datavorous/entrance-exam-dataset huggingface dataset:
What is the value of $\cos 15^{\circ}$ ?

Options:
A $\frac{1}{2}(\sqrt{2-\sqrt{3}})$
B $\frac{1}{2}(\sqrt{2+\sqrt{3}})$
C $\sqrt{2}+\sqrt{3}$
D $\sqrt{2}-\sqrt{3}$

The correct answer is:
$\frac{1}{2}(\sqrt{2+\sqrt{3}})$

Solution:
$\cos 2 \theta=2 \cos ^{2} \theta-1$
Put $\theta=15^{\circ}$
$\therefore \cos 30^{\circ}=2 \cos ^{2} 15^{\circ}-1$
$\Rightarrow \frac{\sqrt{3}}{2}+1=2 \cos ^{2} 15^{\circ}$
$\Rightarrow \cos ^{2} 15^{\circ}=\frac{\sqrt{3}+2}{4}$
$\Rightarrow \cos 15^{\circ}=\frac{1}{2}(\sqrt{2+\sqrt{3}})$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 2)
What is the value of $\cos \left(2 \cos ^{-1}(0.8) ?\right.$

Options:
A $0.81$
B $0.56$
C $0.48$
D $0.28$

The correct answer is:
$0.28$

Solution:
We know that
$2 \cos ^{-1} x=\cos ^{-1}\left(2 x^{2}-1\right)$
here, $x=0.8$
$\therefore \quad 2 \cos ^{-1}(0.8)=\cos ^{-1}\left(2(0.8)^{2}-1\right)$
$=\cos ^{-1}(0.28)$
Now, $\cos \left(\cos ^{-1}(x)\right)=x$
$\therefore \cos \left(\cos ^{-1}(0.28)\right)=0.28$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2016 (Phase 2)
What is the value of $\cos 36^{\circ} ?$

Options:
A $\frac{\sqrt{5}-1}{4}$
B $\frac{\sqrt{5}+1}{4}$
C $\frac{\sqrt{10+2 \sqrt{5}}}{4}$
D $\frac{\sqrt{10-2 \sqrt{5}}}{4}$

The correct answer is:
$\frac{\sqrt{5}+1}{4}$

Solution:
No Solution

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 1)
What is the value of $\cos 46^{\circ} \cos 47^{\circ} \cos 48^{\circ} \cos 49^{\circ} \cos$
$50^{\circ} \ldots \cos 135^{\circ} ?$

Options:
A $-1$
B 0
C 1
D Greater than 1

The correct answer is:
0

Solution:
$\cos 46^{\circ} \cos 47^{\circ} \cos 48^{\circ} \cos 49^{\circ} \ldots . . \cos 135^{\circ}$
We know, $\cos 90^{\circ}=0$
$\therefore$ Given expression has $\cos 90^{\circ}$ and so it's value is 0 .

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
What is the value of $\cos (\pi / 9)+\cos (\pi / 3)+\cos (5 \pi / 9)+\cos (7 \pi / 9) ?$

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

The correct answer is:
$1 / 2$

Solution:
Consider $\cos \frac{\pi}{9}+\cos \frac{\pi}{3}+\cos \frac{5 \pi}{9}+\cos \frac{7 \pi}{9}$
$=\cos \frac{\pi}{9}+\frac{1}{2}+\cos \frac{5 \pi}{9}+\cos \frac{7 \pi}{9}\left(\because \cos \frac{\pi}{3}=\frac{1}{2}\right)$
$=\frac{1}{2}+\left(\cos \frac{\pi}{9}+\cos \frac{5 \pi}{9}\right)+\cos \frac{7 \pi}{9}$
$=\frac{1}{2}+\left[2 \cos \frac{6 \pi}{18} \cos \frac{4 \pi}{18}\right]+\cos \frac{7 \pi}{9}$
$\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=\frac{1}{2}+\left[2 \cos \frac{\pi}{3} \cos \frac{2 \pi}{9}\right]+\cos \frac{7 \pi}{9}$
$=\frac{1}{2}+\left[2 \cdot \frac{1}{2} \cos \frac{2 \pi}{9}\right]+\cos \frac{7 \pi}{9}$
$=\frac{1}{2}+\cos \frac{2 \pi}{9}+\cos \frac{7 \pi}{9}=\frac{1}{2}+2 \cos \left(\frac{9 \pi}{18}\right) \cos \left(\frac{5 \pi}{18}\right)$
$=\frac{1}{2}+2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{18}=\frac{1}{2}\left(\because \cos \frac{\pi}{2}=0\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
What is the value of $\cos 306^{\circ}+\cos 234^{\circ}+\cos 162^{\circ}+\cos 18^{\circ} ?$

