datavorous/entrance-exam-dataset huggingface dataset:
What is the value of $1-\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} ?

Options:
A $1 / 8$
B $3 / 8$
C $5 / 8$
D $7 / 8$

The correct answer is:
$7 / 8$

Solution:
$\begin{aligned} 1 &-\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} \\=& 1-\frac{1}{2}\left[2 \sin 70^{\circ} \sin 10^{\circ} \sin 50^{\circ}\right] \\=& 1-\frac{1}{2}\left[\left(\cos 60^{\circ}-\cos 80^{\circ}\right) \sin 50^{\circ}\right] \\[\because 2 \sin \mathrm{A} \sin \mathrm{B}=\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})] \\=& 1-\frac{1}{2}\left[\frac{1}{2} \sin 50^{\circ}-\frac{1}{2} 2 \cos 80^{\circ} \sin 50^{\circ}\right] \\=& 1-\frac{1}{4}\left[\sin 50^{\circ}-\sin 130^{\circ}+\sin 30^{\circ}\right] \\(\because 2 \cos \mathrm{A} \sin \mathrm{B}=\sin (\mathrm{A}+\mathrm{B})-\sin (\mathrm{A}-\mathrm{B})) \\=& 1-\frac{1}{8}=\frac{7}{8} \quad\left(\because \sin 130^{\circ}=\sin \left(180^{\circ}-50^{\circ}=\sin 50^{\circ}\right)\right.\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
What is the value of $\left(\frac{-1+i \sqrt{3}}{2}\right)^{3 n}+\left(\frac{-1+i \sqrt{3}}{2}\right)^{3 n}$ where
$\mathrm{i}=\sqrt{-1}$ ?

Options:
A 3
B 2
C 1
D 0

The correct answer is:
2

Solution:
From cube root of unity
$\omega=\frac{-1+\sqrt{3} i}{2}$ and
$$
\begin{aligned} \omega^{2} &=\frac{-1-\sqrt{3} i}{2} \\ \omega^{3} &=\left(\frac{-1+\sqrt{3} i}{2}\right)\left(\frac{-1-\sqrt{3} i}{2}\right) \\ &=\frac{(-1)^{2}-(\sqrt{3} i)^{2}}{4}=\frac{4}{4}=1 \\ \text { Now, }(\omega)^{3 n}+\left(\omega^{2}\right)^{3 n} \\=1+1=2 \end{aligned}
$$

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 2)
What is the value of $2 \log _{8} 2-\frac{10 \mathrm{~g}_{3} 7}{2}$ ?

Options:
A 0
B 1
C $8 / 3$
D $16 / 3$

The correct answer is:
0

Solution:
$2 \log _{8} 2-\frac{\log _{3} 9}{3}=2 \log _{2} 32-\frac{\log _{3} 3^{2}}{3}$
$=\frac{2}{3} \log _{2} 2-2 \frac{\log _{3} 3}{3}=\frac{2}{3}-\frac{2}{3}=0 \quad\left(\because \log _{a} a=1\right)$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
What is the value of 2 logg $2-\frac{1}{3} \log _{3}$

Options:
A 0
B 1
C 2
D $1 / 3$

The correct answer is:
0

Solution:
Consider $2 \log _{8} 2-\frac{1}{3} \log _{3} 9$
$=2 \cdot \frac{\log 2}{\log 8}-\frac{1}{3} \cdot \frac{\log 9}{\log 3}=2 \cdot \frac{\log 2}{\log 2^{3}}-\frac{1}{3} \cdot \frac{\log 3^{2}}{\log 3}$
$=2 \cdot \frac{\log 2}{3 \log 2}-\frac{1}{3} 2 \cdot \frac{\log 3}{\log 3}=\frac{2}{3}-\frac{2}{3}=0$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
What is the value of $\sqrt{3}$ cosec $20^{\circ}-\sec 20^{\circ} ? \quad[2009-I I]$

