datavorous/entrance-exam-dataset huggingface dataset:
Let ABC be an acute-angled triangle and let D be the mid-point of BC. If AB=AD, then tanBtanC equal

Options:
A 2
B 3
C 2
D 3

The correct answer is:
3

Solution:
In ∆ABC,D be the mid-point of BC and AB=AD⇒ΔABD is isosceles.

Let AM be the perpendicular drawn from A to BC.

∴ AM is perpendicular bisector of BD

BM=MD

ln ΔABM, tanB=AMBM

ln ΔAMC, tanC=AMCM

∴ tanBtanC=CMBM=3BMBM=3 ∵CM=DM+CD

Topic: Mathematics, Properties of Triangles, BITSAT, BITSAT 2019
Let a,b,c∈R+ and the system of equations (1-a)x+y+z=0,x+(1-b)y+z=0 and x+y+(1-c)z=0 has infinitely many solutions, the minimum value of abc is

Options:
A 33
B 9
C 27
D 3

The correct answer is:
27

Solution:
We have,

1-ax+y+z=0

x+1-by+z=0

x+y+1-cz=0

has many solutions

∴ 1-a1111-b1111-c=0

⇒ ab+ac+bc=abc

∴ ab+bc+ac3≥a2b2c21/3⇒abc1/3≥3

∴ Minimum of abc=33=27

Topic: Mathematics, Determinants, BITSAT, BITSAT 2019
Let $\alpha$ be the solution of $16^{\sin ^2 \theta}+16^{\cos ^2 \theta}=10$ in $\left(0, \frac{\pi}{4}\right)$. If the shadow of a vertical pole is $\frac{1}{\sqrt{3}}$ of its height, then the altitude of the sun is

Options:
A
$\alpha$

B
$\frac{\alpha}{2}$

C
$2 \alpha$

D
$\frac{\alpha}{3}$

The correct answer is:

$2 \alpha$

Solution:
We have,
$16^{\sin ^2 \theta}+16^{\cos ^2 \theta}=10$
$\Rightarrow \quad 16^{\sin ^2 \theta}+16^{1-\sin ^2 \theta}=10$
$\Rightarrow x+\frac{16}{x}=10$, where $x=16^{\sin ^2 \theta}$
$\Rightarrow \quad x^2-10 x+16=0 \Rightarrow x=2,8$
$\therefore \quad 16^{\sin ^2 \theta}=2,8 \Rightarrow 2^{4 \sin ^2 \theta}=2,2^3$
$\Rightarrow \quad 4 \sin ^2 \theta=1,3 \Rightarrow \sin ^2 \theta=\frac{1}{4},\left(\frac{\sqrt{3}}{2}\right)^2$
$\Rightarrow \quad \sin \theta=\frac{1}{2}, \frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6}, \frac{\pi}{3}$

Let the altitude of the sun be $\theta$. Then,
$$
\begin{aligned}
& & \tan \theta & =\frac{h}{h / \sqrt{3}} \Rightarrow \tan \theta=\sqrt{3} \\
\Rightarrow \quad & & \theta & =\pi / 3 \Rightarrow \theta=2 \alpha
\end{aligned}
$$

Topic: Mathematics, Heights and Distances, BITSAT, BITSAT 2022
Let binding energy per nucleon of nucleus is denoted by $\mathrm{E}_{\mathrm{bn}}$ and radius of nucleus is denoted as $r$. If mass number of nuclei $\mathrm{A}, \mathrm{B}$ and 64 and 125 respectively then

Options:
A \(r_{A} < r_{B}, E_{b n A} < E_{b n B}\)
B \(r_{A} < r_{B}, E_{b n A} < E_{b n A}\)
C \(r_{A}=\frac{4}{5} r_{B}, E_{b n A} < E_{b n B}\)
D \(r_{A} < r_{B}, E_{b n A} > E_{b n B}\)

The correct answer is:
\(r_{A} < r_{B}, E_{b n A} > E_{b n B}\)

