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JEE Main 2022 28th July Shift 1 Mathematics Question Paper and 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 the solution curve of the differential equation \(\begin{array}{l}xdy=\left(\sqrt{x^2 + y^2 }+y\right)dx, x>0\end{array} \)
intersect the line x = 1 at y = 0 and the line x = 2 at y = α. Then the value of α is
\(\begin{array}{l}(\text{A})\ \frac{1}{2}\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{3}{2}\end{array} \)
\(\begin{array}{l}(\text{C})\ -\frac{3}{2}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{5}{2}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}\frac{xdy-ydx}{\sqrt{x^2+y^2}}=dx\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=\frac{\sqrt{x^2+y^2}}{x}+\frac{y}{x}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}= \sqrt{1+\frac{y^2}{x^2}} +\frac{y}{x}\end{array} \)
\(\begin{array}{l}\text{Let }\frac{y}{x} = v\end{array} \)
\(\begin{array}{l}\Rightarrow v + x\frac{dv}{dx}=\sqrt{1+v^2}+v\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}\end{array} \)
\(\begin{array}{l}\text{OR}\ \ln \left(v+\sqrt{1+v^2}\right)=\ln x + C\end{array} \)
at x = 1, y = 0
⇒ C = 0
\(\begin{array}{l}\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}=x\end{array} \)
At x = 2,
\(\begin{array}{l}\frac{y}{2}+\sqrt{1+\frac{y^2}{4}}=2\end{array} \)
\(\begin{array}{l}\Rightarrow 1 + \frac{y^2}{4}=4+\frac{y^2}{4}-2y\end{array} \)
\(\begin{array}{l}\text{OR } y = \frac{3}{2}\end{array} \)
2. Considering only the principal values of the inverse trigonometric functions, the domain of the function \(\begin{array}{l}f(x)=\cos^{-1}\left(\frac{x^2-4x+2}{x^2+3}\right)\end{array} \)
is
\(\begin{array}{l}(\text{A})\ \left(-\infty, \frac{1}{4}\right]\end{array} \)
\(\begin{array}{l}(\text{B})\ \left[-\frac{1}{4}, \infty \right)\end{array} \)
\(\begin{array}{l}(\text{C})\ \left(\frac{-1}{3}, \infty \right)\end{array} \)
\(\begin{array}{l}(\text{C})\ \left(-\infty, \frac{1}{3} \right]\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}-1\le \frac{x^2-4x+2}{x^2+3}\le1\end{array} \)
\(\begin{array}{l}\Rightarrow -x^2 – 3 \le x^2 – 4x + 2 \le x^2 + 3 \end{array} \)
⇒ 2x2 – 4x + 5 ≥ 0 & -4x ≤ 1
x ∈ R & x ≥ -1/4 ]
\(\begin{array}{l}\text{So domain is }\left[-\frac{1}{4}, \infty\right)\end{array} \)
3. Let the vectors \(\begin{array}{l}\vec{a}=(1+t)\hat{i}+(1-t)\hat{j}+\hat{k},\ \vec{b}=(1-t)\hat{i}+(1+t)\hat{j}+2\hat{k}\end{array} \)
and \(\begin{array}{l}\vec{c}=t\hat{i}-t\hat{j}+\hat{k}, t\in \mathbf{R}\end{array} \)
be such that for \(\begin{array}{l}\alpha, \beta, \gamma \in \mathbf R,\ \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c} =\vec{0}\ \Rightarrow \alpha =\beta = \gamma =0.\end{array} \)
Then, the set of all values of t is
(A) A non-empty finite set
(B) Equal to N
\(\begin{array}{l}\left(C\right) \text{Equal to}~ \mathbf{R} – \{0\}\end{array} \)
\(\begin{array}{l}\left(C\right) \text{Equal to}~ \mathbf{R}\end{array} \)
Answer (C)
Sol.
\(\begin{array}{l}\text{Clearly }\vec{a}, \vec{b},\vec{c} \text{ are non-coplanar}\end{array} \)
\(\begin{array}{l}\begin{vmatrix} 1+t& 1-t & 1 \\1-t & 1+t & 2 \\t & -t & 1 \\\end{vmatrix} \ne 0\end{array} \)
\(\begin{array}{l}\Rightarrow \left(1 + t\right)\left(1 + t + 2t\right) – \left(1 – t\right)\left(1 – t – 2t\right)+ 1\left(t^2 – t – t – t^2\right) \ne 0\end{array} \)
\(\begin{array}{l}\Rightarrow \left(3t^2 + 4t + 1\right) – \left(1 – t\right) \left(1 – 3t\right) – 2t \neq 0\end{array} \)
\(\begin{array}{l}\Rightarrow \left(3t^2 + 4t + 1\right) – \left(3t^2 – 4t + 1\right) – 2t \neq 0\end{array} \)
\(\begin{array}{l}\Rightarrow t \neq 0\end{array} \)
4. Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation \(\begin{array}{l}cos^{-1}\left(x\right) – 2sin^{-1}\left(x\right) = cos^{-1}\left(2x\right)\end{array} \)
is equal to
\(\begin{array}{l}(\text{A})0\end{array} \)
\(\begin{array}{l}(\text{B})1\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{1}{2}\end{array} \)
\(\begin{array}{l}(\text{D})\ -\frac{1}{2}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}cos^{-1}x – 2sin^{-1}x = cos^{-1}2x\end{array} \)
\(\begin{array}{l}\text{For Domain :}x \in \left[\frac{-1}{2},\frac{1}{2}\right]\end{array} \)
\(\begin{array}{l}\cos^{-1}x-2\left(\frac{\pi}{2}-\cos^{-1}x\right)=\cos^{-1}(2x)\end{array} \)
\(\begin{array}{l} \Rightarrow cos^{-1}x + 2cos^{-1}x = \pi + cos^{-1}2x\end{array} \)
\(\begin{array}{l}\Rightarrow cos\left(3cos^{–1}x\right) = – cos\left(cos^{–1}2x\right)\end{array} \)
\(\begin{array}{l}\Rightarrow 4x^3 = x\end{array} \)
\(\begin{array}{l}\Rightarrow x = 0, \pm \frac{1}{2}\end{array} \)
5. Let the operations *, ◉ ∈ {∧, ∨}. If (p * q) ◉ (p ◉ ~q) is a tautology, then the ordered pair (*, ◉) is
(A) (∨, ∧)
(B) (∨, ∨)
(C) (∧, ∧)
(D) (∧, ∨)
Answer (B)
Sol. *, ◉ ∈ {∧, ∨}.
