Critical Reasoning — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Critical Reasoning MCQs with step-by-step solutions (29 questions). Part of Trigonometric Ratios. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Critical Reasoning · hard
$0 < θ < \frac{π}{2}$ ⇒
A. $\tan θ < θ < \sin θ$
B. $\tan θ > θ > \sin θ$
C. $\tan θ > θ < \sin θ$ ✓ Correct
D. $\tan θ < θ > \sin θ$
Solution: Compare growth rates in the first quadrant.
Q2 — Critical Reasoning · hard
a = sin 1°, b = sin 1 ⇒
A. a = b
B. a < b
C. a > b ✓ Correct
D. a = 2b
Solution: 1 radian ≈ 57.3°, so sin(57.3°) > sin(1°).
Q3 — Critical Reasoning · hard
x = cos 1°, y = cos 1 ⇒
A. x = y
B. x > y
C. x < y ✓ Correct
D. 2x = y
Solution: Since 1 rad = 57.3° and cos is decreasing.
Q4 — Critical Reasoning · hard
p = tan 1°, q = tan 1 ⇒
A. p = q
B. p < q ✓ Correct
C. p > q
D. 2p = 3q
Solution: tan(1°) < tan(57.3°).
Q5 — Critical Reasoning · hard
$(\sqrt{1 - \sin^2 100°}) \sec 100°$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: $\sqrt{1 - \sin^2 100°} = |\cos 100°| = -\cos 100°$ in Q2.
Q6 — Critical Reasoning · hard
$(\sqrt{1 - \sin^2 200°}) \sec 200°$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: $\cos 200° < 0$, so $|\cos 200°| = -\cos 200°$.
Q7 — Critical Reasoning · hard
$(\sqrt{1 - \cos^2 200°}) \cosec 200°$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: $\sin 200° < 0$, so $|\sin 200°| = -\sin 200°$.
Q8 — Critical Reasoning · hard
If $\sqrt{\frac{1 - \sin A}{1 + \sin A}} = \sec A - \tan A$, then A lies in the quadrants
A. I, II
B. II, III
C. I, IV ✓ Correct
D. I, III
Solution: Analyze sign conditions for the equation.
Q9 — Critical Reasoning · hard
If $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$, then A lies in the quadrants
A. I, II
B. II, III
C. I, III ✓ Correct
D. I, IV
Solution: Work through quadrant analysis.
Q10 — Critical Reasoning · hard
If $\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \cosec A - \cot A$, then A lies in the quadrants
A. I, II ✓ Correct
B. II, III
C. I, III
D. I, IV
Solution: Test sign conditions.
Q11 — Critical Reasoning · hard
If $\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \cosec A + \cot A$, then A lies in the quadrants
A. I, II ✓ Correct
B. II, III
C. I, III
D. I, IV
Solution: Verify with quadrant analysis.
Q12 — Critical Reasoning · hard
If $k = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)$, then k =
A. 0
B. ±1 ✓ Correct
C. ±3
D. ±4
Solution: Each product must equal 1.
Q13 — Critical Reasoning · hard
$\sin^6 \frac{49π}{49} + \cos^6 \frac{49π}{49} - 1 + 3\sin^2 \frac{49π}{49} \cos^2 \frac{49π}{49}$ =
A. 1
B. –1
C. 2
D. 0 ✓ Correct
Solution: Use $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ identity.
Q14 — Critical Reasoning · hard
cos A, sin A, cot A are in GP ⇒ $\tan^6 A - \tan^2 A$ =
A. –1
B. 0 ✓ Correct
C. 1
D. 2
Solution: From GP: $\sin^2 A = \cos A \cdot \cot A$.
Q15 — Critical Reasoning · hard
cos A, sin A, cot A are in GP ⇒ $\cot^6 A + \cot^4 A$ =
A. –1
B. 0
C. 1 ✓ Correct
D. 2
Solution: Use GP condition to derive.
Q16 — Critical Reasoning · hard
sin A, cos A, tan A are in GP ⇒ $\cot^6 A - \cot^2 A$ =
A. 2 ✓ Correct
B. 1
C. 0
D. –1
Solution: From GP condition, show $\cot^6 A - \cot^2 A = 2$.
Q17 — Critical Reasoning · hard
If x = cos 10° – sin 10°, then
A. x > 0 ✓ Correct
B. x < 0
C. x = 0
D. x ≥ 0
Solution: $\cos 10° > \sin 10°$ since 10° is acute and < 45°.
