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JEE • MATHEMATICS • TRANSFORMATIONS

Transformations practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 165 questions
31 Easy
31 of 165
Analyzing +4 | -1
If A=tan⁡  6∘  tan⁡  42∘A = \tan \,\,6^\circ \,\,\tan \,\,42^\circ and B=cot⁡  66∘cot⁡  78∘B = \cot \,\,66^\circ \cot \,\,78^\circ, then
View answer and explanation
Correct option: C
A=BA = B
Option C (A=BA = B) is correct because simplifying the expression AB\frac{A}{B} using trigonometric identities results in 1, thus A = B.
32 Medium
32 of 165
Analyzing +4 | -1
If   sin⁡3  x  sin⁡  3x=∑m=0n  cm  cos⁡  mx,   where   c0,  c1  ,  c2,…, cn\;{\sin ^3}\,\,x\,\,\sin \,\,3x = \sum \limits_{m = 0}^n \,\,{c_m}\;\cos \,\,mx,\;\,\mathrm{where}\;\,{c_0},\,\,{c_{1\;}},\,\,{c_2}, \ldots ,\,{c_n} are constants and cn≠0{c_n} \ne 0, then the value of nn is
View answer and explanation
Correct option: B
6\mathrm{6}
Option B (6) is correct because expanding the expression sin⁡3xsin⁡3x\sin^3 x \sin 3x results in a cosine series with the highest multiple of x being 6.
33 Easy
33 of 165
Analyzing +4 | -1
Consider the following statements:

1.If sin⁡A=sin⁡B\sin A = \sin B then we have sin⁡  2A=sin⁡  2B  \sin \,\,2A = \sin \,\,2B\; always

2.The value of   cos  π7  cos⁡  4 π7  cos⁡  5 π7  is   14\mathrm{\;cos}\,\,\frac{\pi }{7}\,\,\cos \,\,\frac{{4\,\pi }}{7}\,\,\cos \,\,\frac{{5\,\pi }}{7}\mathrm{\;is}\,\;\frac{1}{4}

Which of the statements given above is/are correct?
View answer and explanation
Correct option: D
Neither (1)\left( \mathrm{1} \right) nor (2)\left( \mathrm{2} \right)
Option D (Neither (1) nor (2)) is correct because statement 1 is not always true, and the value in statement 2 is 18\frac{1}{8}, not 14\frac{1}{4}.
34 Medium
34 of 165
Analyzing +4 | -1
If   cos A=m  cos⁡  B   and  cot⁡  A+B2=λ  tan⁡  B−A2\mathrm{\;cos}\,A = m\,\,\cos \,\,B\,\;\mathrm{and}\,\,\cot \,\,\frac{{A + B}}{2} = \lambda \,\,\tan \,\,\frac{{B - A}}{2} then λ\lambda is