Options:
A 4
B 3
C 2
D 1

The correct answer is:
4

Solution:
Consider $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$
$=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply and divide by 2 in $N^{r}$.
$=2\left(\frac{\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\right)$
$=2\left(\frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\right)$
$\left(\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right.$ and $\left.\cos 60^{\circ}=\frac{1}{2}\right)$
$=\frac{2 \times 2\left[\sin \left(60^{\circ}-20^{\circ}\right)\right]}{2 \sin 20^{\circ} \cos 20^{\circ}}$
$(\because \sin A \cos B-\cos A \sin B=\sin (A-B)$ and $\sin 2 \theta=$
$2 \sin \theta \cos \theta)$
$=\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}}=4$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
What is the value of $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} ?$

Options:
A 4
B 2
C 1
D 0

The correct answer is:
4

Solution:
$\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$
$=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
$=\frac{4 \cdot \frac{1}{2}\left(\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}\right)}{2 \sin 20^{\circ} \cos 20^{\circ}}$
$=\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)\left(\frac{4}{2 \sin 20^{\circ} \cos 20^{\circ}}\right)$
$=\left(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}\right)\left(\frac{4}{\sin 40^{\circ}}\right)$
$=\sin 40^{\circ} \frac{4}{\sin 40^{\circ}}=4$
$(\because \sin (A-B)=\sin A \cos B-\cos A \sin B)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 1)
What is the value of $\sqrt{3}$ cosec $20^{\circ}-\sec 20^{\circ} ?

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

The correct answer is:
4

Solution:
Consider $\sqrt{3}$ cosec $20^{\circ}-\sec 20^{\circ}$
$=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply and divide by 4 $=\frac{4}{2 \sin 20^{\circ} \cos 20^{\circ}}\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)$
$=\frac{4}{\sin 40^{\circ}}\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)$
$(\because \sin 2 \theta=2 \sin \theta \cos \theta)$
$=\frac{4}{\sin 40^{\circ}}\left(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}\right)$
$=\frac{4}{\sin 40^{\circ}} \sin \left(60^{\circ}-20^{\circ}\right)$
$=4$
$(\because \sin (\mathrm{A}-\mathrm{B})=\sin A \cos B-\cos A \sin B)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 1)
What is the value of $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}(\sin \times \tan x) \mathrm{dx} ? \quad$

Options:
A $-\frac{1}{\sqrt{2}}+\ell \mathrm{n}\left(\frac{1}{\sqrt{2}}\right)$
B $\frac{1}{\sqrt{2}}$
C 0
D $\sqrt{2}$

The correct answer is:
0

Solution:
$\begin{aligned} \text { (c) } & \int_{-\pi}^{\frac{\pi}{4}}(\sin x-\tan x) d x . \\ & \text { Let } f(x)=\sin x-\tan x \\ & f(-x)=\sin (-x)-\tan (-x) \\ &=-\sin x+\tan x=-(\sin x-\tan x) \\ &=-f(x) . \end{aligned}$
So, $f(x)$ is an odd function.
We know $\int_{-a}^{a} f(x) d x=0$, if $f(x)$ is odd function.

Topic: Mathematics, Definite Integration, NDA, NDA 2018 (Phase 1)
What is the value of
$\sqrt{5 \sqrt{5 \sqrt{5 \sqrt{\ldots \infty}}}} ?$

Options:
A 5
B $\sqrt{5}$
C 1
D $(5)^{1 / 4}$

The correct answer is:
5

Solution:
Let $\mathrm{x}=\sqrt{5 \sqrt{5 \sqrt{5 \ldots \ldots \infty}}}$
$\Rightarrow x=\sqrt{5 x} \Rightarrow x^{2}=5 x$
$x^{2}-5 x=0 \Rightarrow x(x-5)=0$
$\Rightarrow x=0$ or 5
5 is given in the option.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 1)
What is the value of $\int_{\pi / 6}^{\pi / 4} \frac{d x}{\sin x \cos x} ? \quad\mathrm{}$

