MCQ 14 Marks
The area (in sq. units) of the region $\left\{(x, y): 0 \leq y \leq 2|x|+1,0 \leq y \leq x^{2}+1,|x| \leq 3\right\}$ is
- A$\frac{80}{3}$
- B$\frac{64}{3}$
- C$\frac{17}{3}$
- D$\frac{32}{3}$
Answer
View full question & answer→Questions
SECTION - A [MATHS - MCQ]
🎯
Test yourself on this topic
20 questions · timed · auto-graded
MCQ 24 Marks
Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $x$ denote the number of defective oranges, then the variance of x is :
- A$28 / 75$
- B$14 / 25$
- C$26 / 75$
- D$18 / 25$
Answer
View full question & answer→
MCQ 34 Marks
Let $O$ be the origin, the point $A$ be $z_{1}=\sqrt{3}+2 \sqrt{2 \mathrm{i}}$, the point $\mathrm{B}\left(\mathrm{z}_{2}\right)$ be such that $\sqrt{3}\left|z_{2}\right|=\left|z_{1}\right|$ and $\arg \left(z_{2}\right)=\arg \left(z_{1}\right)+\frac{\pi}{6}$. Then
- Aarea of triangle ABO is $\frac{11}{\sqrt{3}}$
- BABO is a scalene triangle
- Carea of triangle ABO is $\frac{11}{4}$
- ✓ABO is an obtuse angled isosceles triangle
Answer
View full question & answer→Correct option: D.
ABO is an obtuse angled isosceles triangle
(D)
Sol. $z_{1}=\sqrt{3}+2 \sqrt{2} i \quad \& \frac{\left|z_{2}\right|}{\left|z_{1}\right|}=\frac{1}{\sqrt{3}}$
given $\arg \left(\frac{z_{2}}{z_{1}}\right)=\frac{\pi}{6}$
$z_{2}=\frac{\left|z_{2}\right|}{\left|z_{1}\right|} \cdot z_{1} e^{i\left(\frac{\pi}{6}\right)}$
$z_{2}=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} i)(\sqrt{3}+i)}{2}$
$z_{2}=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}}$
Now,
$z_{1}-z_{2}=\frac{(3+2 \sqrt{2})+i(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}$
$\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|=\left|\mathrm{z}_{2}\right| \Rightarrow \triangle \mathrm{ABO}$ is isosceles with angles $\frac{\pi}{6}, \frac{\pi}{6} \& \frac{2 \pi}{3}$
Sol. $z_{1}=\sqrt{3}+2 \sqrt{2} i \quad \& \frac{\left|z_{2}\right|}{\left|z_{1}\right|}=\frac{1}{\sqrt{3}}$
given $\arg \left(\frac{z_{2}}{z_{1}}\right)=\frac{\pi}{6}$
$z_{2}=\frac{\left|z_{2}\right|}{\left|z_{1}\right|} \cdot z_{1} e^{i\left(\frac{\pi}{6}\right)}$
$z_{2}=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} i)(\sqrt{3}+i)}{2}$
$z_{2}=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}}$
Now,
$z_{1}-z_{2}=\frac{(3+2 \sqrt{2})+i(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}$
$\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|=\left|\mathrm{z}_{2}\right| \Rightarrow \triangle \mathrm{ABO}$ is isosceles with angles $\frac{\pi}{6}, \frac{\pi}{6} \& \frac{2 \pi}{3}$
MCQ 44 Marks
Let ${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56$ and ${ }^{n} C_{r+1}=70$. Let $\mathrm{A}(4 \cos t, 4 \sin t), \mathrm{B}(2 \sin t,-2 \cos t)$ and $\mathrm{C}\left(3 \mathrm{r}-\mathrm{n}, \mathrm{r}^{2}-\mathrm{n}-1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3 x-1)^{2}+(3 y)^{2}=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals :
- ✓20
- B8
- C6
- D18
Answer
View full question & answer→Correct option: A.
