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1

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$f, g :$$ $$\left[ { - 1,2} \right] \to R$$ be continuous functions which are twice differentiable on the interval $$(-1, 2)$$. Let the values of f and g at the points $$-1, 0$$ and $$2$$ be as given in the following table:
X = -1 X = 0 X = 2
f(x) 3 6 0
g(x) 0 1 -1

In each of the intervals $$(-1, 0)$$ and $$(0, 2)$$ the function $$(f-3g)''$$ never vanishes. Then the correct statement(s) is (are)

A
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly three solutions in $$\left( { - 1,0} \right) \cup \left( {0,2} \right)$$
B
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly one solution in $$(-1, 0)$$
C
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly one solution in $$(0, 2)$$
D
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly two solutions in $$(-1, 0)$$ and exactly two solutions in $$(0, 2)$$

Explanation

Let $$F(x) = f(x) - 3g(x)$$

$$\therefore$$ $$F( - 1) = 3$$, $$F(0) = 3$$ and $$F(2) = 3$$

So, $$F'(x)$$ will vanish at least twice in

$$( - 1,0) \cup (0,2)$$.

$$\because$$ $$F''(x) > 0$$ or $$ < 0$$, $$\forall x \in ( - 1,0) \cup (0,2)$$

Hence, $$f'(x) - 3g'(x) = 0$$ has exactly one solution in $$( - 1,0)$$ and one solution in $$(0,2)$$.

2

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

The function $$f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$$ has a local minimum or a local maximum at x =

A
$$-$$2
B
$${{ - 2} \over 3}$$
C
2
D
$${{ 2} \over 3}$$

Explanation

Here, $$f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$$

$$ = \left\{ {\matrix{ { - 2x - (x + 2) + (x - 2),} & {when\,x \le - 2} \cr { - 2x + x + 2 + 3x + 2,} & {when\, - 2 < x \le - {2 \over 3}} \cr { - 4x,} & {when\, - {2 \over 3} < x \le 0} \cr {4x,} & {when\,0 < x \le 2} \cr {2x + 4,} & {when\,x > 2} \cr } } \right.$$

$$ = \left\{ {\matrix{ { - 2x - 4,} & {x \le - 2} \cr {2x + 4,} & { - 2 < x \le - 2/3} \cr { - 4x,} & { - {2 \over 3} < x \le 0} \cr {4x,} & {0 < x \le 2} \cr {2x + 4,} & {x > 2} \cr } } \right.$$

Graph for y = f(x) is shown as

3

JEE Advanced 2013 Paper 1 Offline

MCQ (More than One Correct Answer)
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio $$8:15$$ is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is $$100$$, the resulting box has maximum volume. Then the lengths of the vsides of the rectangular sheet are
A
$$24$$
B
$$32$$
C
$$45$$
D
$$60$$
4

IIT-JEE 2012 Paper 2 Offline

MCQ (More than One Correct Answer)
If $$f\left( x \right) = \int_0^x {{e^{{t^2}}}} \left( {t - 2} \right)\left( {t - 3} \right)dt$$ for all $$x \in \left( {0,\infty } \right),$$ then
A
$$f$$ has a local maximum at $$x=2$$
B
$$f$$ is decreasing on $$(2, 3)$$
C
there exists some $$c \in \left( {0,\infty } \right),$$ such that $$f'(c)=0$$
D
$$f$$ has a local minimum at $$x=3$$

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