Solution:
$\mathrm{r}=\mathrm{r}_{0}(\mathrm{~A})^{1 / 3}$

r increases with increasing A mass number So, $r_{A}
$\mathrm{E}_{\mathrm{bn}}$ decreases with increasing $\mathrm{A}$ for $\mathrm{A}>56$ ${ }^{56} \mathrm{Fe}$ has highest $\mathrm{E}_{\mathrm{bn}}$ value. So, $E_{b n}$ for $A=64$ is larger as compared to $\mathrm{E}_{\mathrm{bn}}$ for nucleus with $\mathrm{A}=125$

$\mathrm{E}_{\mathrm{bnA}}>\mathrm{E}_{\mathrm{bnB}}$

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2014
Let cosα+β=45 and let sinα-β=513, where 0≤α,β≤π4, then tan2α=

Options:
A 207
B 2516
C 5633
D 192

The correct answer is:
5633

Solution:
0≤α,β≤π4; α+β≤π2

cos(α+β)=4/5⇒tan(α+β)=3/4

sin(α-β)=5/13⇒α-β∈θ1

and tan(α-β)=5/12

tan2α=tan(α+β+α-β)

=tan(α+β)+tan(α-β)1-tan(α+β)tan(α-β)

=3/4+5/121-3/4×512=5633

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2019
Let \(E=\{1,2,3,4\}\) and \(F=\{1,2\}\). Then the number of onto functions from \(\mathrm{E}\) to \(\mathrm{F}\) is

Options:
A 14
B 16
C 12
D 8

The correct answer is:
14

Solution:
If set \(A\) has \(m\) elements and set \(B\) has \(\mathrm{n}\) elements then number of onto functions from \(A\) to \(B\) is
\(\sum_{r=1}^n(-1)^{n-r}{ }^n C_r r^m \text { where } 1 \leq n \leq m\)
Here \(E=\{1,2,3,4\}, F=\{1,2\}\)
\(m=4, n=2\)
\(\therefore\) No. of onto functions from \(\mathrm{E}\) to \(\mathrm{F}\)
\(\begin{aligned}
& =\sum_{r=1}^2(-1)^{2-r ~2} C_r(r)^4 \\
& =(-1)^2 C_1+{ }^2 C_2(2)^4=-2+16=14
\end{aligned}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2010
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, BITSAT, BITSAT 2020
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a function defined by

$f(x)=\frac{x-m}{x-n},$ where $m \neq n,$ then

Options:
A $\mathrm{f}$ is one-one onto
B $\mathrm{f}$ is one-one into
C $\mathrm{f}$ is many-one onto
D fis many-one into

The correct answer is:
$\mathrm{f}$ is one-one into

Solution:
Let $f: R \rightarrow R$ be a function defined by

$f(x)=\frac{x-m}{x-n}$

For any $(\mathrm{x}, \mathrm{y}) \in \mathrm{R}$

Let $\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y})$

$\Rightarrow \frac{x-m}{x-n}=\frac{y-m}{y-n} \Rightarrow x=y$

$\therefore \mathrm{f}$ is one $-$ one

Let $\alpha \in \mathrm{R}$ such that $\mathrm{f}(\mathrm{x})=\alpha$

$$

\Rightarrow \alpha=\frac{x-m}{x-n} \Rightarrow(x-n) \alpha=x-m

$$

$\Rightarrow x \alpha-n \alpha=x-m$

$$

\Rightarrow x \alpha-x=n \alpha-m

$$

$\Rightarrow x(\alpha-1)=n \alpha-m$

$\Rightarrow \mathrm{x}=\frac{\mathrm{n} \alpha-\mathrm{m}}{\alpha-1} \cdot$ for $\alpha=1, \mathrm{x} \notin \mathrm{R}$

So, $\mathrm{f}$ is not onto.