Options:
A $2 \ln 3$
B $2 \ln \sqrt{3}$
C $\ln \sqrt{3}$
D $4 \ln 3$

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

Solution:
Let $\mathrm{I}=\int_{\pi / 6}^{\pi / 4} \frac{d x}{\sin x \cos x}$
multiply and divide by 2
$=2 \int_{\pi / 6}^{\pi / 4} \frac{d x}{2 \sin x \cos x}=2 \int_{\pi / 6}^{\pi / 4} \frac{d x}{\sin 2 x}$
$=2 \int_{\pi / 6}^{\pi / 4} \cos e c 2 x d x=2[\log \tan x]_{\pi / 6}^{\pi / 4} \cdot \frac{1}{2}$
$=[\log \tan \pi / 4-\log \tan \pi / 6]$
$=\log 1-\log \frac{1}{\sqrt{3}}=0-\log \frac{1}{\sqrt{3}}$
$=\log \sqrt{3}(\because \log 1=0)$

Topic: Mathematics, Definite Integration, NDA, NDA 2010 (Phase 2)
What is the value of
$\sqrt{8+2 \sqrt{8+2 \sqrt{8+2 \sqrt{8+\ldots .}}}}$

Options:
A 10
B 8
C 6
D 4

The correct answer is:
4

Solution:
Let $y=\sqrt{8+2 y}$
$\Rightarrow y^{2}=8+2 y \Rightarrow y^{2}-2 y-8=0$
$\Rightarrow(y+2)(y-4)=0$
$\Rightarrow y=4,-2$
Hence, required value of given expression is $4 .$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
What is the value of
${ }^{8} C_{0}-{ }^{8} C_{1}+{ }^{8} C_{2}-{ }^{8} C_{3}+{ }^{8} C_{4}-{ }^{8} C_{5}+{ }^{8} C_{6}-{ }^{8} C_{7}+{ }^{8} C_{8}$

Options:
A 0
B 1
C 2
D $2^{8}$

The correct answer is:
0

Solution:
$\begin{array}{ll}(1-x)^{\mathrm{n}}={ }^{n} \mathrm{C}_{0}-{ }^{n} \mathrm{C}_{1}(x)+{ }^{n} \mathrm{C}_{2} x^{2}-{ }^{n} \mathrm{C}_{3} x^{3} & +\ldots \\ & +(-1)^{n n} \mathrm{C}_{n} \\ & \text { Put } x=1 \text { and } n=8 \\ & \therefore(1-1)^{8}={ }^{8} \mathrm{C}_{0}-{ }^{8} \mathrm{C}_{1}+{ }^{8} \mathrm{C}_{2}-{ }^{8} \mathrm{C}_{3}+\ldots+{ }^{8} \mathrm{C}_{8} \\ \Rightarrow & \left({ }^{8} \mathrm{C}_{0}-{ }^{8} \mathrm{C}_{1}+{ }^{8} \mathrm{C}_{2}-{ }^{8} \mathrm{C}_{3}+\ldots+{ }^{8} \mathrm{C}_{8}\right)=0\end{array}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2008 (Phase 2)
What is the value of $9^{1 / 3} .9^{1 / 19} \cdot 9^{1 / 27} \ldots \ldots \infty$ ?

Options:
A 9
B 3
C $9^{1 / 3}$
D 1

The correct answer is:
3

Solution:
The given expression $9^{1 / 3}, 9^{1 / 9} 9^{1 / 27} \ldots \infty$ Can be written as:
$9^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots \infty}=9^{\frac{1}{3}}{1-\frac{1}{3}}=9^{\frac{1}{3} / \frac{2}{3}}=9^{1 / 2}=3$

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
What is the value of
$\frac{(0.101)_{2}^{(11)_{2}}+(0.011)_{2}^{(11)_{2}}}{(0.101)_{2}^{(10)_{2}}-(0.101)_{2}^{(01)_{2}}(0.011)_{2}^{(01)_{2}}+(0.011)_{2}^{(10)_{2}}}$

