Match the following pair:-

1-A, 2-B, 3-C, 4-D

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1-B, 2-D, 3-A, 4-C

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1-A, 2-D, 3-B, 4-C

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1-C, 2-D, 3-B, 4-A

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Solution

The correct option is A
1-A, 2-B, 3-C, 4-D

First Anglo Mysore- war – (1767-1769) Between Hyder Ali and Robert clive.

4th Anglo – Mysore war- (1799) Tipu sultan against wellesely- Tipu sultan died here.

Second Anglo- sikh war- (1848-49) Punjab sikh armies vs. Lord Dalhousie. English were victories. Led to annexation of Punjab.

Third Anglo – Maratha war: (1817-1818) Lord Hesting vs Marathas. (Baji Rao) British were victorious.

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\int_{- 1}^{4} f (x) d x = 4

\int_{2}^{4} [3 - f (x)] d x = 7 t h e n \int_{- 1}^{2} f (x) d x =

List I	List II
I. $\int_{- 1}^{1} x \| x \| d x$	(a) $\frac{π}{2}$
II. $\int_{0}^{\frac{π}{2}} (1 + log (\frac{4 + 3 sin x}{4 + 3 cos x})) d x$	(b) $\int_{0}^{\frac{a}{2}} f (x) d x$
III. $\int_{0}^{a} f (x) d x$	(c) $\int_{0}^{a} [f (x) + f (- x)] d x$
IV. $\int_{- a}^{a} f (x) d x$	(d) $0$
	(e) $\int_{0}^{a} f (a - x) d x$

\int_{0}^{π / 2} \frac{d x}{1 + {tan}^{5} x} = \frac{π}{4}

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a + x) d x = \int_{0}^{π / 2} \frac{d x}{1 + {tan}^{3} x} = \int_{0}^{π / 2} \frac{d_{X}}{1 + {cot}^{3} x} = \frac{π}{4}

$I f \int_{2}^{4} f (x) d x = \frac{1}{2} a n d f (6 - x) = f (x), t h e n \int_{2}^{4} x f (x) d x =$

\int_{0}^{1} e^{2 I n x} d x =

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