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JEE Main 2022 June 26 β Shift 1 Maths Question Paper with Solutions
SECTION – A
Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.
Choose the correct answer :
1. Let
\(\begin{array}{l}f\left ( x \right )=\frac{x-1}{x+1},x\epsilon R-\left\{0,-1,1 \right\}\end{array} \)
If ƒn+1(x) = ƒ(ƒn(x)) for all n∈N, then ƒ6(6) + ƒ7(7) is equal to :
(A)
\(\begin{array}{l}\frac{7}{6}\end{array} \)
(B)
\(\begin{array}{l}-\frac{3}{2}\end{array} \)
(C)
\(\begin{array}{l}\frac{7}{12}\end{array} \)
(D)
\(\begin{array}{l}-\frac{11}{12}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}f\left ( x \right )=\frac{x-1}{x+1}\Rightarrow f\left ( f\left ( x \right ) \right )=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}\end{array} \)
⇒
\(\begin{array}{l}f^3\left ( x \right )=-\frac{x+1}{x-1}\Rightarrow f^4\left (x \right )=-\frac{\frac{x-1}{x+1}+1}{\frac{x-1}{x+1}-1}=x\end{array} \)
So, ƒ6(6) + ƒ7(7) = ƒ2(6) + ƒ3(7)
\(\begin{array}{l}=-\frac{1}{6}-\frac{7+1}{7-1}=-\frac{9}{6}=-\frac{3}{2}\end{array} \)
2. Let
\(\begin{array}{l}A=\left\{\textbf{z}~\epsilon~\textbf{C}:\left|\frac{z+1}{z-1} \right| <1\right\}\end{array} \)
and
\(\begin{array}{l}B=\left\{\textbf{z}~\epsilon~\textbf{C}:arg\left ( \frac{z-1}{z+1} \right ) = \frac{2\pi}{3}\right\}\end{array} \)
Then A∩Bis :
(A) A portion of a circle centred at
\(\begin{array}{l}\left ( 0,~-\frac{1}{\sqrt{3}} \right )\end{array} \)
that lies in the second and third quadrants only
(B) A portion of a circle centred at
\(\begin{array}{l}\left ( 0,~-\frac{1}{\sqrt{3}} \right )\end{array} \)
that lies in the second quadrant only
(C) An empty set
(D) A portion of a circle of radius
\(\begin{array}{l}\frac{2}{\sqrt{3}}\end{array} \)
that lies in the third quadrant only
Answer (B)
Sol.
\(\begin{array}{l}\left| \frac{z+1}{z-1}\right|<1\Rightarrow\left| z+1\right|<\left|z-1 \right|\Rightarrow Re\left ( z \right )<0\end{array} \)
and
\(\begin{array}{l}arg\left ( \frac{z-1}{z+1} \right )=\frac{2\pi}{3}\end{array} \)
is a part of circle as shown.
3. Let A be a 3 × 3 invertible matrix. If |adj (24A)| = |adj (3 adj (2A))|, then |A|2 is equal to :
(A) 66
(B) 212
(C) 26
(D) 1
Answer (C)
Sol.
\(\begin{array}{l}\left|adj\left ( 24A \right ) \right|=\left|adj\left ( 3adj\left ( 2A \right ) \right ) \right|\end{array} \)
⇒
\(\begin{array}{l}\left|24A \right|^2=\left|3~adj\left ( 2A \right ) \right|^2\end{array} \)
⇒
\(\begin{array}{l}\left ( 24^3 \right )^2\cdot\left|A \right|^2=\left ( 3^3 \right )^2\left|adj\left ( 2A \right ) \right|^2\end{array} \)
⇒
\(\begin{array}{l}24^6\cdot\left|A \right|^2=3^6\left|2A \right|^4\end{array} \)
⇒
\(\begin{array}{l}24^6\left|A \right|^2=3^6\cdot\left ( 2^3 \right )^4\left|A \right|^4\end{array} \)
⇒
\(\begin{array}{l}\left|A \right|^2=\frac{24^6}{3^6\cdot2^{12}}=\frac{2^{18}.\cdot3^6}{3^6\cdot2^{12}}=2^6\end{array} \)
4. The ordered pair (a, b), for which the system of linear equations
3x – 2y + z = b
5x – 8y + 9z = 3
2x + y + az = –1
has no solution, is :
(A)
\(\begin{array}{l}\left ( 3,\frac{1}{3} \right )\end{array} \)
(B)
\(\begin{array}{l}\left (- 3,\frac{1}{3} \right )\end{array} \)
(C)
\(\begin{array}{l}\left (- 3,-\frac{1}{3} \right )\end{array} \)
(D)
\(\begin{array}{l}\left ( 3,-\frac{1}{3} \right )\end{array} \)
Answer (C)
Sol.
