KCET 2022 Mathematics Questions with Answers Key Solutions
Solution:
AKMS
A→3!
K→3!
M→3!
SAKM→119
Solution:
a3a1=25
ar2a=25
r2=52
a4a5=ar8ar4=r4=54
Solution:
Given 2x−3y+17=0
is a line perpendicular to line passing through point P(7,17)&Q(15,β) then equation of line passing throught P&Q is
y−17=(β−1715−7)(x−7)
8y−136=βx−7β−17x+119
(7−β)x+8y−255=0………..(2)
Line (1) \& (2) are perpendicular to each other then
m1m2=1
Product of slopes of line (1) \& (2) is −1
Now,
m1=23 and m2=β−78
m1m2=−1⇒23(β−748)=−1
β−7=−12
β=−12+7=−5
Hence [β=−5].
Solution:
\alpha=\displaystyle\lim _{x \rightarrow 2^{-}} f(x)=\displaystyle\lim _{x \rightarrow 2} x^{2}-1=3
\beta=\displaystyle\lim _{x \rightarrow 2^{+}} f(x)=\displaystyle\lim _{x \rightarrow 2} 2 x+3=7
x^{2}-(\alpha+\beta) x+\alpha \beta=0
x^{2}-10 x+21=0
Q6. If 3 x+ i (4 x-y)=6- i where x and y are real numbers, then the values of x and y are respectively,
Solution:
3 x=6
x=2
4 x-y=-1
8-y=-1
9= y
Solution:
\sigma^{2}=5, \overline{ x }=\frac{ k }{4}
\frac{1}{4}\left(1+0+1+ k ^{2}\right)-\frac{ k ^{2}}{16}=5
\frac{ k ^{2}+2}{4} \frac{- k ^{2}}{16}=5
\frac{4 k ^{2}+8- k ^{2}}{16}=5
\Rightarrow 3 k ^{2}+8=80
3 k ^{2}=72
k ^{2}=24
k =\pm \sqrt{24}=2 \sqrt{6}
Solution:
n ( A ) \neq n ( B )
Number of bijections is zero
Solution:
f(-1)=3(-1)=-3
f(2)=2^{2}=4
f(4)=2(4)=8
f(-1)+f(2)+f(4)
=-3+4+8=9
Solution:
2 b =27-3 a
b =\frac{27-3 a }{2}
R =\{(1,2),(3,9),(5,6),(7,3)\}
Solution:
\displaystyle\lim _{y \rightarrow 0} \frac{3+y^{3}-3}{y\left(\sqrt{3+y^{3}}+\sqrt{3}\right)}=\frac{1}{2 \sqrt{3}}
Solution:
\left( A ^{2}\right)^{-1}=\left( A ^{-1}\right)^{2}
Solution:
A=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}
A^{-1}=\frac{1}{-1}\begin{bmatrix}-2 & 1 \\ -3 & 2\end{bmatrix}=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}=A
A^{2}=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}
=\begin{bmatrix}4-3 & -2+2 \\ 6-6 & -3+4\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=I
\therefore A^{3}=A
Solution:
A ^{ T }=- A or A ^{ n } is slow symmetric if n is odd
P = A ^{2021}
P ^{ T }=\left[ A ^{2021}\right]^{ T }=\left[ A ^{ T }\right]^{2021}=(- A )^{2021}=- P
Solution:
A =\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}
{[ aI + bA ]^{1}=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}+\begin{bmatrix}0 & b \\ 0 & 0\end{bmatrix}=\begin{bmatrix} a & b \\ 0 & a \end{bmatrix} }
{[ aI + IA ]^{2}=\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}=\begin{bmatrix} a ^{2} & 2 ab \\ 0 & a ^{2}\end{bmatrix} }
{[ aI + bA ]^{3}=\begin{bmatrix} a ^{2} & 2 ab \\ 0 & a ^{2}\end{bmatrix}\begin{bmatrix}a & b \\ 0 & a \end{bmatrix}=\begin{bmatrix} a ^{3} & 3^{2} b \\ 0 & a ^{3}\end{bmatrix} }
\therefore[ aI + bA ]^{ n }=\begin{bmatrix} a ^{ n } & na a ^{ n -1} b \\ 0 & a ^{ n }\end{bmatrix}
= a ^{ n } I + n \cdot a ^{ n -1} bA
Solution:
A_{3 \times 3} matrix |5 . \operatorname{AdjA}|=5
\Rightarrow 5^{3}| A |^{2}=5
\Rightarrow\left| A ^{2}\right|=\frac{1}{5^{2}}
| A |=\pm \frac{1}{5}
Solution:
\Delta=1\left(2 a^{2}+4\right)+2(4 a-0)+5(8)=86