Now for (q * q)◉(p◉~q) is tautology
(A) (∨, ∧) : (p ∨ q) ∧ (p ∧ ~q) not a tautology
(B) (∨, ∨) : (p ∨ q) ∨ (p ∨ ~q)
= P ∨ T is tautology
(C) (∧ , ∧) : (p ∧ q) ∧ (p ∧ ~q)
= (p ∧ p) ∧ (q ∧ ~q) = p ∧ F not a tautology (Fallasy)
(D) (∧, ∨) : (p ∧ q) ∨ (p ∨ ~q) not a tautology
6. Let a vector \(\begin{array}{l}\vec{a}\ \text{has magnitude}\ 9.\ \text{Let a vector}\ \vec{b}\end{array} \)
be such that for every \(\begin{array}{l}\left(x, y\right) \in \mathbf R \times \mathbf R – \{\left(0, 0\right)\},\ \text{the vector}\ (x\vec{a} + y\vec{b})\end{array} \)
\(\begin{array}{l}\text{is perpendicular to the vector}\ (6y\vec{a} – 18x\vec{b}).\end{array} \)
\(\begin{array}{l}\text{Then the value of}\ |\vec{a}\times \vec{b}|\end{array} \)
is equal to
\(\begin{array}{l}(\text{A})\ 9\sqrt{3}\end{array} \)
\(\begin{array}{l}(\text{B})\ 27\sqrt{3}\end{array} \)
\(\begin{array}{l}(\text{C})\ 9\end{array} \)
\(\begin{array}{l}(\text{D})\ 81\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}(x\vec{a} + y\vec{b})\cdot(6y\vec{a}-18x\vec{b})=0\end{array} \)
\(\begin{array}{l}\Rightarrow \left(6xy|\vec{a}|^2 – 18 xy|\vec{b}|^2\right)+\left(6y^2 -18x^2\right)\vec{a}.\vec{b}=0\end{array} \)
As given equation is identity
Coefficient of x2 = coefficient of y2 = coefficient of
xy = 0
\(\begin{array}{l}\Rightarrow |\vec{a}|^2 = 3|\vec{b}|^2 \Rightarrow |\vec{b}| = 3\sqrt{3}\end{array} \)
\(\begin{array}{l}\text{and } \vec{a}.\vec{b}=0\end{array} \)
\(\begin{array}{l}|\vec{a}\times \vec{b}| = |\vec{a}|| \vec{b}|\sin \theta\end{array} \)
\(\begin{array}{l}=9.3\sqrt{3}.1=27\sqrt{3}\end{array} \)
7. For t ∈ (0, 2π), if ABC is an equilateral triangle with vertices A(sint, – cost), B(cost, sint) and C(a, b) such that its orthocentre lies on a circle with centre (1, 1/3), then (a2 – b2) is equal to
\(\begin{array}{l}(\text{A})\ \frac{8}{3}\end{array} \)
\(\begin{array}{l}(\text{B})\ 8\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{77}{9}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{80}{9}\end{array} \)
Answer (B)
Sol. Let P(h, k) be the orthocentre of ΔABC.
Then
\(\begin{array}{l}h=\frac{\sin t + \cos t + a}{3}, k= \frac{-\cos t + \sin t + b}{3}\end{array} \)
(orthocentre coincide with centroid)
\(\begin{array}{l}\therefore \left(3h – a\right)^2 + \left(3k – b\right)^2 = 2\end{array} \)
\(\begin{array}{l}\therefore \left(h-\frac{a}{3}\right)^2 + \left(k-\frac{b}{3}\right)^2 = \frac{2}{9}\end{array} \)
\(\begin{array}{l}\because\end{array} \)
orthocentre lies on circle with centre \(\begin{array}{l}\left(1, \frac{1}{3}\right)\end{array} \)
∴ a = 3, b = 1
∴ a2 – b2 = 8
8. For α ∈ N, consider a relation R on N given by R = {(x, y) : 3x + αy is a multiple of 7}. The relation R is an equivalence relation if and only if
(A) α = 14
(B) α is a multiple of 4
(C) 4 is the remainder when α is divided by 10
(D) 4 is the remainder when α is divided by 7
Answer (D)
Sol.
\(\begin{array}{l}R = \{(x, y) : 3x + \alpha y~ \text{is a multiple of} ~7\}\end{array} \)
, Now R to be an equivalence relation
\(\begin{array}{l}\left(1\right) \text{R should be reflexive }: \left(a, a\right)\in R~ \forall a \in N\\ \therefore 3a + a \alpha = 7k\end{array} \)
\(\begin{array}{l}\therefore \left(3 + \alpha \right) a = 7k\end{array} \)
\(\begin{array}{l}\therefore 3 + \alpha = 7k_1 \Rightarrow \alpha = 7k_1 – 3\\ = 7k_1 + 4\end{array} \)
\(\begin{array}{l}\left(2\right) R~ \text{should be symmetric} : aRb \Leftrightarrow bRa \end{array} \)
\(\begin{array}{l}aRb : 3a + \left(7k -3\right) b = 7 m \end{array} \)
\(\begin{array}{l}\Rightarrow 3\left(a – b\right) + 7kb = 7 m\\\Rightarrow 3\left(b – a\right) + 7 ka = 7 m\end{array} \)
\(\begin{array}{l}So, aRb \Rightarrow bRa\end{array} \)
\(\begin{array}{l}\therefore \text{R will be symmetric for}~ a = 7k_1 – 3 \end{array} \)
\(\begin{array}{l}\left(3\right) \text{Transitive : Let} \left(a, b\right)\in R, (b, c)\in R\end{array} \)
\(\begin{array}{l}\Rightarrow 3a + \left(7k – 3\right)b = 7k_1 ~\text{and}~\\ 3b + \left(7k_2 – 3\right) c = 7k_3\end{array} \)
Adding 3a + 7kb + (7k2 – 3) c = 7 (k1 + k3)
3a + (7k2 – 3) c = 7 m
\(\begin{array}{l}\therefore \left(a, c\right)\in R\end{array} \)
∴ R is transitive
\(\begin{array}{l}\therefore a = 7k – 3 = 7k + 4\end{array} \)
9. Out of 60% female and 40% male candidates appearing in an exam, 60% of candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability that the chosen candidate is a female, is
\(\begin{array}{l}(\text{A})\ \frac{3}{4}\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{11}{16}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{23}{32}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{13}{16}\end{array} \)
Answer (*) None of the given option is correct
Sol.
\(\begin{array}{l}P(\text{Female})= \frac{60}{100}=\frac{3}{5}\end{array} \)
\(\begin{array}{l}P(\text{Male})= \frac{2}{5}\end{array} \)
\(\begin{array}{l}P(\text{Female/Qualified})= \frac{40}{60} = \frac{2}{3}\end{array} \)
\(\begin{array}{l}P(\text{Male/Qualified})= \frac{20}{60} = \frac{1}{3}\end{array} \)
10. If y = y(x), x ∈ (0, π/2) be the solution curve of the differential equation \(\begin{array}{l}(\sin^22x)\frac{dy}{dx}+ (8\sin^22x + 2\sin 4x)y = 2e^{-4x}(2\sin 2x + \cos 2x),\end{array} \)
\(\begin{array}{l}\text{with}\ y\left(\frac{\pi}{4}\right)=e^{-\pi},\ \text{then}\ y\left(\frac{\pi}{6}\right)\end{array} \)
is equal to
\(\begin{array}{l}(\text{A})\ \frac{2}{\sqrt{3}}e^{-2\pi/3}\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{2}{\sqrt{3}}e^{2\pi/3}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{1}{\sqrt{3}}e^{-2\pi/3}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{1}{\sqrt{3}}e^{2\pi/3}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}(\sin^22x)\frac{dy}{dx}+(8\sin^2 2x + 2\sin 4x)y=2e^{-4x}(2\sin 2x + \cos 2x)\end{array} \)
\(\begin{array}{l}\frac{dy}{dx}+(8+4\cot 2x)y = 2e^{-4x}\left(\frac{2\sin 2x + \cos 2x}{\sin^2 2x}\right)\end{array} \)
\(\begin{array}{l} \text{Integrating factor} (I.F.)=e^{\int (8+4\cot 2x)dx}\end{array} \)
\(\begin{array}{l}=e^{8x+2\ln \sin 2x}\end{array} \)
Solution of differential equation
\(\begin{array}{l}y.e^{8x+2\ln \sin 2x} = \int 2e^{(4x + 2\ln \sin2x)}\frac{(2\sin 2x + \cos 2x)}{\sin^2 2x}dx\end{array} \)
\(\begin{array}{l}=2\int e^{4x}(2 \sin 2x + \cos 2x)dx\end{array} \)
\(\begin{array}{l}y.e^{8x + 2\ln \sin 2x}=e^{4x}\sin 2x + c\end{array} \)
\(\begin{array}{l}y\left(\frac{\pi}{4}\right)=e^{-\pi}\end{array} \)
\(\begin{array}{l}e^{-\pi}.e^{2\pi}=e^{\pi}+c \Rightarrow c=0\end{array} \)
\(\begin{array}{l}y\left(\frac{\pi}{6}\right)=\frac{e^{\frac{2\pi}{3}}\frac{\sqrt{3}}{2}}{e^{\left(\frac{4\pi}{3}+2\ln \frac{\sqrt{3}}{2}\right)}}\end{array} \)
\(\begin{array}{l}=e^{\frac{-2\pi}{3}}.\frac{2}{\sqrt{3}}\end{array} \)
11. If the tangents drawn at the points P and Q on the parabola y2 = 2x – 3 intersect at the point R(0, 1), then the orthocentre of the triangle PQR is :
(A) (0, 1)
(B) (2, –1)
(C) (6, 3)
(D) (2, 1)
Answer (B)
Sol.
Equation of chord PQ
⇒ y × 1 = x – 3
⇒ x – y = 3
For point P & Q
Intersection of PQ with parabola P : (6, 3) Q : (2, –1)
Slope of RQ = –1 & Slope of PQ = 1
\(\begin{array}{l}\text{Therefore}~ \angle PQR = 90 ^\circ \Rightarrow \text{Orthocentre is at} Q : \left(2, -1\right)\end{array} \)
12. Let C be the centre of the circle \(\begin{array}{l}x^2+y^2-x+2y=\frac{11}{4}\end{array} \)
and P be a point on the circle. A line passes through the point C, makes an angle of π/4with the line CP and intersects the circle at the Q and R. Then the area of the triangle PQR (in unit2) is :
\(\begin{array}{l}(\text{A})\ 2\end{array} \)
\(\begin{array}{l}(\text{B})\ 2\sqrt{2}\end{array} \)
\(\begin{array}{l}(\text{C})\ 8\sin\left(\frac{\pi}{8}\right)\end{array} \)
\(\begin{array}{l}(\text{D})\ 8\cos\left(\frac{\pi}{8}\right)\end{array} \)
Answer (B)
Sol.
QR = 2r = 4[
\(\begin{array}{l}P=\left(\frac{1}{2}+2\cos\frac{\pi}{4}, -1 + 2\sin \frac{\pi}{4}\right)\end{array} \)
\(\begin{array}{l}=\left(\frac{1}{2}+\sqrt{2},-1 + \sqrt{2} \right)\end{array} \)
\(\begin{array}{l}\text{Area of } \Delta PQR = \frac{1}{2}\times 4 \times \sqrt{2}\end{array} \)
\(\begin{array}{l}=2\sqrt{2} \text{ sq. units}\end{array} \)
13. The remainder 72022 + 32022 is divided by 5 is:
(A) 0
(B) 2
(C) 3
(D) 4
Answer (C)
Sol. Let E = 72022 + 32022
= (15 – 1)1011 + (10 – 1)1011
= –1 + (multiple of 15) –1 + multiple of 10
= –2 + (multiple of 5)
Hence remainder on dividing E by 5 is 3.
14. Let the matrix \(\begin{array}{l}A= \begin{vmatrix}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0 \\\end{vmatrix}\end{array} \)
and the matrix \(\begin{array}{l}B_0 = A^{49} + 2A^{98}. ~\text{If} ~B_n = Adj(B_{n – 1})~\text{for all}~n \geq 1,\end{array} \)
then det(B4) is equal to :
(A) 328
(B) 330
(C) 332
(D) 336
Answer (C)
Sol.
\(\begin{array}{l}A= \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}\Rightarrow A^2= \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0 \\\end{bmatrix} \times \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0 \\\end{bmatrix} = \begin{bmatrix}0 & 0 & 1 \\1 & 0 & 0 \\0 & 1 & 0 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}\Rightarrow A^3= \begin{bmatrix}0 & 0 & 1 \\1 & 0 & 0 \\0 & 1 & 0 \\\end{bmatrix} \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0 \\\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix} = I\end{array} \)
\(\begin{array}{l}\text{Now } B_0 = A^{49}+ 2A^{98}=\left(A^3\right)^{16}.A+2\left(A^3\right)^{32}.A^2\end{array} \)
\(\begin{array}{l}B_0=A+2A^2= \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0 \\\end{bmatrix}+ \begin{bmatrix}0 & 0 & 2 \\2 & 0 & 0 \\0 & 2 & 0 \\\end{bmatrix} = \begin{bmatrix}0 & 1 & 2 \\2 & 0 & 1 \\1 & 2 & 0 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}\left|B_0\right|= 9 \end{array} \)
\(\begin{array}{l}\text{Since}, B_n = Adj \left|B_{n – 1}\right|\Rightarrow \left|B_n\right| = \left|B_{n – 1}\right|^2 \end{array} \)
\(\begin{array}{l}\text{Hence} \left|B_4\right| = \left|B_3\right|^2 = \left|B_2\right|^4 = \left|B_1\right|^8 = \left|B_0\right|^{16} = \left|3^2\right|^{16} = 3^{32}\end{array} \)
15. Let \(\begin{array}{l}S_1= \left\{z_1 \in C : |z_1 – 3| = \frac{1}{2}\right\}\ \text{and}\ S_2= \left\{z_2 \in C : |z_2 – |z_2 + 1|| = |z_2 + |z_2 – 1||\right\}.\end{array} \)
Then, for z1 ∈ S1 and z2 ∈ S2, the least value of |z2 – z1| is :
\(\begin{array}{l}(\text{A})\ 0\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{1}{2}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{3}{2}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{5}{2}\end{array} \)
Answer (C)
Sol.
\(\begin{array}{l}\because |Z_2 + |Z_2 -1||^2=|Z_2- |Z_2+1| |^2\end{array} \)
\(\begin{array}{l}\Rightarrow (Z_2 + |Z_2-1|)(\bar{Z}_2 + |Z_2-1|)=(Z_2- |Z_2-1|)(\bar{Z}_2-|Z_2+1|)\end{array} \)
\(\begin{array}{l}\Rightarrow Z_2(|Z_2-1|+|Z_2+1|)+ \bar{Z}_2(|Z_2-1| + |Z_2+1|)=|Z_2 + 1|^2 – |Z_2-1|^2\end{array} \)
\(\begin{array}{l}\Rightarrow (Z_2 + \bar{Z}_2)(|Z_2+1|+|Z_2-1|)=2(Z_2 + \bar{Z}_2)\end{array} \)
\(\begin{array}{l}\Rightarrow \text{Either } Z_2 + \bar{Z}_2 = 0 \text{ or } |Z_2 + 1| + |Z_2 + 1|=2\end{array} \)
So, Z2 lies on imaginary axis or on real axis within [–1, 1]
\(\begin{array}{l} \text{Also} |Z_1-3|=\frac{1}{2}\Rightarrow Z_1\end{array} \)
lies on the circle having center 3 and radius ½.
\(\begin{array}{l}\text{Clearly } |Z_1 – Z_2|_{\text{min}} = \frac{3}{2}\end{array} \)
16. The foot of the perpendicular from a point on the circle x2 + y2 = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?
