COLUMN-ICOLUMN-IIA)The area of the ellipse given by 3x2+4xy+3y2=1 is aπb+2 sq.units, where a, b are single digit positive integers. The value of (a+1)b isP)0B)If f:D→R such that f(k)=sin(cosx)+loge(−2cos2x+3cosx−1) then ∫x1x2 [cosx−12] dx is equal to, where x1,x2∈D and [.] denotes the greatest integer functionQ)1C)The area of region, represented by |x|+|y|≤2 and x2≤y is “S” square units then 3S=___R)7D)limy→∞∫1y[tan−1x]dx∫1y[1+1x]dx=_____ ([.]denotes greatest integer function)S)6

=2{32−13}=73a) 3y2+4xy+3x2−1=0 y=−2x±3−5x23⇒−35≤x≤35Area =453∫03535−x2dx=π5=aπb+2 ⇒a=1,b=3b) f(x) is defined when −2cos2x+3cosx−1>0 2cos2x−3cosx+1<0 ⇒12<cosx<1 0<cosx−12<12 ⇒[cosx−12]=0c) required area =2{area OABC−∫01x2dx}d) limy→∞∫1tan10dx+∫tan1y1dx∫1y1dx=limy→∞y−tan1y−1=1

COLUMN-ICOLUMN-IIA)The area of the ellipse given by 3x2+4xy+3y2=1 is aπb+2 sq.units, where a, b are single digit positive integers. The value of (a+1)b isP)0B)If f:D→R such that f(k)=sin(cosx)+loge(−2cos2x+3cosx−1) then ∫x1x2 [cosx−12] dx is equal to, where x1,x2∈D and [.] denotes the greatest integer functionQ)1C)The area of region, represented by |x|+|y|≤2 and x2≤y is “S” square units then 3S=___R)7D)limy→∞∫1y[tan−1x]dx∫1y[1+1x]dx=_____ ([.]denotes greatest integer function)S)6

=2{32−13}=73a) 3y2+4xy+3x2−1=0 y=−2x±3−5x23⇒−35≤x≤35Area =453∫03535−x2dx=π5=aπb+2 ⇒a=1,b=3b) f(x) is defined when −2cos2x+3cosx−1>0 2cos2x−3cosx+1<0 ⇒12<cosx<1 0<cosx−12<12 ⇒[cosx−12]=0c) required area =2{area OABC−∫01x2dx}d) limy→∞∫1tan10dx+∫tan1y1dx∫1y1dx=limy→∞y−tan1y−1=1

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COLUMN-I COLUMN-II
A) The area of the ellipse given by $3 x^{2} + 4 x y + 3 y^{2} = 1$ is $\frac{a π}{\sqrt{b + 2}}$ sq.units, where a, b are single digit positive integers. The value of $(a + 1) b$ is P) 0
B) If $f : D \to R$ such that $f (k) = \sqrt{\sin (\cos x)} + \log_{e} (- 2 \cos^{2} x + 3 \cos x - 1)$ then $\int_{x_{1}}^{x_{2}} [\cos x - \frac{1}{2}] d x$ is equal to, where $x_{1}, x_{2} \in D$ and $[.]$ denotes the greatest integer function Q) 1
C) The area of region, represented by $| x | + | y | \leq 2$ and $x^{2} \leq y$ is “S” square units then $3 S =___$ R) 7
D) $\lim_{y \to \infty} \frac{\int_{1}^{y} [\tan^{- 1} x] d x}{\int_{1}^{y} [1 + \frac{1}{x}] d x} =_____$ ([.]denotes greatest integer function) S) 6

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A-S, B-R, C-P, D-Q

A-R, B-P, C-Q, D-S

A-S, B-Q, C-R, D-P

A-S, B-P, C-R, D-Q

answer is D.

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Detailed Solution

$= 2 {\frac{3}{2} - \frac{1}{3}} = \frac{7}{3}$ a) $3 y^{2} + 4 x y + 3 x^{2} - 1 = 0$
$y = \frac{- 2 x \pm \sqrt{3 - 5 x^{2}}}{3} \Rightarrow - \sqrt{\frac{3}{5}} \leq x \leq \sqrt{\frac{3}{5}}$
Area $= \frac{4 \sqrt{5}}{3} \int_{0}^{\sqrt{\frac{3}{5}}} \sqrt{\frac{3}{5} - x^{2}} d x = \frac{π}{\sqrt{5}} = \frac{a π}{\sqrt{b + 2}}$
$\Rightarrow a = 1, b = 3$
b) $f (x)$ is defined when
$- 2 \cos^{2} x + 3 \cos x - 1 > 0 2 \cos^{2} x - 3 \cos x + 1 < 0 \Rightarrow \frac{1}{2} < \cos x < 1 0 < \cos x - \frac{1}{2} < \frac{1}{2} \Rightarrow [\cos x - \frac{1}{2}] = 0$
c)
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required area $= 2 {a r e a O A B C - \int_{0}^{1} x^{2} d x}$

d)
$\lim_{y \to \infty} \frac{\int_{1}^{\tan 1} 0 d x + \int_{\tan 1}^{y} 1 d x}{\int_{1}^{y} 1 d x} = \lim_{y \to \infty} \frac{y - \tan 1}{y - 1} = 1$

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