Options:
A $(0.001)_{2}$
B $(0.01)_{2}$
C $(0.1)_{2}$
D $(1)_{2}$

The correct answer is:
$(1)_{2}$

Solution:
$(0.101)_{2}=2^{-1} \times 1+2^{-2} \times 0+2^{-3} \times 1$
$=\frac{1}{2}+0+\frac{1}{8}=\frac{5}{8}$
and $(0.011)_{2}=0 \times 2^{-1}+1 \times 2^{-2}+1 \times 2^{-3}$
$=0+\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$
Also, $(11)_{2}=1 \times 2^{1}+1 \times 2^{0}=3$
and $(01)_{2}=0 \times 2^{1} \times 1 \times 2^{0}=1$
$\frac{(0.101)_{2}^{(11)}+(0.011)_{2}^{(11)_{2}}}{(0.101)_{2}^{(10)_{2}}-(0.101)_{2}^{(0)]_{2}}(0.011)_{2}^{(01)}+(0.011)_{2}^{(10)_{2}}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
What is the value of
$\frac{(1001)_{2}^{(11)_{2}}-(101)_{2}^{(11)_{2}}}{(1001)_{2}^{(10)_{2}}+(1001)_{2}^{(0) / 2}(101)_{2}^{(01) / 2}+(101)_{2}^{(10) / 2}} ?$

Options:
A $(1001)_{2}$
B $(101)_{2}$
C $(110)_{2}$
D $(100)_{2}$

The correct answer is:
$(100)_{2}$

Solution:
Converting from binary to decimal $(1001)_{2}=1 \times 2^{3}+2^{0}=8+1=9$
$\begin{array}{l}(11)_{2}=2^{1}+2^{0}=2+1=3 \\ (101)_{2} \\ (10)_{2}=2^{1}=2\end{array} \quad=2^{2}+2^{0}=4+1=5$
and $(01)_{2}=1$
$\frac{(1001)_{2}^{(11)_{2}}-(101)_{2}^{(11)_{2}}}{(1001)_{2}^{(10)_{2}}+(1001)_{2}^{(01)_{2}}(101)_{2}^{(01)_{2}}+(101)_{2}^{(10)} 2}$
$=\frac{9^{3}-5^{3}}{9^{2}+9 \times 5+5^{2}}$
$=\frac{(9-5)\left(9^{2}+9 \times 5+5^{2}\right)}{\left(9^{2}+9 \times 5+5^{2}\right)}=\frac{4 \times(81+45+25)}{(81+45+25)}$
$=4=(100)_{2}$ [Converting from decimal to binary]

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
What is the value of $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ if $a^{3}+b^{3}+c^{3}=0 ?$

Options:
A 0
B 1
C $3 \mathrm{abc}$
D $-3$ abc

The correct answer is:
$3 \mathrm{abc}$

Solution:
$\left|\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{b} & \mathrm{c} & \mathrm{a} \\ \mathrm{c} & \mathrm{a} & \mathrm{b}\end{array}\right|$
$=\mathrm{a}\left(\mathrm{bc}-\mathrm{a}^{2}\right)-\mathrm{b}\left(\mathrm{b}^{2}-\mathrm{ac}\right)+\mathrm{c}\left(\mathrm{ab}-\mathrm{c}^{2}\right)$
$=\mathrm{abc}-\mathrm{a}^{3}-\mathrm{b}^{3}+\mathrm{abc}+\mathrm{abc}-\mathrm{c}^{3}=3 \mathrm{abc}-\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)$
Given that $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=0$
$\Rightarrow\left|\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{b} & \mathrm{c} & \mathrm{a} \\ \mathrm{c} & \mathrm{a} & \mathrm{b}\end{array}\right|=3 \mathrm{abc} .$

Topic: Mathematics, Determinants, NDA, NDA 2006 (Phase 2)
What is the value of a?
Given a function
$f(x)=\left\{\begin{array}{ccc}-1 & \text { If } & x \leq 0 \\ a x+b & \text { If } & 0 < x < 1 \\ 1 & \text { If } & x \geq 1\end{array}\right.$
where a, b are constants. The function is continuous everywhere.