Topic: Mathematics, Functions, BITSAT, BITSAT 2014
Let f:R→R be a function satisfying fx+y=fx+2y2+kxy for all x,y∈R. If f1=2 and f2=8, then fx is equal to

Options:
A 2x2
B 6x-4
C x2+3x-2
D -x2+9x-6

The correct answer is:
2x2

Solution:
We have, fx+y=fx+2y2+kxy for all x,y∈R

⇒fx+y-fxy=2y+kx for all x∈R

⇒limy→0fx+y-fxy=limy→02y+kx

⇒f'x=kx for all x∈R

⇒fx=kx22+C for all x∈R [by integration]

But f1=2 and f2=8

∴ 2=k2+C and 8=2k+C

k=4 and C=0

Hence, fx=2x2 for all x∈R

So, option (a) is correct.

Topic: Mathematics, Functions, BITSAT, BITSAT 2018
Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in \mathbb{R}$

If $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$, then which one of the following is incorrect?

Options:
A $\mathrm{f}(\mathrm{x})$ is continuous, $\forall \mathrm{x} \in \mathrm{R}$
B $\mathrm{f}^{\prime}(\mathrm{x})$ is constant, $\forall \mathrm{x} \in \mathrm{R}$
C $\mathrm{f}(\mathrm{x})$ is differentiable, $\forall \mathrm{x} \in \mathrm{R}$
D $\mathrm{f}(\mathrm{x})$ is differentiable only in a finite interval containing zera.

The correct answer is:
$\mathrm{f}(\mathrm{x})$ is differentiable only in a finite interval containing zera.

Solution:
Let $f(x+y)=f(x)+f(y), \forall x, y \in R$

Put $x=0=y$

$\Rightarrow f(0)=f(0)+f(0)$

$\Rightarrow f(0)=0$

Now, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$f^{\prime \prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)}{h}$

Now, $f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h}$

$$

\begin{array}{l}

\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})}{\mathrm{h}}=\mathrm{f}^{\prime}(0) \\

\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x} \mathrm{f}(0)+\mathrm{C}

\end{array}

$$

But $f(0)=0$

$$

\therefore \mathrm{C}=0

$$

Hence, $f(x)=x f(0), \forall x \in R$

Clearly, $\mathrm{f}(\mathrm{x})$ is everywhere continuous and differentiable and $\mathrm{f}^{\prime}(\mathrm{x})$ is constant.

$\forall \mathrm{x} \in \mathrm{R}$

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2015
Let $f(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{\left.1+x^2\right)}\right.}$ and $f(0)=0$,
then the value of $f(1)$ be

Options:
A
$\log (1+\sqrt{2})$

B
$\log (1+\sqrt{2})-\frac{\pi}{4}$

C
$\log (1+\sqrt{2})+\frac{\pi}{2}$

D
None of these

The correct answer is:

$\log (1+\sqrt{2})-\frac{\pi}{4}$

Solution:
$f(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$
Let $x=\tan \theta$
$\Rightarrow \quad d x=\sec ^2 \theta d \theta=\left(1+x^2\right) d \theta$
$f(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$
$=\int \frac{\tan ^2 \theta \sec ^2 \theta d \theta}{\sec ^2 \theta(1+\sec \theta)}=\int \frac{\tan ^2 \theta d \theta}{1+\sec \theta}$
$=\int \frac{\sin ^2 \theta d \theta}{\cos \theta(1+\cos \theta)}$
$=\int \frac{1-\cos ^2 \theta d \theta}{\cos \theta(1+\cos \theta)}$
$=\int \frac{(1-\cos \theta) d \theta}{\cos \theta}$
$=\int \sec \theta d \theta-\int d \theta$
$=\log \left(x+\sqrt{1+x^2}\right)-\tan ^{-1} x+C$
$\therefore \quad f(0)=\log (0+\sqrt{1+0})-\tan ^{-1}(0)+C$
$0=\log 1-0+C$
$\Rightarrow \quad C=0$
$\therefore \quad f(1)=\log \left(1+\sqrt{1+1^2}\right)-\tan ^{-1}(1)$
$=\log (1+\sqrt{2})-\frac{\pi}{4}$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2022
Let fx=-2x3+21x2-60x+41, then

Options:
A fx is decreasing in 2,5
B fx is decreasing in -∞,2
C fx is increasing in -∞,1
D fx is increasing in -∞,2

The correct answer is:
fx is decreasing in -∞,2

Solution:
Given, fx=-2x3+21x2-60x+41

On differentiating w.r.t. x, we get

f'x=-6x2+42x-60

⇒f'x=-6x2-7x+10

⇒f'x=-6x-2x-5

⇒x<2 or x>5, f'x<0 i.e., fx is decreasing

and 2<x<5, f'x>0 i.e., fx is increasing

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2017
Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\frac{1-\sin ^{3} \mathrm{x}}{3 \cos ^{2} \mathrm{x}}, \mathrm{x}<\frac{\pi}{2} \\ \mathrm{p}, \quad \mathrm{x}=\frac{\pi}{2} \\ \frac{\mathrm{q}(1-\sin \mathrm{x})}{(\pi-2 \mathrm{x})^{2}}, \mathrm{x}>\frac{\pi}{2}\end{array}\right.$

If $f(x)$ is continuous at $x=\frac{\pi}{2},(p, q)=$

Options:
A (1,4)
B $\left(\frac{1}{2}, 2\right)$
C $\left(\frac{1}{2}, 4\right)$
D None of these

The correct answer is:
$\left(\frac{1}{2}, 4\right)$

Solution:
$\mathrm{f}\left[(\pi / 2)^{-}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\sin ^{3}[(\pi / 2)-\mathrm{h}]}{3 \cos ^{2}[(\pi / 2)-\mathrm{h}]}$

$\quad=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos ^{3} \mathrm{~h}}{3 \sin ^{2} \mathrm{~h}}=\frac{1}{2}$

$\mathrm{f}\left[(\pi / 2)^{+}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}[1-\sin \{(\pi / 2)+\mathrm{h}\}]}{[\pi-2\{(\pi / 2)+\mathrm{h}\}]^{2}}$

$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}(1-\cosh )}{4 \mathrm{~h}^{2}}=\frac{\mathrm{q}}{8}$

$\therefore \mathrm{p}=\frac{1}{2}=\frac{\mathrm{q}}{8} \Rightarrow \mathrm{p}=\frac{1}{2}, \mathrm{q}=4$

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2014
Let $f(x)$ be a polynomial of degree three satisfying $\mathrm{f}(0)=-1$ and $\mathrm{f}(1)=0 . \mathrm{Also}, 0$ is a stationary point of $f(x)$ If $f(x)$ does not have an extremum at $x=0$, then the value of $\int \frac{f(x)}{x^{3}-1} d x$ is

Options:
A $\frac{x^{2}}{2}+C$
B $x+C$
C $\frac{x^{3}}{6}+C$
D None of these

The correct answer is:
$x+C$

Solution:
Let $f(x)=a x^{3}+b x^{2}+c x+d$

Put $x=0$ and $x=1$

Then, we get $f(0)=-1$ and $f(1)=0$ $\Rightarrow \mathrm{d}=-1$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0$

$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=1$

It is given that $x=0$ is a stationary point of $\mathrm{f}(\mathrm{x})$, but it is not a point of extremum.

Therefore, $f(0)=0=f^{\prime \prime}(0)$ and $f^{\prime \prime \prime}(0)=0$

Now, $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}$

$\Rightarrow f(x)=3 a x^{2}+2 b x+c$

$$

\begin{array}{l}

f^{\prime \prime}(x)=6 a x+2 b \text { and } f^{\prime \prime}(x)=6 a \\

f^{\prime}=0, f^{\prime \prime}(0)=0 \text { and } f^{\prime \prime \prime}(0)=0 \neq 0

\end{array}

$$

$\Rightarrow c=0, b=0$ and $a \neq 0$

From Eqs. (i) and (ii), we get $\mathrm{a}=1, \mathrm{~b}=\mathrm{c}=0$ and $\mathrm{d}=-1$

