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JEE Advanced · Mathematics · Trigonometry

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1
Considering only the principal values of the inverse trigonometric functions, the value of cot⁡−1(cot⁡(−11))+10sin⁡ ⁣(2cos⁡−1 ⁣(12))+10sin⁡(2tan⁡−1(2))\cot^{-1}(\cot(-11)) + 10\sin\!\left(2\cos^{-1}\!\left(\dfrac{1}{\sqrt2}\right)\right) + 10\sin(2\tan^{-1}(2)) is
Not attempted
A
3π+73\pi+7
B
77
C
4π+74\pi+7
D
3π−53\pi-5
2
α=(1−2cos⁡π11)(1−2cos⁡3π11)(1−2cos⁡9π11)(1−2cos⁡27π11)(1−2cos⁡81π11).\alpha = \left(1-2\cos\dfrac{\pi}{11}\right)\left(1-2\cos\dfrac{3\pi}{11}\right)\left(1-2\cos\dfrac{9\pi}{11}\right)\left(1-2\cos\dfrac{27\pi}{11}\right)\left(1-2\cos\dfrac{81\pi}{11}\right). Then the value of 5−α25-\alpha^2 is
Not attempted
Your answer: —
Correct answer: 4
3
Match each entry in List-I to the correct entry in List-II and choose the correct option. List-I\textbf{List-I} (P) The number of elements in {x∈[−π,π]:sin⁡6x+cos⁡4x=1}\{x\in[-\pi,\pi]: \sin^6x+\cos^4x=1\} (Q) The number of elements in {x∈[−π/2,π/2]:sin⁡2x+cos⁡6x=1}\{x\in[-\pi/2,\pi/2]: \sin^2x+\cos^6x=1\} (R) The number of elements in {x∈[−π,π]:cos⁡2(x/2)−sin⁡2x=1/2}\{x\in[-\pi,\pi]: \cos^2(x/2)-\sin^2x=1/2\} (S) The number of elements in {x∈[−2π,2π]:6sin⁡2(x/2)−cos⁡3x=3}\{x\in[-2\pi,2\pi]: 6\sin^2(x/2)-\cos3x=3\} List-II\textbf{List-II} (1) is 1 (2) is 2 (3) is 3 (4) is 4 (5) is 5
Not attempted
A
P→2,Q→5,R→3,S→4P\to2,Q\to5,R\to3,S\to4
B
P→5,Q→3,R→2,S→4P\to5,Q\to3,R\to2,S\to4
C
P→5,Q→4,R→1,S→3P\to5,Q\to4,R\to1,S\to3
D
P→4,Q→3,R→2,S→5P\to4,Q\to3,R\to2,S\to5
4
The total number of real solutions of the equation θ=tan⁡−1(2tan⁡θ)−12sin⁡−1(6tan⁡θ9+tan⁡2θ)\theta = \tan^{-1}(2\tan\theta) - \dfrac12\sin^{-1}\left(\dfrac{6\tan\theta}{9+\tan^2\theta}\right) is (Here, the inverse trigonometric functions sin⁡−1x\sin^{-1}x and tan⁡−1x\tan^{-1}x assume values in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] and (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right), respectively.)
Not attempted
A
1
B
2
C
3
D
5
5
Let α=1sin⁡60∘sin⁡61∘+1sin⁡62∘sin⁡63∘+⋯+1sin⁡118∘sin⁡119∘\alpha = \dfrac{1}{\sin60^\circ\sin61^\circ} + \dfrac{1}{\sin62^\circ\sin63^\circ} + \cdots + \dfrac{1}{\sin118^\circ\sin119^\circ}. Then the value of (cosec 1∘α)2\left(\dfrac{\text{cosec}\,1^\circ}{\alpha}\right)^2 is ______________.