Options:
A -1
B 0
C 1
D 2

The correct answer is:
-1

Solution:
$F(x)=\left\{\begin{array}{ccc}-1 & , & x \leq 0 \\ a x+b & , & 0 < x < 1 \\ 1 & , & x \geq 1\end{array}\right.$
At $\mathrm{x}=0$,
L.H.L. $=$ R.H.L.
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
$-1=\lim _{x \rightarrow 0}(a x+b)$
$-1=\mathrm{b}$
Since the given function is also continuous at $\mathrm{x}=1$
L.H.L. $=$ R.H.L.
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
$\mathrm{a}+\mathrm{b}=1$
$\mathrm{a}-1=1$
$\mathrm{a}=1+1=2 .$
Function is continuous at $\mathrm{x}=0$
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
$-1=a(0)+b \Rightarrow b=-1$
Now, $\mathrm{f}(\mathrm{x})$ is also continuous at $\mathrm{x}=1$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
$a(1)+b=1$
$a=1-b=1+1$
$\mathrm{a}=2 .$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 1)
What is the value of A?
Given that $\lim _{x \rightarrow \infty}\left(\frac{2+x^{2}}{1+x}-A x-B\right)=3$.

Options:
A -1
B 1
C 2
D 3

The correct answer is:
1

Solution:
$\lim _{x \rightarrow \infty}\left(\frac{2+x^{2}}{1+x}-A x-B\right)=3$
$\Rightarrow \quad \lim _{x \rightarrow \infty}\left(\frac{2+x^{2}-A x-B-A x^{2}-B x}{1+x}\right)=3$
$\Rightarrow \quad \lim _{x \rightarrow \infty}\left[\frac{(1-A) x^{2}-(A+B) x+2-B}{1+x}\right]=3$
Applying L'Hospital rule, $2 x(1-A)-(A+B)=3$
Comparing coefficients $\begin{array}{ll} & 2(1-A)=0 \\ \therefore & A=1 \text { and } \\ & -(A+B)=3 \\ \therefore & B=-3-1=-4 \\ & A+B=-3 \\ & A=1, B=-4\end{array}$

Topic: Mathematics, Limits, NDA, NDA 2015 (Phase 1)
What is the value of b?
Given a function
$f(x)=\left\{\begin{array}{ccc}-1 & \text { If } & x \leq 0 \\ a x+b & \text { If } & 0 < x < 1 \\ 1 & \text { If } & x \geq 1\end{array}\right.$
where a, b are constants. The function is continuous everywhere.

Options:
A -1
B 1
C 0
D 2

The correct answer is:
1

Solution:
$F(x)=\left\{\begin{array}{ccc}-1 & , & x \leq 0 \\ a x+b & , & 0 < x < 1 \\ 1 & , & x \geq 1\end{array}\right.$
At $\mathrm{x}=0$,
L.H.L. $=$ R.H.L.
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
$-1=\lim _{x \rightarrow 0}(a x+b)$
$-1=\mathrm{b}$
Since the given function is also continuous at $\mathrm{x}=1$
L.H.L. $=$ R.H.L.
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
$\mathrm{a}+\mathrm{b}=1$
$\mathrm{a}-1=1$
$\mathrm{a}=1+1=2$.
$\mathrm{b}=-1$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 1)
What is the value of $\mathrm{B}$ ?
Given that $\lim _{x \rightarrow \infty}\left(\frac{2+x^{2}}{1+x}-A x-B\right)=3$.