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

The correct answer is:
0

Solution:
Given expression is:
$\cos 306^{\circ}+\cos 234^{\circ}+\cos 162^{\circ}+\cos 18^{\circ}=\mathrm{x}($ Let $)$
$\Rightarrow \mathrm{x}=\cos \left(360^{\circ}-54^{\circ}\right)+\cos \left(180^{\circ}+54^{\circ}\right)$
$+\cos \left(180^{\circ}-18^{\circ}\right)+\cos 18^{\circ}$
$=\cos 54^{\circ}-\cos 54^{\circ}-\cos 18^{\circ}+\cos 18^{\circ}=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 2)
What is the value of $\cos \left\{\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right\} ?$

Options:
A $63 / 65$
B $33 / 65$
C $22 / 65$
D $11 / 65$

The correct answer is:
$33 / 65$

Solution:
Let $\cos ^{-1} \frac{4}{5}=\mathrm{A} \Rightarrow \cos \mathrm{A}=\frac{4}{5} \Rightarrow \sin \mathrm{A}=\frac{3}{5}$
Let $\cos ^{-1} \frac{12}{13}=\mathrm{B} \Rightarrow \cos \mathrm{B}=\frac{12}{13} \Rightarrow \sin \mathrm{B}=\frac{5}{13}$
Now, $\cos \left(\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right)=\cos (\mathrm{A}+\mathrm{B})$
$=\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}$
$=\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)-\left(\frac{3}{5}\right)\left(\frac{5}{13}\right)=\frac{33}{65}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2012 (Phase 1)
What is the value of :
$\cos \left[\tan ^{-1}\left\{\tan \left(\frac{15 \pi}{4}\right)\right\}\right] ?$

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

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

Solution:
The given trigonometric expression is:
$\cos \left[\tan ^{-1}\left\{\tan \left(\frac{15 \pi}{4}\right)\right\}\right]$
$=\cos \left[\tan ^{-1}\left\{\tan \left(4 \pi-\frac{\pi}{4}\right)\right\}\right]$
$=\cos \left[\tan ^{-1}\left\{-\tan \frac{\pi}{4}\right\}\right]=\cos \left[\tan ^{-1} \tan \left(\frac{-\pi}{4}\right)\right]$
$\left(\right.$ Since $\tan ^{-1} \theta$ is defined for $\left.\frac{-\pi}{2} < \theta < \frac{-\pi}{2}\right)$
$=\cos \left(\frac{-\pi}{4}\right)$
$\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \quad[$ since $\cos (-\theta)=\cos \theta]$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2007 (Phase 2)
What is the value of $\operatorname{cosec}(13 \pi / 12) ?$

Options:
A $\sqrt{6}+\sqrt{2}$
B $-\sqrt{6}+\sqrt{2}$
C $\sqrt{6}-\sqrt{2}$
D $-\sqrt{6}-\sqrt{2}$

The correct answer is:
$-\sqrt{6}-\sqrt{2}$

Solution:
$\operatorname{cosec}\left(\frac{13 \pi}{12}\right)=\operatorname{cosec}\left(\pi+\frac{\pi}{12}\right)$
$=-\operatorname{cosec} \frac{\pi}{12}=-\cos \operatorname{cc} 15^{\circ}$
$=-\sqrt{1+\cot ^{2} 15^{\circ}}$
$=-\sqrt{1+(2+\sqrt{3})^{2}} \quad\left[\because \cot 15^{\circ}=2+\sqrt{3}\right]$
$=-\sqrt{1+4+3+4 \sqrt{3}}$
$=-\sqrt{8+4 \sqrt{3}}=-\sqrt{6+2+2 \sqrt{12}}$
$=-\sqrt{(\sqrt{6})^{2}+(\sqrt{2})^{2}+2(\sqrt{6})(\sqrt{2})}$
$=-\sqrt{(\sqrt{6}+\sqrt{2})^{2}}=-\sqrt{6}-\sqrt{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
What is the value of cot $\left(-870^{\circ}\right) ?$