\(\begin{array}{l}\left(A\right) \left(6x + 5y – 12\right)^2 + 4\left(3x + 7y – 8\right)^2 = 1, z = 6 – 2x – 3y\end{array} \)
\(\begin{array}{l}\left(B\right)\left(5x + 6y – 12\right)^2 + 4\left(3x + 5y – 9\right)^2 = 1, z = 6- 2x – 3y\end{array} \)
\(\begin{array}{l}\left(C\right) \left(6x + 5y – 14\right)^2 + 9\left(3x + 5y – 7\right)^2 = 1, z = 6 – 2x – 3y\end{array} \)
\(\begin{array}{l}\left(D\right) \left(5x + 6y – 14\right)^2 + 9\left(3x + 7y – 8\right)^2 = 1, z = 6 – 2x – 3y\end{array} \)
Answer (B)
Sol. Any point on x2 + y2 = 1, z = 0 is p(cosθ, sinθ, 0)
If foot of perpendicular of p on the plane 2x + 3y + z = 6 is (h, k, l) then
\(\begin{array}{l}\frac{h-\cos \theta}{2}=\frac{k-\sin \theta}{3}=\frac{l-0}{1}\end{array} \)
\(\begin{array}{l}=-\left(\frac{2\cos \theta + 3\sin \theta +0 -6}{2^2 + 3^2 + 1^2}\right)=r(\text{let})\end{array} \)
h = 2r + cosθ, k = 3r + sinθ, l = r
Hence, h – 2l = cosθ and k – 3l = sinθ
Hence (h – 2l)2 + (k – 3l)2 = 1
When l = 6 – 2h – 3k
Hence required locus is
\(\begin{array}{l}\left(x – 2\left(6 – 2x – 3y\right)\right)^2 + \left(y – 3\left(6 – 2x – 3y\right)\right)^2 = 1\end{array} \)
\(\begin{array}{l}\Rightarrow \left(5x + 6y – 12\right)^2 + 4\left(3x + 5y – 9\right)^2 = 1, z = 6 – 2x – 3y\end{array} \)
17. If the minimum value of \(\begin{array}{l}f(x)=\frac{5x^2}{2}+\frac{\alpha}{x^5}, x>0\end{array} \)
is 14, then the value of α is equal to
(A) 32
(B) 64
(C) 128
(D) 256
Answer (C)
Sol.
\(\begin{array}{l}f(x)=\frac{5x^2}{2}+\frac{\alpha}{x^5} \ \ \{x>0\}\end{array} \)
\(\begin{array}{l}f'(x)=5x-\frac{5\alpha}{x^6}=0\end{array} \)
\(\begin{array}{l}\Rightarrow x = (\alpha)^{1/7}\end{array} \)
\(\begin{array}{l}f(x)_{\text{min}}=\frac{5(\alpha)^{2/7}}{2}+\frac{\alpha}{a^{5/7}}=14\end{array} \)
\(\begin{array}{l}\frac{5}{2}\alpha^{2/7}+\alpha^{2/7}=14\end{array} \)
\(\begin{array}{l}\frac{7}{2}\alpha^{2/7} = 14\\ \alpha = 128\end{array} \)
18. Let α, β and γ be three positive real numbers. \(\begin{array}{l}Let f\left(x\right) = \alpha x^5 + \beta x^3 + \gamma x,x \in \mathbb {R} ~\text{and } g:\mathbb{R} \rightarrow \mathbb{R}~\text{be such that}~ g\left(f(x)\right) = x~\text{for all}~ x \in \mathbb{R}. \end{array} \)
If a1, a2, a3, …, an be in arithmetic progression with mean zero, then the value of \(\begin{array}{l}f\left( g\left( \frac{1}{n} \sum_{i=1}^{n}f(a_i)\right)\right)\end{array} \)
is equal to
(A) 0
(B) 3
(C) 9
(D) 27
Answer (A)
Sol.
\(\begin{array}{l}f\left( g\left( \frac{1}{n} \sum_{i=1}^{n}f(a_i)\right)\right)\end{array} \)
\(\begin{array}{l}\frac{a_1+a_2+a_3+…+a_n}{n}=0\end{array} \)
∴ First and last term, second and second last and so on are equal in magnitude but opposite in sign.
\(\begin{array}{l}f(x)=\alpha x^5 + \beta x^3 + \gamma x\end{array} \)
\(\begin{array}{l}\sum_{i=1}^{n}f(a_i)=\alpha\left(a_1^5 + a_2^5 + a_3^5 + … + a_n^5\right) + \beta\left(a_1^3 + a_2^3 + … + a_n^3\right) + \gamma\left(a_1+a_2 + … + a_n\right)\end{array} \)
\(\begin{array}{l}= 0\alpha + 0\beta + 0\gamma= 0\end{array} \)
\(\begin{array}{l}\therefore f\left(g\left(\frac{1}{n}\sum^{n}_{i=1}f\left(a_i\right)\right)\right)= \frac{1}{n}\sum^{n}_{i=1}f\left(a_i\right)=0\end{array} \)
19. Consider the sequence a1, a2, a3, … such that a1 = 1, a2 = 2 and \(\begin{array}{l}a_{n+2}=\frac{2}{a_{n+1}}+a_n \text{ for } n = 1,2,3,…\end{array} \)
. If \(\begin{array}{l}\left(\frac{a_1+\frac{1}{a_2}}{a_3}\right)\left(\frac{a_2+\frac{1}{a_3}}{a_4}\right)\left(\frac{a_3+\frac{1}{a_4}}{a_5}\right)\cdots\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^\alpha (^{61}C_{31}), \end{array} \)
then α is equal to
(A) –30
(B) –31
(C) –60
(D) –61
Answer (C)
Sol.
