Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

(3)

We have,

$f (x) = \int \frac{16 x + 24}{x^{2} + 2 x - 15} d x$

$= \int \frac{16 x + 24}{(x + 5) (x - 3)} d x$

$\frac{16 x + 24}{(x + 5) (x - 3)} = \frac{A}{x + 5} + \frac{B}{x - 3}$

$\Rightarrow$ 16x + 24 = A(x – 3) + B(x + 5)

$∴$ 16 = A + B and 24 = –3A + 5B

$\Rightarrow$ 24 = –3A + 5(16 – A) = –8A + 80

$∴$ $A = \frac{56}{8} = 7$ and $B = 9$

$∴ f (x) = \int (\frac{7}{x + 5} + \frac{9}{x - 3}) d x$

$\Rightarrow f (x) = 7 \log_{e} | x + 5 | + 9 \log_{e} | x - 3 | + c$

$f (4) = 7 \log_{e} | 9 | + 9 \log_{e} (1) + c$

$14 \log_{e} 3 = 7 \log_{e} | 9 | + c \Rightarrow c = 0$

$∴ f (7) = 7 \log_{e} 12 + 9 \log_{e} 4$

$= 32 \log_{e} 2 + 7 \log_{e} 3$

$= \log_{e} (2^{32} \cdot 3^{7})$

$∴ α + β = 39$

(2)

We have, $f'' (x) = g'' (x)$

Integrating on both sides, we get

$f' (x) = g' (x) + c$

Put x = 1

$\Rightarrow 4 = 2 + c \Rightarrow c = 2$

$∴ f' (x) = g' (x) + 2$

Integrating on both sides, we get

f(x) = g(x) + 2x + K

Put x = 2

$\Rightarrow$ f(2) + g(2) + 4 + K

$\Rightarrow$ 3 = 9 + 4 + K

$\Rightarrow$ K = –10

$∴$ f(x) = g(x) + 2x – 10

Put x = 25

$∴$ f(25) – g(25) = 50 – 10 = 40

Want Us to Email you About Special Offers & Updates?