View answer and explanation
Correct option: C
m+1m−1\frac{{m + 1}}{{m - 1}}
Option C ((m+1)/(m-1)) is correct because by applying componendo and dividendo and trigonometric identities, λ is found to be (m+1)/(m-1).
35 Medium
35 of 165
Analyzing +4 | -1
The value of sin⁡  π14  sin⁡  3 π14  sin⁡  5 π14  sin⁡  7π14  sin⁡  9 π14  sin⁡  11 π14  sin⁡  13  π14\sin \,\,\frac{\pi }{{14}}\,\,\sin \,\,\frac{{3\,\pi }}{{14}}\,\,\sin \,\,\frac{{5\,\pi }}{{14}}\,\,\sin \,\,\frac{{7\pi }}{{14}}\,\,\sin \,\,\frac{{9\,\pi }}{{14}}\,\,\sin \,\,\frac{{11\,\pi }}{{14}}\,\,\sin \,\,\frac{{13\;\pi }}{{14}} is
View answer and explanation
Correct option: B
1 / 64\mathrm{1}\,\mathrm{/}\,\mathrm{64}
Option B (1 \/ 64\mathrm{1}\,{\/}\,\mathrm{64}) is correct because by using trigonometric identities and simplification, the expression reduces to 1/64.
36 Medium
36 of 165
Analyzing +4 | -1
The value of cos⁡  2 π15  cos⁡  4 π15  cos⁡  815  cos⁡  14 π15\cos \,\,\frac{{2\,\pi }}{{15}}\,\,\cos \,\,\frac{{4\,\pi }}{{15}}\,\,\cos \,\,\frac{8}{{15}}\,\,\cos \,\,\frac{{14\,\pi }}{{15}} is
View answer and explanation
Correct option: D
1/161/16
Option D (1/161/16) is correct because by applying the product-to-sum trigonometric identity repeatedly, the expression simplifies to 1/16.
37 Medium
37 of 165
Application +4 | -1
The value of cos⁡  2π15  cos⁡  4π15  cos⁡  8π15  cos⁡  14π15\cos \,\,\frac{{2\pi }}{{15}}\,\,\cos \,\,\frac{{4\pi }}{{15}}\,\,\cos \,\,\frac{{8\pi }}{{15}}\,\,\cos \,\,\frac{{14\pi }}{{15}} is
View answer and explanation
Correct option: A
116\frac{1}{{16}}
Option A (116\frac{1}{{16}}) is correct because by repeatedly using the identity 2sin⁡xcos⁡x=sin⁡2x2 \sin x \cos x = \sin 2x, the expression simplifies to 116\frac{1}{{16}}.
38 Easy
38 of 165
Application +4 | -1
The maximum value of 13 sin⁡  θ−4  cos⁡  θ+7\frac{1}{{3\,\sin \,\,\theta - 4\,\,\cos \,\,\theta + 7}}, is
View answer and explanation
Correct option: A
112\frac{1}{{12}}
Option A (112\frac{1}{{12}}) is correct because the maximum value of the given expression is obtained when the denominator is minimized, which occurs when 3sin⁡θ−4cos⁡θ3\sin \theta - 4\cos \theta is at its minimum value of -5.
39 Easy
39 of 165
Application +4 | -1
The maximum value of 5cos⁡  θ+3cos⁡(θ+π3)+35\cos \,\,\theta + 3\cos \left( {\theta + \frac{\pi }{3}} \right) + 3 is
View answer and explanation
Correct option: C
10
Option C (10) is correct because it is derived by simplifying the expression and finding the maximum value of the resulting trigonometric function.
40 Easy
40 of 165
Analyzing +4 | -1
The value of tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the expression simplifies to 4 using trigonometric identities.
41 Easy
41 of 165
Application +4 | -1
sin⁡  47∘−sin⁡  25∘+sin⁡  61∘−sin⁡  11∘=\sin \,\,47^\circ - \sin \,\,25^\circ + \sin \,\,61^\circ - \sin \,\,11^\circ =
View answer and explanation
Correct option: A
cos⁡  7∘\cos \,\,7^\circ
Option A (cos⁡ 7∘\cos \,7^\circ) is correct because it is the simplified form of the given trigonometric expression after applying sum-to-product and product-to-sum identities.
42 Easy
42 of 165
Application +4 | -1
The value of sin⁡  85∘−sin⁡ 35∘cos⁡  65∘\frac{{\sin \,\,85^\circ - \sin \,35^\circ }}{{\cos \,\,65^\circ }} is
View answer and explanation
Correct option: C
1\mathrm{1}
Option C (1) is correct because it is obtained by applying the sine subtraction formula and simplifying the expression.
43 Easy
43 of 165
Application +4 | -1
The maximum of the function 3  cos⁡  x−4  sin⁡  x3\,\,\cos \,\,x - 4\,\,\sin \,\,x is
View answer and explanation
Correct option: D
5
Option D (5) is correct because it represents the maximum value of the function, calculated using the formula a2+b2\sqrt{a^2 + b^2} for functions of the form acos⁡x−bsin⁡xa \cos x - b \sin x.
44 Easy
44 of 165
Application +4 | -1
If n=1,   2,   3,   …n = 1,\;\,2,\,\;3,\,\; \ldots, then cos⁡α  cos⁡  2α  cos⁡  4α…cos⁡  2n−1α\cos \alpha \,\,\cos \,\,2\alpha \,\,\cos \,\,4\alpha \ldots \cos \,\,{2^{n - 1}}\alpha is equal to

View answer and explanation
Correct option: D
sin⁡  2nα2nsin⁡α\frac{{\sin \,\,{2^n}\alpha }}{{{2^n}\sin \alpha }}
Option D is correct because the given expression can be simplified using the identity 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x repeatedly, leading to the result sin⁡ 2nα2nsin⁡ α\frac{{\sin \,{2^n}\alpha }}{{{{2^n}}\sin \,\alpha }}.
45 Easy
45 of 165
Application +4 | -1
The value of 2sin⁡(π8)sin⁡(2π8)sin⁡(3π8)sin⁡(5π8)sin⁡(6π8)sin⁡(7π8)2\sin \left( {\frac{\pi }{8}} \right)\sin \left( {\frac{{2\pi }}{8}} \right)\sin \left( {\frac{{3\pi }}{8}} \right)\sin \left( {\frac{{5\pi }}{8}} \right)\sin \left( {\frac{{6\pi }}{8}} \right)\sin \left( {\frac{{7\pi }}{8}} \right) is
View answer and explanation
Correct option: C
18\frac{1}{8}
Option C (18\frac{1}{8}) is correct because it is the simplified value of the given trigonometric expression using trigonometric identities.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
46 Easy
46 of 165
Application +4 | -1
The period of the function f(θ)=4+4sin⁡3θ−3sin⁡θf\left( \mathrm{\theta } \right) = 4 + 4{\sin ^3}\mathrm{\theta } - 3\sin \mathrm{\theta } is
View answer and explanation
Correct option: A
2π3\frac{{2\pi }}{3}
Option A (2π3\frac{{2\pi }}{3}) is correct because it is the period of the simplified function f(θ) = 4 - sin(3θ).
47 Easy
47 of 165
Application +4 | -1
The projections of a line on the axes are, 9, 12 and 8 . The length of the line is
View answer and explanation
Correct option: A
17
Option A (17) is correct because the length of the line is the square root of the sum of the squares of its projections on the axes.
48 Easy
48 of 165
Application +4 | -1
The value of xx for the maximum value of 3  cos⁡  x+sin⁡  x\sqrt 3 \,\,\cos \,\,x + \sin \,\,x is

