1
NDA Mathematics 13 April 2025
MCQ (Single Correct Answer)
+2.5
-0.833
Change Language
The position vectors of three points A, B and C respectively, where  $\vec{a} ,\vec{b} $ and $\vec{c} $  respectively, where $\vec{c} = (\cos^2 \theta)\vec{a}+(\sin^2 \theta)\vec{b}$. What is $(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})$ equal to?
A
$\vec{0}$
B
$\vec{2c}$
C
$\vec{3c}$
D
Unit vector
2
NDA Mathematics 13 April 2025
MCQ (Single Correct Answer)
+2.5
-0.833
Change Language
Let $\vec{a},\vec{b} ,(\vec{a}\times\vec{b})$ be unit vectors. What is $\vec{a}.\vec{b}$
A
0
B
1/2
C
1
D
3
3
NDA Mathematics 13 April 2025
MCQ (Single Correct Answer)
+2.5
-0.833
Change Language
Let  $x = \sec \theta - \cos \theta \text{ and } y = \sec^4 \theta - \cos^4 \theta \\ $ 
What is $(\frac{dy}{dx})^2$ equal to?
A
$\frac{4(y^2+4)}{(x^2+4)}$
B
$\frac{4(y^2-4)}{(x^2-4)}$
C
$\frac{16(y^2+4)}{(x^2+4)}$
D
$\frac{16(y^2-4)}{(x^2-4)}$
4
NDA Mathematics 13 April 2025
MCQ (Single Correct Answer)
+2.5
-0.833
Change Language
Let  $x = \sec \theta - \cos \theta \text{ and } y = \sec^4 \theta - \cos^4 \theta \\ $ 
What  $\left[ \frac{x^2+4}{y^2+4} \frac{dy}{dx} \left( x^2+4 \frac{d^2y}{dx^2} - 16y \right) \right] $  equal to?
A
$16x$
B
$16y$
C
$-16x$
D
$-16y$
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