NDA II 2026 Mathematics with Solutions
Q.1 [Trigonometry]
Items 1–2: From the top of a building, the angles of depression of the top and bottom of a tower of height $h$ are $\alpha$ and $\beta$ respectively. Let $H$ be the height of the building and $D$ be the horizontal distance between the building and the tower.
What is $H^2+D^2$ equal to?
$\tan\alpha=\dfrac{H-h}{D}$ and $\tan\beta=\dfrac{H}{D}$.
Eliminate $D$: $(H-h)\tan\beta=H\tan\alpha\ \Rightarrow\ H(\tan\beta-\tan\alpha)=h\tan\beta$, so
$H=\dfrac{h\tan\beta}{\tan\beta-\tan\alpha}$ and $D=\dfrac{H}{\tan\beta}=\dfrac{h}{\tan\beta-\tan\alpha}$.
Hence $H^2+D^2=\dfrac{h^2(\tan^2\beta+1)}{(\tan\beta-\tan\alpha)^2}=\dfrac{h^2\sec^2\beta}{(\tan\beta-\tan\alpha)^2}$.
Now $\tan\beta-\tan\alpha=\dfrac{\sin(\beta-\alpha)}{\cos\alpha\cos\beta}$, so
$H^2+D^2=\dfrac{h^2}{\cos^2\beta}\cdot\dfrac{\cos^2\alpha\cos^2\beta}{\sin^2(\beta-\alpha)}=\dfrac{h^2\cos^2\alpha}{\sin^2(\beta-\alpha)}$.
Check the sign of the angle difference: the foot of the tower is always the steeper sighting, so $\beta>\alpha$ and $\sin(\beta-\alpha)>0$.
Q.2 [Trigonometry]
Items 1–2 (continued). Same building and tower.
If $\alpha=15^\circ$ and $\beta=45^\circ$, then which one of the following is correct?
$\dfrac{H}{D}=\tan\beta$.
With $\beta=45^\circ$, $\tan\beta=1$, so $H=D$. Option (d).
The value of $\alpha$ never enters the ratio — the relation $H=D\tan\beta$ holds for any $\alpha$, which is the shortcut worth remembering.
Q.3 [Trigonometry]
Items 3–4: The angles $A$, $B$ and $C$ of a triangle $ABC$ are in the ratio $1:2:7$.
What is the ratio of the side of greatest length to the side of least length?
The greatest side faces the greatest angle, so the ratio wanted is $\dfrac{c}{a}=\dfrac{\sin C}{\sin A}=\dfrac{\sin126^\circ}{\sin18^\circ}$ by the sine rule.
$\sin126^\circ=\sin54^\circ=\cos36^\circ=\dfrac{\sqrt5+1}{4}$ and $\sin18^\circ=\dfrac{\sqrt5-1}{4}$.
$\dfrac{c}{a}=\dfrac{\sqrt5+1}{\sqrt5-1}=\dfrac{(\sqrt5+1)^2}{4}=\dfrac{6+2\sqrt5}{4}=\dfrac{3+\sqrt5}{2}\approx2\cdot618$.
So the ratio is $(3+\sqrt5):2$. Option (c).
Worth memorising: $\sin18^\circ=\dfrac{\sqrt5-1}{4}$ and $\cos36^\circ=\dfrac{\sqrt5+1}{4}$ — they turn up in every paper that uses a $1:2:7$ or pentagon-flavoured triangle.
Q.4 [Trigonometry]
Items 3–4 (continued). Angles in the ratio $1:2:7$.
What is $\sin A\cdot\cos B$ equal to?
$\sin A\cos B=\sin18^\circ\cos36^\circ=\dfrac{\sqrt5-1}{4}\cdot\dfrac{\sqrt5+1}{4}=\dfrac{5-1}{16}=\dfrac{4}{16}=\dfrac14$.
Numerically: $0\cdot30902\times0\cdot80902=0\cdot25$ exactly.
Alternative route without surds: $2\sin18^\circ\cos36^\circ=\dfrac{2\sin18^\circ\cos18^\circ\cos36^\circ}{\cos18^\circ}=\dfrac{\sin72^\circ}{2\cos18^\circ}=\dfrac12$, since $\sin72^\circ=\cos18^\circ$. Halving gives $\tfrac14$.
Note: some circulated keys mark (b) $1/2$. That is the value of $2\sin18^\circ\cos36^\circ$, i.e. twice the quantity asked for.
Q.5 [Trigonometry]
Items 5–6: In a triangle $ABC$, $\dfrac{a+b}{13}=\dfrac{b+c}{11}=\dfrac{c+a}{12}$.
What is $\sin A:\sin B:\sin C$ equal to?
Adding: $2(a+b+c)=36k\Rightarrow a+b+c=18k$.
Subtracting each pair: $c=18k-13k=5k$, $a=18k-11k=7k$, $b=18k-12k=6k$.
By the sine rule $\sin A:\sin B:\sin C=a:b:c=7:6:5$. Option (b).
Note how the largest denominator ($13$, from $a+b$) produces the smallest side $c$ — the reason (c) and (d) are placed as traps.
Q.6 [Trigonometry]
Items 5–6 (continued). Same triangle.
What is $\cos A:\cos B:\cos C$ equal to?
$\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{36+25-49}{60}=\dfrac{12}{60}=\dfrac15$
$\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{49+25-36}{70}=\dfrac{38}{70}=\dfrac{19}{35}$
$\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{49+36-25}{84}=\dfrac{60}{84}=\dfrac57$
Ratio $=\dfrac15:\dfrac{19}{35}:\dfrac57$. Multiply throughout by $35$: $7:19:25$. Option (a).
Q.7 [Trigonometry]
Items 7–8: Given that $\tan\!\left(\dfrac A2+\dfrac B2\right)=p$ and $\tan\!\left(\dfrac A2-\dfrac B2\right)=q$, where $pq\neq\pm1$.
What is $\tan A$ equal to?
$X+Y=A$ and $X-Y=B$.
So $\tan A=\tan(X+Y)=\dfrac{\tan X+\tan Y}{1-\tan X\tan Y}=\dfrac{p+q}{1-pq}$. Option (b).
The condition $pq\neq\pm1$ is exactly what keeps the denominators of this item and the next non-zero.
Q.8 [Trigonometry]
Items 7–8 (continued). Same $p$ and $q$.
What is $\tan B$ equal to?
$\tan B=\tan(X-Y)=\dfrac{\tan X-\tan Y}{1+\tan X\tan Y}=\dfrac{p-q}{1+pq}$. Option (a).
The pair of items is simply the compound-angle formulae read in both directions — get the substitution right and both fall out in one line each.
Q.9 [Trigonometry]
Items 9–10: Let $4(A+B)=\pi$.
What is $(1+\tan A)(1+\tan B)$ equal to?
$\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1\ \Rightarrow\ \tan A+\tan B=1-\tan A\tan B.$
Expand the product:
$(1+\tan A)(1+\tan B)=1+\tan A+\tan B+\tan A\tan B=1+(1-\tan A\tan B)+\tan A\tan B=2.$ Option (b).
Standard result: whenever $A+B=45^\circ$, $(1+\tan A)(1+\tan B)=2$ — e.g. $(1+\tan1^\circ)(1+\tan44^\circ)=2$.
Q.10 [Trigonometry]
Items 9–10 (continued). $4(A+B)=\pi$.
What is $(\cot A-1)(\cot B-1)$ equal to?
$(\cot A-1)(\cot B-1)=\cot A\cot B-(\cot A+\cot B)+1=\dfrac1t-\dfrac st+1$
$=\dfrac{1-s}{t}+1=\dfrac{1-(1-t)}{t}+1=\dfrac tt+1=2.$ Option (d).
Quick check with $A=B=\pi/8$: $\cot(\pi/8)=1+\sqrt2$, so $(\sqrt2)^2=2$ ✓.
Q.11 [Trigonometry]
What is $6\sin(\pi/18)-8\sin^3(\pi/18)$ equal to?
$6\sin\theta-8\sin^3\theta=2\,(3\sin\theta-4\sin^3\theta)=2\sin3\theta.$
With $\theta=10^\circ$: $2\sin30^\circ=2\times\dfrac12=1$. Option (c).
Numerical check: $\sin10^\circ=0\cdot17365$, so $6(0\cdot17365)-8(0\cdot17365)^3=1\cdot04189-0\cdot04188=1\cdot000$ ✓.
Note: some circulated keys mark (a) $1/4$. Substituting the value of $\sin10^\circ$ settles it — the expression is exactly 1.
Q.12 [Trigonometry]
How many values of $\theta$ satisfy the equation $\tan^2\theta+\cot^2\theta=2$, where $0^\circ<\theta<360^\circ$?
Hence $\tan^2\theta=1\Rightarrow\tan\theta=\pm1$.
In $(0^\circ,360^\circ)$ that gives $\theta=45^\circ,\ 135^\circ,\ 225^\circ,\ 315^\circ$ — four values. Option (c).
The AM–GM shortcut is quicker still: $u+\frac1u\ge2$ with equality only at $u=1$, so the equation forces $|\tan\theta|=1$.
Q.13 [Trigonometry]
What is $\sin(\tan^{-1}0\cdot75)$ equal to?
Read it off a $3$–$4$–$5$ right triangle: opposite $=3$, adjacent $=4$, hypotenuse $=\sqrt{3^2+4^2}=5$.
$\sin\theta=\dfrac{3}{5}$. Option (c).
Option (b) $4/5$ is the cosine — the standard trap in inverse-trig items.
Q.14 [Trigonometry]
What is $\tan^2\!\left(\dfrac12\cos^{-1}\dfrac13\right)$ equal to?
Use the half-angle identity $\tan^2\dfrac\theta2=\dfrac{1-\cos\theta}{1+\cos\theta}$:
$\tan^2\dfrac\theta2=\dfrac{1-\frac13}{1+\frac13}=\dfrac{\frac23}{\frac43}=\dfrac12.$ Option (a).
The identity follows from $\cos\theta=\dfrac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$ and is worth carrying into the hall.
Q.15 [Trigonometry]
What is $\tan(1125^\circ)\cot(405^\circ)+\tan(765^\circ)\cot(675^\circ)$ equal to?
$1125^\circ-3(360^\circ)=45^\circ\Rightarrow\tan1125^\circ=\tan45^\circ=1$
$405^\circ-360^\circ=45^\circ\Rightarrow\cot405^\circ=\cot45^\circ=1$
$765^\circ-2(360^\circ)=45^\circ\Rightarrow\tan765^\circ=1$
$675^\circ-360^\circ=315^\circ\Rightarrow\cot315^\circ=\cot(-45^\circ)=-1$
So the expression $=1(1)+1(-1)=0$. Option (c).
The whole item rests on the single fact that $315^\circ$ lies in the fourth quadrant, where the tangent family is negative.