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

The correct answer is:
-4

Solution:
$\lim _{x \rightarrow \infty}\left(\frac{2+x^{2}}{1+x}-A x-B\right)=3$
$\Rightarrow \quad \lim _{x \rightarrow \infty}\left(\frac{2+x^{2}-A x-B-A x^{2}-B x}{1+x}\right)=3$
$\Rightarrow \quad \lim _{x \rightarrow \infty}\left[\frac{(1-A) x^{2}-(A+B) x+2-B}{1+x}\right]=3$
Applying L'Hospital rule, $2 x(1-A)-(A+B)=3$
Comparing coefficients $\begin{array}{ll} & 2(1-A)=0 \\ \therefore & A=1 \text { and } \\ & -(A+B)=3 \\ \therefore & B=-3-1=-4 \\ & A+B=-3 \\ & A=1, B=-4\end{array}$

Topic: Mathematics, Limits, NDA, NDA 2015 (Phase 1)
What is the value of $\mathrm{b}$ such that the scalar product of the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ with the unit vector parallel to the sum of
the vectors $2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ and $\mathrm{b} \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ is unity?

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

The correct answer is:
1

Solution:
Let $\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{B}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$
and $\overrightarrow{\mathrm{C}}=\mathrm{b} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
Now, $\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}+\mathrm{b} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$=(2+\mathrm{b}) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
The unit vector parallel to $\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}$ is
$\hat{n}=\frac{(2+b) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+b)^{2}+6^{2}+(-2)^{2}}}=\frac{(2+b) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{b^{2}+4 b+44}}$
Now, $(\hat{i}+\hat{j}+\hat{k}) \cdot \hat{n}=1 \Rightarrow \frac{2+b+6-2}{\sqrt{b^{2}+4 b+44}}=1$
$\Rightarrow 2+b+6-2=\sqrt{b^{2}+4 b+44}$
$\Rightarrow \quad 8 \mathrm{~b}=8$
$\Rightarrow \quad b=1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
What is the value of $\frac{\cos 10^{\circ}-\sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}}$ ?

Options:
A $\tan 35^{\circ}$
B $\tan 10^{\circ}$
C $\frac{1}{\sqrt{2}}$
D $1

The correct answer is:
$\tan 35^{\circ}$

Solution:
Given expression,
$\frac{\cos 10^{\circ}-\sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}}=\frac{1-\tan 10^{\circ}}{1+\tan 10^{\circ}}=\mathrm{x} \quad$ (let)
$x=\frac{\tan 45-\tan 10^{\circ}}{1+\tan 45 \cdot \tan 10^{\circ}}=\tan (45-10)=\tan 35^{\circ} .$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 1)
What is the value of $\frac{\left(\cos 10^{\circ}+\sin 20^{\circ}\right)}{\left(\cos 20^{\circ}-\sin 10^{\circ}\right)} ?$

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

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

Solution:
The given expression is, $\frac{\cos 10^{\circ}+\sin 20^{\circ}}{\cos 20^{\circ}-\sin 10^{\circ}}$
$=\frac{\cos \left(90^{\circ}-80^{\circ}\right)+\sin 20^{\circ}}{\operatorname{cso}\left(90^{\circ}-70^{\circ}\right)-\sin 10^{\circ}}$
$=\frac{\sin 80^{\circ}+\sin 20^{\circ}}{\sin 70^{\circ}-\sin 10^{\circ}}=\frac{2 \sin \frac{80+20}{2} \cdot \cos \frac{80-20}{2}}{2 \cos \frac{70+10}{2} \cdot \sin \frac{70-10}{2}}$
$=\frac{2 \sin 50^{\circ} \cos 30^{\circ}}{2 \cos 40^{\circ} \sin 30^{\circ}}=\frac{\sin \left(90^{\circ}-40^{\circ}\right) \cot 30^{\circ}}{\cos 40^{\circ}}$
$=\frac{\cos 40^{\circ} \cot 30^{\circ}}{400}=\cot 30^{\circ}=\sqrt{3}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
What is the value of
$\frac{\sin 34^{\circ} \cos 236^{\circ}-\sin 56^{\circ} \sin 124^{\circ}}{\cos 28^{\circ} \cos 88^{\circ}+\cos 178^{\circ} \sin 208^{\circ}} ?$

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

The correct answer is:
$-2$

Solution:
$\begin{aligned} \frac{\sin 34^{\circ} \cos 236^{\circ}-\sin 56^{\circ} \sin 124^{\circ}}{\cos 28^{\circ} \cos 88^{\circ}+\cos 178^{\circ} \sin 208^{\circ}} \\= \frac{\sin 34^{\circ}\left(-\cos 56^{\circ}\right)-\sin 56^{\circ} \cos 34^{\circ}}{\cos 28^{\circ} \sin 2^{\circ}+\cos 2^{\circ} \sin 28^{\circ}} \\= \frac{-\sin 34^{\circ} \cos 56^{\circ}-\sin 56^{\circ} \cos 34^{\circ}}{\sin \left(28^{\circ}+2^{\circ}\right)} \\= \frac{\sin \left(34^{\circ}+56^{\circ}\right)}{\sin 30^{\circ}}=\frac{-\sin 90^{\circ}}{\sin 30^{\circ}}=\frac{-1}{\frac{1}{2}}=-2 \\ \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
What is the value of $\frac{\cos 15^{\circ}+\cos 45^{\circ}}{\cos ^{3} 15^{\circ}+\cos ^{3} 45^{\circ}}$ ?

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{1}{3}$
D None of these

The correct answer is:
None of these

Solution:
Consider $\frac{\cos 15^{\circ}+\cos 45^{\circ}}{\cos ^{3} 15^{\circ}+\cos ^{3} 45^{\circ}}$
$=\frac{\cos 15^{\circ}+\cos 45^{\circ}}{\left(\cos 15^{\circ}+\cos 45^{\circ}\right)\left(\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos 45^{\circ} \cos 15^{\circ}\right)}$
$\left(\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right)$
$=\frac{1}{\left(\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos 45^{\circ} \cos 15^{\circ}\right)}$
$=\frac{1}{\frac{1}{2}+\left(\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}\right)^{2}-\frac{\cos 15^{\circ}}{\sqrt{2}}}$
$\left[\because \cos 15^{\circ}=\cos \left(45^{\circ}-30^{\circ}\right)\right]$
$=\frac{1}{\frac{1}{2}+\left(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right)^{2}-\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}$
$=\frac{1}{\frac{1}{2}+\frac{3+1+2 \sqrt{3}}{8}-\frac{\sqrt{3}+1}{4}}$
$=\frac{1}{\frac{4+4+2 \sqrt{3}-2 \sqrt{3}-2}{8}}=\frac{8}{6}=\frac{4}{3}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 1)
What is the value of
$\frac{\operatorname{cosec}(\pi+\theta) \cot \{(9 \pi / 2-\theta)\} \cos \operatorname{cc}^{2}(2 \pi-\theta)}{\cot (2 \pi-\theta) \sec ^{2}(\pi-\theta) \sec \{(3 \pi / 2)+\theta\}}$

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

The correct answer is:
1

Solution:
The expression
$\frac{\cos \operatorname{cc}(\pi+\theta) \cot \left\{\frac{9 \pi}{2}-\theta\right\} \operatorname{cosec}^{2}(2 \pi-\theta)}{\cot (2 \pi-\theta) \sec ^{2}(\pi-\theta) \sec \left\{\left(\frac{3 \pi}{2}\right)+\theta\right\}}$
$=\frac{-\cos \operatorname{cc} \theta \cdot \tan \theta \cos \operatorname{cc}^{2} \theta}{-\cot \theta \sec ^{2} \theta \operatorname{cosec} \theta}$
$=\frac{\tan ^{2} \theta \cos \operatorname{cc}^{2} \theta}{\sec ^{2} \theta}=\tan ^{2} \theta \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta}$
$=\tan ^{2} \theta \times \frac{1}{\tan ^{2} \theta}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 2)
What is the value of $\alpha ?$
Consider the function $f(x)=\left\{\begin{array}{ccc}\frac{\alpha \cos x}{\pi-2 x} & \text { If } & x \neq \frac{\pi}{2} \\ 3 & \text { If } & x=\frac{\pi}{2}\end{array}\right.$
Which is continuous at $x=\frac{\pi}{2}$, where $\alpha$ is a constant.