\(\begin{array}{l}a_{n+2}=\frac{2}{a_{n+1}}+a_n\end{array} \)
\(\begin{array}{l}\Rightarrow a_na_{n+1}+1 = a_{n+1}a_{n+2}-1\end{array} \)
\(\begin{array}{l}\Rightarrow a_{n+2}a_{n+1}+1 – a_n.a_{n+1}=2\end{array} \)
For n = 1 a3a2 – a1a2 = 2
n = 2 a4a3 – a3a2 = 2
n = 3 a5a4 – a4a3 = 2
:
n = nan + 2an + 1 – anan + 1 = 2 
an + 2 an + 1 = 2n + a1a2
Now,
\(\begin{array}{l}\frac{(a_1a_2+1)}{a_2a_3}.\frac{(a_2a_3+1)}{a_3a_4}.\frac{(a_3a_4+1)}{a_4a_5}.\ … \ . \frac{(a_{30}a_{31}+1)}{a_{31}a_{32}}\end{array} \)
\(\begin{array}{l}=\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.\ … \ . \frac{61}{62}\end{array} \)
\(\begin{array}{l}=2^{-60}\left(^{61}C_{31}\right)\end{array} \)
20. The minimum value of the twice differentiable function \(\begin{array}{l}f(x)=\int_{0}^{x}e^{x-t}f'(t)dt-(x^2-x+1)e^x, x\in \mathbb{R}\end{array} \)
is
\(\begin{array}{l}(\text{A}) \ -\frac{2}{\sqrt{e}}\end{array} \)
\(\begin{array}{l}(\text{B}) \ -2\sqrt{e}\end{array} \)
\(\begin{array}{l}(\text{C}) \ -\sqrt{e}\end{array} \)
\(\begin{array}{l}(\text{D}) \ \frac{2}{\sqrt{e}}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}f(x)=\int_{0}^{x}e^{x-t}f'(t)dt-(x^2-x+1)e^x\end{array} \)
\(\begin{array}{l}f(x)=e^x\int_{0}^{x}e^{-t}f'(t)dt-(x^2-x+1)e^x\end{array} \)
\(\begin{array}{l}e^{-x}f(x)=\int_{0}^{x}e^{-t}f'(t)dt-(x^2-x+1)\end{array} \)
Differentiate on both sides
\(\begin{array}{l}e^{-x}f'(x)+\left(-f(x)e^{-x}\right)=e^{-x}f'(x)-2x+1\end{array} \)
\(\begin{array}{l}f(x)=e^x(2x-1)\end{array} \)
\(\begin{array}{l}f'(x)=e^x(2)+e^x(2x-1)\end{array} \)
\(\begin{array}{l}=e^x(2x+1)\end{array} \)
\(\begin{array}{l}x=-\frac{1}{2}\end{array} \)
\(\begin{array}{l}f”(x)=e^x(2)+(2x+1)e^x\end{array} \)
\(\begin{array}{l}=e^x(2x+3)\end{array} \)
\(\begin{array}{l}\text{For } x= -\frac{1}{2}\ \ f”(x)>0\end{array} \)
\(\begin{array}{l}\Rightarrow {Maxima}\therefore \text{ Max } = e^{-\frac{1}{2}}(-1-1)\end{array} \)
\(\begin{array}{l}\therefore Max.=-\frac{2}{\sqrt{e}}\end{array} \)
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. Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from {A, B, C, D, E} or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from {1, 2, 3, 4, 5} is α × 56, then α is equal to _______.
Answer (7073)
Sol. If password is 6 character long, tehn
Total number of ways having atleast one number = 106 – 56
Similarly, if 7 character long = 107 – 57
and if 8-character long = 108 – 58
\(\begin{array}{l}\text{Number of password = }(10^6 + 10^7 + 10^8) – (5^6 + 5^7 + 5^8) \end{array} \)
\(\begin{array}{l}= 5^6 \left(2^6 + 5.2^7 + 25.2^8 – 1 – 5 – 25\right)\end{array} \)
\(\begin{array}{l}= 56\left(64 + 640 + 6400 – 31\right)\end{array} \)
\(\begin{array}{l}= 7073 \times 5^6\\\therefore \alpha = 7073\end{array} \)
.
2. Let P(–2, –1, 1) and Q(56/17, 43/17, 111/17) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are α, –1, β, where both α and β are integers of minimum absolute values, then α2 + β2 is equal to ___________.
Answer (450)
Sol.
d.r’s of RS = < α, –1, β >
\(\begin{array}{l}\text{d.r’s of }PQ = <\frac{90}{17},\frac{60}{17},\frac{94}{17}>=<45, 30,47>\end{array} \)
as PQ and RS are diagonals of rhombus
\(\begin{array}{l} \alpha\left(45\right) + 30\left(-1\right) + 47\left(\beta\right) = 0\\ \Rightarrow 45\alpha + 47\beta = 30\end{array} \)
\(\begin{array}{l}i.e., \alpha = \frac{30-47\beta}{45}\end{array} \)
for minimum integral value α = – 15 and β = 15
\(\begin{array}{l}\Rightarrow \alpha^2 + \beta^2 = 450.\end{array} \)
3. Let f : [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line
y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1) then the least number of points x ∈ (0, 1) at which f”(x) = 0, is ___________.
Answer (2)
Sol.
If a graph cuts y = 2x + 5 in (0, 1) twice then its concavity changes twice
∴ f”(x) = 0 at atleast two points.