Q.16 [Matrices and Determinants]
If $A$ is a square matrix of order 3 and $|A|=2$, then what is $\operatorname{adj}(\operatorname{adj}A)$ equal to?
Here $n=3$, so $\operatorname{adj}(\operatorname{adj}A)=|A|^{1}A=2A$. Option (c).
The companion results worth memorising for order $n$:
$A(\operatorname{adj}A)=|A|I$, $\ |\operatorname{adj}A|=|A|^{n-1}$, $\ |\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2}$.
Option (a) $8A$ comes from misusing the exponent $n-1$ in place of $n-2$.
Q.17 [Matrices and Determinants]
If $A$ is a square matrix of order 3 such that $A(\operatorname{adj}A)=\begin{bmatrix}64&0&0\\0&64&0\\0&0&64\end{bmatrix}$, then what is $|A|$ equal to?
The matrix given is $64I$, so comparing, $|A|=64$. Option (d).
Traps built into the options: $|\operatorname{adj}A|=|A|^{2}=4096$, and if a candidate instead reads the equation as $|A|^{3}=64$ he gets $4$, which is option (a).
Q.18 [Matrices and Determinants]
Let $A=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}$. What is the least value of $\theta$ for which $A+A^{T}=I$, where $I$ is the identity matrix of order 2?
$A+A^{T}=\begin{bmatrix}2\cos\theta&0\\0&2\cos\theta\end{bmatrix}$.
Setting this equal to $I$ gives $2\cos\theta=1$, i.e. $\cos\theta=\dfrac12$.
The least non-negative solution is $\theta=\dfrac\pi3$. Option (b).
Q.19 [Matrices and Determinants]
Let $A$, $B$ and $C$ be square matrices of order 2 such that $AB=AC$. Which of the statements given below is/are correct?
I. $B$ and $C$ are not necessarily equal if $A$ is a singular matrix.
II. $B$ and $C$ are equal if $A$ is a non-singular matrix.
Statement I is true. Cancellation fails for a singular $A$. A concrete counter-example:
$A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ B=\begin{bmatrix}1&2\\3&4\end{bmatrix},\ C=\begin{bmatrix}1&2\\5&6\end{bmatrix}$
gives $AB=AC=\begin{bmatrix}1&2\\0&0\end{bmatrix}$ although $B\neq C$.
Both hold, so the answer is (c). Matrices form a ring without cancellation — that single idea is what this item tests.
Q.20 [Matrices and Determinants]
If $A$ is a square matrix of order 3 and $|A|=4$, then what is $|\operatorname{adj}A|$ equal to?
With $n=3$ and $|A|=4$: $|\operatorname{adj}A|=4^{2}=16$. Option (b).
Option (a) $64=4^{3}$ is the value of $|A|^{n}$ and is the standard slip.
Q.21 [Matrices and Determinants]
Let $A$ and $B$ be symmetric matrices of the same order. Which of the following statements is/are correct?
I. $(AB-BA)$ is also a symmetric matrix.
II. $(BA-AB)$ is a skew-symmetric matrix.
$(AB-BA)^{T}=B^{T}A^{T}-A^{T}B^{T}=BA-AB=-(AB-BA)$.
So $AB-BA$ is skew-symmetric, not symmetric — Statement I is false (unless the commutator is the zero matrix).
The same computation applied to $BA-AB$ gives $(BA-AB)^{T}=AB-BA=-(BA-AB)$, so Statement II is true.
Answer (b). Result worth carrying: for symmetric $A,B$, the commutator $AB-BA$ is always skew-symmetric and the anticommutator $AB+BA$ is always symmetric.
Q.22 [Matrices and Determinants]
If $A=\begin{bmatrix}-1&2\\3&-4\end{bmatrix}$, then what is $|A^{-1}|$ equal to?
Since $AA^{-1}=I$, taking determinants gives $|A|\,|A^{-1}|=1$, so
$|A^{-1}|=\dfrac1{|A|}=-\dfrac12$. Option (c).
Note that $A^{-1}$ exists precisely because $|A|\neq0$; option (a) is the determinant of $A$ itself.
Q.23 [Sets, Relations and Functions]
Let $S$ be the set of all real numbers and $R$ be a relation on $S$ defined by $xRy\Rightarrow|x|\le y$. Then $R$ is
Transitive? Suppose $xRy$ and $yRz$, i.e. $|x|\le y$ and $|y|\le z$.
Since $y\le|y|$ always, we get $|x|\le y\le|y|\le z$, hence $|x|\le z$, i.e. $xRz$. So $R$ is transitive.
Answer (b). The step that does the work is the inequality $y\le|y|$, which holds for negative $y$ as well.
Q.24 [Permutations and Combinations]
4-digit numbers are formed with 1, 2, 3 and 4. What is the number of 4-digit numbers in which at least one digit is repeated?
Those with no digit repeated use all four digits once each: $4!=24$.
At least one repetition $=256-24=232$. Option (c).
'At least one' almost always means 'total minus none' — attacking it case by case (exactly one pair, two pairs, a triple, all four alike) works but costs four times the effort.
Q.25 [Number Systems]
Given $S=\{(x,y,z)\in R^3:\ xyz=210,\ x
Ordered triples: each prime may go to $x$, $y$ or $z$ independently, giving $3^4=81$ ordered triples with product 210.
Remove the ties. Two coordinates can be equal only if their common value squares into 210; as 210 is square-free the only possibility is $1$, giving the triple $\{1,1,210\}$, which occupies $3$ of the 81 orderings. No triple has all three equal ($210$ is not a cube).
So $81-3=78$ ordered triples have three distinct entries, and each unordered set is counted $3!=6$ times:
$\dfrac{78}{6}=13$ triples with $x
Q.26 [Complex Numbers]
If $x=4+i$, where $i=\sqrt{-1}$, then what is $x^3+2x^2-63x+171$ equal to?
$x-4=i\Rightarrow(x-4)^2=-1\Rightarrow x^2-8x+17=0.$
Now divide the cubic by $x^2-8x+17$:
$x^3+2x^2-63x+171=(x^2-8x+17)(x+10)+1.$
[Check the division: $(x^2-8x+17)(x+10)=x^3+2x^2-63x+170$.]
Since the bracket vanishes, the value is $1$. Option (c).
This 'minimal polynomial' trick turns every such item into one long division and is far safer than expanding $(4+i)^3$.
Q.27 [Quadratic Equations]
Let $\alpha$ and $\beta$ be the roots of the equation $x^2-p(x+1)-q=0$. What is $\dfrac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+q}+\dfrac{\beta^2+2\beta+1}{\beta^2+2\beta+q}$ equal to?
Key step: $(\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=-(p+q)+p+1=1-q.$
Now rewrite each denominator:
$\alpha^2+2\alpha+q=(\alpha+1)^2-(1-q)=(\alpha+1)^2-(\alpha+1)(\beta+1)=(\alpha+1)(\alpha-\beta).$
So the first fraction is $\dfrac{(\alpha+1)^2}{(\alpha+1)(\alpha-\beta)}=\dfrac{\alpha+1}{\alpha-\beta}$.
By symmetry the second is $\dfrac{\beta+1}{\beta-\alpha}=-\dfrac{\beta+1}{\alpha-\beta}$.
Adding: $\dfrac{(\alpha+1)-(\beta+1)}{\alpha-\beta}=\dfrac{\alpha-\beta}{\alpha-\beta}=1$. Option (c).
Q.28 [Quadratic Equations]
If the sum of the roots of the equation $\dfrac1{2x+p}+\dfrac1{2x+q}=\dfrac1r$ is zero, then what is $r$ equal to?
$r\big[(2x+q)+(2x+p)\big]=(2x+p)(2x+q)$
$r(4x+p+q)=4x^2+2x(p+q)+pq$
$4x^2+x\big[2(p+q)-4r\big]+\big[pq-r(p+q)\big]=0.$
Sum of roots $=-\dfrac{2(p+q)-4r}{4}=0\Rightarrow2(p+q)=4r\Rightarrow r=\dfrac{p+q}{2}.$ Option (d).
Only the coefficient of $x$ matters, so the constant term need never be simplified.
Q.29 [Quadratic Equations]
Let $\alpha$ and $\beta$ be the roots of the equation $ax^2+bx+c=0$ and $p_n=\alpha^n+\beta^n$, where $n>1$. What is $ap_{n+1}+bp_n+cp_{n-1}$ equal to?
$a\alpha^{\,n+1}+b\alpha^{\,n}+c\alpha^{\,n-1}=0.$
The identical step for $\beta$ gives $a\beta^{\,n+1}+b\beta^{\,n}+c\beta^{\,n-1}=0$.
Adding the two: $ap_{n+1}+bp_n+cp_{n-1}=0$. Option (a).
This is the Newton recurrence for power sums — the standard way of generating $\alpha^n+\beta^n$ without ever computing the roots.
Q.30 [Sequences and Series]
If $p$, $q$ and $r$ are in AP, then what is $p^3+3pr(p+r)+r^3$ equal to?
So the expression is simply $(p+r)^3$.
In an AP the middle term is the average: $2q=p+r$.
Therefore the value is $(2q)^3=8q^3$. Option (d).
Q.31 [Sequences and Series]
If $\dfrac1{b-a}+\dfrac1{b-c}=\dfrac2b$, then $a$, $b$ and $c$ are in
$b\big[(b-c)+(b-a)\big]=2(b-a)(b-c)$
$2b^2-ab-bc=2\big(b^2-bc-ab+ac\big)$
$2b^2-ab-bc=2b^2-2bc-2ab+2ac$
$ab+bc=2ac.$
Divide throughout by $abc$: $\dfrac1c+\dfrac1a=\dfrac2b$.
That says $\dfrac1a,\dfrac1b,\dfrac1c$ are in AP, i.e. $a$, $b$, $c$ are in HP. Option (c).
The form of the given equation — reciprocals summing to $2/b$ — is itself the signature of a harmonic progression.
Q.32 [Sequences and Series]
Let $P$, $Q$, $R$ and $S$ be the sum of $n$ terms, $2n$ terms, $3n$ terms and $4n$ terms respectively of an AP such that $Q=3P$. Which one of the following is correct?
$Q=3P$ gives $n\big[2a+(2n-1)d\big]=\dfrac{3n}{2}\big[2a+(n-1)d\big]$
$\Rightarrow4a+(4n-2)d=6a+(3n-3)d\Rightarrow(n+1)d=2a.$
Substituting $2a=(n+1)d$:
$R=S_{3n}=\dfrac{3n}{2}\big[(n+1)d+(3n-1)d\big]=\dfrac{3n}{2}(4n)d=6n^2d$
$S=S_{4n}=2n\big[(n+1)d+(4n-1)d\big]=2n(5n)d=10n^2d$
Hence $\dfrac RS=\dfrac{6}{10}=\dfrac35$, i.e. $5R=3S$. Option (a).
(Sanity check: $P=n^2d$ and $Q=3n^2d$, so $Q=3P$ ✓.)
