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JEE • MATHEMATICS • TRANSFORMATIONS

Transformations practice · Easy

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Showing 31-60 of 112 questions
31 Easy
31 of 112
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

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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 }}.
32 Easy
32 of 112
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.
33 Easy
33 of 112
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).
34 Easy
34 of 112
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.
35 Easy
35 of 112
Application +4 | -1
The value of xx for the maximum value of 3  cos⁡  x+sin⁡  x\sqrt 3 \,\,\cos \,\,x + \sin \,\,x is

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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.
36 Easy
36 of 112
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.
37 Easy
37 of 112
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.
38 Easy
38 of 112
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.
39 Easy
39 of 112
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.
40 Easy
40 of 112
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.
41 Easy
41 of 112
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.
42 Easy
42 of 112
Analyzing +4 | -1
If   y=1+4  sin⁡2  x  cos⁡2  x\;y = 1 + 4\,\,{\sin ^2}\,\,x\,\,{\cos ^2}\,\,x then

View answer and explanation
Correct option: A
1≤  y  ≤21 \le \,\,y\,\, \le 2
The correct answer is A (1≤y≤21 \le y \le 2) because by rewriting the equation and using the range of sin⁡22x\sin^2 2x, we can determine the range of y.
43 Easy
43 of 112
Application +4 | -1
If sin⁡  x+sin⁡  y=a\sin \,\,x + \sin \,\,y = a and cos⁡  x+cos⁡  y=b\cos \,\,x + \cos \,\,y = b, then tan⁡(a+y2)\tan \left( {\frac{{a + y}}{2}} \right) is equal to
View answer and explanation
Correct option: B
ab\frac{a}{b}
Option B (ab\frac{a}{b}) is correct because it is derived by dividing the sum-to-product formulas for sine and cosine.
44 Easy
44 of 112
Application +4 | -1
If A+B+C=180∘A+B+C=180^{\circ}, then cot⁡A+cot⁡B+cot⁡Ccot⁡Acot⁡Bcot⁡C\frac{\cot A+\cot B+\cot C}{\cot A \cot B \cot C} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the given expression simplifies to 1 using the trigonometric identity for cot(A+B+C) when A+B+C = 180 degrees.
45 Easy
45 of 112
Application +4 | -1
The value of   cos⁡  π5  cos⁡  2 π5  cos⁡  4 π5  cos⁡  8 π5  \mathrm{\;}\cos \,\,\frac{\mathrm{\pi }}{5}\,\,\cos \,\,\frac{{2\,\mathrm{\pi }}}{5}\,\,\cos \,\,\frac{{4\,\mathrm{\pi }}}{5}\,\,\cos \,\,\frac{{8\,\mathrm{\pi }}}{5}\mathrm{\;} is equal to
View answer and explanation
Correct option: D
−116- \frac{1}{{16}}
Option D (−116- \frac{1}{{16}}) is correct because it is the result of applying the product-to-sine identity repeatedly and simplifying.

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46 Easy
46 of 112
Application +4 | -1
The period of the function sin⁡(πx2)+cos⁡(πx2)\sin \left( {\frac{{\pi x}}{2}} \right) + \cos \left( {\frac{{\pi x}}{2}} \right) is
View answer and explanation
Correct option: A
4
The correct answer is A (4) because the period of both sine and cosine functions with the given argument is 4, and the period of their sum is also 4.
47 Easy
47 of 112
Application +4 | -1
If cos⁡(θ+∅)=mcos  (θ−∅)\cos \left( {\theta + \emptyset } \right) = mcos\;\left( {\theta - \emptyset } \right), then tan⁡θ\tan \theta is equal to
View answer and explanation
Correct option: C
[(1−m)/(1+m)]  cot   ∅\left[ {\left( {1 - m} \right)/\left( {1 + m} \right)} \right]\,\,\mathrm{cot}\;\,\emptyset
Option C [(1−m)/(1+m)]cot   ∅\left[ {\left( {1 - m} \right)/\left( {1 + m} \right)} \right]\mathrm{cot}\;\,\emptyset is correct because it is derived by applying trigonometric identities and simplifying the given equation.
48 Easy
48 of 112
Analyzing +4 | -1
If in a triangle ABC,  a2−b2a2+b2=sin⁡  (A−B)sin⁡  (A+B)ABC,\,\,\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}} = \frac{{\sin \,\,\left( {A - B} \right)}}{{\sin \,\,\left( {A + B} \right)}} , then the triangle is
View answer and explanation
Correct option: A
Right angled or isosceles
Right angled or isosceles is the correct answer because the given equation implies either A = B (isosceles) or a2+b2=c2a^2 + b^2 = c^2 (right-angled).
49 Easy
49 of 112
Application +4 | -1
If A+B+C=3 π2A + B + C = \frac{{3\,\pi }}{2}, then cos⁡  2A+cos⁡  2B+cos⁡  2C=\cos \,\,2A + \cos \,\,2B + \cos \,\,2C =
View answer and explanation
Correct option: D
1−4sin⁡A  sin⁡B  sin⁡C1 - 4\sin A\,\,\sin B\,\,\sin C
Option D (1−4sin⁡Asin⁡Bsin⁡C1 - 4\sin A\sin B\sin C) is correct because it is the simplified form of the given trigonometric expression using the condition A+B+C=3π2A + B + C = \frac{{3\pi }}{{2}}.
50 Easy
50 of 112
Analyzing +4 | -1
The period of the function f(x)=a{tan⁡(πx)+x−[x]}f\left( x \right) = {a^{\left\{ {\tan \left( {\pi x} \right) + x - \left[ x \right]} \right\}}}, where a>0a > 0 denotes the greatest integer function and x is a real number, is
View answer and explanation
Correct option: D
11
Option D 11 is correct because the fractional part of x and the tangent function both have a period of 1.
51 Easy
51 of 112
Application +4 | -1
The expression tan⁡2α+cot⁡2α{\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 tan⁡2α+cot⁡2α{\tan ^2}\alpha + {\cot ^2}\alpha can be rewritten as a squared term plus 2, which is always greater than or equal to 2.
52 Easy
52 of 112
Application +4 | -1
If sin⁡α+sin⁡β=a\sin \alpha+\sin \beta=a and cos⁡α−cos⁡β=b\cos \alpha-\cos \beta=b, then tan⁡(α−β2)\tan \left(\frac{\alpha-\beta}{2}\right) is equal to
View answer and explanation
Correct option: B
−ba-\frac{b}{a}
Option B (−ba-\frac{b}{a}) is correct because it is derived from dividing the two given trigonometric equations after applying sum-to-product formulas.
53 Easy
53 of 112
Analyzing +4 | -1
Let f(x)f\left( x \right) be a real valued function defined by
f(x+λ)=1+[2−5f(x)+10{f(x)}2−10{f(x)}3+5{f(x)}4−{f(x)}5]1/5f\left( {x + \lambda } \right) = 1 + {\left[ {2 - 5f\left( x \right) + 10{{\left\{ {f\left( x \right)} \right\}}^2} - 10{{\left\{ {f\left( x \right)} \right\}}^3} + 5{{\left\{ {f\left( x \right)} \right\}}^4} - {{\left\{ {f\left( x \right)} \right\}}^5}} \right]^{1/5}} for all real x and some positive constant λ\lambda, then f(x)f\left( x \right) is
View answer and explanation
Correct option: B
A periodic function with period 2  λ2\;\lambda
The correct answer is A (A periodic function with period λ\lambda) because the function repeats its values after an interval of 2λ2\lambda, making it periodic.
54 Easy
54 of 112
Application +4 | -1
The maximum value of 3  cos⁡  x+4  sin⁡  x+53\,\,\cos \,\,x + 4\,\,\sin \,\,x + 5 is
View answer and explanation
Correct option: D
None of these
None of these is correct because the maximum value of the expression is 10, which is not among the options.
55 Easy
55 of 112
Analyzing +4 | -1
The period of the function