\(\begin{array}{l}\begin{vmatrix}3 & -2 & 1 \\5 & -8 & 9 \\2 & 1 & a \\\end{vmatrix}=0\Rightarrow-14a-42=0\Rightarrow a=-3\end{array} \)
Now 3(equation (1)) – (equation (2)) – 2(equation (3)) is
3(3x – 2y + z – b) – (5x – 8y + 9z – 3) – 2(2x + y + az + 1) = 0
⇒ –3b + 3 – 2 = 0
⇒
\(\begin{array}{l}b=\frac{1}{3}\end{array} \)
So for no solution a = –3 and
\(\begin{array}{l}b\neq\frac{1}{3}\end{array} \)
5. The remainder when (2021)2023 is divided by 7 is :
(A) 1
(B) 2
(C) 5
(D) 6
Answer (C)
Sol. 2021 ≡ –2 (mod 7)
⇒ (2021)2023 ≡–(2)2023 (mod 7)
≡ –2(8)674 (mod 7)
≡ –2(1)674 (mod 7)
≡ –2(mod 7)
≡ 5(mod 7)
So when (2021)2023 is divided by 7, remainder is 5.
6.
\(\begin{array}{l}\displaystyle \lim_{x\rightarrow\frac{1}{\sqrt{2}}}\frac{\sin\left ( \cos^{-1}x \right )-x}{1-\tan\left ( \cos^{-1}x \right )}\end{array} \)
is equal to :
(A)
\(\begin{array}{l}\sqrt{2}\end{array} \)
(B)
\(\begin{array}{l}-\sqrt{2}\end{array} \)
(C)
\(\begin{array}{l}\frac{1}{\sqrt{2}}\end{array} \)
(D)
\(\begin{array}{l}-\frac{1}{\sqrt{2}}\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l}\lim\limits_{x\rightarrow\frac{1}{\sqrt{2}}}\frac{\sin\left ( \cos^{-1}x \right )-x}{1-\tan\left ( \cos^{-1}x \right )}\end{array} \)
\(\begin{array}{l}\textup{let}~\cos^{-1}x=\frac{\pi}{4}+\theta\end{array} \)
\(\begin{array}{l}=\lim\limits_{\theta\rightarrow0}\frac{\sin\left ( \frac{\pi}{4}+\theta \right )-\cos\left ( \frac{\pi}{4}+\theta \right )}{1-\tan\left ( \frac{\pi}{4}+\theta \right )}\end{array} \)
\(\begin{array}{l}=\lim\limits_{\theta\rightarrow0}\frac{\sqrt{2}\sin\left ( \frac{\pi}{4}+\theta-\frac{\pi}{4} \right )}{1-\frac{1+\tan\theta}{1-\tan\theta}}\end{array} \)
\(\begin{array}{l}=\lim\limits_{\theta\rightarrow0}\frac{\sqrt{2}\sin\theta}{-2\tan\theta}\left ( 1-\tan\theta \right )=-\frac{1}{\sqrt{2}}\end{array} \)
7. g :R→R be two real valued functions defined as
\(\begin{array}{l}f\left ( x \right )=\left\{\begin{matrix}-\left|x+3 \right|, & x<0 \\e^x, & x\geq 0 \\\end{matrix}\right.\end{array} \)
and
\(\begin{array}{l}g\left ( x \right )=\left\{\begin{matrix}x^2+k_1x, & x<0 \\4x+k_2, & x\geq 0 \\\end{matrix}\right.\end{array} \)
where k1 and k2 are real constants. If (goƒ) is differentiable at x = 0, then (goƒ) (–4) + (goƒ) (4) is equal to :
(A) 4(e4 + 1)
(B) 2(2e4 + 1)
(C) 4e4
(D) 2(2e4 – 1)
Answer (D)
Sol. β΅ goƒ is differentiable at x = 0
So R.H.D = L.H.D
\(\begin{array}{l}\frac{d}{dx}\left ( 4e^x+k_2 \right )=\frac{d}{dx}\left ( \left ( -\left|x+3 \right| \right )^2-k_1\left|x+3 \right| \right )\end{array} \)
⇒ 4 = 6 –k1⇒k1 = 2
Also g(ƒ(0+)) = g(ƒ(0–))
⇒ 4 + k2 = 9 – 3k1⇒k2 = –1
Now g(ƒ(–4)) + g(ƒ(4))
= g(–1) + g(e4) = (1 – k1) + (4e4 + k2)
= 4e4 – 2
= 2(2e4 – 1)
8. The sum of the absolute minimum and the absolute maximum values of the function ƒ(x) = |3x – x2 + 2| – x in the interval [–1, 2] is :
(A)
\(\begin{array}{l}\frac{\sqrt{17}+3}{2}\end{array} \)
(B)
\(\begin{array}{l}\frac{\sqrt{17}+5}{2}\end{array} \)
(C) 5
(D)
\(\begin{array}{l}\frac{9-\sqrt{17}}{2}\end{array} \)
Answer (A)
Sol. ƒ(x) = |x2– 3x – 2| – x
\(\begin{array}{l}\forall x\epsilon\left [ -1,2 \right ] \end{array} \)
⇒
\(\begin{array}{l}f\left ( x\right)=\left\{\begin{matrix}x^2-4x-2~~\textup{if}-1\leq x<\frac{3-\sqrt{17}}{2}\\\\ -x^2+2x+2~~\textup{if}~\frac{3-\sqrt{17}}{2}\leq x\leq 2\end{matrix}\right.\end{array} \)
ƒ(x)max = 3
\(\begin{array}{l}f\left ( x \right )_{\textup{min}}=f\left ( \frac{3-\sqrt{17}}{2} \right )\end{array} \)
\(\begin{array}{l}=\frac{\sqrt{17}-3}{2}\end{array} \)
9. Let S be the set of all the natural numbers, for which the line
\(\begin{array}{l}\frac{x}{a}+\frac{y}{b}=2\end{array} \)
is a tangent to the curve \(\begin{array}{l}\left ( \frac{x}{a} \right )^n+\left ( \frac{y}{b} \right )^n=2\end{array} \)
at the point (a, b), ab ≠ 0. Then :
(A) S = ΙΈ
(B) n(S) = 1
(C) S = {2k : k ∈ N }
(D) S = N
Answer (D)
Sol.
\(\begin{array}{l}\left ( \frac{x}{a} \right )^n+\left ( \frac{y}{b} \right )^n=2\end{array} \)
\(\begin{array}{l}\Rightarrow\frac{n}{a}\left ( \frac{x}{a} \right )^{n-1}+\frac{n}{b}\left ( \frac{y}{b} \right )^{n-1}\frac{dy}{dx}=0\end{array} \)
\(\begin{array}{l}\Rightarrow\frac{dy}{dx}=-\frac{b}{a}\left ( \frac{bx}{ay} \right )^{n-1}\end{array} \)
\(\begin{array}{l}\frac{dy}{dx_{\left (a,b \right )}}=-\frac{b}{a}\end{array} \)
So line always touches the given curve.
10. The area bounded by the curve
\(\begin{array}{l}y=\left|x^2-9 \right|\end{array} \)
and the line y = 3 is
(A)
\(\begin{array}{l}4\left ( 2\sqrt{3}+\sqrt{6}-4 \right )\end{array} \)
(B)
\(\begin{array}{l}4\left ( 4\sqrt{3}+\sqrt{6}-4 \right )\end{array} \)
(C)
\(\begin{array}{l}8\left ( 4\sqrt{3}+3\sqrt{6}-9 \right )\end{array} \)
(D)
\(\begin{array}{l}8\left ( 4\sqrt{3}+\sqrt{6}-9 \right )\end{array} \)
Answer (*)
Sol. y = 3 and y = |x2 – 9|
Intersect in first quadrant at
\(\begin{array}{l}x=\sqrt{6}~\textup{and}~x=\sqrt{12}\end{array} \)
Required area
\(\begin{array}{l}=2\left [ \frac{2}{3}\left ( 6\times\sqrt{6} \right )+\displaystyle\int\limits_{\sqrt{6}}^{3}\left ( 3-\left ( 9-x^2 \right ) \right )dx+\displaystyle\int\limits_3^{\sqrt{12}}\left ( 3-\left ( x^2-9 \right ) \right )dx \right ]\end{array} \)
\(\begin{array}{l}=2\left [ 4\sqrt{6}+\left ( \frac{x^3}{3}-6x \right )\left|_{\sqrt{6}}^{3}+\left ( 12x-\frac{x^3}{3} \right ) \right|_3^{\sqrt{12}} \right ]\end{array} \)
\(\begin{array}{l}=2\left [4\sqrt{6}+\left ( 4\sqrt{6}-9 \right )+\left ( 8\sqrt{12}-27 \right ) \right ]\end{array} \)
\(\begin{array}{l}=2\left [ 8\sqrt{6}+16\sqrt{3}-36 \right ]=8\left [ 2\sqrt{6}+4\sqrt{3}-9 \right ]\end{array} \)
11. Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle, Then the area of ΔPQRis :
(A)
\(\begin{array}{l}\frac{25}{4\sqrt{3}}\end{array} \)
(B)
\(\begin{array}{l}\frac{25\sqrt{3}}{2}\end{array} \)
(C)
\(\begin{array}{l}\frac{25}{\sqrt{3}}\end{array} \)
(D)
\(\begin{array}{l}\frac{25}{2\sqrt{3}}\end{array} \)
Answer (D)
Sol.
Altitude of equilateral triangle,
\(\begin{array}{l}\frac{\sqrt{3}l}{2}=\frac{5}{\sqrt{2}}\end{array} \)
\(\begin{array}{l}l=\frac{5\sqrt{2}}{\sqrt{3}}\end{array} \)
Area of triangle
\(\begin{array}{l}=\frac{\sqrt{3}}{4}l^2=\frac{\sqrt3}{4}\cdot\frac{50}{3}=\frac{25}{2\sqrt{3}}\end{array} \)
12. Let C be a circle passing through the points
A(2, –1) and B (3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle
\(\begin{array}{l}\left ( x-5 \right )^2+\left ( y-1 \right )^2=\frac{13}{2}\end{array} \)
then r2 is equal to :
(A) 32
(B)
\(\begin{array}{l}\frac{65}{2}\end{array} \)
(C)
\(\begin{array}{l}\frac{61}{2}\end{array} \)
(D) 30
Answer (B)
Sol. Equation of perpendicular bisector of AB is
\(\begin{array}{l}y-\frac{3}{2}=-\frac{1}{5}\left ( x-\frac{5}{2} \right )\Rightarrow x+5y=10\end{array} \)
Solving it with equation of given circle,
\(\begin{array}{l}\left ( x-5 \right )^2+\left ( \frac{10-x}{5}-1 \right )^2=\frac{13}{2}\end{array} \)
⇒
\(\begin{array}{l}\left ( x-5 \right )^2\left ( 1+\frac{1}{25} \right )=\frac{13}{2}\end{array} \)
⇒
\(\begin{array}{l}x-5=\pm\frac{5}{2}\Rightarrow x=\frac{5}{2}~\textup{or}~\frac{15}{2}\end{array} \)
But
\(\begin{array}{l}x\neq\frac{5}{2}\end{array} \)
because AB is not the diameter.
So, centre will be
\(\begin{array}{l}\left ( \frac{15}{2},\frac{1}{2} \right )\end{array} \)
Now
\(\begin{array}{l}r^2=\left ( \frac{15}{2}-2 \right )^2+\left ( \frac{1}{2}+1 \right )^2\end{array} \)
\(\begin{array}{l}=\frac{65}{2}\end{array} \)
13. Let the normal at the point P on the parabola
y2 = 6x pass through the point (5, –8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :
(A) –3
(B)
\(\begin{array}{l}-\frac{9}{4}\end{array} \)
(C)
\(\begin{array}{l}-\frac{5}{2}\end{array} \)
(D) –2
Answer (B)
Sol. Let P(at2, 2at) where
\(\begin{array}{l}a=\frac{3}{2}\end{array} \)
T :yt = x + at2 So point Q is
\(\begin{array}{l}\left ( -a,~at~-\frac{a}{t} \right )\end{array} \)
N :y = –tx + 2at + at3 passes through (5, –8)
\(\begin{array}{l}-8=-5t+3t+\frac{3}{2}t^3\end{array} \)
⇒ 3t3 – 4t + 16 = 0
⇒ (t + 2)(3t2 – 6t + 8) = 0
⇒ t = –2
So ordinate of point Q is
\(\begin{array}{l}-\frac{9}{4}\end{array} \)
14. If the two lines
\(\begin{array}{l}l_1:\frac{x-2}{3}=\frac{y+1}{-2},z=2\end{array} \)
and \(\begin{array}{l}l_2:\frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}\end{array} \)
are perpendicular, then an angle between the lines l2 and \(\begin{array}{l}l_3:\frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}\end{array} \)
is :
(A)
\(\begin{array}{l}\cos^{-1}\left ( \frac{29}{4}\right )\end{array} \)
(B)
\(\begin{array}{l}\sec^{-1}\left ( \frac{29}{4}\right )\end{array} \)
(C)
\(\begin{array}{l}\cos^{-1}\left ( \frac{2}{29} \right )\end{array} \)
(D)
\(\begin{array}{l}\cos^{-1}\left ( \frac{2}{\sqrt{29}} \right )\end{array} \)
Answer (B)
Sol. β΅ l1 and l2 are perpendicular, so
\(\begin{array}{l}3\times1+\left ( -2 \right )\left ( \frac{\alpha}{2} \right )+0\times2=0\end{array} \)
⇒ α = 3
Now angle between l2 and l3,
\(\begin{array}{l}\cos\theta=\frac{1\left ( -3 \right )+\frac{\alpha}{2}\left ( -2 \right )+2\left ( 4 \right )}{\sqrt{1+\frac{\alpha^2}{4}}+4\sqrt{9+4+16}}\end{array} \)
⇒
\(\begin{array}{l}\cos\theta=\frac{\frac{2}{29}}{2}\Rightarrow\theta=\cos^{-1}\left ( \frac{4}{29} \right )=\sec^{-1}\left ( \frac{29}{4} \right )\end{array} \)
15. Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point
\(\begin{array}{l}\left ( 2,-\frac{1}{2},2 \right )\end{array} \)
in the rotated plane is B( a, b, c),then :
(A)
\(\begin{array}{l}\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}\end{array} \)
(B)
\(\begin{array}{l}\frac{a}{4}=\frac{b}{5}=\frac{c}{-2}\end{array} \)
(C)
\(\begin{array}{l}\frac{a}{8}=\frac{b}{-5}=\frac{c}{4}\end{array} \)
(D)
\(\begin{array}{l}\frac{a}{4}=\frac{b}{5}=\frac{c}{2}\end{array} \)
Answer (A)
Sol. Consider the equation of plane,
P : (2x + 3y + z + 20) + λ(x – 3y + 5z – 8) = 0
P : (2 + λ)x + (3 – 3λ)y + (1 + 5λ)z + (20 – 8λ) = 0
β΅ Plane P is perpendicular to 2x + 3y + z + 20 = 0
So, 4 + 2λ + 9 – 9λ + 1 + 5λ = 0
P :9x – 18y + 36z – 36 = 0
Or P :x – 2y + 4z = 4
If image of
\(\begin{array}{l}\left ( 2,-\frac{1}{2},2 \right )\end{array} \)
in plane P is (a, b, c) then
\(\begin{array}{l}\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}\end{array} \)
and
\(\begin{array}{l}\left ( \frac{a+2}{2} \right )-2\left ( \frac{b-\frac{1}{2}}{2} \right )+4\left ( \frac{c+2}{2} \right )=4\end{array} \)
Clearly
\(\begin{array}{l}a=\frac{4}{3},b=\frac{5}{6}~\textup{and}~c=-\frac{2}{3}\end{array} \)
So, a :b : c = 8 : 5 : – 4
16. If
\(\begin{array}{l}\vec{a}\cdot\vec{b}=1,\vec{b}\cdot\vec{c}=2~\textup{and}~\vec{c}\cdot\vec{a}=3\end{array} \)
, then the value of \(\begin{array}{l}\left [ \vec{a}\times\left ( \vec{b}\times\vec{c} \right ),\vec{b}\times\left ( \vec{c}\times\vec{a} \right ),\vec{c}\times\left ( \vec{b}\times\vec{a} \right ) \right ]\end{array} \)
is :
(A) 0
(B)
\(\begin{array}{l}-6\vec{a}\cdot\left ( \vec{b}\times\vec{c} \right )\end{array} \)
(C)
\(\begin{array}{l}12\vec{c}\cdot\left ( \vec{a}\times\vec{b} \right )\end{array} \)
(D)
\(\begin{array}{l}-12\vec{b}\cdot\left ( \vec{c}\times\vec{a} \right )\end{array} \)
Answer (A)
Sol. β΅
\(\begin{array}{l}\vec{a}\times\left ( \vec{b}\times\vec{c} \right )=3\vec{b}-\vec{c}=\vec{u}\end{array} \)
\(\begin{array}{l}\vec{b}\times\left ( \vec{c}\times\vec{a} \right )=\vec{c}-2\vec{a}=\vec{v}\end{array} \)
\(\begin{array}{l}\vec{c}\times\left ( \vec{b}\times\vec{a} \right )=3\vec{b}-2\vec{a}=\vec{w}\end{array} \)
∴
\(\begin{array}{l}\vec{u}+\vec{v}=\vec{w}\end{array} \)
So vectors
\(\begin{array}{l}\vec{u},\vec{v}~\textup{and}~\vec{w}\end{array} \)
are coplanar, hence their Scalar triple product will be zero.
17. Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:
(A)
\(\begin{array}{l}\frac{275}{6^5}\end{array} \)
(B)
\(\begin{array}{l}\frac{36}{5^4}\end{array} \)
(C)
\(\begin{array}{l}\frac{181}{5^5}\end{array} \)
(D)
\(\begin{array}{l}\frac{46}{6^4}\end{array} \)
Answer (D)
Sol. Let probability of getting head = p
So,
\(\begin{array}{l}^5C_4p^4\left ( 1-p \right )=^5C_5p^5\end{array} \)
⇒
\(\begin{array}{l}p=5\left ( 1-p \right )\Rightarrow p=\frac{5}{6}\end{array} \)
Probability of getting atmost two heads =
\(\begin{array}{l}^5C_0\left ( 1-p \right )^5+^5C_1p\left ( 1-p \right )^4+^5C_2p^2\left ( 1-p \right )^3\end{array} \)
\(\begin{array}{l}=\frac{1+25+250}{6^5}\end{array} \)
\(\begin{array}{l}=\frac{276}{6^5}=\frac{46}{6^4}\end{array} \)
18. The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to:
(A) 60
(B) 55
(C) 50
(D) 45
Answer (A)
Sol. β΅
\(\begin{array}{l}\overline{x}=6=\frac{a+b+8+5+10}{5}\Rightarrow a+b=7\dots\left ( i \right ) \end{array} \)
And
\(\begin{array}{l}\sigma^2=\frac{a^2+b^2+8^2+5^2+10^2}{5}-6^2=6.8\end{array} \)
⇒ a2 + b2 = 25 …(ii)
From (i) and (ii) (a, b) = (3, 4) or (4, 3)
Now mean deviation about mean
\(\begin{array}{l}M=\frac{1}{5}\left ( 3+2+2+1+4 \right )=\frac{12}{5}\end{array} \)
⇒ 25M = 60
19. Let
\(\begin{array}{l}f\left ( x \right )=2\cos^{-1}x+4\cot^{-1}x-3x^2-2x+10,\chi\epsilon\left [ -1,1 \right ]\end{array} \)
If [a, b] is the range of the function,f then 4a – b is equal to :
(A) 11
(B) 11 – π
(C) 11 + π
(D) 15 – π
Answer (B)
Sol.
\(\begin{array}{l}f\left ( x \right )=2\cos^{-1}x+4\cot^{-1}x-3x^2-2x+10\forall x\epsilon\left [ -1,1 \right ]\end{array} \)
⇒
\(\begin{array}{l}f^{‘}\left ( x \right )=-\frac{2}{\sqrt{1-x^2}}-\frac{4}{1+x^2}-6x-2<0~\forall x\epsilon\left [ -1,1 \right ]\end{array} \)
Sof(x) is decreasing function and range of f(x) is
[f(1), f(-1)], which is [π + 5, 5π + 9]
Now 4a – b = 4(π + 5) – (5π + 9)
= 11 – π
20. Let
\(\begin{array}{l}\Delta,\triangledown \epsilon\left\{ \wedge ,\vee \right\}\end{array} \)
be such that \(\begin{array}{l}p\triangledown q\Rightarrow\left ( \left ( p\Delta q \right )\triangledown r \right ) \end{array} \)
is a tautology. Then \(\begin{array}{l}\left ( p\triangledown q \right )\Delta r\end{array} \)
is logically equivalent to :
(A)
\(\begin{array}{l}\left ( p~\Delta~r \right )\vee q\end{array} \)
(B)
\(\begin{array}{l}\left ( p~\Delta~r \right )\wedge q\end{array} \)
(C)
\(\begin{array}{l}\left ( p \wedge r \right )\Delta q\end{array} \)
(D)
\(\begin{array}{l}\left ( p \triangledown r \right )\wedge q\end{array} \)
Answer (A)
Sol. Case-I
If ∇ is same as ∧
Then (p∧q) ⇒ ((pΔq) ∧r) is equivalent to ~ (p∧q) ∨ ((pΔq) ∧r) is equivalent to (~ (p∧q) ∨ (pΔq))∧ (~ (p∧q) ∨r)
Which cannot be a tautology
For both Δ (i.e.∨ or ∧)
Case-II
If ∇ is same as ∨
Then (p∨q) ⇒ ((pΔq) ∨r) is equivalent to
~(p∨q) ∨ (pΔq) ∨r which can be a tautology if Δ is also same as ∨.
Hence both Δ and ∇ are same as ∨.
Now (p∇q) Δr is equivalent to (p∨q∨r).
SECTION – B
Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.
1. The sum of the cubes of all the roots of the equation x4 – 3x3 –2x2 + 3x +1 = 0 is _______.
Answer (36)
Sol. x4 – 3x3 – x2 – x2 + 3x + 1 = 0
(x2 – 1) (x2 – 3x – 1) = 0
Let the root of x2 – 3x – 1 = 0 be α and β and other two roots of given equation are 1 and –1
So sum of cubes of roots = 13 + (–1)3 + α3 + β3
= (α + β)3 – 3αβ(α + β)
= (3)3 – 3(–1)(3)
= 36
2. There are ten boys B1, B2, …, B10 and five girls G1, G2,…, G5 in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B1 and B2 together should not be the members of a group, is ________.
Answer (1120)
Sol. Required number of ways = Total ways of selection – ways in which B1 and B2 are present together.
\(\begin{array}{l}=^{10}C_3\cdot^5C_3-^8C_1\cdot^5C_3=10\left ( 120-8 \right ) \end{array} \)
= 1120
3. Let the common tangents to the curves
4(x2 + y2) = 9 and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and I respectively denote the eccentricity and the length of the latus rectum of this ellipse, then
\(\begin{array}{l}\frac{l}{e^2}\end{array} \)
is equal to ________.
Answer (4)
Sol. Let y = mx + c is the common tangent
So
\(\begin{array}{l}c=\frac{1}{m}=\pm\frac{3}{2}\sqrt{1+m^2}\Rightarrow m^2=\frac{1}{3}\end{array} \)
So equation of common tangents will be
\(\begin{array}{l}y=\pm\frac{1}{\sqrt{3}}x\pm\sqrt{3}\end{array} \)
which intersects at Q(–3, 0)
Major axis and minor axis of ellipse are 12 and 6. So eccentricity
\(\begin{array}{l}e^2=1-\frac{1}{4}=\frac{3}{4}\end{array} \)
and length of latus rectum \(\begin{array}{l}=\frac{2b^2}{a}=3\end{array} \)
Hence
\(\begin{array}{l}\frac{\ell}{e^2}=\frac{3}{3/4}=4\end{array} \)
4. Let
\(\begin{array}{l}f\left ( x \right )=\max\left\{ \left|x+1 \right|,\left|x+2 \right|,\dots\dots,\left|x+5 \right|\right\}\end{array} \)
Then \(\begin{array}{l}\displaystyle\int\limits_{-6}^0f\left ( x \right )dx\end{array} \)
is equal to _______.
Answer (21)
Sol.
\(\begin{array}{l}\displaystyle\int\limits_{-6}^0f\left ( x \right )dx=2\left [ \frac{1}{2}\left ( 2+5 \right )3 \right ]=21\end{array} \)
5. Let the solution curve y = y(x) of the differential equation (4 + x2)dy – 2x(x2 + 3y + 4)dx = 0 pass through the origin. Then y(2) is equal to _______.
Answer (12)
Sol. (4 + x2) dy – 2x(x2 + 3y + 4)dx = 0
⇒
\(\begin{array}{l}\frac{dy}{dx}=\left ( \frac{6x}{x^2+4} \right )y+2x\end{array} \)
⇒
\(\begin{array}{l}\frac{dy}{dx}-\left ( \frac{6x}{x^2+4} \right )y=2x\end{array} \)
I.F. =
\(\begin{array}{l}e^{-3~\textup{In}\left ( x^2+4 \right )}=\frac{1}{\left ( x^2+4 \right )^3}\end{array} \)
So
\(\begin{array}{l}\frac{y}{\left ( x^2+4 \right )^3}=\int\frac{2x}{\left ( x^2+4 \right )^3}dx+c\end{array} \)
⇒
\(\begin{array}{l}y=-\frac{1}{2}\left ( x^2+4 \right )+c\left ( x^2+4 \right )^3\end{array} \)
When x = 0, y = 0 gives
\(\begin{array}{l}c=\frac{1}{32}\end{array} \)
So, for x = 2,y = 12
6. If sin2(10°)sin(20°)sin(40°)sin(50°)sin(70°)
\(\begin{array}{l}=\alpha-\frac{1}{16}\sin\left ( 10^\circ \right )\end{array} \)
, then 16 + α–1 is equal to _______.
Answer (80)
Sol: (sin10° ⋅ sin50° ⋅ sin70°) ⋅ (sin10° ⋅ sin20° ⋅ sin40°)
=
\(\begin{array}{l}\left ( \frac{1}{4}\sin30^\circ \right )\cdot\left [ \frac{1}{2}\sin10^\circ\left ( \cos20^\circ-\cos60^\circ \right ) \right ]\end{array} \)
=
\(\begin{array}{l}\frac{1}{16}\left [ \sin10^\circ\left ( \cos20^\circ-\frac{1}{2} \right ) \right ]\end{array} \)
=
\(\begin{array}{l}\frac{1}{32}[2\sin10^\circ\cdot\cos20^\circ-\sin10^\circ]\end{array} \)
=
\(\begin{array}{l}\frac{1}{32}[\sin30^\circ-\sin10^\circ-\sin10^\circ]\end{array} \)
=
\(\begin{array}{l}\frac{1}{64}-\frac{1}{16}\sin10^\circ\end{array} \)
Clearly
\(\begin{array}{l}\alpha=\frac{1}{64}\end{array} \)
Hence 16 + α–1 = 80
7. Let A = {n∈N : H.C.F. (n, 45) = 1} and
Let B = {2k :k∈ {1, 2, …,100}}. Then the sum of all the elements of A∩B is ___________.
Answer (5264)
Sol: Sum of all elements of A∩B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]
=
\(\begin{array}{l}2\left [ \frac{100\times101}{2}-3\left ( \frac{33\times34}{2} \right )-5\left ( \frac{20\times21}{2} \right )+15\left ( \frac{6\times7}{2} \right ) \right ]\end{array} \)
= 10100 – 3366 – 2100 + 630
= 5264
8. The value of the integral
\(\begin{array}{l}\frac{48}{\pi^4}\displaystyle\int\limits_0^\pi\left ( \frac{3\pi x^2}{2}-x^3 \right )\frac{\sin x}{1+\cos^2x}dx\end{array} \)
is equal to ________.