4. If \(\begin{array}{l}\int_{0}^{\sqrt{3}}\frac{15x^3}{\sqrt{1+x^2 + \sqrt{(1+x^2)^3}}}dx = \alpha \sqrt{2}+\beta\sqrt{3},\end{array} \)
where α, β are integers, then α + β is equal to
Answer (10)
Sol. Put
\(\begin{array}{l}x = tan\theta \Rightarrow dx = sec^2\theta~ d\theta\end{array} \)
\(\begin{array}{l}\Rightarrow I = \int_{0}^{\frac{\pi}{3}}\frac{15\tan^3\theta . \sec^2\theta\ d\theta}{\sqrt{1+\tan^2\theta + \sqrt{\sec^6\theta}}}\end{array} \)
\(\begin{array}{l}\Rightarrow I = \int_{0}^{\frac{\pi}{3}}\frac{15\tan^2\theta \sec^2 \theta \ d\theta}{\sec \theta \sqrt{1+ \sec \theta}}\end{array} \)
\(\begin{array}{l}\Rightarrow I = \int_{0}^{\frac{\pi}{3}}\frac{15(\sec^2 \theta -1)\sec \theta \tan \theta \ d\theta}{\left(\sqrt{1+\sec \theta}\right)}\end{array} \)
Now put 1 + secθ = t2
\(\begin{array}{l}\Rightarrow sec~\theta ~tan~\theta~ d\theta = 2tdt\end{array} \)
\(\begin{array}{l}\Rightarrow I = \int_{\sqrt{2}}^{\sqrt{3}}\frac{15((t^2-1)^2-1)2t \ dt}{t}\end{array} \)
\(\begin{array}{l}\Rightarrow I = 30\int_{\sqrt{2}}^{\sqrt{3}}\left(t^4-2t^2 + 1-1\right)dt\end{array} \)
\(\begin{array}{l}\Rightarrow I = 30\int_{\sqrt{2}}^{\sqrt{3}}\left(t^4-2t^2 \right)dt\end{array} \)
\(\begin{array}{l}\Rightarrow I = 30 \left(\frac{t^5}{5}-\frac{2t^3}{3}\right)_{\sqrt{2}}^{\sqrt{3}}\end{array} \)
\(\begin{array}{l}=30\left[\left(\frac{9}{5}\sqrt{3}-2\sqrt{3}\right)-\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\right]\end{array} \)
\(\begin{array}{l}=(54\sqrt{3}-60\sqrt{3})-(24\sqrt{2}-40\sqrt{2})\end{array} \)
\(\begin{array}{l}=16\sqrt{2}-6\sqrt{3}\end{array} \)
\(\begin{array}{l} \therefore \alpha = 16 ~\text{and}~ \beta = – 6\\ \alpha + \beta = 10.\end{array} \)
5. Let \(\begin{array}{l}A=\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}\ \text{and}\ B=\begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix},\ alpha, \beta \in R.\end{array} \)
Let α1 be the value of α which satisfies \(\begin{array}{l}(A + B)^2 =A^2 + \begin{bmatrix}2 & 2 \\2 & 2 \\\end{bmatrix}\end{array} \)
and α2 be the value of α which satisfies \(\begin{array}{l}\left(A + B\right)^2 = B^2.\end{array} \)
Then |α1 – α2| is equal to _________.
Answer (2)
Sol.
\(\begin{array}{l}\left(A + B\right)^2 = A^2 + B^2 + AB + BA\end{array} \)
\(\begin{array}{l}=A^2 + \begin{bmatrix}2 & 2 \\2 & 2 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}\therefore B^2 + AB + BA =\begin{bmatrix}2 & 2 \\2 & 2 \\\end{bmatrix}\ \ \ …(1)\end{array} \)
\(\begin{array}{l}AB =\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}\begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}= \begin{bmatrix}\beta-1 & 1 \\\alpha + 2\beta & 2 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}BA = \begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}= \begin{bmatrix}\beta+2 & \alpha – \beta \\1 & -1 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}B^2 = \begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}\begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}= \begin{bmatrix}\beta^2 + 1 & \beta \\\beta & 1 \\\end{bmatrix}\end{array} \)
By (1) we get
\(\begin{array}{l}\begin{bmatrix} \beta^2 + 2\beta + 2& \alpha + 1 \\\alpha + 3\beta + 1 & 2 \\\end{bmatrix}=\begin{bmatrix}2 & 2 \\2 & 2 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}\therefore \alpha = 1, \beta = 0 , \Rightarrow \alpha_ 1 = 1\end{array} \)
Similarly If A2 + AB + BA = 0 then
\(\begin{array}{l}\left(A^2 = \begin{bmatrix} 1& -1 \\2 & \alpha \\\end{bmatrix}\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}=\begin{bmatrix}-1 & -1-\alpha \\2+2\alpha & \alpha^2-2 \\\end{bmatrix}\right)\end{array} \)
\(\begin{array}{l}\begin{bmatrix} 2\beta& \alpha – \beta + 1 -1 -\alpha \\\alpha + 2\beta + 1 +2 + 2\alpha & \alpha^2-2+1 \\\end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}\Rightarrow \beta = 0 ~\text{and}~ \alpha = – 1 \Rightarrow \alpha_2 = – 1\end{array} \)
\(\begin{array}{l}\therefore |\alpha_1 – \alpha_2|=|2|=2\end{array} \)
6. For p, q, ∈ R, consider the real valued function f(x) = (x – p)2 – q, x ∈ R and q > 0, Let a1, a2, a3 and a4 be in an arithmetic progression with mean p and positive common difference. If |f(ai)| = 500 for all i = 1, 2, 3, 4, then the absolute difference between the roots of f(x) = 0 is
Answer (50)
Sol. ∵ a1, a2, a3, a4 are in A.P and its mean is p.
∴ a1 = p – 3d, a2 = p –d, a3 = p + d and a4 = p + 3d
Where d > 0
\(\begin{array}{l}\because |f(a_i)|=500\end{array} \)
\(\begin{array}{l}\Rightarrow |9d^2-q|=500\end{array} \)
\(\begin{array}{l}\text{and }|d^2-q|=500\ \ …(i)\end{array} \)
either 9d2 – q = d2 – q
\(\begin{array}{l}\Rightarrow d = 0 ~\text{not acceptable}\\ \therefore 9d^2 – q = q – d^2\end{array} \)
\(\begin{array}{l}\therefore 5d^2 – q = 0 …….\left(ii\right)\end{array} \)
\(\begin{array}{l}\text{Roots of} f\left(x\right) = 0 ~are~ p+\sqrt{q} \text{ and } p-\sqrt{q}\end{array} \)
\(\begin{array}{l}\therefore \text{absolute difference between roots}=\left|2\sqrt{q}\right|= 50 \end{array} \)
7. For the hyperbola H: x2 – y2 = 1 and the ellipse \(\begin{array}{l}E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b > 0,\ \text{let the}\end{array} \)
(1) eccentricity of E be reciprocal of the eccentricity of H, and
(2) the line y = √(5/2) x + K be a common tangent of E and H.