20
(A)
Sol. ${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56$
$\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{28}{56}$
$\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2}$
$\frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2}$
$3 \mathrm{r}=\mathrm{n}+1$
$\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{{ }^{n} \mathrm{C}_{\mathrm{r}+1}}}=\frac{56}{70}$
$\frac{(\mathrm{r}+1)}{(\mathrm{n}-\mathrm{r})}=\frac{56}{70} \Rightarrow 9 \mathrm{r}=4 \mathrm{n}-5$
By (i) \& (ii)
( $\mathrm{r}=3$ ), $(\mathrm{n}=8)$
A (4cost, 4sint) B(2sint, $-2 \cos t) \mathrm{C}\left(3 \mathrm{r}-\mathrm{n}, \mathrm{r}^{2}-\mathrm{n}-1\right)$
A (4cost, 4sint) $\quad \mathrm{B}(2 \sin t,-2 \cos t) \mathrm{C}(1,0)$
$(3 x-1)^{2}+(3 y)^{2}=(4 \cos t+2 \sin t)^{2}+(4 \sin t-\cos t)^{2}$
$(3 \mathrm{x}-1)^{2}+(3 \mathrm{y})^{2}=20 \quad \therefore$
Sol. ${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56$
$\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{28}{56}$
$\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2}$
$\frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2}$
$3 \mathrm{r}=\mathrm{n}+1$
$\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{{ }^{n} \mathrm{C}_{\mathrm{r}+1}}}=\frac{56}{70}$
$\frac{(\mathrm{r}+1)}{(\mathrm{n}-\mathrm{r})}=\frac{56}{70} \Rightarrow 9 \mathrm{r}=4 \mathrm{n}-5$
By (i) \& (ii)
( $\mathrm{r}=3$ ), $(\mathrm{n}=8)$
A (4cost, 4sint) B(2sint, $-2 \cos t) \mathrm{C}\left(3 \mathrm{r}-\mathrm{n}, \mathrm{r}^{2}-\mathrm{n}-1\right)$
A (4cost, 4sint) $\quad \mathrm{B}(2 \sin t,-2 \cos t) \mathrm{C}(1,0)$
$(3 x-1)^{2}+(3 y)^{2}=(4 \cos t+2 \sin t)^{2}+(4 \sin t-\cos t)^{2}$
$(3 \mathrm{x}-1)^{2}+(3 \mathrm{y})^{2}=20 \quad \therefore$
MCQ 54 Marks
Let for some function $y=f(x), \int_{0}^{x} t f(t) d t=x^{2} f(x)$, $x>0$ and $f(2)=3$. Then $f(6)$ is equal to :
- ✓1
- B2
- C3
- D6
Answer
View full question & answer→Correct option: A.
1
(A)
Sol. $\int_{0}^{\mathrm{x}} \mathrm{tf}(\mathrm{t}) \mathrm{dt}=\mathrm{x}^{2}+(\mathrm{x}), \mathrm{x}>0$
Diff. both side w.r. to $x$
$x f(x)=x^{2} f^{\prime}(x)+2 x f(x)$
$-x f(x)=x^{2} f^{\prime}(x)$
$\int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{-1}{x} d x$
$\operatorname{logdf}(x)=-\log x+\log c$
$f(x)=\frac{c}{x}$
$\mathrm{f}(2)=3 \Rightarrow 3=\frac{\mathrm{c}}{2} \Rightarrow \mathrm{c}=6$
$f(x)=\frac{6}{x}$
$f(6)=1$
Sol. $\int_{0}^{\mathrm{x}} \mathrm{tf}(\mathrm{t}) \mathrm{dt}=\mathrm{x}^{2}+(\mathrm{x}), \mathrm{x}>0$
Diff. both side w.r. to $x$
$x f(x)=x^{2} f^{\prime}(x)+2 x f(x)$
$-x f(x)=x^{2} f^{\prime}(x)$
$\int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{-1}{x} d x$
$\operatorname{logdf}(x)=-\log x+\log c$
$f(x)=\frac{c}{x}$
$\mathrm{f}(2)=3 \Rightarrow 3=\frac{\mathrm{c}}{2} \Rightarrow \mathrm{c}=6$
$f(x)=\frac{6}{x}$
$f(6)=1$
MCQ 64 Marks
The sum, of the squares of all the roots of the equation $x^{2}+|2 x-3|-4=0$, is :
- A$3(3-\sqrt{2})$
- B$6(3-\sqrt{2})$
- ✓$6(2-\sqrt{2})$
- D$3(2-\sqrt{2})$
Answer
View full question & answer→Correct option: C.