Q18 — Critical Reasoning · hard
If $\tan θ = -\frac{4}{3}$, then $\sin θ$ =
A. $\frac{4}{5}$ but not $-\frac{4}{5}$
B. $\frac{4}{5}$ (or) $-\frac{4}{5}$ ✓ Correct
C. $-\frac{4}{5}$ but not $\frac{4}{5}$
D. $\frac{3}{5}$
Solution: tan is negative in II and IV quadrants.
Q19 — Critical Reasoning · hard
$\frac{\sqrt{1 - \cos α} + \sqrt{1 + \cos α}}{\sqrt{1 - \cos α} - \sqrt{1 + \cos α}}$ =
A. $-2 \sec α$ ✓ Correct
B. $-2 \sec α$
C. $2 \cosec α$
D. $-2 \cosec α$
Solution: Rationalize and use half-angle formulas.
Q20 — Critical Reasoning · hard
$\frac{\sqrt{1 - \cos θ} - \sqrt{1 + \cos θ}}{\sqrt{1 - \cos θ} + \sqrt{1 + \cos θ}}$ =
A. $-2 \cosec θ$ ✓ Correct
B. $2 \cosec θ$
C. $-2 \cot θ$
D. $2 \cot θ$
Solution: Manipulate using complementary properties.
Q21 — Critical Reasoning · hard
$\frac{1}{2} \sin θ - \sqrt{\cot^2 θ - \cos^2 θ}$ =
A. $\cosec θ$
B. $\sec θ$
C. $\sin θ$ ✓ Correct
D. $\cos θ$
Solution: Simplify the radical first.
Q22 — Critical Reasoning · hard
sin x + $\sin^2 x$ = 1 ⇒ $\cos^{12} x + 3 \cos^{10} x + 3\cos^8 x + \cos^6 x + 2 \cos^4 + \cos^2 x - 2$ =
A. 0 ✓ Correct
B. $\cos^2 x$
C. $\sin^2 x$
D. $-\sin^2 x$
Solution: From sin x + $\sin^2 x$ = 1, derive $\cos^2 x$ relationship.
Q23 — Critical Reasoning · hard
cos x + $\cos^2 x$ = 1 ⇒ $\sin^{12} x + 3\sin^{10}x + 3\sin^8 x + \sin^6 x + \sin^4 x + 2 \sin^2 x - 2$ =
A. 0 ✓ Correct
B. $\sin x$
C. $\cos x$
D. 1
Solution: Use the given constraint to simplify.
Q24 — Critical Reasoning · hard
cos θ + $\cos^2 θ$ = 1, $a \sin^{12} + b \sin^{10} + c \sin^8 + d \sin^6$ = 1 ⇒ $\frac{b + c}{a + d}$ =
A. 2
B. 3
C. 4 ✓ Correct
D. 6
Solution: Identify coefficients from the constraint.
Q25 — Critical Reasoning · hard
sin x + $\sin^2 x$ + $\sin^3 x$ = 1 ⇒ $\cos^6 x - 4 \cos^4x + 8 \cos^2 x$ =
A. 4
B. 2 ✓ Correct
C. 1
D. 0
Solution: Derive from the triple constraint.
Q26 — Critical Reasoning · hard
cos x + $\cos^2 x$ + $\cos^3 x$ = 1, $a \sin^6 x + b \sin^4 x + c \sin^2 x + d$ = 0 ⇒ $a + b + c + d$ =
A. 0 ✓ Correct
B. 1
C. –1
D. 2
Solution: Sum of coefficients.
Q27 — Critical Reasoning · hard
$\sin^2 α = \frac{(x + y)^2}{4xy}$ is possible when
A. x > 0, y > 0, x = 3y
B. x > 0, y > 0, x = y ✓ Correct
C. x = 2y
D. x > 0, y < 0, x = y
Solution: Since $\sin^2 α ≤ 1$, derive constraints.
Q28 — Critical Reasoning · hard
Which of the following is possible?
A. $\cos θ = x + \frac{1}{x}$
B. $\sec θ = \frac{x^2}{1 + x^2}$
C. $\cosec θ = \frac{x}{1 + x^2}$
D. $\tan θ = \frac{x^2 + x - 1}{x^2 - 3x + 5}$ ✓ Correct
Solution: Check bounds for each.
Q29 — Critical Reasoning · hard
If x = a $\cos^2 θ$ sin θ and y = a $\sin^2 θ$ cos θ, then $\frac{(x^2 + y^2)^3}{x^2 y^2}$ =
A. a
B. $a^3$
C. $a^2$ ✓ Correct
D. $a^5$
Solution: Factor and simplify the ratio.