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1

### WB JEE 2020

MCQ (More than One Correct Answer)
English
Bengali
Let $$y = {{{x^2}} \over {{{(x + 1)}^2}(x + 2)}}$$. Then $${{{d^2}y} \over {d{x^2}}}$$ is
A
$$2\left[ {{3 \over {{{(x + 1)}^4}}} - {3 \over {{{(x + 1)}^3}}} + {4 \over {{{(x + 2)}^3}}}} \right]$$
B
$$3\left[ {{2 \over {{{(x + 1)}^3}}} + {4 \over {{{(x + 1)}^2}}} - {5 \over {{{(x + 2)}^3}}}} \right]$$
C
$${6 \over {{{(x + 1)}^3}}} - {4 \over {{{(x + 1)}^2}}} + {3 \over {{{(x + 1)}^3}}}$$
D
$${7 \over {{{(x + 1)}^3}}} - {3 \over {{{(x + 1)}^2}}} + {2 \over {{{(x + 1)}^3}}}$$

## Explanation

Given, $$y = {{{x^2}} \over {{{(x + 1)}^2}(x + 2)}}$$

By using partial fraction,

$${{{x^2}} \over {{{(x + 1)}^2}(x - 2)}} = {4 \over {(x + 2)}} - {3 \over {x + 1}} + {1 \over {{{(x + 1)}^2}}}$$

$$\therefore$$ $${4 \over {x + 2}} - {3 \over {x + 1}} + {1 \over {{{(x + 1)}^2}}}$$

On differentiating both sides w.r.t. x, we get

$${{dy} \over {dx}} = {{ - 4} \over {{{(x + 2)}^2}}} + {3 \over {{{(x + 1)}^2}}} - {2 \over {{{(x + 1)}^3}}}$$

Again differentiating w.r.t. x

$${{{d^2}y} \over {d{x^2}}} = {8 \over {{{(x + 2)}^3}}} - {6 \over {{{(x + 1)}^3}}} + {6 \over {{{(x + 1)}^4}}}$$

$$= 2\left[ {{4 \over {{{(x + 2)}^3}}} - {3 \over {{{(x + 1)}^3}}} + {3 \over {{{(x + 1)}^4}}}} \right]$$

মনে করো $$y = {{{x^2}} \over {{{(x + 1)}^2}(x + 2)}}$$। সেক্ষেত্রে $${{{d^2}y} \over {d{x^2}}}$$ হবে

A
$$2\left[ {{3 \over {{{(x + 1)}^4}}} - {3 \over {{{(x + 1)}^2}}} + {4 \over {{{(x + 2)}^3}}}} \right]$$
B
$$3\left[ {{2 \over {{{(x + 1)}^4}}} + {4 \over {{{(x + 1)}^2}}} - {5 \over {{{(x + 2)}^3}}}} \right]$$
C
$${6 \over {{{(x + 1)}^3}}} - {4 \over {{{(x + 1)}^2}}} + {3 \over {{{(x + 1)}^3}}}$$
D
$${7 \over {{{(x + 1)}^3}}} - {3 \over {{{(x + 1)}^2}}} + {2 \over {{{(x + 1)}^3}}}$$

## Explanation

$${{{x^2}} \over {{{(x + 1)}^2}(x + 2)}} = {A \over {x + 1}} + {B \over {{{(x + 1)}^2}}} + {C \over {x + 2}}$$

$$\therefore$$ $${x^2} = A(x + 1)(x + 2) + B(x + 2) + C{(x + 1)^2}$$

$$x = - 1$$ বসিয়ে

$$\therefore$$ $$1 = B$$

$$x = - 2$$ বসিয়ে, $$4 = C$$

x2-এর সহগ তুলনা করে পাই

$$1 = A + C \Rightarrow A = 1 - C = - 3$$

$$\therefore$$ $$y = {{ - 3} \over {x + 1}} + {1 \over {{{(x + 1)}^2}}} + {4 \over {x + 2}}$$

$$\therefore$$ $${{dy} \over {dx}} = {3 \over {{{(x + 1)}^2}}} - {2 \over {{{(x + 1)}^3}}} - {4 \over {{{(x + 2)}^2}}}$$

$${{{d^2}y} \over {d{x^2}}} = {{ - 6} \over {{{(x + 1)}^3}}} + {6 \over {{{(x + 1)}^4}}} + {8 \over {{{(x + 2)}^3}}}$$

$$= 2\left[ {{3 \over {{{(x + 1)}^4}}} - {3 \over {{{(x + 1)}^3}}} + {4 \over {{{(x + 2)}^3}}}} \right]$$