$6(2-\sqrt{2})$
(C)Sol. $\quad x^{2}+|2 x-3|-4=0$
Case I: $x \geq \frac{3}{2}$
$
\begin{aligned}
& x^{2}+2 x-3-4=0 \\
& x^{2}+2 x-7=0 \\
& x=2 \sqrt{2}-1
\end{aligned}
$
Case II : $\mathrm{x}<\frac{3}{2}$
$
\begin{aligned}
& x^{2}+3-2 x-4=0 \\
& x^{2}-2 x-1=0 \\
& x=1-\sqrt{2}
\end{aligned}
$
Sum of squares $=(2 \sqrt{2}-1)^{2}+(1-\sqrt{2})^{2}$
$
\begin{align*}
& =8-4 \sqrt{2}+1+1-2 \sqrt{2}+2 \\
& =6(2-\sqrt{2}) \quad \therefore(3) \tag{3}
\end{align*}
$
Case I: $x \geq \frac{3}{2}$
$
\begin{aligned}
& x^{2}+2 x-3-4=0 \\
& x^{2}+2 x-7=0 \\
& x=2 \sqrt{2}-1
\end{aligned}
$
Case II : $\mathrm{x}<\frac{3}{2}$
$
\begin{aligned}
& x^{2}+3-2 x-4=0 \\
& x^{2}-2 x-1=0 \\
& x=1-\sqrt{2}
\end{aligned}
$
Sum of squares $=(2 \sqrt{2}-1)^{2}+(1-\sqrt{2})^{2}$
$
\begin{align*}
& =8-4 \sqrt{2}+1+1-2 \sqrt{2}+2 \\
& =6(2-\sqrt{2}) \quad \therefore(3) \tag{3}
\end{align*}
$
MCQ 74 Marks
The sum of all local minimum values of the function $f(x)=\left\{\begin{array}{cc}1-2 x, & x<-1 \\ \frac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x-4)(x-5), & x>2\end{array}\right.$
- A$\frac{171}{72}$
- B$\frac{131}{72}$
- C$\frac{157}{72}$
- D$\frac{167}{72}$
Answer
View full question & answer→
MCQ 84 Marks
If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^{2} \cos ^{2} x}{\left(1+e^{x}\right)} d x=\pi\left(\alpha \pi^{2}+\beta\right), \alpha, \beta \in Z$, then $(\alpha+\beta)^{2}$ equals :
- A144
- B196
- ✓100
- D64
Answer
View full question & answer→Correct option: C.
100
(C)
Sol. $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^{2} \cos ^{2} x}{\left(1+e^{x}\right)} d x$ (Apply King Property)
$
\begin{aligned}
& \int_{0}^{\frac{\pi}{2}} 96 x^{2} \cos ^{2} x=48 \int_{0}^{\frac{\pi}{2}} x^{2}(1+\cos 2 x) d x \\
& 48\left[\left(\frac{x^{3}}{3}\right)_{0}^{\pi / 2}+\int_{0}^{\frac{\pi}{2}} x_{\mathrm{I}}^{2} \cos 2 x d x\right]
\end{aligned}
$
$\Rightarrow$ On solving $\pi\left(2 \pi^{2}-12\right)$
$\Rightarrow \alpha=2, \beta=-12$
$\Rightarrow(\alpha+\beta)^{2}=100$
Sol. $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^{2} \cos ^{2} x}{\left(1+e^{x}\right)} d x$ (Apply King Property)
$
\begin{aligned}
& \int_{0}^{\frac{\pi}{2}} 96 x^{2} \cos ^{2} x=48 \int_{0}^{\frac{\pi}{2}} x^{2}(1+\cos 2 x) d x \\
& 48\left[\left(\frac{x^{3}}{3}\right)_{0}^{\pi / 2}+\int_{0}^{\frac{\pi}{2}} x_{\mathrm{I}}^{2} \cos 2 x d x\right]
\end{aligned}
$
$\Rightarrow$ On solving $\pi\left(2 \pi^{2}-12\right)$
$\Rightarrow \alpha=2, \beta=-12$
$\Rightarrow(\alpha+\beta)^{2}=100$
MCQ 94 Marks
If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to
- A9
- B12
- C8
- D7
MCQ 104 Marks
Let $T_{r}$ be the $r^{\text {th }}$ term of an A.P. If for some $m$,
$\mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}$ and $20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13$, then $5 \mathrm{~m} \sum_{\mathrm{r}=\mathrm{m}}^{2 \mathrm{~m}} \mathrm{~T}_{\mathrm{r}}$ is equal to :
$\mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}$ and $20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13$, then $5 \mathrm{~m} \sum_{\mathrm{r}=\mathrm{m}}^{2 \mathrm{~m}} \mathrm{~T}_{\mathrm{r}}$ is equal to :
- A112
- ✓126
- C98
- D142
Answer
View full question & answer→Correct option: B.