Put these values in $\mathrm{f}(\mathrm{x})$ we get $f(x)=x^{3}-1$

Hence, $\int \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^{3}-1} \mathrm{dx}=\int \frac{\mathrm{x}^{3}-1}{\mathrm{x}^{3}-1} \mathrm{~d} \mathrm{x}=\int 1 \mathrm{~d} \mathrm{x}=\mathrm{x}+\mathrm{C}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2015
Let $f(x)=\frac{a x+b}{c x+d},$ then fof $(x)=x,$ provided that:

Options:
A $\mathrm{d}=-\mathrm{a}$
B $\mathrm{d}=\mathrm{a}$
C $\quad a=b=1$
D $\quad a=b=c=d=1$

The correct answer is:
$\mathrm{d}=-\mathrm{a}$

Solution:
$f(x)=\frac{a x+b}{c x+d}$

fof $(x)=\frac{a\left\{\frac{a x+b}{c x+d}\right\}+b}{c\left\{\frac{a x+b}{c x+d}\right\}+d} \Rightarrow \frac{a^{2} x+a b+b c x+b d}{a c x+b c+c d x+d^{2}}=x$

$\Rightarrow(a c+d c) x^{2}+\left(b c+d^{2}-b c-a^{2}\right) x$

$-a b-b d=0, \quad \forall x \in R$

$\Rightarrow(a+d) c=0, d^{2}-a^{2}=0$ and $(a+d) b=0$

$\Rightarrow a+d=0$

Topic: Mathematics, Functions, BITSAT, BITSAT 2016
Let $f(x)=\left\{\begin{array}{c}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \quad \text { if } x=1\end{array}\right.$

Then which one of the following is true?

Options:
A $f$ is differentiable at $x=0$ and $x=1$
B $f$ is differentiable at $x=0$ but not at $x=1$
C $f$ is differentiable at $x=1$ but not at $x=0$
D $f$ is neither differentiable at $x=0$ nor at $x=1$

The correct answer is:
$f$ is differentiable at $x=0$ but not at $x=1$

Solution:
We have; $f(x)=\left\{\begin{array}{ll}(x-1) \sin \left(\frac{1}{x-1}\right) & \text { if } x \neq 1 \\ 0 & \text { if } x=1\end{array}\right.$

$R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}$

which does not exist.

$\therefore f$ is not differentiable at $x=1$ Also $\left.f^{\prime}(0)=\sin \frac{1}{(x-1)}-\frac{x-1}{(x-1)^{2}} \cos \left(\frac{1}{x-1}\right)\right]_{x=0}$

$=-\sin 1+\cos 1$

$\therefore f$ is differentiable at $x=0$

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2014
$\quad \operatorname{Let} f(x)=\left(x^{5}-1\right)\left(x^{3}+1\right), g(x)=\left(x^{2}-1\right)\left(x^{2}-x+1\right)$

and let $\mathrm{h}(\mathrm{x})$ be such that $\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x}) \mathrm{h}(\mathrm{x})$. Then

$\operatorname{limh}_{x \rightarrow 1} \mathrm{~h}(\mathrm{x})$ is

Options:
A 0
B 1
C 3
D 5

The correct answer is:
5

Solution:
Given $f(x)=g(x) h(x)$

$\Rightarrow h(x)=\frac{f(x)}{g(x)}$

$\Rightarrow \lim _{x \rightarrow 1} h(x)=\lim _{x \rightarrow 1} \frac{f(x)}{g(x)}$

$\Rightarrow \lim _{x \rightarrow 1} \frac{\left(x^{5}-1\right)\left(x^{3}+1\right)}{\left(x^{2}-1\right)\left(x^{2}-x+1\right)}$

$=\lim _{x \rightarrow 1} \frac{x^{5}-1^{5}}{x-1}=5 \times 1^{4}=5$

Topic: Mathematics, Limits, BITSAT, BITSAT 2013
Let fx=xpsin1x,x≠00, x=0 then fx is continuous but not differentiable at x=0, if