Options:
A 6
B 3
C 2
D 1

The correct answer is:
6

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ccc}\frac{\alpha \cos \mathrm{x}}{\pi-2 \mathrm{x}} & \text { if } & \mathrm{x} \neq \frac{\pi}{2} \\ 3 & \text { if } & \mathrm{x}=\frac{\pi}{2}\end{array}\right.$
For continuity at $\mathrm{x}=\frac{\pi}{2}$
L.H.L. $=\lim _{\mathrm{x} \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \frac{\pi^{-}}{2}} \frac{\alpha \cos \mathrm{x}}{\pi-2 \mathrm{x}}$
Put $\mathrm{x}=\frac{\pi}{2}-\mathrm{h}$ where $\mathrm{x} \rightarrow \frac{\pi}{2}$, then $\mathrm{h} \rightarrow 0$
$\therefore \quad$ L.H.L. $=\lim _{\mathrm{h} \rightarrow 0} \frac{\alpha \cos \left(\frac{\pi}{2}-\mathrm{h}\right)}{\pi-2\left(\frac{\pi}{2}-\mathrm{h}\right)}=\lim _{\mathrm{h} \rightarrow 0} \frac{\alpha \sin \mathrm{h}}{\pi-\pi+2 \mathrm{~h}}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 1)
What is the value of $\cos 10^{\circ}+\cos 110^{\circ}+\cos 130^{\circ} ?$

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

The correct answer is:
0

Solution:
Consider $\cos 10^{\circ}+\cos 110^{\circ}+\cos 130^{\circ}$
$=\cos 130^{\circ}+\cos 10^{\circ}+\cos 110^{\circ}$
$=2 \cos \left(\frac{130+10}{2}\right) \cos \left(\frac{130-10}{2}\right)+\cos 110^{\circ}$
$=2 \cos \left(\frac{140}{2}\right) \cos \left(\frac{120}{2}\right)+\cos 110^{\circ}$
$=2 \cos 60^{\circ} \cos 70^{\circ}+\cos 110^{\circ}$
$=\cos 70^{\circ}+\cos 110^{\circ} \quad\left(\because \cos 60^{\circ}=\frac{1}{2}\right)$
$=\cos \left(180^{\circ}-110^{\circ}\right)+\cos 110^{\circ}$
$=-\cos 110^{\circ}+\cos 110^{\circ}=0\left(\because \cos \left(180^{\circ}-\theta\right)=-\cos \theta\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
What is the value of
$\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \cos 45^{\circ} & \sin 45^{\circ}\end{array}\right| \times\left|\begin{array}{ll}\cos 45^{\circ} & \cos 15^{\circ} \\ \sin 45^{\circ} & \sin 15^{\circ}\end{array}\right| ?$

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

The correct answer is:
$-\frac{1}{4}$

Solution:
Consider $\left|\begin{array}{lll}\cos 15^{\circ} & \sin 15^{\circ} \\ \cos 45^{\circ} & \sin 45^{\circ}\end{array}\right| \times\left|\begin{array}{cc}\cos 45^{\circ} & \cos 15^{\circ} \\ \sin 45^{\circ} & \sin 15^{\circ}\end{array}\right|$
$=\left(\sin 45^{\circ} \cos 15^{\circ}-\cos 45^{\circ} \sin 15^{\circ}\right)$
$\times\left(\cos 45^{\circ} \sin 15^{\circ}-\sin 45^{\circ} \cos 15^{\circ}\right)$
$=-\sin \left(45^{\circ}-15^{\circ}\right) \times \sin \left(45^{\circ}-15^{\circ}\right)$
$($ using $\sin (A-B)=-\cos A \sin B+\sin A \cos B)$
$=-\sin 30^{\circ} \times \sin 30^{\circ}=-\frac{1}{2} \times \frac{1}{2}=-\frac{1}{4}$

Topic: Mathematics, Determinants, NDA, NDA 2010 (Phase 1)