126
(B)
Sol. $\mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}, 20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13$
$\mathrm{T}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{25}$
$\mathrm{T}_{25}=\mathrm{a}+24 \mathrm{~d}=\frac{1}{20}$
20. $\frac{25}{2}\left[a+\frac{1}{20}\right]=13 \Rightarrow a=\frac{1}{500}$
also, $20 \mathrm{~S}_{25}=20 \cdot \frac{25}{2}[2 \mathrm{a}+24 \mathrm{~d}]=13 \Rightarrow \mathrm{~d}=\frac{1}{500}$
from (1) $\frac{1}{500}+\frac{m-1}{500}=\frac{1}{25} \Rightarrow m=20$
Now,
$5 \mathrm{~m} \sum_{\mathrm{r}=\mathrm{m}}^{2 \mathrm{~m}} \mathrm{~T}_{\mathrm{r}}=100 \sum_{\mathrm{r}=20}^{40} \mathrm{~T}_{\mathrm{r}}=126$
Sol. $\mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}, 20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13$
$\mathrm{T}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{25}$
$\mathrm{T}_{25}=\mathrm{a}+24 \mathrm{~d}=\frac{1}{20}$
20. $\frac{25}{2}\left[a+\frac{1}{20}\right]=13 \Rightarrow a=\frac{1}{500}$
also, $20 \mathrm{~S}_{25}=20 \cdot \frac{25}{2}[2 \mathrm{a}+24 \mathrm{~d}]=13 \Rightarrow \mathrm{~d}=\frac{1}{500}$
from (1) $\frac{1}{500}+\frac{m-1}{500}=\frac{1}{25} \Rightarrow m=20$
Now,
$5 \mathrm{~m} \sum_{\mathrm{r}=\mathrm{m}}^{2 \mathrm{~m}} \mathrm{~T}_{\mathrm{r}}=100 \sum_{\mathrm{r}=20}^{40} \mathrm{~T}_{\mathrm{r}}=126$
MCQ 114 Marks
$\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right)$ is equal to :
- A1
- ✓0
- C$\frac{33}{65}$
- D$\frac{32}{65}$
Answer
View full question & answer→Correct option: B.
0
(B)
Sol. $\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right)$
$\cos \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{33}{56}\right)$
$\cos \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{5}{12}}{1+\frac{3}{4} \cdot \frac{5}{12}}\right)+\tan ^{-1} \frac{33}{56}\right)$
$\cos \left(\tan ^{-1} \frac{56}{33}+\cot ^{-1} \frac{56}{33}\right)$
$\cos \left(\frac{\pi}{2}\right)=0$
Sol. $\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right)$
$\cos \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{33}{56}\right)$
$\cos \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{5}{12}}{1+\frac{3}{4} \cdot \frac{5}{12}}\right)+\tan ^{-1} \frac{33}{56}\right)$
$\cos \left(\tan ^{-1} \frac{56}{33}+\cot ^{-1} \frac{56}{33}\right)$
$\cos \left(\frac{\pi}{2}\right)=0$
MCQ 124 Marks
Let $$ be a sequence such that $a_{0}=0, a_{1}=\frac{1}{2}$ and $2 a_{n+2}=5 a_{n+1}-3 a_{n}, n=0,1,2,3, \ldots \ldots$ Then $\sum_{k=1}^{100} a_{k}$ is equal to :
- A$3 \mathrm{a}_{99}-100$
- ✓$3 \mathrm{a}_{100}-100$
- C$3 \mathrm{a}_{100}+100$
- D$3 \mathrm{a}_{99}+100$
Answer
View full question & answer→Correct option: B.