Options:
A p<0
B p=0
C 0<p≤1
D p≥1

The correct answer is:
0<p≤1

Solution:
We have,

fx=xpsin1x,x≠00, x=0

For fx to be continuous at x=0

limx→0fx=f0

⇒limx→0xpsin1x=0

⇒limx→0xp×limx→0sin1x=0

⇒limx→0xp×value between -1 to 1=0

∴p>0

Now,

RHD at RHD at x=0=limh→0hpsin1h-0h=limh→0hp-1sin1h

LHD at x=0=limh→0-hpsin-1h-0-h=limh→0-1p-hp-1sin1h

Since, fx is not differentiable at x=0, therefore

p≤1

Hence, 0<p≤1.

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2018
Let f(x)=xx, x∉Z, ([·] denotes greatest integer function), then f'x is equal to

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

The correct answer is:
x

Solution:
We have, f(x)=xx

∴ x is an integer.

⇒fx=px ∵[x]=p

⇒f'x=p=x

∴ f'x=x

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2019
Let $f(x+y)=f(x) \cdot f(y)$ for all $x, y$ where $f(0) \neq 0$ If $f(5)=2$ and $f^{\prime}(0)=3,$ then $f^{\prime}(5)$ is equal to-

Options:
A 6
B 0
C 1
D None of these

The correct answer is:
6

Solution:
$f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5+0)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(5) \cdot f(h)-f(5)+f(0)}{h}$
$(\because f(x+y)=f(x), f(y)$ for all $x, y)$
$=\left(\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\right) \cdot f(5)=f^{\prime}(0) \cdot f(5)$
$=3 \times 2=6$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2012
Let \(f(x)=\frac{\left(e^x-1\right)^2}{\sin \left(\frac{x}{a}\right) \log \left(1+\frac{x}{4}\right)}\) for \(x \neq 0\), and \(f(0)=12\). If \(\mathrm{f}\) is continuous at \(x=0\), then the value of \(a\) is equal to

Options:
A 1
B -1
C 2
D 3

The correct answer is:
3

Solution:
\(\begin{aligned}
& \operatorname{Lt}_{x \rightarrow 0} \frac{\left(e^x-1\right)^2}{\sin \left(\frac{x}{a}\right) \log \left(1+\frac{x}{4}\right)} \\
&=\operatorname{Lt}_{x \rightarrow 0} \frac{\frac{\left(e^x-1\right)^2}{x} \cdot x^2}{\frac{x}{a} \cdot \frac{\sin \left(\frac{x}{a}\right)}{\left(\frac{x}{a}\right)} \cdot \frac{\log \left(1+\frac{x}{4}\right)}{\frac{x}{4}} \cdot \frac{x}{4}} \Rightarrow 4 a=12 \\
& \Rightarrow a=3
\end{aligned}\)

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2010
Let M bea $3 \times 3 \mathrm{n} \times \mathrm{n}$ -singular matrix with $\operatorname{det}(\mathrm{M})$ $=\alpha$. If $\left[\mathrm{M}^{-1}\right.$ adj $(\mathrm{adj} (\mathrm{M})]=\mathrm{KI}$, then the value of $\mathrm{K}$ is

Options:
A 1
B $\alpha$
C $\alpha^{2}$
D $\alpha^{3}$

The correct answer is:
$\alpha$

Solution:
We know that, $M(\operatorname{adj} M)=|M|$ I

Replacing $\mathrm{M}$ by adj $\mathrm{M},$ we get $\operatorname{adj} M[\operatorname{adj}(\operatorname{adj} M)=\operatorname{det}(\operatorname{adj} M) I$

$=\operatorname{det}(M) M^{-1}\left[\operatorname{adj}(\operatorname{adj} M)=\alpha^{2} 1\right.$

$$

\left[\because \mathrm{M}^{-1}=\frac{1}{|\mathrm{M}|} \operatorname{adj}(\mathrm{M})\right]

$$

$\Rightarrow \alpha \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2} \mathrm{I}$