2 a^{2}+8 a+44-86=0
2 a^{2}+8 a-42=0
a^{2}+4 a-21=0
Sum of numbers =-4\left(\therefore-\frac{ b }{ a }=\alpha+\beta\right)
Solution:
\Delta=\frac{1}{2}\begin{vmatrix}-2-3 & -2-1 \\ 6+6 & 6-5\end{vmatrix}=\frac{1}{5}\begin{vmatrix}-5 & -3 \\ 12 & 1\end{vmatrix}
=\frac{1}{2}|-5+36|=\frac{31}{2}=15.5
Solution:
\cos ^{-1}[x]
-1 \leq[x] \leq 1
\Rightarrow[x]=\{-1,0,1\}
x \in[-1,2)
Solution:
y=\left(1+x^{2}\right) \tan ^{-1} x-x
\frac{d y}{d x}=\frac{\left(1+x^{2}\right)}{1+x^{2}}+\tan ^{-1} x \cdot(2 x)-1
=2 x \tan ^{-1} x
Solution:
x = e ^{\theta} \sin \theta=1
y = e ^{\theta} \cos \theta=1, \frac{ x }{ y }=\tan \theta=1
\Rightarrow \theta=\frac{\pi}{4}
\frac{ dy }{ dx }=\left|\frac{ dy / d \theta}{ dx / d \theta}\right|=\frac{- e ^{\theta} \sin \theta+\cos \theta \cdot e ^{\theta}}{ e ^{\theta} \cos \theta+\sin \theta e ^{\theta}}=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}
=\tan \left(\frac{\pi}{4}-\theta\right)=0
Solution:
Y = e^{\sqrt{x\sqrt{x\sqrt{x}}} ........} = e^{x^{\frac{1}{2}} x^{\frac{1}{4}},x^{\frac{1}{8}} ..... = e^{x^{\frac{1}{2}} x^{\frac{1}{4}},x^{\frac{1}{8}} .....}}
e ^{ x ^{\frac{1}{2}\left[1+\frac{1}{2}+\frac{1}{4}+\ldots\right]}}= e ^{ x ^{\frac{1}{2}}}= e ^{ x ^{1}}= e ^{ x }
\frac{d y}{d x}=e^{x}
\frac{ d ^{2} y }{ dx x ^{2}}= e ^{ x } x \log _{ e }^{3}= e ^{\log _{ e }^{3}}=
Solution:
f (1)=1, f ^{\prime}(1)=3
\frac{ d }{ dx }\left[ f ( f ( f ( x )))+( f ( x ))^{2}\right]
{\left[ f ^{\prime}( f ( f ( x ))) \cdot f ^{\prime} f ( x ) \cdot f ^{\prime}( x )+2 f ( x ) \cdot f ^{\prime}( x )\right] }
= f ^{\prime}( f ( f (1))) f ^{\prime}( f (1)) \cdot f ^{\prime}(1)+2 f (1) \cdot f ^{\prime}(1)
f ^{\prime}( f (1)) f ^{\prime}(1) \cdot 3+2 \cdot(1) 3
= f ^{\prime}(1) \cdot 3 \cdot 3+6
=27+6=33
Solution:
y=x^{\sin x}+(\sin x)^{x}
\frac{d y}{d x}=\left[x^{\sin x}\right]\left[\frac{\sin x}{x}+\cos x \cdot \log x\right]+
(\sin x)^{x}[x \cos x+\log \sin x]
x=\frac{\pi}{2}
=\frac{\pi}{2}\left[\frac{2}{\pi}\right]+1[0+0]=1
Solution:
A_{n}=\begin{bmatrix}1-n & n \\ n & 1-n\end{bmatrix}
\left|A_{n}\right|=(1-n)^{2}-n^{2}
=1+ n ^{2}-2 n - n ^{2}
Solution:
f^{\prime}(x)=\frac{x^{2}}{(x+1)(2+x)^{2}}>0
x+1>0 \Rightarrow x>-1
Solution:
\frac{d y}{d x}=-\sqrt{\frac{y}{x}}=-1
y=x
\sqrt{x}+\sqrt{x}=6
x=9, y=9
Solution:
f^{\prime}(x)=\left(12 \sin ^{2} x-12 \sin x+12\right) \cos x
f^{\prime}(x)=12\left(\sin ^{2} x-\sin x+1\right) \cos x
\sin ^{2} x-\sin x+1>0
x \in\left(\frac{\pi}{2}, \pi\right) \cos x < 0
Solution:
\int\limits_{0}^{8}[x] d x=1+2+3+\ldots .+7
=\frac{7(7+1)}{2}=28
Solution:
Put \sin \theta= t
\int\limits_{0}^{1} t^{1 / 2}\left(1-t^{2}\right) d t=\frac{8}{21}
Solution:
x=0 \Rightarrow y=1
\frac{d y}{d x}=\frac{-y}{e^{y}+x}
\left(\frac{d y}{d x}\right)_{(0,1)}=-\frac{1}{e}
\left(\frac{d^{2} y}{d x^{2}}\right)_{(0,1)}=\frac{1}{e^{2}}
Solution:
2 \int \frac{\cos ^{2} x-\cos ^{2} \alpha}{\cos x-\cos \alpha} d x=2 \int(\cos x+\cos \alpha) d x
=2[\sin x+x \cos \alpha]
Solution:
\int\limits_{0}^{1} e^{x}\left[\frac{1}{(x+2)^{2}}-\frac{2}{(x+2)^{3}}\right] d x