Q.33 [Sequences and Series]
In a GP, the 3rd, 5th and 7th terms are $x$, $x^2+2$ and $x^3+10$ respectively, where $x>1$. What is the 8th term?
$(x^2+2)^2=x(x^3+10)$
$x^4+4x^2+4=x^4+10x\Rightarrow4x^2-10x+4=0\Rightarrow2x^2-5x+2=0.$
Roots $x=2$ and $x=\tfrac12$; since $x>1$, $x=2$.
So the 3rd, 5th, 7th terms are $2,\ 6,\ 18$, giving $r^2=\dfrac62=3$, i.e. $r=\sqrt3$.
8th term $=$ 7th term $\times r=18\sqrt3$. Option (b).
Q.34 [Complex Numbers]
If $(1+i)^n-16=0$, where $i=\sqrt{-1}$ and $n$ is a positive integer, then what is the least value of $n$?
Now confirm the argument as well:
$(1+i)^2=2i\Rightarrow(1+i)^8=(2i)^4=16\,i^4=16$ ✓.
So the least such $n$ is $8$. Option (c).
Matching the modulus alone would have allowed $n=8$ only; had the modulus condition admitted several $n$, the argument condition $\dfrac{n\pi}{4}=2k\pi$ would have picked the right one.
Q.35 [Complex Numbers]
What is $i\times i^4\times i^9\times i^{16}\times\cdots\times i^{576}$, where $i=\sqrt{-1}$, equal to?
So the product is $i^{\,S}$ where $S=\displaystyle\sum_{k=1}^{24}k^2=\dfrac{24\cdot25\cdot49}{6}=4900.$
Powers of $i$ repeat with period 4, and $4900=4\times1225$, so $i^{4900}=\left(i^4\right)^{1225}=1$. Option (c).
Only the remainder of the exponent on division by 4 matters — the full value 4900 need not even be written out once you see it is a multiple of 4.
Q.36 [Complex Numbers]
If $z=x+iy$ be such that $\left|\dfrac{z+\lambda i}{z-\lambda i}\right|=1$, where $i=\sqrt{-1}$ and $\lambda$ is a positive real number, then which of the following statements is/are correct?
I. $z$ lies on the line $y=x$.
II. The amplitude of $z$ is $\dfrac\pi4$.
The locus is the perpendicular bisector of the segment joining those two points, namely the real axis $y=0$.
Algebraic confirmation: $|x+i(y+\lambda)|=|x+i(y-\lambda)|\Rightarrow(y+\lambda)^2=(y-\lambda)^2\Rightarrow4\lambda y=0\Rightarrow y=0$ (as $\lambda>0$).
I is false — $z$ lies on $y=0$, not $y=x$.
II is false — on the real axis the amplitude is $0$ (or $\pi$), never $\pi/4$.
Answer (d).
Q.37 [Matrices and Determinants]
What is $\begin{vmatrix}\frac1a&bc&a^3\\[2pt]\frac1a+\frac1b&c(a+b)&a^3+b^3\\[2pt]\frac1a+\frac1b+\frac1c&ab+bc+ca&a^3+b^3+c^3\end{vmatrix}$ equal to?
$R_3\to R_3-R_2$ and $R_2\to R_2-R_1$ give
$\begin{vmatrix}\frac1a&bc&a^3\\ \frac1b&ca&b^3\\ \frac1c&ab&c^3\end{vmatrix}$
(using $c(a+b)-bc=ca$ and $ab+bc+ca-c(a+b)=ab$).
Now multiply $R_1$ by $a$, $R_2$ by $b$, $R_3$ by $c$ — this multiplies the determinant by $abc$:
$abc\cdot D=\begin{vmatrix}1&abc&a^4\\1&abc&b^4\\1&abc&c^4\end{vmatrix}.$
Column 2 is $abc$ times column 1, so the determinant is zero; hence $D=0$. Option (a).
Q.38 [Matrices and Determinants]
If $ae+bg=cf+dh=p$ and $af+bh=ce+dg=q$, then what is $\begin{vmatrix}a&b\\c&d\end{vmatrix}\times\begin{vmatrix}e&f\\g&h\end{vmatrix}$ equal to?
$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}e&f\\g&h\end{bmatrix}=\begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix}=\begin{bmatrix}p&q\\q&p\end{bmatrix}.$
So the required product of determinants is $\begin{vmatrix}p&q\\q&p\end{vmatrix}=p^2-q^2$. Option (d).
The four given equalities are placed exactly so that the product matrix comes out symmetric — spotting that is the whole item.
Q.39 [Matrices and Determinants]
If $x+ay+a^2z=1$, $x+by+b^2z=1$, $x+cy+c^2z=1$, where $a\neq b\neq c$, then what is $x+y+z$ equal to?
Each given equation says exactly that $f(a)=0$, $f(b)=0$, $f(c)=0$.
A quadratic cannot have three distinct roots unless it is identically zero, so
$z=0,\quad y=0,\quad x-1=0\Rightarrow x=1.$
Therefore $x+y+z=1$. Option (b).
(The coefficient matrix is a Vandermonde matrix, whose determinant $(b-a)(c-a)(c-b)\ne0$ guarantees the solution is unique — so the one found above is the only one.)
Q.40 [Matrices and Determinants]
Let $A=\begin{pmatrix}1&-\tan\theta\\ \tan\theta&1\end{pmatrix}$ and $B=\begin{pmatrix}1&\tan\theta\\ -\tan\theta&1\end{pmatrix}$. If $C=AB^{-1}$, then what is $\det(C)$ equal to?
$\det C=\dfrac{\det A}{\det B}.$
$\det A=1\cdot1-(-\tan\theta)(\tan\theta)=1+\tan^2\theta=\sec^2\theta$
$\det B=1\cdot1-(\tan\theta)(-\tan\theta)=1+\tan^2\theta=\sec^2\theta$
$\det C=\dfrac{\sec^2\theta}{\sec^2\theta}=1.$ Option (d).
Note $B=A^{T}$, and a matrix always has the same determinant as its transpose — which is the one-line way to see the answer must be 1.
Q.41 [Number Systems]
What is the LCM of $(11110)_2$, $(10100)_2$ and $(11011)_2$?
$(11110)_2=16+8+4+2=30$
$(10100)_2=16+4=20$
$(11011)_2=16+8+2+1=27$
Factorise: $30=2\cdot3\cdot5$, $20=2^2\cdot5$, $27=3^3$.
$\mathrm{LCM}=2^2\cdot3^3\cdot5=4\times27\times5=540.$
Back to binary: $540=512+16+8+4=2^9+2^4+2^3+2^2$, so bits 9, 4, 3, 2 are set:
$540=(1000011100)_2.$ Option (b).
Quick filter: the answer must be even (30 and 20 are even), which kills option (d) at a glance.
Q.42 [Binomial Theorem]
What is the coefficient of $x^{2n}$ in the expansion of $(1+x)^{2n}\left(1+\dfrac1x\right)^{2n}$?
So the product is $\dfrac{(1+x)^{4n}}{x^{2n}}$.
The coefficient of $x^{2n}$ in that equals the coefficient of $x^{4n}$ in $(1+x)^{4n}$, which is $\binom{4n}{4n}=1$. Option (a).
Term-by-term check: a general term is $\binom{2n}{j}x^{j}\cdot\binom{2n}{k}x^{-k}$, and $j-k=2n$ with $j,k\le2n$ forces $j=2n,\ k=0$ — exactly one term, of coefficient $\binom{2n}{2n}\binom{2n}{0}=1$.
Q.43 [Binomial Theorem]
Consider the following statements in respect of the expansion of $(x^2+x^{-2}+2)^4$ :
I. The number of terms in the expansion is 5.
II. One of the coefficients in the expansion has a maximum value equal to 70.
$(x^2+x^{-2}+2)^4=\left(x+\dfrac1x\right)^8.$
Statement I. The expansion of $\left(x+\frac1x\right)^8$ has terms in $x^{8},x^{6},x^{4},x^{2},x^{0},x^{-2},x^{-4},x^{-6},x^{-8}$ — nine terms, not 5. False.
Statement II. The coefficients are $\binom8k$ for $k=0,\dots,8$, the largest being the middle one, $\binom84=70$. True.
Answer (b).
Q.44 [Sets, Relations and Functions]
Consider the following relations from $A$ to $B$, where $A=\{1,3,5\}$ and $B=\{2,4,6,8\}$ :
I. $\{(1,2),(3,2),(3,6),(5,8)\}$
II. $\{(3,4),(5,8),(1,6),(3,2)\}$
III. $\{(1,2),(3,6)\}$
IV. $\{(1,6),(3,2),(5,2)\}$
Which of the above is/are function(s) from $A$ to $B$?
I — the element 3 is paired with both 2 and 6. Not a function.
II — 3 is paired with both 4 and 2. Not a function.
III — the element 5 has no image at all; the domain is $\{1,3\}\ne A$. Not a function.
IV — $1\mapsto6$, $3\mapsto2$, $5\mapsto2$: all three elements covered, each exactly once. A function (two elements sharing an image is perfectly allowed — that is a many-one function).
Answer (b).
Q.45 [Complex Numbers]
If $x^2-2x\cos\theta+1=0$, then what is the magnitude of $x$?
$x=\dfrac{2\cos\theta\pm\sqrt{4\cos^2\theta-4}}{2}=\cos\theta\pm\sqrt{\cos^2\theta-1}=\cos\theta\pm i\sin\theta.$
Hence $|x|=\sqrt{\cos^2\theta+\sin^2\theta}=1$. Option (c).
Shortcut: the product of the roots is $1$ and the roots are conjugates, so $|x|^2=1$. Option (a) can be dismissed on sight — a magnitude is never negative.
Q.46 [Sets, Relations and Functions]
Consider the following statements in respect of sets $A$ and $B$ :
I. If $x\in A$ and $A\in B$, then $x\in B$.
II. If $A\subset B$ and $x\notin B$, then $x\notin A$.
Statement II is true. It is the contrapositive of $A\subset B$: since every element of $A$ lies in $B$, anything outside $B$ cannot be inside $A$.
Answer (b). The item is testing the difference between $\in$ (membership) and $\subset$ (inclusion).
Q.47 [Sets, Relations and Functions]
A research group conducted a survey of 500 consumers and reported that 370 consumers preferred product $X$ and 240 consumers preferred product $Y$. What is the least number that must have preferred both the products?
$n(X\cap Y)=n(X)+n(Y)-n(X\cup Y)\ \ge\ 370+240-500=110.$
The least possible overlap is therefore 110, attained when every consumer prefers at least one product. Option (c).
The maximum overlap, for contrast, would be $\min(370,240)=240$.
Q.48 [Sets, Relations and Functions]
The Cartesian product $A\times A$ has 25 elements among which are found $(3,1)$, $(6,2)$, $(5,3)$. Which of the following statements is/are correct?