Then 4(a2 + b2) is equal to _______.
Answer (03.00)
Sol. The equation of tangent to hyperbola x2 – y2 = 1 within slope m is equal to
\(\begin{array}{l}y=mx \pm \sqrt{m^2+1}\ …(i)\end{array} \)
And for same slope m, equation of tangent to ellipse
\(\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\end{array} \)
is \(\begin{array}{l}y=mx \pm \sqrt{a^2m^2+b^2}\ …(ii)\end{array} \)
∵ Equation (i) and (ii) are identical
∴ a2m2 + b2 = m2 – 1
\(\begin{array}{l}\therefore m^2 = \frac{1+b^2}{1-a^2}\end{array} \)
But equation of common tangent is
\(\begin{array}{l}y=\sqrt{\frac{5}{2}}x+K\end{array} \)
\(\begin{array}{l}\therefore m = \sqrt{\frac{5}{2}}\Rightarrow \frac{5}{2}=\frac{1+b^2}{1-a^2}\end{array} \)
∴ 5a2 + 2b2 = 3 …(i)
\(\begin{array}{l} \text{Eccentricity of ellipse} =\frac{1}{\sqrt{2}}\end{array} \)
\(\begin{array}{l}\therefore 1-\frac{b^2}{a^2}=\frac{1}{2}\end{array} \)
⇒a2 = 2b2 …(ii)
From equation (i) and (ii):
\(\begin{array}{l}a^2=\frac{1}{2}, b^2 = \frac{1}{4}\end{array} \)
∴ 4(a2 + b2) = 3
8. Let x1, x2, x3, …, x20 be in geometric progression with x1 = 3 and the common ratio 1/2. A new data is constructed replacing each xi by (xi – i)2. If x̄ is the mean of new data, then the greatest integer less than or equal to x̄ is _________.
Answer (142)
Sol. x1, x2, x3, …, x20 are in G.P.
\(\begin{array}{l}x_1=3, r=\frac{1}{2}\end{array} \)
\(\begin{array}{l}\bar{x}=\frac{\sum x_1^2-2x_ii + i^2}{20}\end{array} \)
\(\begin{array}{l}=\frac{1}{20}\left[12\left(1-\frac{1}{2^{40}}\right)-6\left(4-\frac{11}{2^{18}}\right) + 70 \times 41\right] \end{array} \)
\(\begin{array}{l}\left\{\begin{matrix}S=1+2.\frac{1}{2}+3.\frac{1}{2^2}+… \\\frac{S}{2} =\frac{1}{2}+ \frac{2}{2^2}+…\end{matrix}\right.\end{array} \)
\(\begin{array}{l}\frac{S}{2}=2\left(1-\frac{1}{2^{20}}\right)\left.-\frac{20}{2^{20} }= 4 – \frac{11}{2^{18}}\right\}\end{array} \)
\(\begin{array}{l}\therefore[\bar{x}]=\left[\frac{2858}{20}-\left(\frac{12}{240}-\frac{66}{2^{18}}\right).\frac{1}{20}\right] = 142 \end{array} \)
9. \(\begin{array}{l}\displaystyle \lim_{ x\to 0}\left(\frac{(x+2\cos x)^3 + 2(x+2\cos x)^2 + 3 \sin(x+2\cos x)}{(x+2)^3+2(x+2)^2+3\sin(x+2)}\right)^{\frac{100}{x}} \end{array} \)
is equal to _________.
Answer (01)
Sol. Let x + 2cosx = a
x + 2 = b
as x → 0, a → 2 and b → 2
\(\begin{array}{l}\displaystyle \lim_{ x\to 0}\left(\frac{a^3+2a^2+3\sin a}{b^3 + 2b^2 + 3\sin b}\right)^{\frac{100}{x}} \end{array} \)
\(\begin{array}{l}=e^{\displaystyle \lim_{x \to 0}\frac{100}{x}.\frac{(a^3-b^3)+2(a^2-b^2)+3(\sin a – \sin b)}{b^3+2b^2 + 3\sin b}}\end{array} \)
\(\begin{array}{l}\because \displaystyle \lim_{x\to 0}\frac{a-b}{x}=\displaystyle \lim_{x\to 0}\frac{2(\cos x -1)}{x}=0\end{array} \)
= e0
= 1
10. The sum of all real value of x for which \(\begin{array}{l}\frac{3x^2-9x+17}{x^2 +3x +10}=\frac{5x^2 -7x + 19}{3x^2 +5x + 12}\end{array} \)
is equal to ________.
Answer (06)
Sol.
\(\begin{array}{l}\frac{3x^2-9x+17}{x^2 +3x +10}=\frac{5x^2 -7x + 19}{3x^2 +5x + 12}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{3x^2-9x+17}{5x^2 -7x +19}=\frac{x^2 +3x + 10}{3x^2 +5x + 12}\end{array} \)
\(\begin{array}{l}\frac{-2x^2-2x-2}{5x^2 – 7x + 19}=\frac{-2x^2-2x-2}{3x^2+5x+12}\end{array} \)
Either x2 + x + 1 = 0 or 5x2 – 7x + 19 = 3x2 + 5x + 12 ⇒ 2x2 – 12x + 7 = 0 ⇒ sum of roots = 6
For x2 + x + 1 = 0, no real roots.
JEE Main 2022 July 28th Shift 1 Paper Analysis
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