$3 \mathrm{a}_{100}-100$
(B)
Sol. $\mathrm{a}_{0}=0, \mathrm{a}_{1}=\frac{1}{2}$
$2 \mathrm{a}_{\mathrm{n}+2}=5 \mathrm{a}_{\mathrm{n}+1}-3 \mathrm{a}_{\mathrm{n}}$
$2 \mathrm{x}^{2}-5 \mathrm{x}+3=0 \Rightarrow \mathrm{x}=1,3 / 2$
$\therefore \mathrm{a}_{\mathrm{n}}=\mathrm{A}(1)^{\mathrm{n}}+\mathrm{B}\left(\frac{3}{2}\right)^{\mathrm{n}}$
$\left.\begin{array}{ll}\mathrm{n}=0 & 0=\mathrm{A}+\mathrm{B} \\ \mathrm{n}=1 & \frac{1}{2}=\mathrm{A}+\frac{3}{2} \mathrm{~B}\end{array}\right] \begin{aligned} & \mathrm{A}=-1 \\ & \mathrm{~B}=1\end{aligned}$
$\Rightarrow \mathrm{a}_{\mathrm{n}}=-1+\left(\frac{3}{2}\right)^{\mathrm{n}}$
$\sum_{k=1}^{100} a_{k}=\sum_{k=1}^{100}(-1)+\left(\frac{3}{2}\right)^{k}$
$=-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1}$
$=-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right)$
$=3 .\left(\mathrm{a}_{100}\right)-100$
Sol. $\mathrm{a}_{0}=0, \mathrm{a}_{1}=\frac{1}{2}$
$2 \mathrm{a}_{\mathrm{n}+2}=5 \mathrm{a}_{\mathrm{n}+1}-3 \mathrm{a}_{\mathrm{n}}$
$2 \mathrm{x}^{2}-5 \mathrm{x}+3=0 \Rightarrow \mathrm{x}=1,3 / 2$
$\therefore \mathrm{a}_{\mathrm{n}}=\mathrm{A}(1)^{\mathrm{n}}+\mathrm{B}\left(\frac{3}{2}\right)^{\mathrm{n}}$
$\left.\begin{array}{ll}\mathrm{n}=0 & 0=\mathrm{A}+\mathrm{B} \\ \mathrm{n}=1 & \frac{1}{2}=\mathrm{A}+\frac{3}{2} \mathrm{~B}\end{array}\right] \begin{aligned} & \mathrm{A}=-1 \\ & \mathrm{~B}=1\end{aligned}$
$\Rightarrow \mathrm{a}_{\mathrm{n}}=-1+\left(\frac{3}{2}\right)^{\mathrm{n}}$
$\sum_{k=1}^{100} a_{k}=\sum_{k=1}^{100}(-1)+\left(\frac{3}{2}\right)^{k}$
$=-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1}$
$=-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right)$
$=3 .\left(\mathrm{a}_{100}\right)-100$
MCQ 134 Marks
Let the equation of the circle, which touches $x$-axis at the point $(a, 0), a>0$ and cuts off an intercept of length $b$ on $y$-axis be $x^{2}+y^{2}-\alpha x+\beta y+\gamma=0$. If the circle lies below $x$-axis, then the ordered pair ( $2 \mathrm{a}, \mathrm{b}^{2}$ ) is equal to :
- A$\left(\alpha, \beta^{2}+4 \gamma\right)$
- B$\left(\gamma, \beta^{2}-4 \alpha\right)$
- C$\left(\gamma, \beta^{2}+4 \alpha\right)$
- ✓$\left(\alpha, \beta^{2}-4 \gamma\right)$
Answer
View full question & answer→Correct option: D.
$\left(\alpha, \beta^{2}-4 \gamma\right)$
(D)
By pytogorus $\mathrm{r}^{2}=\mathrm{a}^{2}+\frac{\mathrm{b}^{2}}{4}=\mathrm{P}^{2}$
$r=\sqrt{\frac{4 \mathrm{a}^{2}+\mathrm{b}^{2}}{4}}$
Equation of circle is $(x-\alpha)^{2}+(y-\beta)^{2}=r^{2}$
$x^{2}+y^{2}-2 a x-2 p y+\alpha^{2}+p^{2}-r^{2}=0$
comparision $x^{2}+y^{2}-\alpha x+\beta y+r=0$
$-\alpha=-2 a, \beta=-2 p, r=a^{2}$
$\Rightarrow 2 \mathrm{a}=\alpha, 4 \mathrm{a}^{2}+\mathrm{b}^{2}=4 \mathrm{p}^{2}$
$
\begin{aligned}
& \alpha^{2}+b^{2}=4 p^{2} \\
& \alpha^{2}+b^{2}=\beta^{2}
\end{aligned}
$So, $\left(2 \mathrm{a}, \mathrm{b}^{2}\right)=\left(\alpha, \beta^{2}-4 \mathrm{r}\right)$
By pytogorus $\mathrm{r}^{2}=\mathrm{a}^{2}+\frac{\mathrm{b}^{2}}{4}=\mathrm{P}^{2}$
$r=\sqrt{\frac{4 \mathrm{a}^{2}+\mathrm{b}^{2}}{4}}$
Equation of circle is $(x-\alpha)^{2}+(y-\beta)^{2}=r^{2}$
$x^{2}+y^{2}-2 a x-2 p y+\alpha^{2}+p^{2}-r^{2}=0$
comparision $x^{2}+y^{2}-\alpha x+\beta y+r=0$
$-\alpha=-2 a, \beta=-2 p, r=a^{2}$
$\Rightarrow 2 \mathrm{a}=\alpha, 4 \mathrm{a}^{2}+\mathrm{b}^{2}=4 \mathrm{p}^{2}$
$
\begin{aligned}
& \alpha^{2}+b^{2}=4 p^{2} \\
& \alpha^{2}+b^{2}=\beta^{2}
\end{aligned}
$So, $\left(2 \mathrm{a}, \mathrm{b}^{2}\right)=\left(\alpha, \beta^{2}-4 \mathrm{r}\right)$
MCQ 144 Marks
The relation $R=\{(x, y): x, y \in z$ and $x+y$ is even $\}$ is :
- Areflexive and transitive but not symmetric
- Breflexive and symmetric but not transitive
- ✓an equivalence relation
- Dsymmetric and transitive but not reflexive
Answer
View full question & answer→Correct option: C.
an equivalence relation
(C)
Sol. $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}, \mathrm{y} \in \mathrm{z}$ and $\mathrm{x}+\mathrm{y}$ is even $\}$
reflexive $x+x=2 x$ even
symmetric of $x+y$ is even, then $(y+x)$ is also even
transitive of $\mathrm{x}+\mathrm{y}$ is even $\& \mathrm{y}+\mathrm{z}$ is even then $x+z$ is also even
So, relation is an equivalence relation.
Sol. $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}, \mathrm{y} \in \mathrm{z}$ and $\mathrm{x}+\mathrm{y}$ is even $\}$
reflexive $x+x=2 x$ even
symmetric of $x+y$ is even, then $(y+x)$ is also even
transitive of $\mathrm{x}+\mathrm{y}$ is even $\& \mathrm{y}+\mathrm{z}$ is even then $x+z$ is also even
So, relation is an equivalence relation.
MCQ 154 Marks
Let $\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be a point in xy-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and $(0,0,1)$.
Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements
(S1): $\triangle \mathrm{ABC}$ is an isosceles right angled triangle and
(S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$.
Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements
(S1): $\triangle \mathrm{ABC}$ is an isosceles right angled triangle and
(S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$.
- Aboth are true
- ✓only (S1) is true
- Conly (S2) is true
- Dboth are false
Answer
View full question & answer→Correct option: B.
only (S1) is true
(B)
Sol. $\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ Let $\mathrm{P}(0,3,2), \mathrm{Q}(2,0,3), \mathrm{R}(0,0,1)$
$\mathrm{AP}=\mathrm{AQ}=\mathrm{AR}$
$x^{2}+(y-3)^{2}+(z-2)^{2}=(x-2)^{2}+y^{2}+(z-3)^{2}=x^{2}+$
$y^{2}+(z-1)^{2}$
In $x y$ plane $z=0$
So, $x^{2}-4 x+4+y^{2}+9=x^{2}+y^{2}+1$
$\Rightarrow \mathrm{y}=2$
$\mathrm{x}=3$
$9+y^{2}-6 y+9+4=x^{2}+y^{2}+1$
So, $\mathrm{A}(3,2,0)$ also $\mathrm{B}(1,4,-1) \& \mathrm{C}(2,0,-2)$
Now $A B=\sqrt{4+4+1}=3$
$
\begin{aligned}
& \mathrm{AC}=\sqrt{1+4+4}=3 \\
& \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18}
\end{aligned}
$
$\mathrm{AB}=\mathrm{AC}$
isosceles $\Delta \& \mathrm{AB}^{2}+\mathrm{AC}^{2}=\mathrm{BC}^{2}$
right angle $\Delta$
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times$ base.height
$\frac{1}{2} \times 3 \times 3=\frac{9}{2}$
So only $\mathrm{S}_{1}$ is true
Sol. $\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ Let $\mathrm{P}(0,3,2), \mathrm{Q}(2,0,3), \mathrm{R}(0,0,1)$
$\mathrm{AP}=\mathrm{AQ}=\mathrm{AR}$
$x^{2}+(y-3)^{2}+(z-2)^{2}=(x-2)^{2}+y^{2}+(z-3)^{2}=x^{2}+$
$y^{2}+(z-1)^{2}$
In $x y$ plane $z=0$
So, $x^{2}-4 x+4+y^{2}+9=x^{2}+y^{2}+1$
$\Rightarrow \mathrm{y}=2$
$\mathrm{x}=3$
$9+y^{2}-6 y+9+4=x^{2}+y^{2}+1$
So, $\mathrm{A}(3,2,0)$ also $\mathrm{B}(1,4,-1) \& \mathrm{C}(2,0,-2)$
Now $A B=\sqrt{4+4+1}=3$
$
\begin{aligned}
& \mathrm{AC}=\sqrt{1+4+4}=3 \\
& \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18}
\end{aligned}
$
$\mathrm{AB}=\mathrm{AC}$
isosceles $\Delta \& \mathrm{AB}^{2}+\mathrm{AC}^{2}=\mathrm{BC}^{2}$
right angle $\Delta$
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times$ base.height
$\frac{1}{2} \times 3 \times 3=\frac{9}{2}$
So only $\mathrm{S}_{1}$ is true
MCQ 164 Marks
Let $f: R \rightarrow R$ be a function defined by $f(x)=(2+3 a) x^{2}+\left(\frac{a+2}{a-1}\right) x+b, a \neq 1$. If $f(x+y)=f(x)+f(y)+1-\frac{2}{7} x y$, then the value of $28 \sum_{i=1}^{5}|f(i)|$ is:
- A715
- B735
- C545
- ✓675
Answer
View full question & answer→Correct option: D.