I. It is possible to determine other elements of $A\times A$.
II. $(5,5)\in A\times A$ and $(1,3)\notin A\times A$.
The three given pairs expose the elements $3,1,6,2,5$ — that is already five distinct elements, so
$A=\{1,2,3,5,6\}$ and $A$ is completely determined.
Statement I is true: with $A$ known, all 25 ordered pairs can be written down.
Statement II is false: $(5,5)\in A\times A$ is correct, but $1\in A$ and $3\in A$, so $(1,3)$ does belong to $A\times A$. One false half sinks the statement.
Answer (a).
Q.49 [Sets, Relations and Functions]
Consider the following statements in respect of a relation $R$ from a set $A$ to a set $B$ :
I. The set $B$ is called codomain of the relation $R$.
II. The range of the relation is always equal to codomain of the relation $R$.
III. The domain of the relation $R$ must be equal to the set $A$.
II is false — the range (the set of second coordinates actually used) is a subset of the codomain. Equality holds only for onto relations.
III is false — this is where the item bites. For a relation, the domain is the set of first coordinates actually used, which may be a proper subset of $A$. The requirement that every element of $A$ be used applies to functions, not to relations.
Answer (a).
Q.50 [Analytical Geometry — 2D]
Consider the following system of linear inequalities : $x-y\le3$ and $x+y\ge5$. The solution of the inequalities lies in
First quadrant: $(4,1)$ gives $4-1=3\le3$ ✓ and $4+1=5\ge5$ ✓. Points exist.
Second quadrant ($x<0,y>0$): $(-1,7)$ gives $-8\le3$ ✓ and $6\ge5$ ✓. Points exist.
Third quadrant ($x<0,y<0$): then $x+y<0$, which can never reach $5$. Impossible.
Fourth quadrant ($x>0,y<0$): $x-y\le3$ with $y<0$ forces $x\le3+y<3$; then $x+y\ge5$ forces $y\ge5-x>2$, contradicting $y<0$. Impossible.
So the solution set lies in the first and second quadrants only. Option (c).
Q.51 [Analytical Geometry — 2D]
A circle touches each of the lines $x-y=0$ and $x+y=0$ at unit distance from the origin. The centre of the circle may be at
The point of contact with $y=x$ is the foot of the perpendicular from $(a,0)$ to that line, namely $\left(\dfrac a2,\dfrac a2\right)$.
Its distance from the origin is $\sqrt{\dfrac{a^2}{4}+\dfrac{a^2}{4}}=\dfrac{a}{\sqrt2}$.
Setting this equal to 1 gives $a=\sqrt2$, so the centre may be at $(\sqrt2,0)$. Option (a).
(The radius is then $\dfrac{|a|}{\sqrt2}=1$ — consistent, since for perpendicular tangents from a point the contact distance and the radius coincide.)
Q.52 [Analytical Geometry — 2D]
A line $(\sin\theta)x+(\cos\theta)y=\sin2\theta$ cuts the coordinate axes at points $P$ and $Q$. Let $M$ be the midpoint of the line segment $PQ$. What is the distance of $M$ from the origin?
Put $y=0$: $x=\dfrac{2\sin\theta\cos\theta}{\sin\theta}=2\cos\theta$, so $P=(2\cos\theta,\,0)$.
Put $x=0$: $y=\dfrac{2\sin\theta\cos\theta}{\cos\theta}=2\sin\theta$, so $Q=(0,\,2\sin\theta)$.
Midpoint $M=(\cos\theta,\ \sin\theta)$, whose distance from the origin is
$\sqrt{\cos^2\theta+\sin^2\theta}=1.$ Option (b).
So $M$ traces the unit circle as $\theta$ varies — a neat way to see the answer cannot depend on $\theta$, which rules out options (c) and (d) immediately.
Q.53 [Analytical Geometry — 2D]
The centre and one of the foci $(F)$ of an ellipse are at $(0,0)$ and $(-c,0)$ respectively. If $P(x,y)$ is any point on the ellipse and $2a$ is the length of major axis, then what is $PF$ equal to?
$r_1=a+ex$ (from the left focus $(-c,0)$) and $r_2=a-ex$ (from the right focus $(c,0)$).
Here $F$ is the left focus, so
$PF=a+ex=a+\dfrac{c}{a}x.$ Option (d).
Sanity check at the right-hand vertex $x=a$: $PF=a+c$, which is indeed the distance from $(-c,0)$ to $(a,0)$ ✓. (Note $r_1+r_2=2a$ — the defining property of the ellipse.)
Q.54 [Analytical Geometry — 2D]
What is the equation of the line which is equidistant from the lines $2x-4y-7=0$ and $6x-12y+1=0$?
$2x-4y-7=0$ and $2x-4y+\dfrac13=0$ — parallel lines.
The line midway between $2x-4y+c_1=0$ and $2x-4y+c_2=0$ is $2x-4y+\dfrac{c_1+c_2}{2}=0$:
$c=\dfrac{-7+\frac13}{2}=\dfrac{-\frac{20}{3}}{2}=-\dfrac{10}{3}.$
So the line is $2x-4y-\dfrac{10}{3}=0$; multiplying by $\dfrac32$ gives
$3x-6y-5=0.$ Option (c).
The trap is averaging $-7$ and $+1$ without first scaling the second equation — that produces options (a)/(b).
Q.55 [Analytical Geometry — 2D]
A ray of light passing through the point $P(1,2)$ reflects on the $x$-axis at point $N$ and the reflected ray passes through the point $Q(5,3)$. What is the distance of the point $N$ from the origin?
Line $P'Q$: slope $=\dfrac{3-(-2)}{5-1}=\dfrac54$, so $y+2=\dfrac54(x-1)$.
$N$ is where this meets $y=0$: $2=\dfrac54(x-1)\Rightarrow x-1=\dfrac85\Rightarrow x=\dfrac{13}{5}$.
$N=\left(\dfrac{13}{5},0\right)$, so $ON=\dfrac{13}{5}$ units. Option (c).
Reflecting $Q$ instead of $P$ gives the same $N$ — either image works.
Q.56 [Analytical Geometry — 2D]
A line cuts off intercept $p$ on the $x$-axis and intercept $q$ on the $y$-axis, where $p>q$. The sum of the intercepts is 2 and the product of the intercepts is $-15$. What is the equation of the line?
Since $p>q$: $p=5$, $q=-3$.
Intercept form: $\dfrac xp+\dfrac yq=1\Rightarrow\dfrac x5-\dfrac y3=1$.
Multiply by 15: $3x-5y=15$, i.e. $3x-5y-15=0$. Option (a).
Verify the intercepts: $y=0\Rightarrow x=5$ ✓; $x=0\Rightarrow y=-3$ ✓.
Q.57 [Analytical Geometry — 2D]
Let $P$ and $Q$ be the points on the positive $x$-axis and positive $y$-axis respectively. A point $N(2,1)$ divides the line segment $PQ$ in the ratio $1:2$. What is the equation of the line?
$N=\left(\dfrac{1\cdot0+2\cdot p}{1+2},\ \dfrac{1\cdot q+2\cdot0}{1+2}\right)=\left(\dfrac{2p}{3},\ \dfrac q3\right).$
Matching with $N(2,1)$: $\dfrac{2p}{3}=2\Rightarrow p=3$ and $\dfrac q3=1\Rightarrow q=3$.
Intercept form: $\dfrac x3+\dfrac y3=1\Rightarrow x+y=3$, i.e. $x+y-3=0$. Option (b).
Getting the ratio the wrong way round gives $P=(6,0)$, $Q=(0,\tfrac32)$ — a line not among the options, which is a useful self-check.
Q.58 [Analytical Geometry — 2D]
$ABCD$ is a square. The equations of $AB$, $AD$ and $BD$ are $y=0$, $x=0$ and $x+y-4=0$ respectively. What is the equation of $AC$?
$B$ lies on $AB$ ($y=0$) and on the diagonal $BD$: $x+0=4\Rightarrow B=(4,0)$.
$D$ lies on $AD$ ($x=0$) and on $BD$: $0+y=4\Rightarrow D=(0,4)$.
In the square $ABCD$ the diagonals bisect each other, so $C=B+D-A=(4,4)$.
Line $AC$ joins $(0,0)$ to $(4,4)$: $y=x$, i.e. $x-y=0$. Option (b).
A quick sanity check: $AC$ must be perpendicular to $BD$ ($x+y=4$, slope $-1$) — and the slope of $AC$ is $+1$ ✓.
Q.59 [Analytical Geometry — 2D]
$ABC$ is a triangle, where the vertices $A$ and $B$ are fixed points both lying on the $x$-axis. Let $AB=10$ cm. The vertex $C$ moves such that $\dfrac1{\tan A}+\dfrac1{\tan B}=\dfrac1k$, where $k\neq0$. What is the equation of the locus of the point $C$?
$\cot A=\dfrac{x}{y},\qquad \cot B=\dfrac{10-x}{y}.$
Adding, the $x$ cancels:
$\cot A+\cot B=\dfrac{x+(10-x)}{y}=\dfrac{10}{y}.$
The condition $\cot A+\cot B=\dfrac1k$ therefore gives $\dfrac{10}{y}=\dfrac1k$, i.e. $y=10k$. Option (a).
So $C$ moves on a line parallel to $AB$ — the locus keeps the height, and hence the area, of the triangle constant.
Q.60 [Analytical Geometry — 2D]
What is the eccentricity $(e)$ of the parabola $4x^2+y=0$?
But the numbers never matter here: every parabola has eccentricity exactly 1, by definition, since a point on it is equidistant from the focus and the directrix.
Answer (c). For reference: $e=0$ for a circle, $0
Q.61 [Analytical Geometry — 3D]
The vertices of a triangle are $A(2,0,0)$, $B(0,6,0)$ and $C(0,0,4)$. If $AD$, $BE$ and $CF$ are the medians of the triangle, then what is $AD^2+BE^2+CF^2$ equal to?
$AB^2=(2-0)^2+(0-6)^2=4+36=40$
$BC^2=6^2+4^2=36+16=52$
$CA^2=4^2+2^2=16+4=20$
Sum of squares of sides $=40+52+20=112$.
$AD^2+BE^2+CF^2=\dfrac34\times112=84.$ Option (b).
(Direct check: $D$, the midpoint of $BC$, is $(0,3,2)$, so $AD^2=4+9+4=17$; $E=(1,0,2)$ gives $BE^2=1+36+4=41$; $F=(1,3,0)$ gives $CF^2=1+9+16=26$. Total $17+41+26=84$ ✓.)
Q.62 [Analytical Geometry — 3D]
A line makes angles $\alpha$, $\beta$ and $\gamma$ with the positive directions of $x$-axis, $y$-axis and $z$-axis respectively such that $\alpha+\beta=90^\circ$. Which of the following statements is/are correct?
