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JEE • MATHEMATICS • TRANSFORMATIONS

Transformations practice · Mixed difficulty

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Practice format and chapter quiz

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 151-165 of 165 questions
151 Easy
151 of 165
Application +4 | -1
If   sin⁡  A−sin⁡  Ccos⁡  C−cos⁡  A=cot⁡  B,   then   A, B,C\;\frac{{\sin \,\,A - \sin \,\,C}}{{\cos \,\,C - \cos \,\,A}} = \cot \,\,B,\,\;\mathrm{then}\;\,A,\,B,C are in

View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the given condition implies that BB is the arithmetic mean of AA and CC, which is the definition of an arithmetic progression.
152 Easy
152 of 165
Application +4 | -1
16sin⁡(20∘)sin⁡(40∘)sin⁡(80∘)16\sin \left( {20^\circ } \right)\sin \left( {40^\circ } \right)\sin \left( {80^\circ } \right) is equal to:
View answer and explanation
Correct option: B
232\sqrt 3
Option B (232\sqrt 3) is correct because the expression simplifies using the trigonometric identity 4sin⁡θsin⁡(60°−θ)sin⁡(60°+θ)=sin⁡3θ4\sin \theta \sin(60°-\theta)\sin(60°+\theta) = \sin 3\theta.
153 Easy
153 of 165
Application +4 | -1
The value of 2  cos⁡  x−cos⁡  3x−cos⁡  5x  2\,\,\cos \,\,x - \cos \,\,3x - \cos \,\,5x\mathrm{\;} is equal to
View answer and explanation
Correct option: A
16  cos⁡3 x  sin⁡2 x16\,\,{\cos ^3}\,x\,\,{\sin ^2}\,x
Option A (16  cos⁡3 x  sin⁡2 x16\,\,{\cos ^3}\,x\,\,{\sin ^2}\,x) is correct because by using trigonometric identities and factoring, the expression simplifies to 16  cos⁡3 x  sin⁡2 x16\,\,{\cos ^3}\,x\,\,{\sin ^2}\,x.
154 Easy
154 of 165
Application +4 | -1
The least value of cosec2 x+25  sec⁡2 x\mathrm{cose}{\mathrm{c}^2}\,x + 25\,\,{\sec ^2}\,x is
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Correct option: D
38
Option D (38) is correct because the minimum value of the expression is obtained by completing the square and finding the minimum value of the squared term.
155 Easy
155 of 165
Understanding +4 | -1
The function f(x)=∣cos⁡x∣f\left( x \right) = \left| {\cos x} \right| is periodic with period
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Correct option: B
π\pi
Option B (π\pi) is correct because the period of ∣cos⁡x∣|\cos x| is π\pi.
156 Medium
156 of 165
Application +4 | -1
If f:R→Rf:R \to R and is defined by f(x)=12−cos⁡3xf\left( x \right) = \frac{1}{{2 - \cos 3x}} for each x∈Rx \in R, then the range ff is
View answer and explanation
Correct option: B
[1/3,1]\left[ {1/3,\mathrm{ }1} \right]
Option B [1/3,1]\left[ {1/3,\mathrm{ }1} \right] is correct because the range is found by considering the bounds of the cosine function and its effect on the given function.
157 Easy
157 of 165
Analyzing +4 | -1
In aΔ ABC,  cos⁡(B+2C+3A2)+cos⁡(A−B2)\Delta \,ABC,\,\,\cos \left( {\frac{{B + 2C + 3A}}{2}} \right) + \cos \left( {\frac{{A - B}}{2}} \right) is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because by using the identity A+B+C=πA+B+C = \pi and simplifying the expression, we can show that the sum of the cosines equals 0.
158 Medium
158 of 165
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 is equal to
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, with the given condition, the expression simplifies to this form using trigonometric identities.
159 Medium
159 of 165
Analyzing +4 | -1
If y=cos⁡2x+sec⁡2xy = {\cos ^2}x + {\sec ^2}x, then
View answer and explanation
Correct option: C
y≥2y \ge 2
Option C (y≥2y \ge 2) is correct because cos⁡2x+sec⁡2x{\cos ^2}x + {\sec ^2}x can be rewritten as (cos⁡x−sec⁡x)2+2(\cos x - \sec x)^2 + 2, which is always greater than or equal to 2.
160 Easy
160 of 165
Application +4 | -1
The sum   S=sin  θ+sin⁡  2θ+...+sin⁡  nθ  \mathrm{\;S} = \mathrm{sin}\,\,\mathrm{\theta } + \sin \,\,2\mathrm{\theta } + ... + \sin \,\,n\mathrm{\theta \;} equals
View answer and explanation
Correct option: A
sin⁡  12  (n+1)  θ  sin⁡  nθ2/sin⁡  θ2\sin \,\,\frac{1}{2}\,\,\left( {n + 1} \right)\,\,\mathrm{\theta }\,\,\sin \,\,\frac{{n\mathrm{\theta }}}{2}/\sin \,\,\frac{\mathrm{\theta }}{2}
Option A (sin⁡12(n+1)θsin⁡nθ2/sin⁡θ2\sin \frac{1}{2}\left( {n + 1} \right)\theta \sin \frac{{n\theta }}{2}/\sin \frac{{\theta }}{2}) is correct because it represents the sum of a sine series using the appropriate trigonometric formula.
161 Easy
161 of 165
Application +4 | -1
The maximum value of cos⁡2A+cos⁡2B−cos⁡2C{\cos ^2}A + {\cos ^2}B - {\cos ^2}C, is
View answer and explanation
Correct option: D
2
Option D (2) is correct because the maximum value is attained when cos²A and cos²B are maximized (equal to 1) and cos²C is minimized (equal to 0).
162 Medium
162 of 165
Analyzing +4 | -1
If   A+B+C=π    and  cos⁡  A=cos⁡  B  cos⁡  C\;A + B + C = \pi \,\,\;\mathrm{and}\,\,\cos \,\,A = \cos \,\,B\,\,\cos \,\,C then tan⁡  B  tan⁡  C\tan \,\,B\,\,\tan \,\,C is equal to