2

### WB JEE 2019

MCQ (More than One Correct Answer)
English
Bengali
Let f and g be differentiable on the interval I and let a, b $$\in$$ I, a < b. Then,
A
If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 is soluble in (a, b)
B
If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 may not be soluble in (a, b)
C
If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 is soluble in (a, b), k $$\in$$ R
D
If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 may not be soluble in (a, b), k $$\in$$ R

## Explanation

From option (a)

We have,

$$f(a) = f(b) = 0$$

$$\Rightarrow f'(a)\,.\,f'(b) < 0$$

Again, let $$h(x) = f'(x) + f(x)g'(x)$$

$$\Rightarrow h(a) = f'(a) + f(a)g'(a) = f'(a)$$

and $$h(b) = f'(b) + f(b)g'(b) = f'(b)$$

$$\therefore$$ $$h(a)\,.\,h(b) = f'(a)\,.\,f'(b) < 0$$

$$\therefore$$ h(x) = 0 has root between (a, b)

From option (c)

We have, $$g(a) = g(b) = 0$$

$$\Rightarrow g'(a)\,.\,g'(b) < 0$$

Again, let $$m(x) = g'(x) + kg(x)$$

$$\Rightarrow m(a) = g'(a) + kg(a) = g'(a)$$

$$\Rightarrow m(b) = g'(b) + kg(b) = g'(b)$$

$$\therefore$$ $$m(a)\,.\,m(b) = g'(a)\,.\,g'(b) < 0$$

$$\therefore$$ m(x) = 0 has root between (a, b).

So, option (a) and (c) are correct.

মনে করো f ও g অন্তরাল I তে অন্তরকলনযোগ্য এবং a, b $$\in$$ I, a < b । সেক্ষেত্রে -

A
যদি f (a) = 0 = f (b) হয়, তবে f' (x) + f (x) g' (x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য
B
যদি f (a) = 0 = f (b) হয়, তবে f' (x) + f (x) g' (x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য নাও হতে পারে
C
যদি g(a) = 0 = g(b) হয়, তবে g'(x) + kg(x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য, k $$\in$$ R
D
যদি g(a) = 0 = g(b) হয়, তবে g'(x) + kg(x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য নাও হতে পারে k $$\in$$ R
3

### WB JEE 2017

MCQ (More than One Correct Answer)
English
Bengali
If f(x) = xn, being a non-negative integer, then the values of n for which f'($$\alpha$$ + $$\beta$$) = f'($$\alpha$$) + f'($$\beta$$) for all $$\alpha$$, $$\beta$$ > 0 is
A
1
B
2
C
0
D
5

## Explanation

We have,

f(x) = xn

$$\Rightarrow$$ f'(x) = nxn$$-$$1

Now, f'($$\alpha$$ + $$\beta$$) = f'($$\alpha$$) + f'($$\beta$$)

$$\Rightarrow$$ n($$\alpha$$ + $$\beta$$)n$$-$$1 = n$$\alpha$$n$$-$$1 + n$$\beta$$n$$-$$1

$$\Rightarrow$$ ($$\alpha$$ + $$\beta$$)n$$-$$1 = $$\alpha$$n$$-$$1 + $$\beta$$n$$-$$1

From options we see that n = 2, satisfy the above equation.

$$\therefore$$ n = 2

যদি $$f(x) = {x^n}$$ হয় যেখানে $$n$$-অঋণাত্মক পূর্ণসংখ্যা, তবে সব $$\alpha ,\beta > 0$$-এর জন্য $$f'(\alpha + \beta ) = f'(\alpha ) + f'(\beta )$$ হলে

A
$$1$$
B
$$2$$
C
$$0$$
D
$$5$$

## Explanation

$$n = 1$$ হলে $$f(x) = x \Rightarrow f'(x) = 1$$

$$f'(\alpha + \beta ) = 1$$ এবং $$f'(\alpha ) + f'(\beta ) = 1 + 1 = 2$$

$$\therefore$$ $$n \ne 1$$

$$n = 2$$ হলে $$f(x) = {x^2}$$

$$\therefore$$ $$f'(x) = 2x$$

$$f'(\alpha + \beta ) =2(\alpha+\beta)= f'(\alpha ) + f'(\beta )$$

$$\therefore$$ $$n = 2$$ সম্ভব।

$$n = 0$$ হলে $$f(1) = 1 \Rightarrow f'(x) = 0$$

$$f'(\alpha + \beta ) = f'(\alpha ) + f'(\beta ) = 0$$

$$\therefore$$ $$n = 0$$ হতে পারে

$$n = 5$$ হলে $$f(x) = {x^5} \Rightarrow f'(x) = 5{x^4}$$

$$f'(\alpha + \beta ) = 5{(\alpha + \beta )^4}$$

$$f'(\alpha ) + f'(\beta ) = 5({\alpha ^4} + {\beta ^4})$$

$$\therefore$$ $$n \ne 5$$

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