$\Rightarrow \mathrm{M}^{-1}[\mathrm{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}$

But $M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=K I$

Hence, $\mathrm{K}=\alpha$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2015
Let \(\mathrm{A}=\{x: x \in \mathrm{R},|x| < 1\}\);
\(\mathrm{B}=\{x: x \in \mathrm{R},|x-1| \geq 1\} \text { and } \mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D} \text {, }\)
then the set \(\mathrm{D}\) is

Options:
A \(\{x: 1 < x \leq 2\}\)
B \(\{x: 1 \leq x < 2\}\)
C \(\{x: 1 \leq x \leq 2\}\)
D None of these

The correct answer is:
\(\{x: 1 \leq x < 2\}\)

Solution:
\(\begin{aligned}
& \mathrm{A}=[x: x \in \mathrm{R},-1 < x < 1] \\
& \mathrm{B}=[x: x \in \mathrm{R}: x-1 \leq-1 \text { or } x-1 \geq 1] \\
&=[x: x \in R: x \leq 0 \text { or } x \geq 2] \\
& \therefore A \cup B=R-D,
\end{aligned}\)
where \(D=[x: x \in R, 1 \leq x < 2]\)

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2010
Let \(\mathrm{f}(\mathrm{x})=\) \(\left\{\begin{array}{l}a x^2+1, x > 1 \\ x+a, x \leq 1\end{array}\right.\). Then \(f(x)\) is derivable at \(x=1\), if

Options:
A \(a=2\)
B \(a=1\)
C \(a=0\)
D \(a=1 / 2\)

The correct answer is:
\(a=1 / 2\)

Solution:
\(\begin{aligned}
& \operatorname{Lf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\
& =\lim _{h \rightarrow 0} \frac{(1-h+a)-(1+a)}{-h}=\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \\
& \operatorname{Rf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left[a(1+h)^2+1\right]-(1+a)}{h} \\
& =\lim _{h \rightarrow 0}(a h+2 a)=2 a
\end{aligned}\)
Since f' (1) exists, \(\therefore L f^{\prime}(1)=R f^{\prime}(1) \Rightarrow a=1 / 2\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2010
Let \(\mathrm{I}_1=\int_1^2 \frac{1}{\sqrt{1+\mathrm{x}^2}} \mathrm{dx}\) and \(\mathrm{I}_2=\int_1^2 \frac{1}{x} d x\), then

Options:
A \(\mathrm{I}_1 > \mathrm{I}_2\)
B \(\mathrm{I}_2 > \mathrm{I}_1\)
C \(\mathrm{I}_1=\mathrm{I}_2\)
D None of these

The correct answer is:
\(\mathrm{I}_2 > \mathrm{I}_1\)

Solution:
If \(I_1=\int_1^2 \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^2}}, I_2=\int_1^2 \frac{\mathrm{dx}}{\mathrm{x}}\)
\(I_1=\ell n \left(\frac{2+\sqrt{5}}{1+\sqrt{2}}\right), \quad I_2=\ell n 2 \Rightarrow I_1 < I_2\)

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2011
Let p and q be roots of the equation x2-2x+A=0 and let r and s be the roots of the equation x2-18x+B=0. If p<q<r<s are in arithmetic progression, then A and B are

Options:
A -3,-77
B 3,-77
C -3,77
D 3,77

The correct answer is:
-3,77

Solution:
x2-2x+A=0

⇒ x=1±1-A

∴ p=1-1-A,q=1+1-A

Similarly, r=9-81-B,s=9+81-B

p,q,r,s are in AP.