View answer and explanation
Correct option: B
1\mathrm{1}
Option B (1) is correct because by using the given condition A+B+C=πA + B + C = \pi and the trigonometric identity, we can derive that tan⁡Btan⁡C=1\tan B \tan C = 1.
163 Medium
163 of 165
Analyzing +4 | -1
If the equation sin⁡  θ(sin⁡  θ+2  cos⁡  θ)=a\sin \,\,\theta \left( {\sin \,\,\theta + 2\,\,\cos \,\,\theta } \right) = a has a real solution, then the shortest interval containing ′a′'a' is
View answer and explanation
Correct option: A
[1−52,  1+52]\left[ {\frac{{1 - \sqrt 5 }}{2},\,\,\frac{{1 + \sqrt 5 }}{2}} \right]
Option A is correct because the range of 'a' is determined by finding the maximum and minimum values of the trigonometric expression.
164 Easy
164 of 165
Analyzing +4 | -1
The equation sin⁡4θ+cos⁡4θ=a{\sin ^4}\theta + {\cos ^4}\theta = a has a real solution if
View answer and explanation
Correct option: A
a∈[1/2,1]a \in \left[ {1/2,1} \right]
Option A is correct because it finds the range of 'a' for which the equation has real solutions by using trigonometric identities and the range of cosine function.
165 Medium
165 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 is equal to
View answer and explanation
Correct option: A
4  sin⁡A  sin⁡B  sin⁡C4\,\,\sin A\,\,\sin B\,\,\sin C
Option A (4  sin⁡A  sin⁡B  sin⁡C4\,\,\sin A\,\,\sin B\,\,\sin C) is correct because it is the result of applying trigonometric identities to simplify the expression sin⁡2A+sin⁡2B+sin⁡2C\sin 2A + \sin 2B + \sin 2C when A+B+C=πA + B + C = \pi.
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