Not attempted
Your answer: —
Correct answer: 3
6
Considering only the principal values of the inverse trigonometric functions, the value of tan⁡(sin⁡−1(35)−2cos⁡−1(25))\tan\left(\sin^{-1}\left(\dfrac{3}{5}\right) - 2\cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)\right) is
Not attempted
A
724\dfrac{7}{24}
B
−724-\dfrac{7}{24}
C
−524-\dfrac{5}{24}
D
524\dfrac{5}{24}
7
Let π2<x<π\dfrac{\pi}{2}<x<\pi be such that cot⁡x=−511\cot x=\dfrac{-5}{\sqrt{11}}. Then (sin⁡11x2)(sin⁡6x−cos⁡6x)+(cos⁡11x2)(sin⁡6x+cos⁡6x)\left(\sin\dfrac{11x}{2}\right)\left(\sin6x-\cos6x\right)+\left(\cos\dfrac{11x}{2}\right)\left(\sin6x+\cos6x\right) is equal to
Not attempted
A
11−123\dfrac{\sqrt{11}-1}{2\sqrt3}
B
11+123\dfrac{\sqrt{11}+1}{2\sqrt3}
C
11+132\dfrac{\sqrt{11}+1}{3\sqrt2}
D
11−132\dfrac{\sqrt{11}-1}{3\sqrt2}
8
Consider an obtuse-angled triangle ABCABC in which the difference between the largest and the smallest angle is π2\dfrac{\pi}{2} and whose sides are in arithmetic progression. Suppose the vertices of this triangle lie on a circle of radius 11. Let aa denote the area of the triangle ABCABC. Then the value of (64a)2(64a)^2 is
Not attempted
Your answer: —
Correct answer: 1008
9
Consider an obtuse-angled triangle ABCABC in which the difference between the largest and the smallest angle is π2\dfrac{\pi}{2} and whose sides are in arithmetic progression. Suppose the vertices of this triangle lie on a circle of radius 11. Then the inradius of the triangle ABCABC is
Not attempted
Your answer: —
Correct answer: 0.25
10
For any y∈Ry\in\mathbb{R}, let cot⁡−1(y)∈(0,π)\cot^{-1}(y)\in(0,\pi) and tan⁡−1(y)∈(−π2,π2)\tan^{-1}(y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right). Then the sum of all the solutions of the equation tan⁡−1 ⁣(6y9−y2)+cot⁡−1 ⁣(9−y26y)=2π3\tan^{-1}\!\left(\dfrac{6y}{9-y^2}\right)+\cot^{-1}\!\left(\dfrac{9-y^2}{6y}\right)=\dfrac{2\pi}{3} for 0<∣y∣<30<|y|<3 is equal to
Not attempted
A
23−32\sqrt3-3
B
3−233-2\sqrt3
C
43−64\sqrt3-6
D
6−436-4\sqrt3
11
Let tan⁡−1(x)∈(−π2,π2)\tan^{-1}(x)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right), for x∈Rx\in\mathbb{R}. Then the number of real solutions of the equation 1+cos⁡(2x)=2 tan⁡−1(tan⁡x)\sqrt{1+\cos(2x)}=\sqrt2\,\tan^{-1}(\tan x) in the set (−3π2,−π2)∪(−π2,π2)∪(π2,3π2)\left(-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right)\cup\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right) is equal to
Not attempted
Your answer: —
Correct answer: 3
12
Let PQRSPQRS be a quadrilateral in a plane, where QR=1QR=1, ∠PQR=∠QRS=70∘\angle PQR=\angle QRS=70^\circ, ∠PQS=15∘\angle PQS=15^\circ and ∠PRS=40∘\angle PRS=40^\circ. If ∠RPS=θ∘\angle RPS=\theta^\circ, PQ=αPQ=\alpha and PS=βPS=\beta, then the interval(s) that contain(s) the value of 4αβsin⁡θ∘4\alpha\beta\sin\theta^\circ is/are
Not attempted
A
(0,2)(0,\sqrt2)
B
(1,2)(1,2)
C
(2,3)(\sqrt2,3)
D
(22,32)(2\sqrt2,3\sqrt2)
13
Let α\alpha and β\beta be real numbers such that −π4<β<0<α<π4-\dfrac{\pi}{4} < \beta < 0 < \alpha < \dfrac{\pi}{4}. If sin⁡(α+β)=13\sin(\alpha+\beta) = \dfrac{1}{3} and cos⁡(α−β)=23\cos(\alpha-\beta) = \dfrac{2}{3}, then the greatest integer less than or equal to (sin⁡αcos⁡β+cos⁡βsin⁡α+cos⁡αsin⁡β+sin⁡βcos⁡α)2\left(\dfrac{\sin\alpha}{\cos\beta} + \dfrac{\cos\beta}{\sin\alpha} + \dfrac{\cos\alpha}{\sin\beta} + \dfrac{\sin\beta}{\cos\alpha}\right)^2 is _______.
Not attempted
Your answer: —
Correct answer: 1
14
Consider the following lists: List-I (I) {x∈[−2π3,2π3]:cos⁡x+sin⁡x=1}\left\{x\in\left[-\dfrac{2\pi}{3},\dfrac{2\pi}{3}\right]: \cos x+\sin x=1\right\} (II) {x∈[−5π18,5π18]:3tan⁡3x=1}\left\{x\in\left[-\dfrac{5\pi}{18},\dfrac{5\pi}{18}\right]: \sqrt3\tan3x=1\right\} (III) {x∈[−6π5,6π5]:2cos⁡(2x)=3}\left\{x\in\left[-\dfrac{6\pi}{5},\dfrac{6\pi}{5}\right]: 2\cos(2x)=\sqrt3\right\} (IV) {x∈[−7π4,7π4]:sin⁡x−cos⁡x=1}\left\{x\in\left[-\dfrac{7\pi}{4},\dfrac{7\pi}{4}\right]: \sin x-\cos x=1\right\} List-II (P) has two elements (Q) has three elements (R) has four elements (S) has five elements (T) has six elements The correct option is:
Not attempted
A
(I)→(P); (II)→(S); (III)→(P); (IV)→(S)(I)\to(P);\ (II)\to(S);\ (III)\to(P);\ (IV)\to(S)
B
(I)→(P); (II)→(P); (III)→(T); (IV)→(R)(I)\to(P);\ (II)\to(P);\ (III)\to(T);\ (IV)\to(R)
C
(I)→(Q); (II)→(P); (III)→(T); (IV)→(S)(I)\to(Q);\ (II)\to(P);\ (III)\to(T);\ (IV)\to(S)
D
(I)→(Q); (II)→(S); (III)→(P); (IV)→(R)(I)\to(Q);\ (II)\to(S);\ (III)\to(P);\ (IV)\to(R)
15
Considering only the principal values of the inverse trigonometric functions, the value of 32cos⁡−122+π2+14sin⁡−122 π2+π2+tan⁡−12π\dfrac{3}{2}\cos^{-1}\sqrt{\dfrac{2}{2+\pi^2}} + \dfrac{1}{4}\sin^{-1}\dfrac{2\sqrt{2}\,\pi}{2+\pi^2} + \tan^{-1}\dfrac{\sqrt{2}}{\pi} is __________.
Not attempted
Your answer: —
Correct answer: 2.36
16
Consider a triangle PQRPQR having sides of lengths p,qp, q and rr opposite to the angles P,QP, Q and RR, respectively. Then which of the following statements is (are) TRUE?
Not attempted
A
cos⁡P≥1−p22qr\cos P \ge 1 - \frac{p^2}{2qr}
B
cos⁡R≥(q−rp+q)cos⁡P+(p−rp+q)cos⁡Q\cos R \ge \left(\frac{q-r}{p+q}\right)\cos P + \left(\frac{p-r}{p+q}\right)\cos Q
C
q+rp<2sin⁡Qsin⁡Rsin⁡P\frac{q+r}{p} < 2\frac{\sqrt{\sin Q \sin R}}{\sin P}
D
If p<qp<q and p<rp<r, then cos⁡Q>pr\cos Q > \frac{p}{r} and cos⁡R>pq\cos R > \frac{p}{q}
17
In a triangle ABCABC, let AB=23AB=\sqrt{23}, BC=3BC=3 and CA=4CA=4. Then the value of cot⁡A+cot⁡Ccot⁡B\frac{\cot A+\cot C}{\cot B} is ____.
Not attempted
Your answer: —
Correct answer: 2
18
For any positive integer nn, let Sn:(0,∞)→RS_n:(0,\infty)\to\mathbb{R} be defined by Sn(x)=∑k=1ncot⁡−1(1+k(k+1)x2x)S_n(x)=\sum_{k=1}^{n}\cot^{-1}\left(\frac{1+k(k+1)x^2}{x}\right), where for any x∈Rx\in\mathbb{R}, cot⁡−1(x)∈(0,π)\cot^{-1}(x)\in(0,\pi) and tan⁡−1(x)∈(−π2,π2)\tan^{-1}(x)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). Then which of the following statements is (are) TRUE?
Not attempted
A
S10(x)=π2−tan⁡−1(1+11x210x)S_{10}(x)=\frac{\pi}{2}-\tan^{-1}\left(\frac{1+11x^2}{10x}\right), for all x>0x>0
B
lim⁡n→∞cot⁡(Sn(x))=x\lim\limits_{n\to\infty}\cot(S_n(x))=x, for all x>0x>0
C
The equation S3(x)=π4S_3(x)=\frac{\pi}{4} has a root in (0,∞)(0,\infty)
D
tan⁡(Sn(x))≤12\tan(S_n(x))\le\frac{1}{2}, for all n≥1n\ge1 and x>0x>0
19
Let f:[0,2]→Rf: [0, 2] \to \mathbb{R} be the function defined by f(x)=(3−sin⁡(2πx))sin⁡(πx−π4)−sin⁡(3πx+π4)f(x) = (3 - \sin(2\pi x))\sin\left(\pi x - \frac{\pi}{4}\right) - \sin\left(3\pi x + \frac{\pi}{4}\right). If α,β∈[0,2]\alpha, \beta \in [0, 2] are such that {x∈[0,2]:f(x)≥0}=[α,β]\{x \in [0,2] : f(x) \geq 0\} = [\alpha, \beta], then the value of β−α\beta - \alpha is
Not attempted
Your answer: —
Correct answer: 1
20
Let x,yx, y and zz be positive real numbers. Suppose x,yx, y and zz are the lengths of the sides of a triangle opposite to its angles X,YX, Y and ZZ, respectively. If tan⁡X2+tan⁡Z2=2yx+y+z\tan\frac{X}{2} + \tan\frac{Z}{2} = \frac{2y}{x+y+z}, then which of the following statements is/are TRUE?
Not attempted
A
2Y=X+Z2Y = X+Z
B
Y=X+ZY = X+Z
C
tan⁡X2=xy+z\tan\frac{X}{2} = \frac{x}{y+z}
D
x2+z2−y2=xzx^2+z^2-y^2 = xz
21
Let f(x)=sin⁡(πcos⁡x)f(x)=\sin(\pi\cos x) and g(x)=cos⁡(2πsin⁡x)g(x)=\cos(2\pi\sin x) be two functions defined for x>0x>0. Define the following sets, whose elements are written in increasing order: X={x:f(x)=0}X=\{x: f(x)=0\}, Y={x:f′(x)=0}Y=\{x: f'(x)=0\}, Z={x:g(x)=0}Z=\{x: g(x)=0\}, W={x:g′(x)=0}W=\{x: g'(x)=0\}. List-I contains the sets X,Y,Z,WX,Y,Z,W and List-II contains some information regarding these sets. List-I: (I) XX (II) YY (III) ZZ (IV) WW List-II: (P) ⊇{π2,3π2,4π,7π}\supseteq\left\{\dfrac{\pi}{2},\dfrac{3\pi}{2},4\pi,7\pi\right\} (Q) an arithmetic progression (R) NOT an arithmetic progression (S) ⊇{π6,7π6,13π6}\supseteq\left\{\dfrac{\pi}{6},\dfrac{7\pi}{6},\dfrac{13\pi}{6}\right\} (T) ⊇{π3,2π3,π}\supseteq\left\{\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi\right\} (U) ⊇{π6,3π4}\supseteq\left\{\dfrac{\pi}{6},\dfrac{3\pi}{4}\right\} Which of the following is the only CORRECT combination?
Not attempted
A
(II), (R), (S)
B
(I), (P), (R)
C
(II), (Q), (T)
D
(I), (Q), (U)
22
Let f(x)=sin⁡(πcos⁡x)f(x)=\sin(\pi\cos x) and g(x)=cos⁡(2πsin⁡x)g(x)=\cos(2\pi\sin x) be two functions defined for x>0x>0. Define the following sets, whose elements are written in increasing order: X={x:f(x)=0}X=\{x: f(x)=0\}, Y={x:f′(x)=0}Y=\{x: f'(x)=0\}, Z={x:g(x)=0}Z=\{x: g(x)=0\}, W={x:g′(x)=0}W=\{x: g'(x)=0\}. List-I contains the sets X,Y,Z,WX,Y,Z,W and List-II contains some information regarding these sets. List-I: (I) XX (II) YY (III) ZZ (IV) WW List-II: (P) ⊇{π2,3π2,4π,7π}\supseteq\left\{\dfrac{\pi}{2},\dfrac{3\pi}{2},4\pi,7\pi\right\} (Q) an arithmetic progression (R) NOT an arithmetic progression (S) ⊇{π6,7π6,13π6}\supseteq\left\{\dfrac{\pi}{6},\dfrac{7\pi}{6},\dfrac{13\pi}{6}\right\} (T) ⊇{π3,2π3,π}\supseteq\left\{\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi\right\} (U) ⊇{π6,3π4}\supseteq\left\{\dfrac{\pi}{6},\dfrac{3\pi}{4}\right\} Which of the following is the only CORRECT combination?
Not attempted
A
(IV), (Q), (T)
B
(IV), (P), (R), (S)
C
(III), (R), (U)
D
(III), (P), (Q), (U)
23
The value of sec⁡−1(14∑k=010sec⁡ ⁣(7π12+kπ2)sec⁡ ⁣(7π12+(k+1)π2))\sec^{-1}\left(\dfrac14\displaystyle\sum_{k=0}^{10}\sec\!\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\!\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right) in the interval [−π4,3π4]\left[-\dfrac{\pi}{4},\dfrac{3\pi}{4}\right] equals ____
Not attempted
Your answer: —
Correct answer: 0
24
For non-negative integers nn, let f(n)=∑k=0nsin⁡ ⁣(k+1n+2π)sin⁡ ⁣(k+2n+2π)∑k=0nsin⁡2 ⁣(k+1n+2π).f(n)=\dfrac{\displaystyle\sum_{k=0}^{n}\sin\!\left(\dfrac{k+1}{n+2}\pi\right)\sin\!\left(\dfrac{k+2}{n+2}\pi\right)}{\displaystyle\sum_{k=0}^{n}\sin^2\!\left(\dfrac{k+1}{n+2}\pi\right)}. Assuming cos⁡−1x\cos^{-1}x takes values in [0,π][0,\pi], which of the following options is/are correct?
Not attempted
A
sin⁡(7cos⁡−1f(5))=0\sin(7\cos^{-1}f(5))=0
B
f(4)=32f(4)=\dfrac{\sqrt3}{2}
C
lim⁡n→∞f(n)=12\lim_{n\to\infty} f(n) = \dfrac12
D
If α=tan⁡(cos⁡−1f(6))\alpha=\tan(\cos^{-1}f(6)), then α2+2α−1=0\alpha^2+2\alpha-1=0
25
In a non-right-angled triangle △PQR\triangle PQR, let p,q,rp,q,r denote the lengths of the sides opposite to the angles at P,Q,RP,Q,R respectively. The median from RR meets the side PQPQ at SS, the perpendicular from PP meets the side QRQR at EE, and RSRS and PEPE intersect at OO. If p=3p=\sqrt3, q=1q=1, and the radius of the circumcircle of the △PQR\triangle PQR equals 11, then which of the following options is/are correct?
Not attempted
A
Area of △SOE=312\triangle SOE=\dfrac{\sqrt3}{12}
B
Radius of incircle of △PQR=32(2−3)\triangle PQR=\dfrac{\sqrt3}{2}(2-\sqrt3)
C
Length of RS=72RS=\dfrac{\sqrt7}{2}
D
Length of OE=16OE=\dfrac16
26
For any positive integer nn, define fn:(0,∞)→Rf_n: (0, \infty) \to \mathbb{R} as fn(x)=∑j=1ntan⁡−1(11+(x+j)(x+j−1))f_n(x) = \sum_{j=1}^{n} \tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) for all x∈(0,∞)x \in (0, \infty). (Here, the inverse trigonometric function tan⁡−1x\tan^{-1}x assumes values in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).) Then, which of the following statement(s) is (are) TRUE?
Not attempted
A
∑j=15tan⁡2(fj(0))=55\sum_{j=1}^{5} \tan^2(f_j(0)) = 55
B
∑j=110(1+fj′(0))sec⁡2(fj(0))=10\sum_{j=1}^{10} \left(1+f_j'(0)\right)\sec^2(f_j(0)) = 10
C
For any fixed positive integer nn, lim⁡x→∞tan⁡(fn(x))=1n\lim_{x\to\infty}\tan(f_n(x)) = \frac{1}{n}
D
For any fixed positive integer nn, lim⁡x→∞sec⁡2(fn(x))=1\lim_{x\to\infty}\sec^2(f_n(x)) = 1
27
Let a,b,ca,b,c be three non-zero real numbers such that the equation 3acos⁡x+2bsin⁡x=c\sqrt3 a\cos x + 2b\sin x = c, x∈[−π2,π2]x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], has two distinct real roots α\alpha and β\beta with α+β=π3\alpha+\beta=\dfrac{\pi}{3}. Then, the value of ba\dfrac ba is ______.
Not attempted
Your answer: —
Correct answer: 0.5
28
The number of real solutions of the equation sin⁡−1(∑i=1∞xi+1−x∑i=1∞(x2)i)=π2−cos⁡−1(∑i=1∞(−x2)i−∑i=1∞(−x)i)\sin^{-1}\left(\displaystyle\sum_{i=1}^{\infty}x^{i+1} - x\sum_{i=1}^{\infty}\left(\dfrac{x}{2}\right)^{i}\right) = \dfrac{\pi}{2} - \cos^{-1}\left(\displaystyle\sum_{i=1}^{\infty}\left(-\dfrac{x}{2}\right)^{i} - \sum_{i=1}^{\infty}(-x)^{i}\right) lying in the interval (−12,12)\left(-\dfrac12,\dfrac12\right) is ______. (Here, sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x assume values in [−π2,π2]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] and [0,π][0,\pi], respectively.)
Not attempted
Your answer: —
Correct answer: 2
29
In a triangle PQRPQR, let ∠PQR=30∘\angle PQR = 30^\circ and the sides PQPQ and QRQR have lengths 10310\sqrt{3} and 1010, respectively. Then, which of the following statement(s) is (are) TRUE?
Not attempted
A
∠QPR=45∘\angle QPR = 45^\circ
B
The area of the triangle PQRPQR is 25325\sqrt{3} and ∠QRP=120∘\angle QRP = 120^\circ
C
The radius of the incircle of the triangle PQRPQR is 103−1510\sqrt{3}-15
D
The area of the circumcircle of the triangle PQRPQR is 100π100\pi
30
Let α\alpha and β\beta be nonzero real numbers such that 2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=12(\cos\beta - \cos\alpha) + \cos\alpha\cos\beta = 1. Then which of the following is/are true?
Not attempted
A
tan⁡(α2)+3tan⁡(β2)=0\tan\left(\frac{\alpha}{2}\right) + \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0
B
3tan⁡(α2)+tan⁡(β2)=0\sqrt{3}\tan\left(\frac{\alpha}{2}\right) + \tan\left(\frac{\beta}{2}\right) = 0
C
tan⁡(α2)−3tan⁡(β2)=0\tan\left(\frac{\alpha}{2}\right) - \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0
D
3tan⁡(α2)−tan⁡(β2)=0\sqrt{3}\tan\left(\frac{\alpha}{2}\right) - \tan\left(\frac{\beta}{2}\right) = 0
31
The value of ∑k=1131sin⁡(π4+(k−1)π6)sin⁡(π4+kπ6)\displaystyle\sum_{k=1}^{13}\dfrac{1}{\sin\left(\dfrac{\pi}{4}+\dfrac{(k-1)\pi}{6}\right)\sin\left(\dfrac{\pi}{4}+\dfrac{k\pi}{6}\right)} is equal to
Not attempted
A
3−33-\sqrt3
B
2(3−3)2(3-\sqrt3)
C
2(3−1)2(\sqrt3-1)
D
2(2+3)2(2+\sqrt3)
32
In a triangle XYZXYZ, let x,y,zx,y,z be the lengths of sides opposite to the angles X,Y,ZX,Y,Z, respectively, and 2s=x+y+z2s=x+y+z. If s−x4=s−y3=s−z2\dfrac{s-x}{4}=\dfrac{s-y}{3}=\dfrac{s-z}{2} and area of incircle of the triangle XYZXYZ is 8π3\dfrac{8\pi}{3}, then
Not attempted
A
area of the triangle XYZXYZ is 666\sqrt6
B
the radius of circumcircle of the triangle XYZXYZ is 3566\dfrac{35}{6}\sqrt6
C
sin⁡X2sin⁡Y2sin⁡Z2=435\sin\dfrac X2\sin\dfrac Y2\sin\dfrac Z2=\dfrac{4}{35}
D
sin⁡2 ⁣(X+Y2)=35\sin^2\!\left(\dfrac{X+Y}{2}\right)=\dfrac35
33
Let S={x∈(−π,π):x≠0,±π2}S=\left\{x\in(-\pi,\pi):x\ne0,\pm\dfrac{\pi}{2}\right\}. The sum of all distinct solutions of the equation 3sec⁡x+cosec⁡x+2(tan⁡x−cot⁡x)=0\sqrt3\sec x+\operatorname{cosec}x+2(\tan x-\cot x)=0 in the set SS is equal to
Not attempted
A
−7π9-\dfrac{7\pi}{9}
B
−2π9-\dfrac{2\pi}{9}
C
0
D
5π9\dfrac{5\pi}{9}
34
Let −π6<θ<−π12-\dfrac{\pi}{6} < \theta < -\dfrac{\pi}{12}. Suppose α1\alpha_1 and β1\beta_1 are the roots of the equation x2−2xsec⁡θ+1=0x^2 - 2x\sec\theta + 1 = 0 and α2\alpha_2 and β2\beta_2 are the roots of the equation x2+2xtan⁡θ−1=0x^2 + 2x\tan\theta - 1 = 0. If α1>β1\alpha_1 > \beta_1 and α2>β2\alpha_2 > \beta_2, then α1+β2\alpha_1 + \beta_2 equals
Not attempted
A
2(sec⁡θ−tan⁡θ)2(\sec\theta-\tan\theta)
B
2sec⁡θ2\sec\theta
C
−2tan⁡θ-2\tan\theta
D
00
35
If α=3sin⁡−1(611)\alpha=3\sin^{-1}\left(\dfrac{6}{11}\right) and β=3cos⁡−1(49)\beta=3\cos^{-1}\left(\dfrac{4}{9}\right), where the inverse trigonometric functions take only the principal values, then the correct option(s) is(are)
Not attempted
A
cos⁡β>0\cos\beta>0
B
sin⁡β<0\sin\beta<0
C
cos⁡(α+β)>0\cos(\alpha+\beta)>0
D
cos⁡α<0\cos\alpha<0
36
This question is part of a Column-I to Column-II matching set (JEE Advanced 2015 Section 3, Q.60). Column II lists the values 1,2,3,5,61,2,3,5,6; select every value that correctly matches this Column-I entry. Entry (A)\textbf{Entry (A)}: In a triangle △XYZ\triangle XYZ, let a,b,ca,b,c be the lengths of the sides opposite to the angles X,Y,ZX,Y,Z respectively. If 2(a2−b2)=c22(a^2-b^2)=c^2 and λ=sin⁡(X−Y)sin⁡Z\lambda=\dfrac{\sin(X-Y)}{\sin Z}, then possible values of nn for which cos⁡(nπλ)=0\cos(n\pi\lambda)=0 is (are)
Not attempted
A
11
B
22
C
33
D
55
E
66
37
This question is part of a Column-I to Column-II matching set (JEE Advanced 2015 Section 3, Q.60). Column II lists the values 1,2,3,5,61,2,3,5,6; select every value that correctly matches this Column-I entry. Entry (B)\textbf{Entry (B)}: In a triangle △XYZ\triangle XYZ, let a,b,ca,b,c be the lengths of the sides opposite to the angles X,Y,ZX,Y,Z respectively. If 1+cos⁡2X−2cos⁡2Y=2sin⁡Xsin⁡Y1+\cos2X-2\cos2Y=2\sin X\sin Y, then possible value(s) of ab\dfrac{a}{b} is (are)
Not attempted
A
11
B
22
C
33
D
55
E
66
38
The number of distinct solutions of the equation 54cos⁡22x+cos⁡4x+sin⁡4x+cos⁡6x+sin⁡6x=2\dfrac{5}{4}\cos^2 2x+\cos^4 x+\sin^4 x+\cos^6 x+\sin^6 x=2 in the interval [0,2π][0,2\pi] is
Not attempted
Your answer: —
Correct answer: 8
39
Match each entry of List I\textbf{List I} to its value in List II\textbf{List II}. List I\textbf{List I} P.\textbf{P.} Let y(x)=cos⁡(3cos⁡−1x)y(x)=\cos(3\cos^{-1}x), x∈[−1,1]x\in[-1,1], x≠±32x\ne\pm\dfrac{\sqrt3}{2}. Then 1y(x){(x2−1)d2y(x)dx2+xdy(x)dx}\dfrac{1}{y(x)}\left\{(x^2-1)\dfrac{d^2y(x)}{dx^2}+x\dfrac{dy(x)}{dx}\right\} equals Q.\textbf{Q.} Let A1,A2,…,AnA_1,A_2,\dots,A_n (n>2n>2) be the vertices of a regular polygon of nn sides with centre at the origin. Let a⃗k\vec{a}_k be the position vector of AkA_k. If ∣∑k=1n−1(a⃗k×a⃗k+1)∣=∣∑k=1n−1(a⃗k⋅a⃗k+1)∣\left|\displaystyle\sum_{k=1}^{n-1}(\vec a_k\times\vec a_{k+1})\right|=\left|\displaystyle\sum_{k=1}^{n-1}(\vec a_k\cdot\vec a_{k+1})\right|, the minimum value of nn is R.\textbf{R.} If the normal from the point P(h,1)P(h,1) on the ellipse x26+y23=1\dfrac{x^2}{6}+\dfrac{y^2}{3}=1 is perpendicular to the line x+y=8x+y=8, then the value of hh is S.\textbf{S.} Number of positive solutions satisfying the equation tan⁡−1 ⁣(12x+1)+tan⁡−1 ⁣(14x+1)=tan⁡−1 ⁣(2x2)\tan^{-1}\!\left(\dfrac{1}{2x+1}\right)+\tan^{-1}\!\left(\dfrac{1}{4x+1}\right)=\tan^{-1}\!\left(\dfrac{2}{x^2}\right) is List II\textbf{List II} 1.\textbf{1.} 11 2.\textbf{2.} 22 3.\textbf{3.} 88 4.\textbf{4.} 99 Which of the following is the correct combination?
Not attempted
A
P→4, Q→3, R→2, S→1P{\to}4,\ Q{\to}3,\ R{\to}2,\ S{\to}1
B
P→2, Q→4, R→3, S→1P{\to}2,\ Q{\to}4,\ R{\to}3,\ S{\to}1
C
P→4, Q→3, R→1, S→2P{\to}4,\ Q{\to}3,\ R{\to}1,\ S{\to}2
D
P→2, Q→4, R→1, S→3P{\to}2,\ Q{\to}4,\ R{\to}1,\ S{\to}3
40
For x∈(0,π)x\in(0,\pi), the equation sin⁡x+2sin⁡2x−sin⁡3x=3\sin x+2\sin2x-\sin3x=3 has
Not attempted
A
Infinitely many solutions
B
Three solutions
C
One solution
D
No solution
41
In a triangle the sum of two sides is xx and the product of the same two sides is yy. If x2−c2=yx^2-c^2=y, where cc is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is
Not attempted
A
3y2x(x+c)\dfrac{3y}{2x(x+c)}
B
3y2c(x+c)\dfrac{3y}{2c(x+c)}
C
3y4x(x+c)\dfrac{3y}{4x(x+c)}
D
3y4c(x+c)\dfrac{3y}{4c(x+c)}
42
Let f:[0,4π]→[0,π]f:[0,4\pi]\to[0,\pi] be defined by f(x)=cos⁡−1(cos⁡x)f(x)=\cos^{-1}(\cos x). The number of points x∈[0,4π]x\in[0,4\pi] satisfying the equation f(x)=10−x10f(x)=\dfrac{10-x}{10} is
Not attempted
Your answer: —
Correct answer: 3
43
In a triangle PQRPQR, PP is the largest angle and cos⁡P=13\cos P=\dfrac13. Further the incircle of the triangle touches the sides PQPQ, QRQR and RPRP at NN, LL and MM respectively, such that the lengths of PNPN, QLQL, RMRM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
Not attempted
A
1616
B
1818
C
2424
D
2222
44
Match List I with List II and select the correct answer using the code given below the lists: List I: P. (1y2(cos⁡(tan⁡−1y)+ysin⁡(tan⁡−1y)cot⁡(sin⁡−1y)+tan⁡(sin⁡−1y))2+y4)1/2\left(\dfrac1{y^2}\left(\dfrac{\cos(\tan^{-1}y)+y\sin(\tan^{-1}y)}{\cot(\sin^{-1}y)+\tan(\sin^{-1}y)}\right)^2+y^4\right)^{1/2} takes value Q. If cos⁡x+cos⁡y+cos⁡z=0=sin⁡x+sin⁡y+sin⁡z\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z then a possible value of cos⁡x−y2\cos\dfrac{x-y}{2} is R. If cos⁡(π4−x)cos⁡2x+sin⁡xsin⁡2xsec⁡x=cos⁡xsin⁡2xsec⁡x+cos⁡(π4+x)cos⁡2x\cos\left(\dfrac{\pi}{4}-x\right)\cos2x+\sin x\sin2x\sec x=\cos x\sin2x\sec x+\cos\left(\dfrac{\pi}{4}+x\right)\cos2x then a possible value of sec⁡x\sec x is S. If cot⁡(sin⁡−11−x2)=sin⁡(tan⁡−1(x6))\cot\left(\sin^{-1}\sqrt{1-x^2}\right)=\sin\left(\tan^{-1}(x\sqrt6)\right), x≠0x\neq0, then a possible value of xx is List II: 1. 1253\dfrac12\sqrt{\dfrac53} 2. 2\sqrt2 3. 12\dfrac12 4. 11 Codes:
Not attempted
A
(A) P→4, Q→3, R→1, S→2P\to4,\ Q\to3,\ R\to1,\ S\to2
B
(B) P→4, Q→3, R→2, S→1P\to4,\ Q\to3,\ R\to2,\ S\to1
C
(C) P→3, Q→4, R→2, S→1P\to3,\ Q\to4,\ R\to2,\ S\to1
D
(D) P→3, Q→4, R→1, S→2P\to3,\ Q\to4,\ R\to1,\ S\to2
45
The value of cot⁡(∑n=123cot⁡−1(1+∑k=1n2k))\cot\left(\displaystyle\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right) is
Not attempted
A
2325\dfrac{23}{25}
B
2523\dfrac{25}{23}
C
2324\dfrac{23}{24}
D
2423\dfrac{24}{23}
∑k=1n2k=n(n+1)\displaystyle\sum_{k=1}^n 2k=n(n+1), and cot⁡−1(1+n(n+1))=tan⁡−1(n+1)−tan⁡−1(n)\cot^{-1}(1+n(n+1))=\tan^{-1}(n+1)-\tan^{-1}(n) — a telescoping identity.
Deconstruction\textbf{Deconstruction}: Evaluate the cotangent of a sum of inverse-cotangents whose arguments are built from an inner sum. Strategy\textbf{Strategy}: Simplify the inner sum, recognize the resulting cot⁡−1\cot^{-1} term as a telescoping difference of tan⁡−1\tan^{-1} terms, sum the telescope, then take cot⁡\cot of the result. Execution\textbf{Execution}: ∑k=1n2k=n(n+1)\displaystyle\sum_{k=1}^n 2k=n(n+1), so each term is cot⁡−1(1+n(n+1))=tan⁡−1 ⁣(11+n(n+1))\cot^{-1}\big(1+n(n+1)\big)=\tan^{-1}\!\left(\dfrac{1}{1+n(n+1)}\right). - Since tan⁡−1(n+1)−tan⁡−1(n)=tan⁡−1 ⁣((n+1)−n1+n(n+1))=tan⁡−1 ⁣(11+n(n+1))\tan^{-1}(n+1)-\tan^{-1}(n)=\tan^{-1}\!\left(\dfrac{(n+1)-n}{1+n(n+1)}\right)=\tan^{-1}\!\left(\dfrac{1}{1+n(n+1)}\right), each term telescopes. - ∑n=123cot⁡−1(1+n(n+1))=∑n=123[tan⁡−1(n+1)−tan⁡−1(n)]=tan⁡−1(24)−tan⁡−1(1)\displaystyle\sum_{n=1}^{23}\cot^{-1}(1+n(n+1)) = \sum_{n=1}^{23}\big[\tan^{-1}(n+1)-\tan^{-1}(n)\big] = \tan^{-1}(24)-\tan^{-1}(1). - cot⁡(tan⁡−124−tan⁡−11)=cot⁡(tan⁡−124−11+24⋅1)=cot⁡(tan⁡−12325)=2523\cot\big(\tan^{-1}24-\tan^{-1}1\big) = \cot\left(\tan^{-1}\dfrac{24-1}{1+24\cdot1}\right)=\cot\left(\tan^{-1}\dfrac{23}{25}\right)=\dfrac{25}{23}. Verification\textbf{Verification}: The telescoping sum is a standard, easily re-checked pattern (each internal tan⁡−1\tan^{-1} term cancels except the first and last); re-deriving via the tangent subtraction formula independently confirms tan⁡(tan⁡−124−tan⁡−11)=23/25\tan(\tan^{-1}24-\tan^{-1}1)=23/25, whose reciprocal is 25/2325/23. Final Answer\textbf{Final Answer}: (B) 2523\dfrac{25}{23}