I. The maximum value of $\cos\alpha+\cos\beta$ is $\sqrt2$.
II. The minimum value of $\cos\alpha+\cos\beta+\cos\gamma$ is 1.
Since $\beta=90^\circ-\alpha$, $\cos\beta=\sin\alpha$, so
$\cos^2\alpha+\sin^2\alpha+\cos^2\gamma=1\Rightarrow\cos^2\gamma=0\Rightarrow\cos\gamma=0.$
The line is therefore always perpendicular to the $z$-axis, and both statements reduce to studying $f(\alpha)=\cos\alpha+\sin\alpha=\sqrt2\sin(\alpha+45^\circ)$ on $0^\circ\le\alpha\le90^\circ$ (the range forced by $\beta=90^\circ-\alpha\ge0$).
I. $\alpha+45^\circ$ runs over $[45^\circ,135^\circ]$, so $\sin$ reaches 1 at $\alpha=45^\circ$ and $f$ attains $\sqrt2$. True.
II. On the same interval $\sin$ is least at the endpoints, where $f=\sqrt2\cdot\dfrac{1}{\sqrt2}=1$; adding $\cos\gamma=0$ leaves 1. True (attained at $\alpha=0^\circ$, the line along the $x$-axis).
Answer (c).
Note: some circulated keys mark (a) I only. The endpoint $\alpha=0^\circ$ gives direction cosines $(1,0,0)$ — a perfectly valid line — so the minimum 1 is attained and II stands.
Q.63 [Analytical Geometry — 3D]
What is the area of the triangle whose vertices are $(0,7,10)$, $(-1,6,6)$ and $(-4,9,6)$?
$\vec{AB}=(-1,-1,-4)$ and $\vec{AC}=(-4,2,-4)$.
$\vec{AB}\times\vec{AC}=\big((-1)(-4)-(-4)(2),\ \ -[(-1)(-4)-(-4)(-4)],\ \ (-1)(2)-(-1)(-4)\big)$
$=(4+8,\ -[4-16],\ -2-4)=(12,\ 12,\ -6).$
$|\vec{AB}\times\vec{AC}|=\sqrt{144+144+36}=\sqrt{324}=18.$
Area $=\dfrac12\times18=9$ square units. Option (a).
(As it happens $AB=\sqrt{18}$, $AC=\sqrt{36}$ and $BC=\sqrt{18}$, so $AB^2+BC^2=CA^2$ — the triangle is right-angled at $B$, and $\tfrac12\sqrt{18}\sqrt{18}=9$ ✓.)
Q.64 [Analytical Geometry — 3D]
If $O$ is the origin and $P$ is the point $(2,-4,6)$, then what is the equation of the plane through $P$ and perpendicular to $OP$?
Point-normal form through $P(2,-4,6)$:
$2(x-2)-4(y+4)+6(z-6)=0$
$2x-4y+6z-4-16-36=0\Rightarrow2x-4y+6z-56=0.$
Divide by 2: $x-2y+3z-28=0$. Option (b).
Faster still: for a plane through $P$ with normal $\vec{OP}$, the equation is $\vec r\cdot\vec{OP}=|\vec{OP}|^2$, and $|\vec{OP}|^2=4+16+36=56$, giving $2x-4y+6z=56$ directly.
Q.65 [Analytical Geometry — 3D]
The image of the point $P(1,3,4)$ with respect to the plane $Ax+By+Cz+D=0$ is $(-3,5,2)$. If $A+B+C=2$, then what is $D$ equal to?
$P'-P=(-3-1,\ 5-3,\ 2-4)=(-4,2,-2)\ \parallel\ (2,-1,1).$
Write $(A,B,C)=t(2,-1,1)$. Then $A+B+C=2t=2\Rightarrow t=1$, so $(A,B,C)=(2,-1,1)$.
The midpoint of $PP'$ lies on the plane:
$M=\left(\dfrac{1-3}{2},\dfrac{3+5}{2},\dfrac{4+2}{2}\right)=(-1,4,3).$
Substituting: $2(-1)-1(4)+1(3)+D=0\Rightarrow-3+D=0\Rightarrow D=3.$ Option (b).
The condition $A+B+C=2$ is what fixes the scale — without it the plane equation would only be known up to a multiple, which is why option (d) is tempting but wrong.
Q.66 [Vectors]
Let $\vec a$, $\vec b$, $\vec c$ and $\vec d$ be the vectors. Consider the following :
I. $(\vec a\times\vec b)\cdot(\vec c\times\vec d)$
II. $(\vec a\times\vec b)\times(\vec c\times\vec d)$
III. $(\vec a\cdot\vec b)\cdot(\vec c\cdot\vec d)$
IV. $(\vec a\cdot\vec b)\times(\vec c\cdot\vec d)$
V. $\{(\vec a\times\vec b)\cdot\vec c\}\times\vec d$
where '$\cdot$' represents dot product and '$\times$' represents cross product of vectors. How many of the above are not well-defined?
I. $\vec a\times\vec b$ and $\vec c\times\vec d$ are vectors; their dot product is fine. Well-defined.
II. Same two vectors, cross product. Well-defined.
III. $\vec a\cdot\vec b$ and $\vec c\cdot\vec d$ are scalars; a dot product of two scalars is meaningless. Not defined.
IV. Cross product of two scalars. Not defined.
V. $\{(\vec a\times\vec b)\cdot\vec c\}$ is the scalar triple product — a scalar; crossing a scalar with a vector is meaningless. Not defined.
Three expressions fail. Option (c).
Q.67 [Vectors]
The scalar projection of $\vec a=\lambda\hat i+\hat j-2\hat k$ on $\vec b=2\hat i-\hat j-\lambda\hat k$ is $7/3$. What is the value of $\lambda$?
$\vec a\cdot\vec b=2\lambda-1+2\lambda=4\lambda-1$ and $|\vec b|=\sqrt{4+1+\lambda^2}=\sqrt{5+\lambda^2}$.
$\dfrac{4\lambda-1}{\sqrt{5+\lambda^2}}=\dfrac73\Rightarrow3(4\lambda-1)=7\sqrt{5+\lambda^2}.$
Square: $144\lambda^2-72\lambda+9=49\lambda^2+245\Rightarrow95\lambda^2-72\lambda-236=0.$
$\Delta=72^2+4(95)(236)=5184+89680=94864=308^2$, so $\lambda=\dfrac{72\pm308}{190}$, giving $\lambda=2$ or $\lambda=-\dfrac{118}{95}$.
Only $\lambda=2$ is offered, and it checks: $\dfrac{8-1}{\sqrt{9}}=\dfrac73$ ✓. Option (b).
With four clean options, substitution is faster than the quadratic — but squaring can introduce a false root, so the check at the end is not optional.
Q.68 [Vectors]
How much angle does $\vec r=(\cos\theta)\hat i+(\sin\theta)\hat j+\hat k$ make with the positive direction of $z$-axis?
The angle $\phi$ with $\hat k$ satisfies
$\cos\phi=\dfrac{\vec r\cdot\hat k}{|\vec r|}=\dfrac{1}{\sqrt2}\Rightarrow\phi=\dfrac\pi4.$ Option (c).
Geometrically the vector sweeps a cone of semi-vertical angle $45^\circ$ about the $z$-axis as $\theta$ varies.
Q.69 [Vectors]
Let $\vec a$ and $\vec b$ be unit vectors inclined at $30^\circ$. What is the area of the parallelogram whose sides are represented by the vectors $\vec a+3\vec b$ and $3\vec a+\vec b$?
$(\vec a+3\vec b)\times(3\vec a+\vec b)=\vec a\times\vec b+9(\vec b\times\vec a)=\vec a\times\vec b-9\,\vec a\times\vec b=-8(\vec a\times\vec b).$
$|\vec a\times\vec b|=|\vec a||\vec b|\sin30^\circ=1\cdot1\cdot\dfrac12=\dfrac12.$
Area $=8\times\dfrac12=4$ square units. Option (d).
Q.70 [Vectors]
Let $\vec a$, $\vec b$ and $\vec c$ be unit vectors such that $\vec a\cdot\vec b=\vec a\cdot\vec c=0$. If the angle between $\vec b$ and $\vec c$ is $\pi/6$, then what is $\vec a$ equal to?
$\vec a=\mu(\vec b\times\vec c).$
$|\vec b\times\vec c|=|\vec b||\vec c|\sin\dfrac\pi6=1\cdot1\cdot\dfrac12=\dfrac12.$
Since $\vec a$ is a unit vector, $1=|\mu|\cdot\dfrac12\Rightarrow|\mu|=2$, giving $\vec a=\pm2(\vec b\times\vec c)$.
The option offered is (c) $2(\vec b\times\vec c)$.
The sign choice simply fixes which of the two unit normals is taken — both are legitimate, so only the magnitude 2 is being tested.
Q.71 [Integral Calculus]
What is $\displaystyle\int\dfrac{\sqrt x}{\sqrt{1-x^3}}\,dx$ equal to?
$u=x^{3/2}\Rightarrow du=\dfrac32x^{1/2}dx\Rightarrow\sqrt x\,dx=\dfrac23\,du.$
The integral becomes
$\dfrac23\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}=\dfrac23\sin^{-1}u+c=\dfrac23\sin^{-1}\!\left(x^{3/2}\right)+c.$ Option (d).
Differentiating back is the quickest check: $\dfrac{d}{dx}\left[\dfrac23\sin^{-1}x^{3/2}\right]=\dfrac23\cdot\dfrac{\frac32x^{1/2}}{\sqrt{1-x^3}}=\dfrac{\sqrt x}{\sqrt{1-x^3}}$ ✓.
Q.72 [Integral Calculus]
What is $\displaystyle\int_0^1\dfrac{dx}{(ax+bx+c)^2}$ equal to?
$\displaystyle\int_0^1\frac{dx}{(mx+c)^2}=\left[\frac{-1}{m(mx+c)}\right]_0^1=\frac{-1}{m(m+c)}+\frac{1}{mc}$
$=\frac1m\left(\frac1c-\frac1{m+c}\right)=\frac1m\cdot\frac{m}{c(m+c)}=\frac1{c(m+c)}.$
Restoring $m=a+b$: the value is $\dfrac1{c(a+b+c)}$. Option (d).
A definite integral of a positive integrand must be positive, which rules out options (a) and (b) before any work is done.
Q.73 [Integral Calculus]
What is $\displaystyle\int e^{x\ln10}\,e^x\,dx$ equal to?
So the integrand is $10^x e^x=(10e)^x$, a plain exponential with base $10e$.
$\displaystyle\int a^x dx=\frac{a^x}{\ln a}+c$ with $a=10e$, and $\ln(10e)=\ln10+\ln e=1+\ln10$:
$\displaystyle\int(10e)^x dx=\frac{(10e)^x}{1+\ln10}+c.$ Option (a).
The whole item turns on reading $e^{x\ln10}$ as $10^x$ rather than as $e^{x}\ln10$.
