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Jee Main Advanced Math Chapter Wise Mock Test
Quiz 1
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Q.1
if $2 \sin^2 (x) – 5 \sin (x) + 2 > 0$, $x \in ( 0 ,2 \pi)$ then $x \in $
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$(0, \frac {\pi}{6} ) \cup (\frac {5\pi}{6} , 2 \pi)$
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$(\frac {\pi}{80}, \frac {\pi}{6} )$
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$(0, \frac {\pi}{6} )$
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$(\frac {5\pi}{6} , 2 \pi)$
Explanation
Solution:
Let s = sin x. The inequality 2s²-5s+2 > 0 factors as (2s-1)(s-2) > 0.
Since sin x never exceeds 1, the factor (s-2) is always negative, so the product is positive exactly when the other factor (2s-1) is also negative, i.e. s < 1/2, i.e. sin x < 1/2.
On (0, 2π), sin x ≥ 1/2 exactly on [π/6, 5π/6], so sin x < 1/2 on the remaining (0, π/6) ∪ (5π/6, 2π).
Q.2
for a positive integer n , let $f_n \theta = ( tan \frac {\theta}{2}) ( 1+ sec \theta) (1 + sec 2 \theta) .....(1+ sec 2^n \theta)$. Now four values are given as (i) $f_2 ( \frac {\pi}{16})=1$ (ii) $f_3 ( \frac {\pi}{32})=1$ (iii) $f_4 ( \frac {\pi}{64})=1$ (iv) $f_5 ( \frac {\pi}{128})=1$ Which of these are correct
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(i) and (ii)
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(i) and (iv)
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All are correct
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(i) ,(ii) and (iii)
Explanation
Solution:
Key identity: tanθ·(1+sec2θ) = tan2θ. Proof: 1+sec2θ = (cos2θ+1)/cos2θ = 2cos²θ/cos2θ, so tanθ·(1+sec2θ) = (2sinθcosθ)/cos2θ = sin2θ/cos2θ = tan2θ.
Applying this repeatedly: tan(θ/2)·(1+secθ) = tanθ (using θ/2 in place of θ above), then ×(1+sec2θ) gives tan2θ, then ×(1+sec4θ) gives tan4θ, and so on -- each extra factor doubles the angle. So f
n
(θ) = tan(2
n
θ).
f
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(π/16)=tan(4·π/16)=tan(π/4)=1, f
3
(π/32)=tan(8·π/32)=tan(π/4)=1, f
4
(π/64)=tan(16·π/64)=tan(π/4)=1, f
5
(π/128)=tan(32·π/128)=tan(π/4)=1 -- all four equal 1.
Q.3
The minimum value of 3 cosx + 4 sinx + 5 is
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0
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-5
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5
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4
Explanation
Solution:
For an expression A cos x + B sin x, the amplitude is √(A²+B²), so it ranges over [-√(A²+B²), √(A²+B²)]. Here √(3²+4²) = √25 = 5, so 3cos x + 4sin x ranges over [-5, 5].
Adding the constant 5 shifts this to [0, 10], so the minimum value of the whole expression is 0.
Q.4
which of these is not the solution for the equation $7 cos^2 x + 3 sin^2 x =4$
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$\frac {4 \pi}{5}$
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$\frac {4 \pi}{3}$
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$\frac {2 \pi}{3}$
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$\frac {11 \pi}{3}$
Explanation
Solution:
7cos²x + 3sin²x = 4 ⇒ 7cos²x + 3(1-cos²x) = 4 ⇒ 4cos²x = 1 ⇒ cos²x = 1/4 ⇒ cos x = ±1/2, giving the general solution x = nπ ± π/3.
2π/3 = π - π/3, 4π/3 = π + π/3, and 11π/3 = 2π + 5π/3 = 2π + (2π-π/3), which reduces to the same family -- all three fit nπ±π/3.
4π/5 does not fit that family for any integer n, so it is not a solution.
Q.5
The number of integral values of k for which equation 7cos(x) + 5 sin(x) =2k +1 has a solution
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4
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10
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12
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8
Explanation
Solution:
An equation of the form A cos x + B sin x = C has a real solution exactly when |C| ≤ √(A²+B²). Here A=7, B=5, so √(49+25) = √74 ≈ 8.60.
So we need |2k+1| ≤ √74, i.e. -8.60 ≤ 2k+1 ≤ 8.60, i.e. -4.80 ≤ k ≤ 3.80.
The integers in that range are -4, -3, -2, -1, 0, 1, 2, 3 -- 8 values.