View answer and explanation
Correct option: A
30∘{30^ \circ }
Option A (30∘{30^ \circ }) is correct because the maximum value of the function occurs when x+π3=π2x + \frac{\pi}{3} = \frac{\pi}{2}, which gives x=π6=30∘x = \frac{\pi}{6} = 30^\circ.
49 Easy
49 of 165
Application +4 | -1
A+B=C  ⇒cos⁡2A+cos⁡2  B+cos⁡2  C−2  cos⁡  A  cos⁡  B  cos⁡  CA + B = C\; \Rightarrow {\cos ^2}A + {\cos ^2}\,\,B + {\cos ^2}\,\,C - 2\,\,\cos \,\,A\,\,\cos \,\,B\,\,\cos \,\,C is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because substituting C=A+BC = A + B and using trigonometric identities simplifies the expression to 1.
50 Medium
50 of 165
Application +4 | -1
If 3sin⁡(A+B)=4sin⁡(A−B)3 \sin (A+B)=4 \sin (A-B) and If tan⁡A=ktan⁡B\tan A=k \tan B, then value of kk is
View answer and explanation
Correct option: C
7
Option C (7) is correct because it is derived by using the componendo and dividendo rule on the given trigonometric equation.
51 Easy
51 of 165
Application +4 | -1
In a right angled Δ   ABC  sin⁡2A+sin⁡2  B+sin⁡2  C=\Delta \,\;ABC\,\,{\sin ^2}A + {\sin ^2}\,\,B + {\sin ^2}\,\,C =
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression evaluates to 2 when C = 90 degrees, which is not among the given options.
52 Medium
52 of 165
Application +4 | -1
The maximum value of   sin⁡  (x+π6)+cos⁡  (x+π6)  \mathrm{\;}\sin \,\,\left( {x + \frac{\pi }{6}} \right) + \cos \,\,\left( {x + \frac{\pi }{6}} \right)\mathrm{\;} in the interval (0,  π2)\left( {0,\,\,\frac{\pi }{2}} \right) is attained at
View answer and explanation
Correct option: A
x=π12x = \frac{\pi }{{12}}
Option A (x=π12x = \frac{\pi }{{12}}) is correct because it is the value of x where the given trigonometric function attains its maximum value in the specified interval.
53 Easy
53 of 165
Application +4 | -1
The expression tan⁡2  α+cot⁡2  α{\tan ^2}\,\,\alpha + {\cot ^2}\,\,\alpha is
View answer and explanation
Correct option: A
≥2\ge 2
Option A (≥2\ge 2) is correct because the square of any real number is non-negative, leading to the given inequality.
54 Medium
54 of 165
Application +4 | -1
If in △ABC,cos⁡B+cos⁡C=2−2cos⁡A\triangle A B C, \cos B+\cos C=2-2 \cos A, then
View answer and explanation
Correct option: B
b,a,cb, a, c are in AP
Option B (b,a,cb, a, c are in AP) is correct because the given condition leads to the relationship sin⁡B+sin⁡C=2sin⁡A\sin B + \sin C = 2 \sin A, which implies that b,a,cb, a, c are in arithmetic progression.
55 Easy
55 of 165
Application +4 | -1
If A+C=2BA + C = 2B, then cos⁡  C−cos⁡  Asin⁡  A−sin⁡  C\frac{{\cos \,\,C - \cos \,\,A}}{{\sin \,\,A - \sin \,\,C}} is equal to
View answer and explanation
Correct option: D
tan⁡  B\tan \,\,B
Option D (tan⁡ B\tan \,B) is correct because by using the sum-to-product formulas and the given condition A+C=2BA + C = 2B, the expression simplifies to tan⁡ B\tan \,B.
56 Medium
56 of 165
Analyzing +4 | -1
The values of α\alpha for which the equation sin⁡4x+cos⁡4x+sin⁡  2x+α=0{\sin ^4}x + {\cos ^4}x + \sin \,\,2x + \alpha = 0 may be valid, are
View answer and explanation
Correct option: B
0≤α≤120 \le \alpha \le \frac{1}{2}
Option C (−32≤α≤12-\frac{3}{2} \le \alpha \le \frac{1}{2}) is incorrect. The correct answer is B (0≤α≤120 \le \alpha \le \frac{1}{2}) because the equation is valid only when 0≤3+2α≤40 \le 3 + 2\alpha \le 4.
57 Easy
57 of 165
Analyzing +4 | -1
If cos⁡  θ=cos⁡  α  cos⁡  β\cos \,\,\theta = \cos \,\,\alpha \,\,\cos \,\,\beta, then tan⁡(θ+α2)  tan⁡  (θ−α2)\tan \left( {\frac{{\theta + \alpha }}{2}} \right)\,\,\tan \,\,\left( {\frac{{\theta - \alpha }}{2}} \right) is equal to
View answer and explanation
Correct option: B
tan⁡2  β2{\tan ^2}\,\,\frac{\beta }{2}
Option B (tan⁡2β2{\tan ^2}\frac{\beta }{2}) is correct because it is derived from applying trigonometric identities to the given equation cos⁡θ=cos⁡αcos⁡β\cos \theta = \cos \alpha \cos \beta.
58 Easy
58 of 165
Application +4 | -1
If A={x:π6≤x≤π3}A = \left\{ {x:\frac{\pi }{6} \le x \le \frac{\pi }{3}} \right\} and f(x)=cos⁡x−x(1+x)f\left( x \right) = \cos x - x\left( {1 + x} \right), then f(A)f\left( A \right) is equal to
View answer and explanation
Correct option: C
[12−π3(1+π3),  32−π6(1+π6)]\left[ {\frac{1}{2} - \frac{\pi }{3}\left( {1 + \frac{\pi }{3}} \right),\,\,\frac{{\sqrt 3 }}{2} - \frac{\pi }{6}\left( {1 + \frac{\pi }{6}} \right)} \right]
Option C ([12−π3(1+π3), 32−π6(1+π6)\frac{1}{2} - \frac{\pi }{3}\left( {1 + \frac{\pi }{3}} \right),\,\frac{{\sqrt 3 }}{2} - \frac{\pi }{6}\left( {1 + \frac{\pi }{6}} \right)]) is correct because it correctly evaluates the function at the endpoints of the interval, considering that the function is decreasing.
59 Hard
59 of 165
Analyzing +4 | -1
The equation y=sin⁡xsin⁡(x+2)−sin⁡2(x+1)y = \sin x\sin \left( {x + 2} \right) - {\sin ^2}\left( {x + 1} \right) represents a straight line lying in
View answer and explanation
Correct option: D
Third and fourth quadrants only
Option D (Third and fourth quadrants only) is correct because the simplified equation represents a horizontal line with a negative y-value.
60 Medium
60 of 165
Application +4 | -1
If A+B+C=πA + B + C = \pi, then sin⁡  2A+sin⁡  2B+sin⁡  2C=\sin \,\,2A + \sin \,\,2B + \sin \,\,2C =
View answer and explanation
Correct option: A
4  sin⁡  A  sin⁡ B  sin⁡  C4\,\,\sin \,\,A\,\,\sin \,B\,\,\sin \,\,C
Option A (4sin⁡Asin⁡Bsin⁡C4 \sin A \sin B \sin C) is correct because by using the sum-to-product trigonometric identities and the fact that A+B+C=πA + B + C = \pi, the expression sin⁡2A+sin⁡2B+sin⁡2C\sin 2A + \sin 2B + \sin 2C simplifies to 4sin⁡Asin⁡Bsin⁡C4 \sin A \sin B \sin C.
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