Q.74 [Integral Calculus]
What is $\displaystyle\int_1^2 x^x(1+\ln x)\,dx$ equal to?
$\dfrac{y'}{y}=\ln x+1\Rightarrow\dfrac{d}{dx}\left(x^x\right)=x^x(1+\ln x).$
Therefore
$\displaystyle\int_1^2x^x(1+\ln x)dx=\Big[x^x\Big]_1^2=2^2-1^1=4-1=3.$ Option (b).
Option (a) 4 is what a candidate gets by forgetting the lower limit.
Q.75 [Integral Calculus]
If $f(x+y)=f(x)+f(y)$, then what is $\displaystyle\int_{-1}^{1}f(x)\,dx$ equal to?
Now put $y=-x$: $f(0)=f(x)+f(-x)\Rightarrow f(-x)=-f(x)$, so $f$ is an odd function.
The integral of an odd function over an interval symmetric about the origin vanishes:
$\displaystyle\int_{-1}^{1}f(x)dx=0.$ Option (a).
(Cauchy's functional equation, with any mild regularity, forces $f(x)=cx$ — visibly odd.)
Q.76 [Integral Calculus]
What is the area bounded by $y=e^{|x|}$ and the lines $|x|=1$ and $y=0$?
The curve is symmetric about the $y$-axis, so
$A=\displaystyle\int_{-1}^{1}e^{|x|}dx=2\int_0^1e^x dx=2\Big[e^x\Big]_0^1=2(e-1)=2e-2.$ Option (b).
Option (a) is the area of just the right half — the commonest slip in $|x|$ area questions.
Q.77 [Integral Calculus]
The slope of the tangent to the curve $y=f(x)$ at $(x,f(x))$ is $2x$. If the curve passes through the origin, then what is the area bounded by the curve, the $x$-axis and the line $x=1$?
$A=\displaystyle\int_0^1x^2dx=\left[\frac{x^3}{3}\right]_0^1=\frac13$ square unit. Option (a).
The two steps — integrate the slope to get the curve, then integrate the curve to get the area — are what the item is really checking.
Q.78 [Differential Equations]
What is the solution of $y\,dx-x\,dy-y^2\cos x\,dx=0$?
$\dfrac{y\,dx-x\,dy}{y^2}=\cos x\,dx.$
The left side is an exact differential: $d\!\left(\dfrac xy\right)=\dfrac{y\,dx-x\,dy}{y^2}$.
Integrating both sides: $\dfrac xy=\sin x+c$, so
$y=\dfrac{x}{c+\sin x}.$ Option (a).
Spotting $\dfrac{y\,dx-x\,dy}{y^2}$ as $d(x/y)$ — and $\dfrac{x\,dy-y\,dx}{x^2}$ as $d(y/x)$ — converts a whole family of these problems into one-line integrations.
Q.79 [Differential Equations]
What is the differential equation of the family of straight lines passing through the origin?
$\dfrac{dy}{dx}=m$, and from the equation itself $m=\dfrac yx$.
Equating: $\dfrac{dy}{dx}=\dfrac yx\Rightarrow x\dfrac{dy}{dx}-y=0.$ Option (c).
Option (a) $y\,y'=x$ is the differential equation of the rectangular hyperbolas $y^2-x^2=c$, and option (d) $x\,y'+y=0$ that of the hyperbolas $xy=c$.
Q.80 [Differential Equations]
What is the degree of the differential equation representing the family of curves $y^2=\sqrt c\,x$, where $c$ is a positive parameter?
$2y\dfrac{dy}{dx}=\sqrt c.$
Substituting $\sqrt c$ back into the original equation:
$y^2=2xy\dfrac{dy}{dx}\Rightarrow y-2x\dfrac{dy}{dx}=0\quad(y\ne0).$
The equation is polynomial in $\dfrac{dy}{dx}$ and the highest power of the highest derivative is 1, so the degree is 1 (the order is also 1). Option (a).
The radical sits on the parameter, not on a derivative, so it never raises the degree — that is the trap the question is built around.
Q.81 [Sets, Relations and Functions]
What is the domain of the function $f(x)=\log_x e$?
Two conditions must hold:
• the base of a logarithm must be positive: $x>0$;
• the denominator must not vanish: $\ln x\ne0\Rightarrow x\ne1$.
So the domain is $(0,\infty)-\{1\}$. Option (b).
$x=1$ is excluded for the same reason that 1 is barred as a logarithm base — $1^y$ is always 1 and can never equal $e$.
Q.82 [Limits and Continuity]
If $f(x)$ is the integral of $\dfrac{1+\cos^2x-2\cos x}{x\sin^2x\tan x}$, then what is $\displaystyle\lim_{x\to0}f'(x)$ equal to?
The numerator is a perfect square: $1-2\cos x+\cos^2x=(1-\cos x)^2$.
Now use the small-angle equivalents as $x\to0$:
$1-\cos x\sim\dfrac{x^2}{2}\Rightarrow(1-\cos x)^2\sim\dfrac{x^4}{4}$;
$\sin^2x\sim x^2$ and $\tan x\sim x$, so the denominator $x\sin^2x\tan x\sim x\cdot x^2\cdot x=x^4$.
$\displaystyle\lim_{x\to0}f'(x)=\lim_{x\to0}\frac{x^4/4}{x^4}=\frac14.$ Option (d).
Nothing here needs integrating — reading $f'$ straight off the integrand is the whole trick.
Q.83 [Limits and Continuity]
Consider the following statements in respect of the function $f(x)=[3x]$, where $[\cdot]$ is the greatest integer function :
I. $f(x)$ is continuous at $x=1/3$.
II. $f(x)$ is differentiable at $x=1/4$.
I. At $x=\dfrac13$, $3x=1$ — a jump point. Left limit $=[3\cdot\tfrac13^-]=0$, value $=[1]=1$. Not continuous, so I is false.
II. At $x=\dfrac14$, $3x=0\cdot75$ is not an integer; on a neighbourhood of $\tfrac14$ the function is the constant 0, so $f'(1/4)=0$ exists. II is true.
Answer (b). A step function is differentiable everywhere except at its jumps — where it is not even continuous.
Q.84 [Limits and Continuity]
What is $\displaystyle\lim_{x\to0}\frac{\sin^2x}{x|x|}$ equal to?
Right ($x\to0^+$): $|x|=x$, so the expression is $\dfrac{\sin^2x}{x^2}\to1$.
Left ($x\to0^-$): $|x|=-x$, so the expression is $\dfrac{\sin^2x}{-x^2}\to-1$.
The one-sided limits are $1$ and $-1$; they differ, so the limit does not exist. Option (d).
Whenever $|x|$ or $[x]$ appears at the point of approach, the two-sided limit must be split before anything else is done.
Q.85 [Differential Calculus]
Let $f(x)=p|x|+q$, where $p$ and $q$ are real constants. If $f'(x)$ exists at $x=0$, then what is the value of $p$?
Right: $\displaystyle\lim_{h\to0^+}\frac{p|h|+q-(0+q)}{h}=\lim_{h\to0^+}\frac{ph}{h}=p.$
Left: $\displaystyle\lim_{h\to0^-}\frac{p|h|}{h}=\lim_{h\to0^-}\frac{-ph}{h}=-p.$
Differentiability requires these to agree: $p=-p\Rightarrow p=0$. Option (b).
With $p=0$ the function is the constant $q$ — the corner of $|x|$ is removed only by flattening it entirely. The value of $q$ is irrelevant, which is why option (d) is wrong.
Q.86 [Differential Calculus]
Consider the following statements in respect of the given function : $f(x)=\begin{cases}\dfrac{x^3}{|x|},&x\ne0\\[4pt]0,&x=0\end{cases}$
I. $f(x)$ is continuous everywhere.
II. $f(x)$ is differentiable everywhere.
$f(x)=x|x|=\begin{cases}x^2,&x\ge0\\-x^2,&x<0\end{cases}$
I. Both pieces are polynomials and they agree at 0, so $f$ is continuous everywhere. True.
II. $f'(x)=2x$ for $x>0$ and $-2x$ for $x<0$; both one-sided derivatives at 0 equal 0, so $f'(0)=0$ exists. In fact $f'(x)=2|x|$ throughout. True.
Answer (c). Note that $f''$ does not exist at 0 — the function is $C^1$ but not $C^2$, which is the subtlety behind the item.
Q.87 [Limits and Continuity]
What is $\displaystyle\lim_{x\to\infty}\left(\sqrt{x+\sqrt x}-\sqrt x\right)$ equal to?
$\sqrt{x+\sqrt x}-\sqrt x=\dfrac{\left(x+\sqrt x\right)-x}{\sqrt{x+\sqrt x}+\sqrt x}=\dfrac{\sqrt x}{\sqrt{x+\sqrt x}+\sqrt x}.$
Divide numerator and denominator by $\sqrt x$:
$=\dfrac{1}{\sqrt{1+\frac1{\sqrt x}}+1}\ \longrightarrow\ \dfrac{1}{1+1}=\dfrac12.$ Option (c).
The same rationalising step handles the whole family $\sqrt{x+a}-\sqrt{x}\to0$ and $\sqrt{x^2+ax}-x\to\dfrac a2$.
Q.88 [Limits and Continuity]
If $m$ and $n$ are the roots of the equation $x^2-px+q=0$, then what is $\displaystyle\lim_{x\to m}\frac{e^{x^2-px+q}-1}{(x-m)(x-n)}$ equal to?
Write $u=(x-m)(x-n)$. As $x\to m$, $u\to0$, and the expression is exactly
$\dfrac{e^{u}-1}{u}.$
Using the standard limit $\displaystyle\lim_{u\to0}\frac{e^u-1}{u}=1$, the answer is 1. Option (c).
No expansion of the quadratic, and no L'Hôpital, is needed once the numerator's exponent is recognised as the denominator itself.
Q.89 [Sets, Relations and Functions]
Let $f(x)=x^n+k$, where $n$ is a natural number and $k$ is a positive real constant such that $f(x)+f\!\left(\dfrac1x\right)=f(x)\,f\!\left(\dfrac1x\right)$ and $f(2)=9$. What is $f(-1)$ equal to?
LHS $=x^n+x^{-n}+2k$
RHS $=(x^n+k)(x^{-n}+k)=1+k\left(x^n+x^{-n}\right)+k^2$
Equating and collecting: $(k-1)\left(x^n+x^{-n}\right)=k^2-2k+1=(k-1)^2$.
This must hold for all $x$, and $x^n+x^{-n}$ is not constant, so $k-1=0$, i.e. $k=1$ (and then both sides are 0).
So $f(x)=x^n+1$. From $f(2)=2^n+1=9$ we get $2^n=8$, i.e. $n=3$.
$f(-1)=(-1)^3+1=0.$ Option (c).
The relation $f(x)+f(1/x)=f(x)f(1/x)$ is the standard signature of $f(x)=x^n+1$.