=\left[\frac{e^{x}}{(x+2)^{2}}\right]_{0}^{1}=\frac{e}{9}-\frac{1}{4}
Solution:
\frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1}
A=\frac{1}{5}, B=\frac{-1}{5}, C=\frac{2}{5}
Solution:
A=\int\limits_{0}^{\pi / 3} \tan x d x=\log |\sec x|_{0}^{\pi / 3}=\log 2
Solution:
I=\left[\frac{x^{3}}{3}\right]_{2}^{3}=\frac{1}{3}(27-8)=\frac{19}{3}
Solution:
\sin x=t
\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos x \sin x}{1+\sin x} d x
\Rightarrow \int\limits_{0}^{1} \frac{t}{1+t} d t=1-\log 2
Solution:
y \cdot x=\frac{x^{4}}{4}+C
2 y(2)-y(1)=\frac{15}{4}
Solution:
x+y=z \Rightarrow \frac{d z}{d x}=1+z^{2}
\int \frac{1}{1+z^{2}} d z=\int 1 d x
\operatorname{Tan}^{-1}(x+y)=x+c
Solution:
I . F=\log x,
y \log x=2 x(\log x-1)+c
If x=e then y=c then y(e)=2 e
Solution:
\left(1+ y _{1}^{2}\right)=\left( y _{2}\right)^{3}
2+3=5
Q45. The co-ordinates of foot of the perpendicular drawn from the origin to the plane 2 x-3 y+4 z=29 are
Solution:
verification (2,-3,4)
Solution:
\cos \theta=\frac{3 \times 1+5 \times 4+4 \times 2}{\sqrt{3^{2}+5^{2}+4^{2}} \times \sqrt{1^{2}+4^{2}+2^{2}}}=\frac{31}{5 \sqrt{42}}
\theta=\cos ^{-1} \frac{31}{5 \sqrt{42}}
Solution:
At (0,2),(3,0), z=12
Hence minimum at 2 points.
Solution:
Distance =\frac{|2-2-4-4|}{\sqrt{1+4+16}}=\frac{(8)}{\sqrt{21}}
Solution:
X
0
1
2
3
P(x)
\frac{1}{8}
\frac{3}{8}
\frac{3}{8}
\frac{1}{8}
Mean =\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=\frac{3}{2}=1.5
X | 0 | 1 | 2 | 3 |
P(x) | \frac{1}{8} | \frac{3}{8} | \frac{3}{8} | \frac{1}{8} |
Solution:
P ( A )=\frac{1}{2}, P ( B )=\frac{1}{3}, P ( A \cap B )=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}
P ( A \cap B )= P (\overline{ A \cup B })
P ( A \cap B )=1- P ( A \cup B )
P ( A \cap B )=1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\right]
P ( A \cap B )=1-\left[\frac{6+4-1}{12}\right]
P ( A \cap B )=1-\frac{9}{12}
P ( A \cap B )=\frac{1}{4}
Solution:
P \left( E _{1}\right)= probability of there is lockdown =0.7
P\left(E_{2}\right)= probability of there is lockdown =0.3
A is the event controlled in one month
P \left( A / E _{1}\right)=0.8, P \left( A / E _{2}\right)=0.3
P(A)=0.7(0.8)+(0.3)(0.3)
=0.56+0.09=0.65
Solution:
P ( A )=\frac{1}{4}, P ( B )=\frac{3}{4}, P ( A \cup B )=\frac{13}{20}
\frac{1}{4}+ x -\frac{1}{4} \cdot x =\frac{13}{20}
\frac{3}{4} x-\frac{13}{20}-\frac{5}{20}=\frac{8}{20}
x=\frac{8}{20} \times \frac{4}{3}=\frac{8}{15}
Solution:
n ( A )= p , n ( B )= q
n ( A \times B )=7
pq =7
p ^{2}+ q ^{2}=7^{2}+1^{2} or 1^{2}+7^{2}
p ^{2}+ q ^{2}=50
Solution:
1 - x > 0 , 1 - x \neq 1
x -1<0 \,\,\,\,\,\,x \neq 0 \,\,\,\,\,\,x +2 \geq 0
x <1 \,\,\,\,\,\,x \geq-2
\therefore x \in[-2,0) \cup(0,1)
Solution:
\frac{\pi}{32}=\frac{180^{\circ}}{32}=5^{0} 37^{\prime} 30^{\prime \prime}
Solution:
\sin \frac{5 \pi}{12} \cdot \sin \frac{\pi}{12}
=\frac{1}{2} \sin \frac{\pi}{6}
=\frac{1}{2}+\frac{1}{2}=\frac{1}{4}
Solution:
1+\cos \theta=2 \cos ^{2}\left(\frac{\theta}{2}\right)
\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}=2 \cos \theta
Solution: