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(i) and (ii):
(i) If F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}, show that F(x)F(y)=F(x+y)F(x)F(y) = F(x+y).
(ii) Prove that ∫02af(x) dx=∫0af(x) dx+∫0af(2a−x) dx\displaystyle\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx. OR Answer
(a) and (b):
(a) If ∣x218x∣=∣62186∣\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}, then find xx.
(b) If x=a(cos⁡t+tsin⁡t)x = a(\cos t + t\sin t), y=a(sin⁡t−tcos⁡t)y = a(\sin t - t\cos t), find dydx\dfrac{dy}{dx}.
Assam AhsecAHSEC Higher Secondary (HS) Final Examination 2023Subjective· 4mImportance★★★★★
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(i) Matrix multiplication + angle-sum identities; (ii) split the integral and substitute x=2a−tx=2a-t in the second piece; (OR-a) equate determinants; (OR-b) differentiate the parametric equations and divide.

(i) F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]F(x) = \begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0&0&1\end{bmatrix}. Compute F(x)F(y)F(x)F(y) by matrix multiplication:

Row 1: [cos⁡xcos⁡y−sin⁡xsin⁡y, −cos⁡xsin⁡y−sin⁡xcos⁡y, 0]=[cos⁡(x+y), −sin⁡(x+y), 0]\big[\cos x\cos y - \sin x\sin y,\ -\cos x\sin y - \sin x\cos y,\ 0\big] = [\cos(x+y),\ -\sin(x+y),\ 0]

Row 2: [sin⁡xcos⁡y+cos⁡xsin⁡y, −sin⁡xsin⁡y+cos⁡xcos⁡y, 0]=[sin⁡(x+y), cos⁡(x+y), 0]\big[\sin x\cos y + \cos x\sin y,\ -\sin x\sin y + \cos x\cos y,\ 0\big] = [\sin(x+y),\ \cos(x+y),\ 0]

Row 3: [0,0,1][0,0,1]

using cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)\cos x\cos y - \sin x\sin y = \cos(x+y) and sin⁡xcos⁡y+cos⁡xsin⁡y=sin⁡(x+y)\sin x\cos y+\cos x\sin y = \sin(x+y). This is exactly F(x+y)F(x+y). Hence F(x)F(y)=F(x+y)F(x)F(y) = F(x+y).

(ii) ∫02af(x)dx=∫0af(x)dx+∫a2af(x)dx\int_0^{2a}f(x)dx = \int_0^a f(x)dx + \int_a^{2a}f(x)dx (splitting the interval).

In ∫a2af(x)dx\int_a^{2a}f(x)dx, substitute x=2a−tx = 2a-t, so dx=−dtdx=-dt; when x=a,t=ax=a, t=a; when x=2a,t=0x=2a, t=0:

∫a2af(x)dx=∫a0f(2a−t)(−dt)=∫0af(2a−t) dt=∫0af(2a−x) dx\int_a^{2a}f(x)dx = \int_a^0 f(2a-t)(-dt) = \int_0^a f(2a-t)\,dt = \int_0^a f(2a-x)\,dx (renaming the dummy variable).

Therefore ∫02af(x)dx=∫0af(x)dx+∫0af(2a−x)dx\displaystyle\int_0^{2a}f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a-x)dx, as required.

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