Q.90 [Differential Calculus]
If $f(x)=\cos\left\{\dfrac\pi3[x]+x\right\}$ for $1
$\dfrac\pi2\approx1\cdot571$ does lie in $(1,2)$, so
$f\!\left(\dfrac\pi2\right)=\cos\left(\dfrac\pi3+\dfrac\pi2\right)=\cos\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{2}.$ Option (b).
Checking that the given point actually lies inside the stated interval is part of the question — outside $(1,2)$ the formula for $[x]$ would change.
Q.91 [Differential Calculus]
Consider the following statements :
I. $\dfrac{d}{dx}\ln|x|=-\dfrac1x$ if $x<0$
II. $\dfrac{d}{dx}\ln|x|=\dfrac1x$ if $x>0$
$\dfrac{d}{dx}\ln(-x)=\dfrac{1}{-x}\cdot(-1)=\dfrac1x.$
So Statement I is false — the derivative is $+\dfrac1x$ (itself a negative number when $x<0$), not $-\dfrac1x$.
For $x>0$, $|x|=x$ and the derivative is plainly $\dfrac1x$, so Statement II is true.
Answer (b). The point of the item: $\dfrac{d}{dx}\ln|x|=\dfrac1x$ holds on both sides of the origin, which is exactly why $\displaystyle\int\frac{dx}{x}=\ln|x|+c$ is written with the modulus.
Q.92 [Differential Calculus]
If $y=\left|\sin\!\left(\dfrac\pi4-x\right)\right|$, then what is $\dfrac{dy}{dx}$ at $x=\dfrac\pi4$ equal to?
Put $t=\dfrac\pi4-x$, so near $t=0$, $y=|\sin t|\approx|t|=\left|\dfrac\pi4-x\right|$.
Right-hand derivative at $x=\dfrac\pi4$: $+1$. Left-hand derivative: $-1$.
They differ, so $\dfrac{dy}{dx}$ does not exist there. Option (d).
(Formally, $\dfrac{dy}{dx}=-\cos\!\left(\dfrac\pi4-x\right)\operatorname{sgn}\!\left[\sin\!\left(\dfrac\pi4-x\right)\right]$, and the sign factor is undefined at the crossing.)
Q.93 [Differential Calculus]
Let $f$ be a differentiable function such that $f(x+y)=f(x)+f(y)$ for all $x,y\in R$. Which of the following statements is/are correct?
I. If $f(x)=x\,g(x)$, then the derivative of $f(x)$ with respect to $x$ is equal to $g(0)$.
II. If $f(x)=x^2g(x)$, then the derivative of $f(x)$ with respect to $x$ is equal to 0.
(Here $g$ is a continuous function)
Statement I. If $f(x)=x\,g(x)$ then for $x\ne0$, $g(x)=\dfrac{cx}{x}=c$; continuity of $g$ forces $g(0)=c$ too. Since $f'(x)=c$, indeed $f'(x)=g(0)$. True.
Statement II. If $f(x)=x^2g(x)$ then for $x\ne0$, $g(x)=\dfrac{c}{x}$. For $g$ to be continuous at 0 this must not blow up, which forces $c=0$. Then $f\equiv0$ and $f'(x)=0$. True.
Answer (c). The word 'continuous' in the bracket is doing all the work in Statement II — without it, $g$ could be defined arbitrarily at 0 and the conclusion would fail.
Q.94 [Differential Calculus]
If $f(x)=x^{n-1}+x^{n-2}+x^{n-3}+\cdots+1$, then what is $f'(2)$ equal to?
$f'(x)=(n-1)x^{n-2}+(n-2)x^{n-3}+\cdots+2x+1=\displaystyle\sum_{k=1}^{n-1}k\,x^{k-1}.$
At $x=2$: $f'(2)=\displaystyle\sum_{k=1}^{n-1}k\,2^{k-1}$.
Use the standard sum $\displaystyle\sum_{k=1}^{m}k\,2^{k-1}=(m-1)2^{m}+1$, with $m=n-1$:
$f'(2)=(n-2)2^{\,n-1}+1=n2^{\,n-1}-2\cdot2^{\,n-1}+1=n2^{\,n-1}-2^{\,n}+1.$ Option (a).
Check with $n=3$: $f(x)=x^2+x+1$, $f'(2)=2(2)+1=5$; the formula gives $3\cdot4-8+1=5$ ✓.
Q.95 [Applications of Derivatives]
$ABC$ is a triangle right angled at $B$. If $AC=8$ units, then what is the area of the triangle of maximum area?
By AM–GM, $ac\le\dfrac{a^2+c^2}{2}=32$, with equality when $a=c$.
Maximum area $=\dfrac12(32)=16$ square units, attained by the isosceles right triangle with legs $4\sqrt2$. Option (c).
Calculus route: with $a=8\sin\theta$, $c=8\cos\theta$, area $=32\sin\theta\cos\theta=16\sin2\theta$, maximal at $\theta=45^\circ$ — the same answer, 16.
A geometric way to see it: $B$ lies on the circle with $AC$ as diameter (radius 4), so the greatest height above $AC$ is the radius 4, giving area $\tfrac12\cdot8\cdot4=16$.
Q.96 [Applications of Derivatives]
For the curve $y=xe^{2x}$,
Since $e^{2x}>0$ always, the only stationary point is $1+2x=0$, i.e. $x=-\dfrac12$.
$\dfrac{d^2y}{dx^2}=2e^{2x}(1+2x)+2e^{2x}=e^{2x}(4x+4)$; at $x=-\dfrac12$ this is $e^{-1}(2)>0$, so the point is a minimum.
Answer (c). (The minimum value is $-\dfrac{1}{2e}$, and the curve has no maximum — it rises without bound as $x\to\infty$.)
Q.97 [Applications of Derivatives]
Let $f(x)=\displaystyle\int xe^x\,dx$. Then $f(x)$ decreases in the interval
$f$ decreases where $f'(x)<0$. Since $e^x>0$ for every $x$, the sign of $f'$ is the sign of $x$:
$xe^x<0\iff x<0.$
So $f$ decreases on $(-\infty,0)$. Option (a).
Options (c) and (d) are placed for candidates who integrate to $f(x)=e^x(x-1)+c$ and then differentiate that, mistakenly landing on the turning point of $f'$ rather than of $f$.
Q.98 [Differential Equations]
What is the solution of the differential equation $\dfrac{dy}{dx}=1+x\cot(y-x)$?
The equation becomes $\dfrac{dv}{dx}=x\cot v$, which separates:
$\tan v\,dv=x\,dx.$
Integrate: $-\ln|\cos v|=\dfrac{x^2}{2}+c_1$, i.e. $\ln|\sec v|=\dfrac{x^2}{2}+c_1$.
Exponentiating: $\sec v=c\,e^{x^2/2}$, that is
$\sec(y-x)=c\,e^{x^2/2}.$ Option (a).
Option (b) is the same family written upside down and would need $c\,e^{-x^2/2}$ to be correct — the sign in the exponent is the discriminator.
Q.99 [Sets, Relations and Functions]
Consider the following statements in respect of the function $f:R-\left\{\dfrac35\right\}\to R-\left\{\dfrac35\right\}$ such that $f(x)=\dfrac{3x+2}{5x-3}$ :
I. $f(x)$ is a bijective function.
II. $f^{-1}(x)=f(x)$
$f(f(x))=\dfrac{3\left(\frac{3x+2}{5x-3}\right)+2}{5\left(\frac{3x+2}{5x-3}\right)-3}=\dfrac{3(3x+2)+2(5x-3)}{5(3x+2)-3(5x-3)}=\dfrac{9x+6+10x-6}{15x+10-15x+9}=\dfrac{19x}{19}=x.$
So $f\circ f=\mathrm{id}$, which proves both statements at once: a function that is its own inverse must be one-one and onto (I true), and $f^{-1}=f$ (II true).
Answer (c). Such maps are called involutions; for $f(x)=\dfrac{ax+b}{cx+d}$ the condition is simply $a+d=0$, which holds here since $3+(-3)=0$.
Q.100 [Sets, Relations and Functions]
Let $A$ and $B$ be the sets having only 2 and 3 elements respectively. What is the total number of mappings from $A$ to $B$?
Each of the 2 elements of $A$ has 3 possible images, so the count is
$3\times3=3^2=9.$ Option (d).
The general rule is $n(B)^{\,n(A)}$ — note which set supplies the base and which the exponent; reversing them gives $2^3=8$, which is option (c).
Q.101 [Statistics]
The variance of 10 observations is 25. If 2 is multiplied to each of the observations and subsequently 7 is subtracted from each multiplied observation, then what is the new variance?
$\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X).$
Here the transformation is $Y=2X-7$, so $a=2$ and $b=-7$:
$\operatorname{Var}(Y)=2^2\times25=100.$ Option (b).
The subtraction of 7 changes the mean (from $\mu$ to $2\mu-7$) but leaves the spread untouched — that is the idea being tested.
Q.102 [Statistics]
If the random variable $X$ has mean 5 and standard deviation 4, then what is the standard deviation of the random variable $Y=3X+4$?
$\sigma_Y=|a|\,\sigma_X=3\times4=12.$ Option (b).
The mean of 5 is a decoy — it plays no part. (For the record, $E[Y]=3(5)+4=19$.)
Q.103 [Probability]
For a binomial distribution with mean 4 and standard deviation $\sqrt3$, what is the value of $P(X=0)$?
$np=4$ and $npq=\left(\sqrt3\right)^2=3$, so dividing, $q=\dfrac34$ and hence $p=\dfrac14$.
From $np=4$: $n\cdot\dfrac14=4\Rightarrow n=16$.
$P(X=0)=\binom{16}{0}p^0q^{16}=\left(\dfrac34\right)^{16}.$ Option (b).
Option (a) is the trap for anyone who writes $p^{16}$ instead of $q^{16}$ — $X=0$ means no successes, so every trial must be a failure.
Q.104 [Statistics]
If the correlation coefficient between the variables $X$ and $Y$ is zero, then the two lines of regression are
$Y-\bar Y=r\dfrac{\sigma_Y}{\sigma_X}\left(X-\bar X\right)$ and $X-\bar X=r\dfrac{\sigma_X}{\sigma_Y}\left(Y-\bar Y\right).$
With $r=0$ both regression coefficients vanish, leaving
$Y=\bar Y$ and $X=\bar X$ — a horizontal line and a vertical line. Option (b).
They meet at $(\bar X,\bar Y)$, as regression lines always do, and the angle between them is $90^\circ$ — the geometric statement of 'no linear relationship'. At the other extreme, $r=\pm1$ makes the two lines coincide.
Q.105 [Probability]
Four dice are rolled. What is the probability of getting a total of the numbers on the dice as 7?
Put $y_i=x_i-1\ge0$; then $y_1+y_2+y_3+y_4=3$, and the upper bound $y_i\le5$ cannot bite.
