)\right\}
$$
$=0^2 \times$ a finite quantity between $-1$ and $1=0$
$$
\text { RHL }=\lim _{h \rightarrow 0^{+}} h^2 \sin \left(\frac{1}{h}\right)=0
$$
Also $g(0)=0$
$\therefore g(x)$ is continuous at $x=0$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2014 (12 Apr Online)
Let fix=sin2pix for i=1,2,3 & piâN. It is given that the fundamental periods of f1x+f2x+f3x, f1x+f2x and f1x+f3x are Ï,Ï2 and Ï3 respectively, then the minimum value of p1+p2+p3 is

Options:

The correct answer is:
11

Solution:
f1x=sinâ¡2p1x
f2x=sinâ¡2p2x
f3x=sinâ¡2p3x
Now, Ï=LCM2Ï2p1,2Ï2p2,2Ï2p3âHCFp1,p2,p3=1
Ï2=LCM2Ï2p1,2Ï2p2âHCFp1,p2=2âp1=2ap2=2b
Ï3=LCM2Ï2p1,2Ï2p3âHCFp1,p3=3âp1=3xp3=3y
Hence, p1=6, p2=2, p3=3 (for minimum value)

Topic: Mathematics, Functions, JEE Main
Let fkx=1ksink x+cosk x where x∈R and k≥1. Then f4x-f6x equals

Options:
A 1 4
B 1 1 2
C 1 6
D 1 3

The correct answer is:
1 1 2

Solution:
f4x=14sin4x+cos4x
=14sin2x+cos2x2-2sin2x cos2x
=141-2sin2x cos2x
f6x=16sin6x+cos6x
=16sin2x+cos2x3-3sin2x cos2xsin2x+cos2x
=161-3sin2x cos2x1
f4x-f6x=14-16
=112

Topic: Mathematics, Trigonometric Equations, JEE Main
Let fkx=1ksink x+cosk x where x∈R and k≥1. Then f4x-f6x equals

Options:
A 1 4
B 1 1 2
C 1 6
D 1 3

The correct answer is:
1 1 2

Solution:
f4x=14sin4x+cos4x
=14sin2x+cos2x2-2sin2x cos2x
=141-2sin2x cos2x
f6x=16sin6x+cos6x
=16sin2x+cos2x3-3sin2x cos2xsin2x+cos2x
=161-3sin2x cos2x1
f4x-f6x=14-16
=112

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2014 (06 Apr)
Let $f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$ for $\mathrm{k}=1,2,3, \ldots$ Then for all $\mathrm{x} \in \mathrm{R},$ the value of $f_{4}(x)-f_{6}(x)$ is equal to :

Options:
A $\frac{1}{12}$
B $\frac{1}{4}$
C $\frac{-1}{12}$
D $\frac{5}{12}$

The correct answer is:
$\frac{1}{12}$

Solution:
$f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$

$f_{4}(x)=\frac{1}{4}\left[\sin ^{4} x+\cos ^{4} x\right]$

$=\frac{1}{4}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-\frac{(\sin 2 x)^{2}}{2}\right]$

$=\frac{1}{4}\left[1-\frac{(\sin 2 x)^{2}}{2}\right]$

$f_{6}(x)=\frac{1}{6}\left[\sin ^{6} x+\cos ^{6} x\right]$

$=\frac{1}{6}\left[\left(\sin ^{2} x+\left(\cos ^{2} x\right)-\frac{3}{4}\left(\sin ^{2} x\right)^{2}\right]\right.$

$=\frac{1}{6}\left[1-\frac{3}{4}(\sin 2 x)^{2}\right]$

Now $f_{4}(x)-f_{(6)}(x)=\frac{1}{4}-\frac{1}{6}-\frac{(\sin 2 x)^{2}}{8}+\frac{1}{8}(\sin 2 x)^{2}$

$=\frac{1}{12}$

Topic: Mathematics, Trigonometric Equations, JEE Main