Answer (6)
Sol:
\(\begin{array}{l}I=\frac{48}{\pi^4}\int_{0}^{\pi}\left [ \left ( \frac{\pi}{2}-x \right )^3-\frac{3\pi^2}{4}\left ( \frac{\pi}{2}-x \right )+\frac{\pi^3}{4} \right ]\frac{\sin x~dx}{1+\cos^2x}\end{array} \)
Using
\(\begin{array}{l}\int_a^bf\left ( x \right )dx=\int_a^bf\left ( a+b-x \right )dx\end{array} \)
we get
\(\begin{array}{l}I=\frac{48}{\pi^4}\int_0^\pi\left [ -\left ( \frac{\pi}{2}-x \right )^3+\frac{3\pi^2}{4}\left ( \frac{\pi}{2}-x \right )+\frac{\pi^3}{4} \right ]\frac{\sin~x~dx}{1+\cos^2~x}\end{array} \)
Adding these two equations, we get
\(\begin{array}{l}2I=\frac{48}{\pi^4}\int_0^\pi\frac{\pi^3}{2}\cdot\frac{\sin~x~dx}{1+\cos^2~x}\end{array} \)
⇒
\(\begin{array}{l}I=\frac{12}{\pi}\left [ -\tan^{-1}\left ( \cos x \right ) \right ]_0^\pi=\frac{12}{\pi}\cdot\frac{\pi}{2}=6\end{array} \)
9. Let
\(\begin{array}{l}A=\displaystyle\sum\limits_{i=1}^{10}\displaystyle\sum\limits_{j=1}^{10}\textup{min}\left\{i,j \right\}\end{array} \)
and\(\begin{array}{l}B=\displaystyle\sum\limits_{i=1}^{10}\displaystyle\sum\limits_{j=1}^{10}\textup{max}\left\{i,j \right\}\end{array} \)
. Then A + B is equal to ____________.
Answer (1100)
Sol: Each element of ordered pair {i, j} is either present in A or in B.
So,A + B =Sum of all elements of all ordered pairs {i, j} for 1 ≤i≤ 10 and 1 ≤j≤ 10
= 20 (1 + 2 + 3+…. + 10)
= 1100
10. Let
\(\begin{array}{l}S=\left ( 0,2\pi \right )-\left\{\frac{\pi}{2},\frac{3\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4} \right\}\end{array} \)
. Let y = y(x), x∈S, be the solution curve of the differential equation \(\begin{array}{l}\frac{dy}{dx}=\frac{1}{1+\sin2x},y\left ( \frac{\pi}{4} \right )=\frac{1}{2}\end{array} \)
. If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve\(\begin{array}{l}y=\sqrt{2}\sin x\end{array} \)
is \(\begin{array}{l}\frac{k\pi}{12}\end{array} \)
, then k is equal to _________.
Answer (42)
Sol:
\(\begin{array}{l}\frac{dy}{dx}=\frac{1}{1+\sin 2x}\end{array} \)
⇒
\(\begin{array}{l}dy=\frac{\sec^2x~dx}{\left ( 1+\tan~x \right )^2}\end{array} \)
⇒
\(\begin{array}{l}y=-\frac{1}{1+\tan x}+c\end{array} \)
When
\(\begin{array}{l}x=\frac{\pi}{4},~~y=\frac{1}{2}~\textup{gives}~~c=1\end{array} \)
So
\(\begin{array}{l}y=\frac{\tan x}{1+\tan x}\Rightarrow y=\frac{\sin x}{\sin x+\cos x}\end{array} \)
Now,
\(\begin{array}{l}y=\sqrt{2}\sin x~~\Rightarrow~~\sin x=0\end{array} \)
or
\(\begin{array}{l}\sin x+\cos x=\frac{1}{\sqrt{2}}\end{array} \)
sinx = 0 gives x = π only.
and
\(\begin{array}{l}\sin x+\cos x=\frac{1}{\sqrt{2}}\Rightarrow\sin\left ( x+\frac{\pi}{4} \right )=\frac{1}{2}\end{array} \)
So
\(\begin{array}{l}x+\frac{\pi}{4}=\frac{5\pi}{6}~~\textup{or}~~\frac{13\pi}{6}\Rightarrow x=\frac{7\pi}{12}~~\textup{or}~~\frac{23\pi}{12}\end{array} \)
Sum of all solutions =
\(\begin{array}{l}\pi+\frac{7\pi}{12}+\frac{23\pi}{12}=\frac{42\pi}{12}\end{array} \)
Hence k = 42.
β β β
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