Options:
A $\sqrt{3}$
B $\frac{1}{\sqrt{3}}$
C $-\sqrt{3}$
D $-\frac{1}{\sqrt{3}}

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

Solution:
$\cot \left(-870^{\circ}\right)=-\cot \left(2 \times 360^{\circ}+150^{\circ}\right)$
$=-\cot 150^{\circ}=-\cot \left(180^{\circ}-30^{\circ}\right)=\cot 30^{\circ}=\sqrt{3} .$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 2)
What is the value of $\lambda$ for which
$(\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \times(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=(2 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})$ ?

Options:
A 2
B $-2$
C 1
D 7

The correct answer is:
2

Solution:
Given $(\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \times(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=(2 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})$
$\Rightarrow \quad 2 \hat{\mathrm{i}}-\hat{\mathrm{j}}(4 \lambda+3)+\hat{\mathrm{k}}(-2 \lambda-3)=2 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
$\Rightarrow \quad-4 \lambda-3=-11$
$\Rightarrow \quad 4 \lambda=8 \Rightarrow \lambda=2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
What is the value of $\lambda$ for which the vectors
$3 \hat{i}+4 \hat{j}-\hat{k}$ and $-2 \hat{i}+\lambda \hat{j}+10 \hat{k}$ are perpendicular?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Given vectors are $3 \hat{i}+4 \hat{j}-\hat{k}$ and $-2 \hat{i}+\lambda \hat{j}+10 \hat{k}$.
If these are perpendicular, dot product is $0 .$
$(3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(-2 \hat{i}+\lambda \hat{j}+10 \hat{k})=0$
$\Rightarrow(3)(-2)+(4)(\lambda)+(-10)=0$
$\Rightarrow-6+4 \lambda-10=0$
$\Rightarrow 4 \lambda=16$
$\Rightarrow \lambda=4$

Topic: Mathematics, Vector Algebra, NDA, NDA 2019 (Phase 1)
What is the value of $\lambda$ for which the vectors
$\hat{i}-\hat{j}+\hat{k}, 2 \hat{i}+\hat{j}-\hat{k}$ and $\lambda \hat{i}-\hat{j}+\lambda \hat{k}$ are co-planar?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
Given vectors are $\hat{i}-\hat{j}+\hat{k}, 2 \hat{i}+\hat{j}-\hat{k}$ and $\lambda \hat{i}-\hat{j}+\lambda \hat{k}$
We know given vectors are coplanar, if $\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda\end{array}\right|=0$
$\Rightarrow 1(\lambda-1)+1(2 \lambda+\lambda)+1(-2-\lambda)=0$
$\Rightarrow \lambda-1+3 \lambda-2-\lambda=0$
$\Rightarrow 3 \lambda=3 \Rightarrow \lambda=1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 1)
What is the value of $\left(\frac{i+\sqrt{3}}{-i+\sqrt{3}}\right)^{200}+\left(\frac{i-\sqrt{3}}{i+\sqrt{3}}\right)^{200}+1$ ?