f(x)=sin⁡8x  cosx−sin⁡6x  cos3xcos⁡2xcos⁡x−sin⁡3xsin⁡4xf\left( x \right) = \frac{{\sin 8\mathrm{x\;cosx} - \sin 6\mathrm{x\;cos}3\mathrm{x}}}{{\cos 2\mathrm{x}\cos \mathrm{x} - \sin 3\mathrm{x}\sin 4\mathrm{x}}} is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{{2}}) is correct because the given function simplifies to tan⁡2x\tan 2x, and the period of tan⁡kx\tan kx is π∣k∣.\frac{\pi}{{|k|}}.
56 Easy
56 of 112
Application +4 | -1
The value of cos⁡  π15  cos⁡  2 π15  cos⁡  2 π15  cos⁡  8 π15\cos \,\,\frac{\pi }{{15}}\,\,\cos \,\,\frac{{2\,\pi }}{{15}}\,\,\cos \,\,\frac{{2\,\pi }}{{15}}\,\,\cos \,\,\frac{{8\,\pi }}{{15}} is
View answer and explanation
Correct option: B
−116- \frac{1}{{16}}
Option B (-116\frac{1}{{16}}) is correct because by repeatedly using the product-to-sum trigonometric identities, the expression simplifies to -116\frac{1}{{16}}.
57 Easy
57 of 112
Application +4 | -1
The range of f(x)=cos⁡x−sin⁡xf\left( x \right) = \cos x - \sin x is
View answer and explanation
Correct option: D
[−2,  2]\left[ { - \sqrt 2 ,\;\sqrt 2 } \right]
Option D ([−2,;2]\left[ { - \sqrt 2 ,\\;\sqrt 2 } \right]) is correct because it represents the range of the function, which is a sinusoidal function with amplitude 2\sqrt{2}.
58 Easy
58 of 112
Analyzing +4 | -1
The maximum value of 12  sin⁡  θ−9  sin⁡2  θ12\,\,\sin \,\,\theta - 9\,\,{\sin ^2}\,\,\theta is
View answer and explanation
Correct option: B
4\mathrm{4}
Option B (4) is correct because the maximum value of the given expression is 4, obtained by completing the square.
59 Easy
59 of 112
Application +4 | -1
For α,β∈(0,π/2)\alpha, \beta \in(0, \pi / 2), let 3sin⁡(α+β)=2sin⁡(α−β)3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta) and a real number kk be such that tan⁡α=ktan⁡β\tan \alpha=k \tan \beta. Then, the value of kk is equal to
View answer and explanation
Correct option: B
−5-5
Option B (-5) is correct because it involves using trigonometric identities to simplify the given equation and find the relationship between tan⁡α\tan \alpha and tan⁡β\tan \beta.
60 Easy
60 of 112
Application +4 | -1
sin⁡  65∘+sin⁡  43∘−sin⁡  29∘−sin⁡  7∘\sin \,\,65^\circ + \sin \,\,43^\circ - \sin \,\,29^\circ - \sin \,\,7^\circ is equal to
View answer and explanation
Correct option: D
None of these
Option C (3/2) is correct because it is the result of applying the sine rule and componendo and dividendo.
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