675
(D)
Sol. $f(x)=(3 a+2) x^{2}+\left(\frac{a+2}{a-1}\right) x+b$
$f\left(x+\frac{1}{2}\right)=f(x)+f(y)+1-\frac{2}{7} x y$
In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$ So, $f(0)=0+0+b=-1 \Rightarrow b=-1$ In (1) Put $y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^{2}$
$-1=2(3 a+2) x^{2}+2 b+1+\frac{2}{7} x^{2}$
$-1=\left(2(3 a+2)+\frac{2}{7}\right) x^{2}+1-2$
$\Rightarrow 6 \mathrm{a}+4+\frac{2}{7}=0$
$a=-\frac{5}{7}$
So $\mathrm{f}(\mathrm{x})=-\frac{1}{7} \mathrm{x}^{2}-\frac{3}{4} \mathrm{x}-1$
$\Rightarrow|\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4 \mathrm{x}^{2}+21 \mathrm{x}+28\right|$
Now, $28 \sum_{i=1}^{5}|f(i)|=28(|f(1)|+|f(2)|+\ldots+|f(5)|)$
28. $\frac{1}{28} \cdot 675=675$
Sol. $f(x)=(3 a+2) x^{2}+\left(\frac{a+2}{a-1}\right) x+b$
$f\left(x+\frac{1}{2}\right)=f(x)+f(y)+1-\frac{2}{7} x y$
In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$ So, $f(0)=0+0+b=-1 \Rightarrow b=-1$ In (1) Put $y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^{2}$
$-1=2(3 a+2) x^{2}+2 b+1+\frac{2}{7} x^{2}$
$-1=\left(2(3 a+2)+\frac{2}{7}\right) x^{2}+1-2$
$\Rightarrow 6 \mathrm{a}+4+\frac{2}{7}=0$
$a=-\frac{5}{7}$
So $\mathrm{f}(\mathrm{x})=-\frac{1}{7} \mathrm{x}^{2}-\frac{3}{4} \mathrm{x}-1$
$\Rightarrow|\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4 \mathrm{x}^{2}+21 \mathrm{x}+28\right|$
Now, $28 \sum_{i=1}^{5}|f(i)|=28(|f(1)|+|f(2)|+\ldots+|f(5)|)$
28. $\frac{1}{28} \cdot 675=675$
MCQ 174 Marks
If $f(x)=\frac{2^{x}}{2^{x}+\sqrt{2}}, x \in R$, then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to :
- A41
- ✓$\frac{81}{2}$
- C82
- D$81 \sqrt{2}$
Answer
View full question & answer→Correct option: B.