Options:
A -1
B 0
C 1
D 2

The correct answer is:
0

Solution:
Consider,
$\frac{i+\sqrt{3}}{-i+\sqrt{3}}=\frac{(i+\sqrt{3})^{2}}{(\sqrt{3}-i)(\sqrt{3}+i)}($ By Rationalizing $)$
$=\frac{i^{2}+3+2 \sqrt{3} i}{3+1}\left(\right.$ using $\left.(a-b)(a+b)=a^{2}-b^{2}\right)$
$=\frac{-1+3+2 \sqrt{3} i}{4}=\frac{1+\sqrt{3} i}{2}$
$=\frac{-(-1-\sqrt{3} i)}{2}=-\omega^{2} \quad\left(\because i^{2}=-1\right)$
and consider
$\frac{i-\sqrt{3}}{i+\sqrt{3}}=\frac{(i-\sqrt{3})^{2}}{i^{2}-(\sqrt{3})^{2}}($ By Rationalizing)
$=\frac{i^{2}+3-2 i \sqrt{3}}{-4}=\frac{2-2 i \sqrt{3}}{-4}\left(\because i^{2}=-1\right)$
$=\frac{-1+i \sqrt{3}}{2}=\omega$
$\therefore\left(\frac{i+\sqrt{3}}{-i+\sqrt{3}}\right)^{200}+\left(\frac{i-\sqrt{3}}{i+\sqrt{3}}\right)^{200}+1$
$=\left(-\omega^{2}\right)^{200}+\omega^{200}+1$
$=\omega^{400}+\omega^{200}+1=\omega^{3 \times 133+1}+\omega^{3 \times 66+2}+1$
$=\left(\omega^{3}\right)^{133} \omega+\left(\omega^{3}\right)^{66} \omega^{2}+1$
$=\omega+\omega^{2}+1 \quad\left(\because \omega^{3}=1\right)$
$=0$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 1)
What is the value of $\lambda$ if the function is continuous at
$x=\frac{\pi}{2} ?$
Consider the function $f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}$
Where $x \neq \frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right)=\lambda$

Options:
A $1 / 8$
B $1 / 4$
C $1 / 2$
D 1

The correct answer is:
$1 / 8$

Solution:
Function is continuous at $\mathrm{x}=\frac{\pi}{2}$
$\mathrm{f}\left(\frac{\pi}{2}\right)=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{1-\sin \mathrm{x}}{(\pi-2 \mathrm{x})^{2}}=\frac{1}{8}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 1)
What is the value of $\lambda$ if the straight line $(2 x+3 y+4)+\lambda$ $(6 \mathrm{x}-\mathrm{y}+12)=0$ is parallel to y-axis?

Options:
A 3
B $-6$
C 4
D $-3$

The correct answer is:
3

Solution:
The given line is, $(2 x+3 y+4)+\lambda(6 x-y+12)=0$
$\Rightarrow(2+6 \lambda) x+(3-\lambda) y+4+12 \lambda=0 \quad \ldots .(1)$
Since line (i) is parallel to y-axis. $\therefore \quad$ Coefficient of $\mathrm{y}=0$
$3-\lambda=0 \Rightarrow \lambda=3$

Topic: Mathematics, Pair of Lines, NDA, NDA 2012 (Phase 2)
What is the value of $\lambda$ if the triangle whose vertices are
$\hat{i}, \hat{j}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ will be right angled?

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

The correct answer is:
0

Solution:
Let us consider triangle $\mathrm{ABC}$. Suppose $\hat{i}, \hat{j}$ and
$\hat{i}+\hat{j}+\lambda \hat{k}$ are the position vector of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$.
Then $\overrightarrow{A B}=\hat{j}-\hat{i}, \overrightarrow{A C}=\hat{j}+\lambda \hat{k}, \overrightarrow{B C}=\hat{i}+\lambda \hat{k}$
$|\overrightarrow{A B}|=\sqrt{(-1)^{2}+(1)^{2}}=\sqrt{2}$
$|\overrightarrow{B C}|=\sqrt{(1)^{2}+(\lambda)^{2}}=\sqrt{1+\lambda^{2}}$
$|\overrightarrow{A C}|=\sqrt{(1)^{2}+(\lambda)^{2}}=\sqrt{1+\lambda^{2}}$
To be $\Delta \mathrm{ABC}$ is a right angled triangle, $\angle C$ should be right angle,
i.e., $\overrightarrow{B C} \cdot \overrightarrow{A C}=0$
$\Rightarrow(\hat{i}+\lambda \hat{k}) \cdot(\hat{j}+\lambda \hat{k})=0$
$\Rightarrow 0+0+\lambda^{2}=0$
$\therefore \quad \lambda=0$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
What is the value of integral $\mathrm{I}=\int_{0}^{1} \mathrm{x}(1-\mathrm{x})^{9} \mathrm{dx}$ ?

Options:
A $\frac{1}{110}$
B $\frac{1}{111}$
C $\frac{1}{112} \quad$
D $\frac{1}{119} \quad$

The correct answer is:
$\frac{1}{110}$

Solution:
Let the given integral be,
$\mathrm{I}=\int_{0}^{1} \mathrm{x}(1-\mathrm{x})^{9} \mathrm{~d} \mathrm{x}$
Put $1-\mathrm{x}=\mathrm{t} \Rightarrow \mathrm{dx}=-\mathrm{dt}$ and $\mathrm{x}=1-\mathrm{t}$
when $x=0, t=1$ and when $x=1, t=0$
$\Rightarrow \quad I=\int_{1}^{0}(1-i) i^{9}(-d l)$
$=-\int_{1}^{0}\left(-\mathrm{t}^{10}+\mathrm{t}^{9}\right) \mathrm{dt}=-\int_{1}^{0}\left(-\mathrm{t}^{10}+\mathrm{t}^{9}\right) \mathrm{dt}$
$\left[\frac{-\mathrm{t}^{11}}{11}+\frac{\mathrm{t}^{10}}{10}\right]_{0}^{1}=\frac{-1}{11}+\frac{1}{10}=\frac{-10+11}{110}=\frac{1}{110}$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 2)
What is the value of $k$ for which the following function $f(x)$ is continuous for all $x ?$
$f(x)=\left\{\begin{array}{ll}\frac{x^{3}-3 x+2}{(x-1)^{2}} & , \text { for } x \neq 1 \\ k & , \text { for } \quad x=1\end{array}\right.$

Options:
A 3
B 2
C 1
D -1

The correct answer is:
3

Solution:
Let $f(x)=\left\{\begin{array}{cl}\frac{x^{3}-3 x+2}{(x-1)^{2}}, & \forall x \neq 1 \\ k, & \forall x=1\end{array}\right.$
and $f(x)$ is continuous.
$\therefore \quad \lim _{x \rightarrow 1} f(x)=k$
$\Rightarrow \quad \lim _{x \rightarrow 1} \frac{x^{3}-3 x+2}{(x-1)^{2}}=k$
$\Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{3 x^{2}-3}{2(x-1)} \quad[$ By $L$ 'Hospitals rule $]$
$\Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{6 x}{2} \quad[$ ByL'Hospitals rule $]$
$\Rightarrow k=3$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2011 (Phase 1)
What is the value of $k$, if
$\left|\begin{array}{lll}k & b+c & b^{2}+c^{2} \\ k & c+a & c^{2}+a^{2} \\ k & a+b & a^{2}+b^{2}\end{array}\right|=(a-b)(b-c)(c-a) ?$

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

The correct answer is:
1

Solution:
Let $\left|\begin{array}{ccc}k & b+c & b^{2}+c^{2} \\ k & c+a & c^{2}+a^{2} \\ k & a+b & a^{2}+b^{2}\end{array}\right|=\Delta$
$=k\left|\begin{array}{ccc}1 & 1 & 1 \\ b+c & c+a & a+b \\ b^{2}+c^{2} & c^{2}+a^{2} & a^{2}+b^{2}\end{array}\right|$
Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$
$=k\left|\begin{array}{ccc}1 & 0 & 0 \\ b+c & a-b & a-c \\ b^{2}+c^{2} & a^{2}-b^{2} & a^{2}-c^{2}\end{array}\right|$
$=k\left|\begin{array}{ccc}1 & 0 & 0 \\ b+c & a-b & a-c \\ b^{2}+c^{2} & (a-b)(a+b) & (a-c)(a+c)\end{array}\right|$
$=k(a-b)(a-c)\left|\begin{array}{cc}1 & 1 \\ a+b & a+c\end{array}\right|$
$=k(a-b)(a-c)(a+c-a-b)$
$=k(a-b)(b-c)(c-a)$
But given $\Delta=(a-b)(b-c)(c-a)$
Thus, $k=1$

Topic: Mathematics, Determinants, NDA, NDA 2009 (Phase 1)
What is the value of $k$ if the area bounded by the curve $y=\sin k x, y=0, x=\pi / k, x=\pi /(3 k)$ is 3 sq unit?