∵ q-p=s-r

⇒ 21-A=281-B

⇒ B=80+A

and q-p=r-q

⇒ 31-A=8-81-B

⇒ A=-3,B=77

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2019
Let Px=∫dxex+8e-x+4e-3x, Qx=∫dxe3x+8ex+4e-x and Rx=Px-2Qx. If Rx=12AB+2e-xC+K, then the value of A,B,C is

Options:
A tan-1,2,ex
B tan-1,ex,2
C tan-1,12,1ex
D tan-1,1ex,12

The correct answer is:
tan-1,ex,2

Solution:
We have, Rx=∫dxex+8e-x+4e-3x-2∫dxe3x+8ex+4e-x

⇒Rx=∫exe2x-2e4x+8e2x+4dx

we get Rt=∫t2-2dtt4+8t2+4=∫1-2t-2dtt+2t-12+4

=12tan-1t+2t-12+K

⇒Rx=12tan-1ex+2e-x2+K

Hence, A,B,C=tan-1,ex,2

∴ Option (b) is correct.

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2018
Let $R=\{(3,3),(6,6),(9,9),(12,12),(6,12), (3,9),(3,12),(3,6)$ be a relation on the set $A=\{3,6,9,12\}$. Then, the relation is

Options:
A an equivalence relation
B reflexive and symmetric
C reflexive and transitive
D onlyreflexive

The correct answer is:
reflexive and transitive

Solution:
$(3,3),(6,6),(9,9),(12,12), \in R$.

$R$ isnot symmetricas (6,12)$\notin R$ but (12,6)$\in R$ $R$ is transitive as the only pair which needs verification is (3,6) and (6,12)$\in R$.

$$

\Rightarrow \quad(3,12) \in R

$$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2014
Let $R$ be the relation on the set $R$ of all real numbers, defined by $a R b$ If $|a-b| \leq 1$. Then, $R$ is

Options:
A reflexive and symmetric only
B reflexive and transitive only
C equivalence
D None of the above

The correct answer is:
reflexive and symmetric only

Solution:
Since, $|a-a|=0 \leq 1,$ so $a R a, \forall a \in R$ $\therefore R$ is reflexive. Now, $a R b \Rightarrow|a-b| \leq 1 \Rightarrow|b-a| \leq 1 \Rightarrow b R a$

$\therefore R$ is symmetric. But $R$ is not transitive as

$1 R 2,2 R 3$ but $1 \not R^{\prime} 3$

$$

[\because|1-3|=2>1]

$$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2013
Let $\mathrm{S}$ be the focus of the parabola $\mathrm{y}^{2}=8 \mathrm{x}$ and $\mathrm{PQ}$ be the common chord of the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-$ $4 y=0$ and the given parabola. The area of $\Delta P Q S$ is

Options:
A $4 \mathrm{sq}$ units
B $3 \mathrm{sq}$ units
C 2 sq units
D $8 s q$ units

The correct answer is:
$4 \mathrm{sq}$ units

Solution:
The parametric equations of the parabola $y^{2}=8 x$ are $x=2 t^{2}$ and $y=4 t$

and the given equation of circle is

$x^{2}+y^{2}-2 x-4 y=0$

On putting $x=2 t^{2}$ and $y=4$ t in circle

we get

$$

\begin{array}{l}

4 t^{4}+16 t^{2}-4 t^{2}-16 t=0 \\

\Rightarrow 4 t^{2}+12 t^{2}-16 t=0 \\

\Rightarrow 4 t\left(t^{3}+3 t-4\right)=0 \\

\Rightarrow t(t-1)\left(t^{2}+t+4\right)=0 \\

\Rightarrow t=0, t=1 \\

\quad\left[\because t^{2}+t+4 \neq 0\right]

\end{array}

$$

Thus the coordinates of points of intersection of the circle and the parabola are $\mathrm{Q}(0,0)$ and $\mathrm{P}(2,4) .$ Clearly these are diametrically opposite points on the circle.

The coordinates of the focus $\mathrm{S}$ of the parabola are (2,0) which lies on the circle.

$$

\begin{aligned}

\therefore \text { Area of } \Delta \mathrm{PQS} &=\frac{1}{2} \times \mathrm{QS} \times \mathrm{SP}=\frac{1}{2} \times 2 \times 4 \\

&=4 \mathrm{sq} . \text { units. }

\end{aligned}

$$

Topic: Mathematics, Parabola, BITSAT, BITSAT 2015