Q.6
if $\alpha + \beta = \frac {\pi}{2}$ and $ \beta + \gamma = \alpha$,then $tan \alpha$ equals to
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$ tan \beta + tan \gamma$
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$2 tan \beta + tan \gamma$
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$2( tan \beta + tan \gamma)$
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$ tan \beta + 2 tan \gamma$
Explanation
Solution:
From β+γ=α, we get α-β=γ, so tan(α-β) = tanγ:
(tanα - tanβ) / (1 + tanα tanβ) = tanγ.
From α+β=π/2, tanβ = cotα = 1/tanα. Writing t = tanα and substituting tanβ=1/t:
(t - 1/t) / (1 + t·(1/t)) = tanγ ⇒ (t - 1/t)/2 = tanγ ⇒ t - 1/t = 2tanγ.
Since 1/t = tanβ, this is t - tanβ = 2tanγ, i.e. tanα = tanβ + 2tanγ.
Q.7
Minimum value of the expression $ \frac {1}{cos^2 ( \pi/4 +x) + sin^2( \pi/4 -x)}$ equals is
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$\frac {1}{\sqrt 2}$
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$\sqrt 2$
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$\frac {1}{2}$
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2
Explanation
Solution:
sin(π/4 - x) = cos(π/2 - (π/4-x)) = cos(π/4+x), so sin²(π/4-x) = cos²(π/4+x).
The denominator becomes cos²(π/4+x) + cos²(π/4+x) = 2cos²(π/4+x).
The whole expression is 1 / (2cos²(π/4+x)), which is smallest when cos²(π/4+x) is largest, i.e. equal to 1 -- giving a minimum value of 1/2.
Q.8
The solutions of the trigonometric equation tan x + tan 2x + tan x tan 2x=1
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$\frac {n \pi}{2} + \frac {\pi}{12}, n \in Z$
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$\frac {n \pi}{3} + \frac {\pi}{12}, n \in Z$
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$\frac {n \pi}{3} + \frac {\pi}{4}, n \in Z$
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$\frac {n \pi}{3} + \frac {\pi}{8}, n \in Z$
Explanation
Solution:
Rearranging: tan x + tan 2x = 1 - tan x tan 2x, i.e.
(tan x + tan 2x) / (1 - tan x tan 2x) = 1, which is exactly tan(x+2x) = tan(3x).
So tan(3x) = 1, giving 3x = nπ + π/4, i.e. x = nπ/3 + π/12, for integer n.
Q.9
which of the following number is rational?
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$ cos 15^0$
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$ sin 15^0$
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$ sin 15^0 cos 75^0$
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$ sin 15^0 cos 15^0$
Explanation
Solution:
sin15°cos15° = ½sin(2×15°) = ½sin30° = ½×½ = 1/4, a rational number.
cos15° and sin15° individually involve √6 and √2 terms (irrational), and sin15°cos75° = sin15°sin15° = sin²15° = (1-cos30°)/2, which involves √3 and is irrational too.
Q.10
if y = |sin (x) | + |cos (x)| and $x \in R$ then
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$y \in [0,2]$
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$y \in [0,1]$
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$y \in [1, \sqrt 2]$
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$y \in [0,\sqrt 2]$
Explanation
Solution:
y = |sin x| + |cos x|. Squaring: y² = sin²x + cos²x + 2|sinx cosx| = 1 + |sin2x|.
|sin2x| ranges over [0,1], so y² ranges over [1,2], meaning y itself ranges over [1, √2] (y is never negative, being a sum of absolute values).
The minimum, y=1, happens when one of |sinx|,|cosx| is 0 (e.g. x=0); the maximum, y=√2, happens when |sinx|=|cosx| (e.g. x=π/4).
Q.11
The maximum value of sin (cos x) is equal to
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0
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sin 1
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$sin ( \frac {1}{\sqrt 2}) $
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1
Explanation
Solution:
cos x always lies in [-1, 1], and since 1 radian is well inside [-π/2, π/2] (where sine is increasing), sin(cos x) is largest exactly when cos x is largest, i.e. cos x = 1, giving sin(1).
Q.12
If sin A and cos A are the roots of the equation $ax^2 – bx + c = 0$, then a, b and c satisfy the relation
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$a^2 – b^2 – 2ac = 0$
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$a^2 + b^2 + 2ac = 0$
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$a^2 + c^2 + 2ab = 0$
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$a^2 – b^2 + 2ac = 0$
Explanation
Solution:
For ax²-bx+c=0, the sum of roots is sinA+cosA = b/a and the product is sinA·cosA = c/a.
Squaring the sum: (sinA+cosA)² = sin²A+cos²A+2sinAcosA = 1+2sinAcosA, so (b/a)² = 1 + 2(c/a).
Multiplying through by a²: b² = a² + 2ac, which rearranges to a²-b²+2ac = 0.
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