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

The correct answer is:
$1 / 2$

Solution:
Given $x=\pi / k, x=\frac{\pi}{3 k}$ and $y=\sin k x$
Let the required area be $A$ So, $\mathrm{A}=3$ (Given)
ThereforeArea, $A=\int_{\pi / 3 k}^{\pi / k} \sin k x d x$
$\Rightarrow 3=-\left[\frac{\cos k x}{k}\right]_{\pi / 3 k}^{\pi / k}$
$\Rightarrow 3=-\frac{1}{k}\left[\cos \pi-\cos \frac{\pi}{3}\right]$
$\Rightarrow 3=-\frac{1}{k}\left[-1-\frac{1}{2}\right]$
$(\because \cos \pi=\cos (\pi / 2+\pi / 2)$
$\quad=-\sin \pi / 2$ and $\left.\cos 60^{\circ}=\frac{1}{2}\right)$
$\Rightarrow 3=\frac{3}{2 k} \Rightarrow k=\frac{1}{2}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2009 (Phase 2)
What is the value of
$\frac{\log _{27} 9 \times \log _{16} 64}{\log _{4} \sqrt{2}} ?$

Options:
A $\frac{1}{6}$
B $\frac{1}{4}$
C 8
D 4

The correct answer is:
4

Solution:
The given logarithm expression
$\frac{\log _{27} 9 \log _{16} 64}{\log _{4} \sqrt{2}}$
is simplified as:
$\frac{\log 9}{\log 27} \times \frac{\log 64}{\log 16} \times \frac{\log 4}{\log \sqrt{2}}$
$=\frac{2 \log 3}{3 \log 3} \times \frac{6 \log 2}{4 \log 2} \times \frac{2 \log 2}{\frac{1}{2} \log 2}$
$=\frac{2}{3} \times \frac{6}{4} \times 4=4$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
What is the value of $\frac{\left(\log _{27} 9\right)\left(\log _{16} 64\right)}{\log _{4} \sqrt{2}}$ ?

Options:
A 1
B 2
C 4
D 8

The correct answer is:
4

Solution:
Consider $\frac{\left(\log _{27} 9\right)\left(\log _{16} 64\right)}{\log _{4} \sqrt{2}}$
$=\frac{\log _{3^{3}}\left(3^{2}\right) \log _{4^{2}}(4)^{3}}{\log _{2^{2}}\left(2^{1 / 2}\right)}$
$=\frac{\frac{2}{3} \log _{3} 3 \times \frac{3}{2} \log _{4} 4}{\frac{1}{2 \times 2} \log _{2} 2}=\frac{1}{\frac{1}{4}}=4$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
What is the value of $\frac{\log \sqrt{\alpha \beta}(H)}{\log \sqrt{\alpha \beta \gamma}(H)}$ ?

Options:
A $\log _{\alpha \beta}(\alpha)$
B $\log _{\alpha \beta \gamma}(\alpha \beta)$
C $\log _{\alpha \beta}(\alpha \beta \gamma)$
D $\log _{\alpha \beta}(\beta)$

The correct answer is:
$\log _{\alpha \beta}(\alpha \beta \gamma)$

Solution:
Consider $\frac{\log _{\sqrt{\alpha \rho}} H}{\log _{\sqrt{a \gamma}} H}=\frac{\log _{H} \sqrt{a_{n} r}}{\log _{H} \sqrt{x \beta}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
What is the value of $\lim _{x \rightarrow 0} \frac{x \sin 5 x}{\sin ^{2} 4 x}$

Options:
A 0
B $\frac{5}{4}$
C $\frac{5}{16}$
D $\frac{25}{4}$

The correct answer is:
$\frac{5}{16}$

Solution:
$\lim _{x \rightarrow 0} \frac{x \sin 5 x}{\sin ^{2} 4 x}$
[multiply denominator and number with $\mathrm{x}]$ We get,
$\lim _{x \rightarrow 0} \frac{x^{2} \sin 5 x}{x \sin ^{2} 4 x}=\lim _{x \rightarrow 0} \frac{\sin 5 x}{x} \cdot \frac{x^{2}}{\sin ^{2} 4 x}$
Rearranging to bring a standard form, we get,
$\lim _{x \rightarrow 0} \frac{5 \sin 5 x}{5 x} \cdot \frac{(4 x)^{2}}{16 \sin ^{2} 4 x}$
$=\frac{5}{16}\left(\lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}\right) \cdot \frac{1}{\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{4 x}\right)^{2}}=\frac{5}{16}$

Topic: Mathematics, Limits, NDA, NDA 2006 (Phase 2)
What is the value of $\operatorname{Lim}_{x \rightarrow 0} \frac{\cos (a x)-\cos (b x)}{x^{2}}$ ?

Options:
A $a-b$
B $\mathrm{a}+\mathrm{b}$
C $\frac{b^{2}-a^{2}}{2}$
D $\frac{b^{2}+a^{2}}{2}$

The correct answer is:
$\frac{b^{2}-a^{2}}{2}$

Solution:
Consider $\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{-a \sin a x+b \sin b x}{2 x}$
(using L' Hospital's rule)
$=\lim _{x \rightarrow 0} \frac{-a^{2} \cos a x+b^{2} \cos b x}{2}$
(using L' Hospital's rule)
$=\frac{b^{2}-a^{2}}{2}(\because \cos 0=1)$

Topic: Mathematics, Limits, NDA, NDA 2010 (Phase 2)
What is the value of $\lim _{x \rightarrow 0} \frac{\sin |x|}{x}$ ?

Options:
A 1
B $-1$
C $\infty$
D Limit does not exist

The correct answer is:
Limit does not exist

Solution:
$\lim _{\mathrm{x} \rightarrow 0} \frac{\sin |\mathrm{x}|}{\mathrm{x}}, \mathrm{LHL}$ is limit when $\mathrm{x} < 0$
$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 0} \frac{\sin (-\mathrm{x})}{\mathrm{x}}=-\lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=-1$
RHL is limit when $\mathrm{x}>0$
$\mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 0} \frac{\sin (\mathrm{x})}{\mathrm{x}}=1$
So, LHL $\neq$ RHL Hence, Limit does not exist.

Topic: Mathematics, Limits, NDA, NDA 2006 (Phase 1)
What is the value of $\lim _{x \rightarrow 1} \frac{(x-1)^{2}}{|x-1|} ?$

Options:
A 0
B 1
C -1
D The limit does not exist

The correct answer is:
0

Solution:
Let $f(x)=\frac{(x-1)^{2}}{|x-1|}=\left\{\begin{array}{l}(x-1), x \geq 1 \\ -(x-1), x < 1\end{array}\right.$
Now, LHL $\begin{aligned} &=\lim _{h \rightarrow 0} f(1-h) \\=& \lim _{h \rightarrow 0}[-(1-h-1)]=\lim _{h \rightarrow 0} h=0 \end{aligned}$
and $\mathrm{RHL}=\lim _{h \rightarrow 0} f(1+h)$
$\quad=\lim _{h \rightarrow 0}(1+h-1)=\lim _{h \rightarrow 0} h=0$
$\therefore \mathrm{LHL}=\mathrm{RHL}$
$\therefore \quad \lim _{x \rightarrow 0} f(x)=\mathrm{LHL}=\mathrm{RHL}$
$\Rightarrow \quad \lim _{x \rightarrow 0} \frac{(x-1)^{2}}{|x-1|}=0$

Topic: Mathematics, Limits, NDA, NDA 2010 (Phase 1)
What is the value of $\lim _{x \rightarrow a} \frac{\sqrt{\alpha+2 x}-\sqrt{3 x}}{\sqrt{3 \alpha+x}-2 \sqrt{x}}$ ?

Options:
A $\frac{2}{\sqrt{3}}$
B $\frac{1}{(3 \sqrt{3})}$
C $\frac{2}{(3 \sqrt{3})}$
D $\frac{1}{\sqrt{3}}$

The correct answer is:
$\frac{1}{\sqrt{3}}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\sqrt{\alpha+2 x}-\sqrt{3 x}}{\sqrt{3 \alpha+x}-2 \sqrt{x}} \\ &=\lim _{x \rightarrow \alpha} \frac{(\sqrt{\alpha+2 x})^{2}-(\sqrt{3 x})^{2}}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \times \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{(\sqrt{3 \alpha+x})^{2}-(2 \sqrt{x})^{2}} \\ &=\lim _{x \rightarrow \alpha} \frac{\alpha+2 x-3 x}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \times \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{3 \alpha+x-4 x} \\ &=\lim _{x \rightarrow \alpha} \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \times \frac{(\alpha-x)}{3(\alpha-x)} \\ &=\frac{1}{3} \lim _{x \rightarrow \alpha} \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \\ &=\frac{1}{3} \frac{\sqrt{4 \alpha}+2 \sqrt{\alpha}}{\sqrt{3 \alpha}+\sqrt{3 \alpha}}=\frac{1}{2}\left(\frac{4 \sqrt{\alpha}}{2 \sqrt{3} \sqrt{\alpha}}\right)=\frac{1}{\sqrt{3}} \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2008 (Phase 2)
What is the value of $\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$