Number of non-negative solutions $=\binom{3+3}{3}=\binom63=20.$
Total outcomes $=6^4=1296$.
$P=\dfrac{20}{1296}=\dfrac{5}{324}.$ Option (d).
Option (a) $5/1296$ is what a candidate gets by forgetting to reduce the fraction incorrectly; option (c) keeps the unsimplified numerator.
Q.106 [Probability]
If $P(A\cap B)=1/2$ and $P(\bar A\cap\bar B)=1/2$, and $2P(A)=P(B)=k$, then what is the value of $k$?
$P(A\cup B)=1-P(\bar A\cap\bar B)=1-\dfrac12=\dfrac12.$
From $2P(A)=P(B)=k$: $P(A)=\dfrac k2$ and $P(B)=k$.
Addition rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$\dfrac12=\dfrac k2+k-\dfrac12\Rightarrow1=\dfrac{3k}{2}\Rightarrow k=\dfrac23.$ Option (d).
(Check: $P(A)=\tfrac13$, $P(B)=\tfrac23$, $P(A\cup B)=\tfrac13+\tfrac23-\tfrac12=\tfrac12$ ✓.)
Q.107 [Probability]
A committee has 6 men and 4 women. One member is selected randomly which is woman only. What is the probability that she belongs to a subgroup of 2 senior women?
Of those 4, exactly 2 are senior:
$P=\dfrac24=\dfrac12.$ Option (b).
The 6 men are irrelevant once the conditioning is applied; including them would wrongly give $\dfrac2{10}$.
Q.108 [Statistics]
The standard deviation of $X$ is 4 and the standard deviation of $Y$ is 5. If the correlation coefficient $r$ is $0\cdot8$, then what is the regression coefficient of $Y$ on $X$?
$b_{YX}=r\,\dfrac{\sigma_Y}{\sigma_X}=0\cdot8\times\dfrac54=1.$ Option (d).
Two useful checks: a regression coefficient always carries the sign of $r$ (so the negative options are impossible here), and $b_{YX}\times b_{XY}=r^2=0\cdot64$ — indeed $b_{XY}=0\cdot8\times\dfrac45=0\cdot64$, and $1\times0\cdot64=0\cdot64$ ✓.
Q.109 [Probability]
In a class, the probability of passing students in Mathematics is $0\cdot7$ and the probability of passing both Mathematics and Statistics is $0\cdot5$. What is the probability that a student passes Statistics given that the student passed Mathematics?
$P(S\mid M)=\dfrac{P(S\cap M)}{P(M)}=\dfrac{0\cdot5}{0\cdot7}=\dfrac57.$ Option (a).
Note the order of the conditioning: dividing by the probability of the given event. Reversing it would give $P(M\mid S)$, which cannot even be computed here since $P(S)$ is not supplied.
Q.110 [Statistics]
If the mean of 50 observations is 40 and the standard deviation is 8, then what is the coefficient of variation?
The number of observations (50) is not needed. The coefficient of variation is a relative measure of dispersion, which is what makes it the right tool for comparing spread between data sets measured in different units.
Q.111 [Probability]
If two fair dice are tossed, then what is the probability that the sum of the numbers on the faces of the dice is neither 5 nor 7?
Sum 5: $(1,4),(2,3),(3,2),(4,1)$ — 4 ways.
Sum 7: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — 6 ways.
These are disjoint, so $P(\text{5 or 7})=\dfrac{10}{36}=\dfrac5{18}$.
$P(\text{neither})=1-\dfrac5{18}=\dfrac{13}{18}.$ Option (c).
Q.112 [Probability]
Consider the following statements for three events $A$, $B$ and $C$ :
I. $P(A\cap B\cap C)\le P(A)+P(B)+P(C)-2$
II. $P(A\cup B\cup C)\ge P(A)+P(B)+P(C)$
$P(A\cap B\cap C)\ \ge\ P(A)+P(B)+P(C)-2.$
Counter-example to the version printed: take $A=B=C$ with $P=0\cdot5$. Then the left side is $0\cdot5$ and the right side is $1\cdot5-2=-0\cdot5$, so $0\cdot5\le-0\cdot5$ fails. False.
Statement II. Inclusion–exclusion gives
$P(A\cup B\cup C)=\Sigma P(A)-\Sigma P(A\cap B)+P(A\cap B\cap C)\ \le\ P(A)+P(B)+P(C),$
by sub-additivity — the union can never exceed the sum. Equality holds only for pairwise disjoint events, so the strict '$\ge$' as a general claim is false.
Answer (d).
Q.113 [Probability]
A Mathematics problem is given to two students $X$ and $Y$ to solve. The odds in favour of $X$ solving the problem are 6 to 9 and the odds against $Y$ in solving the problem are 6 to 5. What is the probability that the problem will be solved if both $X$ and $Y$ try to solve the problem?
Odds in favour of $X$ are $6:9$, so $P(X)=\dfrac{6}{6+9}=\dfrac{6}{15}=\dfrac25$ and $P(\bar X)=\dfrac35$.
Odds against $Y$ are $6:5$, so $P(\bar Y)=\dfrac{6}{11}$ and $P(Y)=\dfrac{5}{11}$.
The problem is solved unless both fail:
$P(\text{solved})=1-P(\bar X)P(\bar Y)=1-\dfrac35\cdot\dfrac6{11}=1-\dfrac{18}{55}=\dfrac{37}{55}.$ Option (c).
Reading the second set of odds as 'in favour' gives $\dfrac{39}{55}$, which is exactly option (d).
Q.114 [Probability]
The events $A$ and $B$ are independent among three events $A$, $B$ and $D$. If $P(A\cap B\cap D)=0\cdot04$, $P(D\mid A\cap B)=0\cdot25$ and $P(B)=4P(A)$, then what is the value of $P(A\cup B)$?
$P(A\cap B\cap D)=P(D\mid A\cap B)\cdot P(A\cap B)\Rightarrow0\cdot04=0\cdot25\times P(A\cap B),$
so $P(A\cap B)=0\cdot16$.
$A$ and $B$ are independent, so $P(A)P(B)=0\cdot16$. With $P(B)=4P(A)$:
$4P(A)^2=0\cdot16\Rightarrow P(A)^2=0\cdot04\Rightarrow P(A)=0\cdot2,\ P(B)=0\cdot8.$
$P(A\cup B)=0\cdot2+0\cdot8-0\cdot16=0\cdot84.$ Option (b).
Event $D$ serves only to deliver $P(A\cap B)$ — once that is extracted it plays no further part.
Q.115 [Probability]
Items 115–117: Let $A$, $B$, $C$ and $D$ be mutually exclusive and exhaustive events such that $\dfrac{P(A)}{6}=\dfrac{P(B)}{3}=\dfrac{P(C)}{4}=\dfrac{P(D)}{2}$.
What is $\dfrac{P(A)+2P(B)}{3P(C)+P(D)}$ equal to?
Mutually exclusive and exhaustive means the four probabilities add to 1:
$6k+3k+4k+2k=15k=1\Rightarrow k=\dfrac1{15}.$
So $P(A)=\dfrac{6}{15},\ P(B)=\dfrac{3}{15},\ P(C)=\dfrac{4}{15},\ P(D)=\dfrac{2}{15}.$
$\dfrac{P(A)+2P(B)}{3P(C)+P(D)}=\dfrac{6k+6k}{12k+2k}=\dfrac{12k}{14k}=\dfrac67.$ Option (b).
The $k$ cancels, so this part could have been answered without even finding $k$.
Q.116 [Probability]
Items 115–117 (continued). Same four events.
If $G$ is the geometric mean of $P(A)$, $P(B)$, $P(C)$ and $P(D)$, then what is $G$ equal to?
$G=\left(6k\cdot3k\cdot4k\cdot2k\right)^{1/4}=\left(144\,k^4\right)^{1/4}=k\cdot144^{1/4}.$
$144^{1/4}=\left(12^2\right)^{1/4}=\sqrt{12}=2\sqrt3.$
$G=\dfrac{2\sqrt3}{15}.$ Rationalising the other way, $\dfrac{2\sqrt3}{15}=\dfrac{2\sqrt3}{5\sqrt3\cdot\sqrt3}=\dfrac{2}{5\sqrt3}.$ Option (b).
(Numerically $G\approx0\cdot2309$, which sits sensibly between the smallest probability $\tfrac2{15}\approx0\cdot133$ and the largest $\tfrac6{15}=0\cdot4$.)
Q.117 [Probability]
Items 115–117 (continued). Same four events.
If $H$ is the harmonic mean of $P(A)$, $P(B)$, $P(C)$ and $P(D)$, then what is $H$ equal to?
$\dfrac1{6k}+\dfrac1{3k}+\dfrac1{4k}+\dfrac1{2k}=\dfrac1k\left(\dfrac16+\dfrac13+\dfrac14+\dfrac12\right)=\dfrac1k\cdot\dfrac{2+4+3+6}{12}=\dfrac{15}{12k}=\dfrac{5}{4k}.$
$H=\dfrac{4}{\frac{5}{4k}}=\dfrac{16k}{5}=\dfrac{16}{5\times15}=\dfrac{16}{75}.$ Option (c).
Consistency check: $H\le G\le A$. Here $H=0\cdot213$, $G=0\cdot231$ and the arithmetic mean is $\tfrac{15k}{4}=0\cdot25$ ✓.
Q.118 [Probability]
Items 118–120: Five numbers are randomly picked from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and arranged them in ascending order $[x_1
What is $P(x_1=4)$ equal to?
$x_1=4$ means 4 is the smallest chosen, so the other four must all come from $\{5,6,7,8,9,10\}$:
Favourable $=\binom64=15.$
$P(x_1=4)=\dfrac{15}{252}=\dfrac{5}{84}.$ Option (a).
The ascending arrangement adds nothing to the counting — every 5-element subset corresponds to exactly one arrangement.
Q.119 [Probability]
Items 118–120 (continued). Same selection of five numbers from 1 to 10.
What is $P(x_3=6)$ equal to?
two from $\{1,2,3,4,5\}$: $\binom52=10$ ways
two from $\{7,8,9,10\}$: $\binom42=6$ ways
Favourable $=10\times6=60$, out of $\binom{10}{5}=252$.
$P(x_3=6)=\dfrac{60}{252}=\dfrac{5}{21}.$ Option (a).
Q.120 [Probability]
Items 118–120 (continued). Same selection of five numbers from 1 to 10.
What is $P(x_1=2,\ x_2=3,\ x_3=8)$ equal to?
Numbers greater than 8 available: $\{9,10\}$ — exactly two of them, so there is precisely one favourable selection, $\{2,3,8,9,10\}$.
$P=\dfrac1{\binom{10}{5}}=\dfrac1{252}.$ Option (c).
Option (d) tempts anyone who thinks the requirement is impossible — it is not, but only just: had $x_3$ been 9, no valid completion would exist and the answer really would be 0.