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1

### JEE Main 2019 (Online) 12th January Morning Slot

Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then $$\int\limits_0^a \,$$f(x) g(x) dx is equal to :
A
4$$\int\limits_0^a \,$$f(x)dx
B
$$-$$ 3$$\int\limits_0^a \,$$f(x)dx
C
$$\int\limits_0^a \,$$f(x)dx
D
2$$\int\limits_0^a \,$$f(x)dx

## Explanation

$${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx}$$

$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx}$$

$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx}$$

$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}}$$

$$\Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx}$$
2

### JEE Main 2019 (Online) 11th January Evening Slot

$$\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}$$ is equal to :
A
0
B
4
C
1
D
2

## Explanation

$$\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1$$
3

### JEE Main 2019 (Online) 11th January Evening Slot

Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\pi$$) cos |x| is not differentiable. Then the set K is equal to :
A
{0, $$\pi$$}
B
$$\phi$$ (an empty set)
C
{ r }
D
{0}

## Explanation

f(x) = sin$$\left| x \right| - \left| x \right|$$ + 2(x $$-$$ $$\pi$$) cosx

$$\because$$  sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0

$$\therefore$$  k = $$\phi$$
4

### JEE Main 2019 (Online) 11th January Morning Slot

Let [x] denote the greatest integer less than or equal to x. Then $$\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$
A
equals $$\pi$$ + 1
B
equals 0
C
does not exist
D
equals $$\pi$$

## Explanation

R.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$

(as x $$\to$$ 0+ $$\Rightarrow$$  [x] $$=$$ 0)

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$$

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$$

L.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$$

(as x $$\to$$ 0$$-$$ $$\Rightarrow$$  [x] $$=$$ $$-$$1)

$$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi$$

R.H.L.  $$\ne$$  L.H.L.

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