$\frac{81}{2}$
(B)
Sol. $\mathrm{f}(\mathrm{x})=\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}+\sqrt{2}}$
$f(x)+f(1-x)=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}}$
$=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^{x}}=\frac{2^{x}+\sqrt{2}}{2^{x}+\sqrt{2}}=1$
Now, $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)=f\left(\frac{1}{82}\right)+f\left(\frac{2}{82}\right)+\ldots \ldots .+f\left(\frac{81}{82}\right)$
$=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)$
$
=\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots . .40 \text { cases }+\mathrm{f}\left(\frac{41}{82}\right)
$
$=(1+1+\ldots 40$ times $)+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}}$
$40+\frac{1}{2}=\frac{81}{2}$
Sol. $\mathrm{f}(\mathrm{x})=\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}+\sqrt{2}}$
$f(x)+f(1-x)=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}}$
$=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^{x}}=\frac{2^{x}+\sqrt{2}}{2^{x}+\sqrt{2}}=1$
Now, $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)=f\left(\frac{1}{82}\right)+f\left(\frac{2}{82}\right)+\ldots \ldots .+f\left(\frac{81}{82}\right)$
$=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)$
$
=\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots . .40 \text { cases }+\mathrm{f}\left(\frac{41}{82}\right)
$
$=(1+1+\ldots 40$ times $)+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}}$
$40+\frac{1}{2}=\frac{81}{2}$
MCQ 184 Marks
Two number $k_{1}$ and $k_{2}$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_{1}}+i^{k_{2}},(i=\sqrt{-1})$ is non-zero, equals
- A$\frac{1}{2}$
- B$\frac{1}{4}$
- ✓$\frac{3}{4}$
- D$\frac{2}{3}$
Answer
View full question & answer→Correct option: C.
$\frac{3}{4}$
(C)
Sol. $\quad i^{k_{1}}+i^{k_{2}} \neq 0 \quad i^{k_{1}} \rightarrow 4$ option for $i,-1,-i, 1$
Total cases $\Rightarrow 4 \times 4=16$
Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_{1}}+\mathrm{i}^{\mathrm{k}_{2}}=0$
$\left\{\begin{array}{c}1,-1 \\ -1,1 \\ i,-i \\ -i, i\end{array}\right\}$
4 Cases $\Rightarrow$ Probability $=\frac{16-4}{16}=\frac{3}{4}$
Sol. $\quad i^{k_{1}}+i^{k_{2}} \neq 0 \quad i^{k_{1}} \rightarrow 4$ option for $i,-1,-i, 1$
Total cases $\Rightarrow 4 \times 4=16$
Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_{1}}+\mathrm{i}^{\mathrm{k}_{2}}=0$
$\left\{\begin{array}{c}1,-1 \\ -1,1 \\ i,-i \\ -i, i\end{array}\right\}$
4 Cases $\Rightarrow$ Probability $=\frac{16-4}{16}=\frac{3}{4}$
MCQ 194 Marks
Let $A B C D$ be a trapezium whose vertices lie on the parabola $y^{2}=4 x$. Let the sides $A D$ and $B C$ of the trapezium be parallel to $y$-axis. If the diagonal AC is of length $\frac{25}{4}$ and it passes through the point $(1,0)$, then the area of ABCD is :
- ✓$\frac{75}{4}$
- B$\frac{25}{2}$
- C$\frac{125}{8}$
- D$\frac{75}{8}$
Answer
View full question & answer→Correct option: A.
$\frac{75}{4}$
(A)
$A\left(\mathrm{at}_{1}^{2}, 2 \mathrm{at}_{1}\right) \& C\left(\frac{\mathrm{a}}{\mathrm{t}_{1}^{2}},-\frac{2 \mathrm{a}}{\mathrm{t}_{1}}\right)$
Length $A C=a\left(t_{1}+\frac{1}{t_{1}}\right)^{2}=\frac{25}{4}, t_{1}+\frac{1}{t_{1}}= \pm \frac{5}{2}$
$\Rightarrow \mathrm{t}_{1}=2$ or $\frac{1}{2}, \mathrm{~A}\left(\frac{1}{2}, 1\right), \mathrm{D}\left(\frac{1}{4},-1\right), \mathrm{B}(4,4), \mathrm{C}(4,-4)$
So, area of trapezium $=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4}$
$A\left(\mathrm{at}_{1}^{2}, 2 \mathrm{at}_{1}\right) \& C\left(\frac{\mathrm{a}}{\mathrm{t}_{1}^{2}},-\frac{2 \mathrm{a}}{\mathrm{t}_{1}}\right)$
Length $A C=a\left(t_{1}+\frac{1}{t_{1}}\right)^{2}=\frac{25}{4}, t_{1}+\frac{1}{t_{1}}= \pm \frac{5}{2}$
$\Rightarrow \mathrm{t}_{1}=2$ or $\frac{1}{2}, \mathrm{~A}\left(\frac{1}{2}, 1\right), \mathrm{D}\left(\frac{1}{4},-1\right), \mathrm{B}(4,4), \mathrm{C}(4,-4)$
So, area of trapezium $=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4}$
MCQ 204 Marks
The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0,1 , $2,3,4,5,6,7$, such that the sum of their first and last digits should not be more than 8 , is
- A4608
- B5720
- C5719
- D4607
Answer
View full question & answer→