Options:
A e
B $e^{2}$
C $e^{4}$
D $e^{5}$

The correct answer is:
$e^{5}$

Solution:
$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=\lim _{x \rightarrow \infty}\left(\frac{x+5+1}{x+1}\right)^{x+4}=\lim _{x \rightarrow \infty}\left(\frac{x+1}{x+1}+\frac{5}{x+1}\right)^{x+4}$
$=\lim _{x \rightarrow \infty}\left(1+\frac{5}{x+1}\right)^{\frac{x+4}{5} \times \frac{5(x+1)}{x+1}}$
$=\lim _{x \rightarrow \infty}\left[\left(1+\frac{5}{x+1}\right)^{\frac{x+1}{5}}\right]^{5\left(\frac{x+4}{x+1}\right)}$
$=e^{5 \lim _{x \rightarrow \infty} \frac{1+\frac{4}{x}}{x+\frac{1}{x}}}=e^{5} \quad\left(\because \lim _{x \rightarrow \infty} \frac{1+4 / x}{1+1 / x}=1\right)$

Topic: Mathematics, Limits, NDA, NDA 2010 (Phase 1)
What is the value of $\lim _{x \rightarrow \infty}\left(\frac{x-2}{x+2}\right)^{x+2}$ ?

Options:
A 0
B $\mathrm{e}^{4}$
C $\mathrm{e}^{-2}$
D $\mathrm{e}^{-4}$

The correct answer is:
$\mathrm{e}^{-4}$

Solution:
$\lim _{x \rightarrow 0}\left(\frac{x-2}{x+2}\right)^{x+2}$
can be written as
$\lim _{x \rightarrow 0}\left(\frac{x+2-4}{x+2}\right)^{x+2}$
or, $\lim _{x \rightarrow \infty}\left\{1-\frac{4}{(x+2)}\right\}^{x+2}$
Putting $\mathrm{x}+2=\mathrm{t}$
when $\mathrm{x} \rightarrow \infty, \mathrm{t} \rightarrow \infty$
So, $\lim _{t \rightarrow \infty}\left\{1-\frac{4}{t}\right\}^{t}$
or, $\lim _{t \rightarrow \infty}\left\{1-\frac{4}{t}\right\}^{\frac{t}{4} \times 4}$
or, $\lim _{x \rightarrow \infty}\left\{\left(1-\frac{4}{t}\right)^{\frac{t}{4}}\right\}^{4}=\left(e^{-1}\right)^{4}=e^{-4}$

Topic: Mathematics, Limits, NDA, NDA 2007 (Phase 1)
What is the value of $\lim _{x \rightarrow \infty} \frac{\sin x}{x}$ ?

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

The correct answer is:
0

Solution:
Given limit is: $\lim _{x \rightarrow \infty} \frac{\sin x}{x}$
Let, $\mathrm{x}=\frac{1}{\mathrm{~h}}$ and as $\mathrm{x} \rightarrow \infty, \mathrm{h} \rightarrow 0$
$\therefore \lim _{x \rightarrow \infty} \frac{\sin x}{x}$ change to : $\lim _{h \rightarrow 0} h \sin \frac{1}{h}$
$=0$ (value of $\sin \frac{1}{\mathrm{~h}}$ is finite as it lies between $-1$ and 1 )

Topic: Mathematics, Limits, NDA, NDA 2008 (Phase 1)
What is the value of $\log _{2}\left(\log _{3} 81\right)$ ?

Options:
A 2
B 3
C 4
D 9

The correct answer is:
2

Solution:
Let $\log _{2}\left(\log _{3} 81\right)=x$
$\Rightarrow \log _{2}\left(\log _{3} 3^{4}\right)=x$
$\Rightarrow \log _{2}\left(4 \log _{3} 3\right)=x$
$\Rightarrow \log _{2}(4)=x$
$\left(\because \log _{a} a=1\right)$
$\Rightarrow 4=2$
$\Rightarrow 2^{2}=2^{x} \Rightarrow x=2$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 2)